The representation theory of finite sets and correspondences

Serge Bouc and Jacques Thévenaz

Abstract. We investigate correspondence functors, namely the functors from the category of finite sets and correspondences to the category of k-modules, where k is a commutative ring. They have various specific properties which do not hold for other types of functors. In particular, if k is a field and if F is a correspondence functor, then F is finitely generated if and only if the dimension of F (X) grows exponentially in terms of the cardinality of the finite set X. In such a case, F has finite length. Also, if k is noetherian, then any subfunctor of a finitely generated functor is finitely generated. When k is a field, we give a description of all the simple functors and we determine the dimension of their evaluations at any finite set.

A main tool is the construction of a functor associated to any finite lattice T. We prove for instance that this functor is projective if and only if the lattice T is distributive. Moreover, it has quotients which play a crucial role in the analysis of simple functors. The special case of total orders yields some more specific results. Several other properties are also discussed, such as projectivity, duality, and symmetry. In an appendix, all the lattices associated to a given poset are described.

AMS Subject Classification: 06B05, 06B15, 06D05, 06D50, 16B50, 18B05, 18B10, 18B35, 18E05

Keywords: finite set, correspondence, relation, functor category, simple functor, order, lattice, poset

Contents

1. Introduction 4

Part 1. CORRESPONDENCE FUNCTORS 7 2. The representation theory of categories 8 3. Correspondence functors 10 4. Some basic examples and results 15 5. Posets and lattices 20 6. Finite generation 23 7. Bounded type 26 8. Exponential behavior and finite length 28 9. Projective functors and duality 34 10. The noetherian case 38

Part 2. FUNCTORS AND LATTICES 47 11. Functors associated to lattices 48 12. Quotients of functors associated to lattices 53 13. The fundamental functor associated to a poset 57 14. Duality and opposite lattices 62 15. The case of a total order 73 16. On the structure of lattices 83 17. Evaluations of fundamental functors and simple functors 87 18. Properties of fundamental functors 98 19. Tensor product and internal hom 105 20. Examples of decompositions 114

Part 3. APPENDICES 119 21. Lattices associated to posets 120 22. Forests associated to lattices 128

Bibliography 139

1. Introduction

Representations of categories have been used by many authors in different contexts. The purpose of the present memoir is to develop the theory in the case of the category of finite sets, when morphisms are all correspondences between finite sets. For representing the category of finite sets, there are several possible choices. Pirashvili [Pi] treats the case of pointed sets and maps, while Church, Ellenberg and Farb [CEF] consider the case where the morphisms are all injective maps. Putman and Sam [PS] use all k-linear splittable injections between finite-rank free k-modules (where k is a commutative ring). Here, we move away from such choices by using all correspondences as morphisms. The cited papers are concerned with applications to cohomological stability, while we develop our theory without any specific application in mind. The main motivation is provided by the fact that finite sets are basic objects in mathematics. Moreover, the theory turns out to have many quite surprising results, which justify the development presented here.

Let C be the category of finite sets and correspondences. A correspondence functor over a commutative ring k is a functor from C to the category k-Mod of all k-modules. As much as possible, we develop the theory for an arbitrary commutative ring k. Let us start however with the case when k is a field. If F is a correspondence functor over a field k, then we prove that F is finitely generated if and only if the dimension of F (X) grows exponentially in terms of the cardinality of the finite set X (Theorem 8.4). In such a case, we also obtain that F has finite length (Theorem 8.6). This result was obtained independently by Gitlin [Gi], using a criterion proved by Wiltshire-Gordon [WG]. Moreover, for finitely generated correspondence functors, we show that the Krull-Remak-Schmidt theorem holds (Proposition 6.6) and that projective functors coincide with injective functors (Theorem 9.6).

Suppose that k is a field. By well-known results about representations of categories, simple correspondence functors can be easily classified. In our case, they are parametrized by triples (E, R, V), where E is a finite set, R is a partial order relation on E, and V is a simple k Aut(E, R)-module (Theorem 3.12). This is the first indication of the importance of posets in our work. However, if S_{E,R,V} is a simple functor parametrized by (E, R, V), then it is quite hard to describe the evaluation S_{E,R,V} (X) at a finite set X. We achieve this in Section 17 by giving a closed formula for its dimension (Theorem 17.19).

Simple functors have precursors depending only on a poset (E, R). There is a correspondence functor 𝕊_{E,R}, which we call a fundamental correspondence functor (Definition 4.7), from which we recover in a straightforward fashion every simple correspondence functor S_{E,R,V} by means of a suitable tensor product with V. Actually, 𝕊_{E,R} can be defined over any commutative ring k and a large part of our work is concerned with proving properties of the fundamental correspondence functors 𝕊_{E,R}. In particular, we prove in Section 17 that the evaluation 𝕊_{E,R}(X) at a finite set X is a free k-module, by describing a k-basis of 𝕊_{E,R}(X) and giving a closed formula for its rank (Corollary 17.17). This is the key which is used, when k is a field, to obtain the formula for the dimension of S_{E,R,V} (X), for any simple k Aut(E, R)-module V.

A natural question when dealing with a commutative ring k is to obtain specific results when k is noetherian. We follow this track in Section 10 and show for instance that any subfunctor of a finitely generated correspondence functor is again finitely generated (Corollary 10.5). Also, we obtain stabilization results for Hom and Ext between correspondence functors evaluated at large enough finite sets (Theorem 10.10).

In Section 19, we also introduce a natural tensor structure on the category of correspondence functors, and show that there is an adjoint internal hom construction.

A main tool in this work is the construction of a correspondence functor F_{T} associated to a finite lattice T (Section 11). This is the second indication of the importance of posets and lattices in our work. Part 2 of this memoir describes the interplay between lattices and functors. For instance, one of our first results asserts that the functor F_{T} is projective if and only if the lattice T is distributive (Theorem 11.11).

The fundamental functors can be analyzed by using lattices: if (E, R) is the subposet of irreducible elements in a lattice T, then the functor F_{T} has a fundamental functor as a quotient, which turns out to be 𝕊_{E,R^{op}} where R^{op} denotes the opposite order relation (Theorem 13.1). We show that there is a duality between F_{T} and F_{T^{op}} over any commutative ring k (Theorem 14.9) and the fundamental functor 𝕊_{E,R} also appears as a subfunctor of F_{T^{op}} (Theorem 14.16). The special case of a totally ordered lattice T yields some specific results (Section 15), connected to the fact that F_{T} and its associated fundamental functor are projective functors.

A surprising byproduct of our work is concerned with the theory of finite lattices. In any lattice T, we define some operations r^{∞} and s^{∞}, as well as some special elements which we call bulbs (Section 16). Also, a canonical forest (disjoint union of trees) can be constructed in T, from which some idempotents can be defined in F_{T} (T) (Theorem 17.7 and Theorem 17.9). All this plays a crucial role in the description of a k-basis of 𝕊_{E,R}(X) in Section 17.

Some specific results about lattices have been postponed to two appendices (Section 21 and Section 22) because they are not used directly in our analysis of correspondence functors.

There are a few basic results in Section 2 which have been imported from elsewhere, but this memoir is almost self-contained. The main exception is the algebra E_{E} of essential relations on a finite set E. This algebra has been analyzed in [BT] and all its simple modules have been classified there. This uses the fundamental module P_{E}f_{R} and there is an explicit description of the action of relations on P_{E}f_{R}. All the necessary background on this algebra E_{E} of essential relations is recalled in Section 3. Note that the fundamental module P_{E}f_{R} is a main ingredient for the definition of the fundamental functor 𝕊_{E,R}.

Part 1

CORRESPONDENCE FUNCTORS

2. The representation theory of categories

Before introducing the category C of finite sets and correspondences, we first recall some standard facts from the representation theory of categories. Let D be a category and let X and Y be two objects of D. We adopt a slightly unusual notation by writing D(Y, X) for the set of all morphisms from X to Y. We reverse the order of X and Y in view of having later a left action of morphisms behaving nicely under composition.

We assume that D is small (or more generally that a skeleton of D is small). This allows us to talk about the set of natural transformations between two functors starting from D.

Throughout this paper, k denotes a commutative ring. It will sometimes be noetherian and sometimes a field, but we shall always emphasize when we make additional assumptions.

2.1. Definition. The k-linearization of a category D, where k is any commutative ring, is defined as follows :

• The objects of kD are the objects of D.

• For any two objects X and Y, the set of morphisms from X to Y is the free k-module kD(Y, X) with basis D(Y, X).

• The composition of morphisms in kD is the k-bilinear extension

kD(Z, Y) × kD(Y, X) ⟶ kD(Z, X)

of the composition in D.

2.2. Definition. Let D be a category and k a commutative ring. A k-representation of the category D is a k-linear functor from kD to the category k-Mod of k-modules.

We could have defined a k-representation of D as a functor from D to k-Mod, but it is convenient to linearize first the category D (just as for group representations, where one can first introduce the group algebra).

If F : kD → k-Mod is a k-representation of D and if X is an object of D, then F (X) will be called the evaluation of F at X. Morphisms in kD act on the left on the evaluations of F by setting, for every m ∈ F (X) and for every morphism α ∈ kD(Y, X),

α · m := F (α)(m) ∈ F (Y) .

We often use a dot for this action of morphisms on evaluation of functors. With our choice of notation, if β ∈ kD(Z, Y), then

(βα) · m = β · (α · m) .

The category F(kD, k-Mod) of all k-representations of D is an abelian category. (We need to restrict to a small skeleton of D in order to have sets of natural transformations, which are morphisms in F(kD, k-Mod), but we will avoid this technical discussion). A sequence of functors

0 ⟶ F_{1} ⟶ F_{2} ⟶ F_{3} ⟶ 0

is exact if and only if, for every object X, the evaluation sequence

0 ⟶ F_{1}(X) ⟶ F_{2}(X) ⟶ F_{3}(X) ⟶ 0

is exact. Also, a k-representation of D is called simple if it is nonzero and has no proper nonzero subfunctor.

For any object X of D, consider the representable functor kD(−, X) (which is a projective functor). Its evaluation at an object Y is the k-module kD(Y, X), which has a natural structure of a (kD(Y, Y), kD(X, X))-bimodule by composition.

For any kD(X, X)-module W, we define, as in [Bo1], the functor

L_{X,W} = kD(−, X) ⊗_{kD(X,X)} W ,

which satisfies the following adjunction property.

2.3. Lemma. Let F = F(kD, k- Mod) be the category of all k-representations of D and let X be an object of D.

(a) The functor

kD(X, X)− Mod ⟶ F , W ↦ L_{X,W}

is left adjoint of the evaluation functor

F ⟶ kD(X, X)− Mod , F ↦ F (X) .

In other words, for any k-representation F : kD → k-Mod and any kD(X, X)-module W, there is a natural isomorphism

Hom_{F}(L_{X,W} , F) ≅ Hom_{kD(X,X)}(W, F (X)) .

Moreover L_{X,W} (X) ≅ W as kD(X, X)-modules.

(b) This functor kD(X, X)− Mod ⟶ F is right exact. It maps projective modules to projective functors, and indecomposable projective modules to indecomposable projective functors.

Proof : Part (a) is straightforward and is proved in Section 2 of [Bo1]. Part (b) follows because this functor is left adjoint of an exact functor.

Our next result is a slight extension of the first lemma of [Bo1].

2.4. Lemma. Let X be an object of D and let W be a kD(X, X)-module. For any object Y of D, let

J_{X,W} (Y) = { ∑_{i} ϕ_{i} ⊗ w_{i} ∈ L_{X,W} (Y) | ∀ψ ∈ kD(X, Y), ∑_{i}(ψϕ_{i}) · w_{i} = 0} .

(a) J_{X,W} is the unique subfunctor of L_{X,W} which is maximal with respect to the condition that it vanishes at X.

(b) If W is a simple module, then J_{X,W} is the unique maximal subfunctor of L_{X,W} and L_{X,W} /J_{X,W} is a simple functor.

Proof : The proof is sketched in Lemma 2.3 of [BST] in the special case of biset functors for finite groups, but it extends without change to representations of an arbitrary category D.

2.5. Notation. Let X be an object of D and let W be a kD(X, X)-module. We define

S_{X,W} = L_{X,W} /J_{X,W} .

If W is a simple kD(X, X)-module, then S_{X,W} is a simple functor.

We emphasize that L_{X,W} and S_{X,W} are defined for any kD(X, X)-module W and any commutative ring k. Note that we always have J_{X,W} (X) = {0} because if

a = ∑_{i} ϕ_{i} ⊗ w_{i} ∈ J_{X,W} (X), then a = id_{X} ⊗(∑_{i} ϕ_{i} · w_{i}) = 0.

Therefore, we have isomorphisms of kD(X, X)-modules

L_{X,W} (X) ≅ S_{X,W} (X) ≅ W .

2.6. Proposition. Let S be a simple k-representation of D and let Y be an object of D such that S(Y) ≠ 0.

(a) S is generated by S(Y), that is, S(X) = kD(X, Y)S(Y) for all objects X. More precisely, if 0 ≠ u ∈ S(Y), then S(X) = kD(X, Y) · u.

(b) S(Y) is a simple kD(Y, Y)-module.

(c) S ≅ S_{Y,S(Y)}.

Proof : (a) Given 0 ≠ u ∈ S(Y), let S′(X) = kD(X, Y) · u for all objects X. This clearly defines a nonzero subfunctor S′ of S, so S′ = S by simplicity of S.

(b) This follows from (a).

(c) By the adjunction of Lemma 2.3, the identity id : S(Y) → S(Y) corresponds to a non-zero morphism θ : L_{Y,S(Y)} → S. Since S is simple, θ must be surjective. But S_{Y,S(Y)} is the unique simple quotient of L_{Y,S(Y)}, by Lemma 2.4 and Notation 2.5, so S ≅ S_{Y,S(Y)}.

It should be noted that S has many realizations S ≅ S_{Y,W} as above, where W = S(Y) ≠ 0. However, if there is a notion of unique minimal object, then one can parametrize simple functors S by setting S ≅ S_{Y,W}, where Y is the unique minimal object such that S(Y) ≠ 0 (see Theorem 3.7 for the case of correspondence functors).

Our next proposition is Proposition 3.5 in [BST] in the case of biset functors, but it holds more generally and we just recall the proof of [BST].

2.7. Proposition. Let S be a simple k-representation of D and let Y be an object of D such that S(Y) ≠ 0. Let F be any k-representation of D. Then the following are equivalent:

(a) S is isomorphic to a subquotient of F.

(b) The simple kD(Y, Y)-module S(Y) is isomorphic to a subquotient of the kD(Y, Y)-module F (Y).

Proof : It is clear that (a) implies (b). Suppose that (b) holds and let W_{1}, W_{2} be submodules of F (Y) such that W_{2} ⊂ W_{1} and W_{1}/W_{2} ≅ S(Y). For i ∈ {1, 2}, let F_{i} be the subfunctor of F generated by W_{i}. Explicitly, for any object X of D, F_{i}(X) = kD(X, Y) · W_{i} ⊆ F (X). Then F_{i}(Y) = W_{i} and (F_{1}/F_{2})(Y) = W_{1}/W_{2} ≅ S(Y). The isomorphism S(Y) → (F_{1}/F_{2})(Y) induces, by the adjunction of Lemma 2.3, a nonzero morphism θ : L_{Y,S(Y)} → F_{1}/F_{2}. Since S(Y) is simple, L_{Y,S(Y)} has a unique maximal subfunctor J_{Y,S(Y)}, by Lemma 2.4, and L_{Y,S(Y)}/J_{Y,S(Y)} ≅ S_{Y,S(Y)} ≅ S, by Proposition 2.6. Let F′_{1} = θ(L_{Y,S(Y)}) and F′_{2} = θ(J_{Y,S(Y)}). Since θ ≠ 0, we obtain

F′_{1}/F′_{2} ≅ L_{Y,S(Y)}/J_{Y,S(Y)} ≅ S_{Y,S(Y)} ≅ S ,

showing that S is isomorphic to a subquotient of F.

3. Correspondence functors

Leaving the general case, we now prepare the ground for the category C we are going to work with.

3.1. Definition. Let X and Y be sets.

(a) A correspondence from X to Y is a subset of the cartesian product Y ×X. Note that we have reversed the order of X and Y for the reasons mentioned at the beginning of Section 2.

(b) A correspondence is often called a relation but we use this terminology only when X = Y, in which case we say that a subset of X × X is a relation on X.

(c) If σ is a permutation of X, then there is a corresponding relation on X which we write

Δ_{σ} = {(σ(x), x) ∈ X × X | x ∈ X} .

In particular, when σ = id, we also write

Δ_{id} = Δ_{X} = {(x, x) ∈ X × X | x ∈ X} .

3.2. Definition. Let C denote the following category :

• The objects of C are the finite sets.

• For any two finite sets X and Y, the set C(Y, X) is the set of all correspondences from X to Y.

• The composition of correspondences is as follows. If R ⊆ Z × Y and S ⊆ Y × X, then RS is defined by

RS = { (z, x) ∈ Z × X | ∃ y ∈ Y such that (z, y) ∈ R and (y, x) ∈ S } .

The identity morphism id_{X} is the diagonal subset Δ_{X} ⊆ X ×X (in other words the equality relation on X).

3.3. Definition. Let kC be the linearization of the category C, where k is any commutative ring.

(a) A correspondence functor (over k) is a k-representation of the category C, that is, a k-linear functor from kC to the category k-Mod of k-modules.

(b) We let F_{k} = F(kC, k-Mod) be the category of all such correspondence functors (an abelian category).

In part (b), we need to restrict to a small skeleton of C in order to have sets of natural transformations, which are morphisms in F_{k}, but we avoid this technical discussion. It is clear that C has a small skeleton, for instance by taking the full subcategory having one object for each cardinality.

3.4. Proposition. Let X and Y be finite sets. If R ⊆ Y × X, let R^{op} denote the opposite correspondence, defined by

R^{op} = {(x, y) ∈ X × Y | (y, x) ∈ R} ⊆ X × Y .

Then the assignment R ↦ R^{op} induces an isomorphism from C to the opposite category C^{op}, which extends to an isomorphism from kC to kC^{op}.

Proof : One checks easily that if X, Y, Z are finite sets, if R ⊆ Y × X and S ⊆ Z × Y, then (SR)^{op} = R^{op}S^{op}.

We use opposite correspondences to define dual functors. The notion will be used in Section 9 and Section 14.

3.5. Definition. Let F be a correspondence functor over k. The dual F^{♮} of F is the correspondence functor defined on a finite set X by

F^{♮}(X) = Hom_{k} (F (X), k) .

If Y is a finite set and R ⊆ Y ×X, then the map F^{♮}(R) : F^{♮}(X) → F^{♮}(Y) is defined by

∀α ∈ F^{♮}(X), F^{♮}(R)(α) = α ∘ F (R^{op}) .

In order to study simple modules or simple functors, it suffices to work over a field k, by standard commutative algebra. If we assume that k is a field, then the evaluation at a finite set X of a representable functor or of a simple functor is always a finite-dimensional k-vector space.

3.6. Definition. A minimal set for a correspondence functor F is a finite set X of minimal cardinality such that F (X) ≠ 0.

Clearly, for a nonzero functor, such a minimal set always exists and is unique up to bijection.

Our next task is to describe the parametrization of simple correspondence functors, assuming that k is a field. This uses the finite-dimensional algebra

R_{X} := kC(X, X)

of all relations on X, which was studied in [BT]. A relation R on X is called essential if it does not factor through a set of cardinality strictly smaller than |X|. The k-submodule generated by set of inessential relations is a two-sided ideal

I_{X} = ∑_{|Y |<|X|} kC(X, Y)kC(Y, X)

and the quotient

E_{X} := kC(X, X)/I_{X}

is called the essential algebra. A large part of its structure has been elucidated in [BT].

The following parametrization theorem is similar to Proposition 2 in [Bo1] or Theorem 4.3.10 in [Bo2]. The context here is different, but the proof is essentially the same.

3.7. Theorem. Assume that k is a field.

(a) Let S be a simple correspondence functor, let E be a minimal set for S, and let W = S(E). Then W is a simple module for the essential algebra E_{E} (with I_{E} acting by zero) and S ≅ S_{E,W}.

(b) Let E be a finite set and let W be a simple module for the essential algebra E_{E}, viewed as a module for the algebra R_{E} by making I_{E} act by zero on W. Then E is a minimal set for S_{E,W}. Moreover, S_{E,W} (E) ≅ W (as E_{E}-modules).

(c) The set of isomorphism classes of simple correspondence functors is parametrized by the set of isomorphism classes of pairs (E, W) where E is a finite set and W is a simple E_{E}-module.

Proof : (a) Since S(Y) = {0} if |Y | < |E|, we have

I_{E} · S(E) = ∑_{|Y |<|E|} kC(E, Y)kC(Y, E) · S(E) ⊆ ∑_{|Y |<|E|} kC(E, Y) · S(Y) = {0} ,

so S(E) is a module for the essential algebra E_{E}. Now the identity of S(E) corresponds by adjunction to a nonzero homomorphism L_{E,W} → S, where W = S(E)

(see Lemma 2.3). This homomorphism is surjective since S is simple. But L_{E,W} has a unique simple quotient, namely S_{E,W}, hence S ≅ S_{E,W}.

(b) Suppose that S_{E,W} (Y) ≠ {0}. Then L_{E,W} (Y) ≠ J_{E,W} (Y), so there exists a correspondence ϕ ∈ C(Y, E) and v ∈ W such that ϕ ⊗ v ∈ L_{E,W} (Y) − J_{E,W} (Y). By definition of J_{E,W}, this means that there exists a correspondence ψ ∈ C(E, Y) such that ψϕ · v ≠ 0. Since W is a module for the essential algebra E_{E} = kC(E, E)/I_{E}, we have ψϕ ∉ I_{E}. But ψϕ factorizes through Y, so we must have |Y | ≥ |E|. Thus E is a minimal set for S_{E,W}. The isomorphism S_{E,W} (E) ≅ W is a general fact mentioned before.

(c) This follows from (a) and (b).

Theorem 3.7 reduces the classification of simple correspondence functors to the question of classifying all simple modules for the essential algebra E_{E}. Fortunately, this has been achieved in [BT]. The simple E_{E}-modules are actually modules for a quotient P_{E} = E_{E}/N where N is a nilpotent ideal defined in [BT]. We call P_{E} the algebra of permuted orders, because it has a k-basis consisting of all relations on E of the form Δ_{σ}R, where σ runs through the symmetric group Σ_{E} of all permutations of E, and R is an order on E. By an order, we always mean a partial order relation. We let O be the set of all orders on E and Aut(E, R) the stabilizer of the order R in the symmetric group Σ_{E}. For the description of simple E_{E}-modules, we need the following new basis of P_{E} (see Theorem 6.2 in [BT] for details). 3.8. Lemma.

(a) There is a set {f_{R} | R ∈ O} of orthogonal idempotents whose sum is 1, such that P_{E} has a k-basis consisting of all elements of the form Δ_{σ}f_{R}, where σ ∈ Σ_{E} and R ∈ O.

(b) For any σ ∈ Σ_{E}, we have ^{σ}f_{R} = f_{^{σ}R}, where ^{σ}x = Δ_{σ}xΔ_{σ^{−1}} for any x ∈ P_{E}. In particular, Δ_{σ}f_{R}Δ_{σ^{−1}} = f_{R} if σ ∈ Aut(E, R).

(c) For any order Q on E, we have :

Qf_{R} ≠ 0 ⟺ Qf_{R} = f_{R} ⟺ Q ⊆ R .

For the description of simple E_{E}-modules and then simple correspondence functors, we will make use of the left E_{E}-module P_{E}f_{R}. This module is actually defined without assuming that k is a field. 3.9. Definition. Let E be a finite set and R an order on E. We call P_{E}f_{R} the fundamental module for the algebra E_{E}, associated with the poset (E, R).

We now describe its structure. 3.10. Proposition. Let E be a finite set and R an order on E.

(a) The fundamental module P_{E}f_{R} is a left module for the algebra P_{E}, hence also a left module for the essential algebra E_{E} and for the algebra of relations R_{E}.

(b) P_{E}f_{R} is a (P_{E}, k Aut(E, R))-bimodule and the right action of k Aut(E, R) is free.

(c) P_{E}f_{R} is a free k-module with a k-basis consisting of the elements Δ_{σ}f_{R}, where σ runs through the group Σ_{E} of all permutations of E.

(d) The action of the algebra of relations R_{E} on the module P_{E}f_{R} is given as follows. For any relation Q ∈ C(E, E),

Q · Δ_{σ}f_{R} = \{ \begin{matrix}Δ_{τσ}f_{R} & if ∃τ ∈ Σ_{E} such that Δ_{E} ⊆ Δ_{τ^{−1}}Q ⊆ ^{σ}R, \\ 0 & otherwise , \end{matrix}

where Δ_{E} is the diagonal of E × E, and ^{σ}R = {(σ(e), σ(f)) | (e, f) ∈ R} (recall that τ is unique in the first case).

Proof : See Corollary 7.3 and Proposition 8.5 in [BT].

The description of all simple E_{E}-modules is as follows (see Theorem 8.1 in [BT] for details).

3.11. Theorem. Assume that k is a field.

(a) Let R be an order on E and let P_{E}f_{R} be the corresponding fundamental module. If V is a simple k Aut(E, R)-module, then

T_{R,V} := P_{E}f_{R} ⊗_{k Aut(E,R)} V

is a simple P_{E}-module (hence also a simple E_{E}-module).

(b) Every simple E_{E}-module is isomorphic to a module T_{R,V} as in (a).

(c) For any permutation σ ∈ Σ_{E}, we have T_{^{σ}R,^{σ}V} ≅ T_{R,V}, where ^{σ}R = Δ_{σ}RΔ_{σ^{−1}} is the conjugate order and ^{σ}V is the conjugate module.

(d) The set of isomorphism classes of simple E_{E}-modules is parametrized by the set of conjugacy classes of pairs (R, V) where R is an order on E and V is a simple k Aut(E, R)-module.

Putting together Theorem 3.7 and Theorem 3.11, we finally obtain the following parametrization, which is essential for our purposes.

3.12. Theorem. Assume that k is a field. The set of isomorphism classes of simple correspondence functors is parametrized by the set of isomorphism classes of triples (E, R, V) where E is a finite set, R is an order on E, and V is a simple k Aut(E, R)-module.

3.13. Notation. Let E be a finite set and R an order on E.

(a) If V is a simple k Aut(E, R)-module, we denote by S_{E,R,V} the simple correspondence functor parametrized by the triple (E, R, V).

(b) More generally, for any commutative ring k and any k Aut(E, R)-module V, we define

S_{E,R,V} = (L_{E,P_{E}f_{R}}/J_{E,P_{E}f_{R}}) ⊗_{k Aut(E,R)} V ,

by which we mean that S_{E,R,V} (X) = (L_{E,P_{E}f_{R}}/J_{E,P_{E}f_{R}})(X)⊗_{k Aut(E,R)}V for any finite set X.

We end this section with a a basic result concerning the correspondence functors S_{E,R,V}, where k is any commutative ring and V is any k Aut(E, R)-module.

3.14. Lemma. Let E be a finite set, let R be an order on E, and let V be any k Aut(E, R)-module.

(a) E is a minimal set for S_{E,R,V}.
(b) S_{E,R,V} (E) ≅ P_{E}f_{R} ⊗_{k Aut(E,R)} V.

Proof : Both P_{E}f_{R} and P_{E}f_{R} ⊗_{k Aut(E,R)} V are left modules for the essential algebra E_{E}. Therefore, the argument given in part (b) of Theorem 3.7 shows again that E is a minimal set for S_{E,R,V}. Moreover, since J_{E,P_{E}f_{R}} vanishes on evaluation at E, we have

S_{E,R,V} (E) = L_{E,P_{E}f_{R}}(E) ⊗_{k Aut(E,R)} V = P_{E}f_{R} ⊗_{k Aut(E,R)} V ,

as required.

When k is a field and V is simple, we obtain S_{E,R,V} (E) = T_{R,V}, so we recover the module T_{R,V} of Theorem 3.11.

4. Some basic examples and results

In this section we present a few important examples of correspondence functors and prove some of their basic properties.

For any finite set E, the representable functor kC(−, E) (sometimes called Yoneda functor) is the very first example of a correspondence functor. By definition, it is actually isomorphic to the functor L_{E,kC(E,E)}. If W is a kC(E, E)-module generated by a single element w (for instance a simple module), then the functor L_{E,W} is isomorphic to a quotient of kC(−, E) via the surjective homomorphism

kC(−, E) ⟶ L_{E,W} = kC(−, E) ⊗_{kC(E,E)} W , ϕ ↦ ϕ ⊗ w .

The representable functor kC(−, E) is projective (by Yoneda’s lemma).

The next result is basic. It has several important corollaries which are often used.

4.1. Lemma. Let E and F be finite sets with |E| ≤ |F |. There exist correspondences i_{∗} ∈ C(F, E) and i^{∗} ∈ C(E, F) such that i^{∗}i_{∗} = id_{E}.

Proof : Since |E| ≤ |F |, there exists an injective map i : E ↪ F. Let i_{∗} ⊆ (F ×E) denote the correspondence

i_{∗} = {(i(e), e) | e ∈ E} ,

and i^{∗} ⊆ (E × F) denote the opposite correspondence

i^{∗} = {(e, i(e)) | e ∈ E} .

As i is injective, one checks easily that i^{∗}i_{∗} = Δ_{E}, that is, i^{∗}i_{∗} = id_{E}.

In other words, this lemma says that the object E of C is a direct summand of the object F whenever |E| ≤ |F |.

4.2. Corollary. Let E and F be finite sets with |E| ≤ |F |. The representable functor kC(−, E) is isomorphic to a direct summand of the representable functor kC(−, F).

Proof : Right multiplication by i^{∗} defines a homomorphism of correspondence functors

kC(−, E) ⟶ kC(−, F) ,

and right multiplication by i_{∗} defines a homomorphism of correspondence functors

kC(−, F) ⟶ kC(−, E) .

Their composite is the identity of kC(−, E), because i^{∗}i_{∗} = id_{E}.

4.3. Corollary. Let E and F be finite sets with |E| ≤ |F |. The left C(F, F)- module C(F, E) is projective.

Proof : By Corollary 4.2, kC(F, E) is isomorphic to a direct summand of kC(F, F), which is free.

4.4. Corollary. Let E and F be finite sets with |E| ≤ |F |. Let M be a correspondence functor. If M(F) = 0, then M(E) = 0.

Proof : For any m ∈ M(E), we have m = i^{∗}i_{∗}·m. But i_{∗}·m ∈ M(F), so i_{∗}·m = 0. Therefore m = 0.

4.5. Corollary. Let E and F be finite sets with |E| ≤ |F |. For every finite set X, composition in the category kC

μ : kC(X, F) ⊗_{R_{F}} kC(F, E) ⟶ kC(X, E)

is an isomorphism.

Proof : The inverse of μ is given by

ϕ : kC(X, E) ⟶ kC(X, F) ⊗_{R_{F}} kC(F, E) , ϕ(α) = αi^{∗} ⊗ i_{∗} .

Composing with μ, we obtain μϕ(α) = μ(αi^{∗} ⊗ i_{∗}) = αi^{∗}i_{∗} = α, so μϕ = id. On the other hand, if β ∈ kC(X, F) and γ ∈ kC(F, E), then γi^{∗} belongs to R_{F} and therefore

ϕμ(β ⊗ γ) = ϕ(βγ) = βγi^{∗} ⊗ i_{∗} = β ⊗ γi^{∗}i_{∗} = β ⊗ γ ,

showing that ϕμ = id.

Now we move to direct summands of representable functors, given by some idempotent. If R is an idempotent in kC(E, E), then kC(−, E)R is a direct summand of kC(−, E), hence projective again. In particular, if R is a preorder on E, that is, a relation which is reflexive and transitive, then R is idempotent (because R ⊆ R^{2} by reflexivity and R^{2} ⊆ R by transitivity). There is an equivalence relation ∼ associated with R, defined by

x ∼ y ⟺ (x, y) ∈ R and (y, x) ∈ R .

Then R induces an order relation \overline{R} on the quotient set \overline{E} = E/ ∼ such that

(x, y) ∈ R ⟺ (\overline{x}, \overline{y}) ∈ \overline{R} ,

where \overline{x} denotes the equivalence class of x under ∼.

4.6. Lemma. Let E be a finite set and let R be a preorder on E. Let \overline{R} be the corresponding order on the quotient set \overline{E} = E/ ∼ and write e ↦ \overline{e} for the quotient map E → \overline{E}. The correspondence functors kC(−, E)R and kC(−, \overline{E})\overline{R} are isomorphic via the isomorphism

kC(−, E)R ⟶ kC(−, \overline{E})\overline{R} , S ↦ \overline{S} ,

where for any correspondence S ⊆ X × E, the correspondence \overline{S} ⊆ X × \overline{E} is defined by

(x, \overline{e}) ∈ \overline{S} ⟺ (x, e) ∈ S .

Proof : It is straightforward to check that \overline{S} is well-defined. If S ∈ C(X, E)R, then S = SR and it follows that \overline{S} = \overline{S} \overline{R}. Surjectivity is easy and injectivity follows from the definition of \overline{S}.

This shows that it is relevant to consider the functors kC(−, E)R where R is an order on E. We will see later that they turn out to be connected with the simple functors parametrized by R^{op}.

We now introduce the fundamental functors 𝕊_{E,R} which will play a crucial role in the sequel.

4.7. Definition. If E is a finite set and R is an order on E, we denote by 𝕊_{E,R} the correspondence functor

𝕊_{E,R} = L_{E,P_{E}f_{R}}/J_{E,P_{E}f_{R}} ,

where P_{E}f_{R} is the fundamental module associated with the poset (E, R) (see Proposition 3.10). We call it the fundamental correspondence functor associated with the poset (E, R). In other words 𝕊_{E,R} ≅ S_{E,R,kAut(E,R)} .

By Lemma 3.14, E is a minimal set for 𝕊_{E,R} and 𝕊_{E,R}(E) = P_{E}f_{R}.

4.8. Remark. Since P_{E}f_{R} has a bimodule structure (see Proposition 3.10), we can tensor on the right with a k Aut(E, R)-module V. Then we recover the functor

S_{E,R,V} = 𝕊_{E,R} ⊗_{k Aut(E,R)} V .

This is because 𝕊_{E,R} is a quotient of the functor L_{E,P_{E}f_{R}} and there is a compatible right k Aut(E, R)-module structure on each of its evaluations

L_{E,P_{E}f_{R}}(X) = kC(X, E) ⊗_{kC(E,E)} P_{E}f_{R} .

Whenever k is a field and V is a simple k Aut(E, R)-module, then we recover in this way the simple functor S_{E,R,V}.

Evaluations of functors are not easy to control. One of our main tasks will be to show that the evaluations of the fundamental correspondence functors 𝕊_{E,R} are free k-modules and to find their rank. This will be completely achieved in Corollary 17.17. We start here by giving a first result in this direction. It holds over an arbitrary commutative ring k.

4.9. Lemma. Let (E, R) be a finite poset.

(a) There is a unique morphism of correspondence functors

ω_{E,R} : kC(−, E) ⟶ 𝕊_{E,R}

such that ω_{E,R}(Δ_{E}) = f_{R} ∈ P_{E}f_{R} = 𝕊_{E,R}(E).

(b) The morphism ω_{E,R} is surjective. For any finite set X, the kernel of the

map

ω_{E,R,X} : kC(X, E) ⟶ 𝕊_{E,R}(X)

consists of the set of linear combinations ∑_{S⊆X×E} λ_{S}S with coefficients λ_{S}

in k, such that

∀U ⊆ E × X, ∑_{Δ⊆US⊆R} λ_{S} = 0 ,

where Δ = Δ_{E} for simplicity.

(c) The morphism ω_{E,R} factorises as

\xymatrix{kC(−, E) \ar@{->>}[r] & kC(−, E)R \ar@{->>}[r]^{^{\overline{ω}^{E,R}}} & 𝕊_{E,R} ,}

where the morphism on the left hand side is right multiplication by R. For any finite set X, the kernel of \overline{ω}_{E,R,X} is equal to the set of linear combinations ∑_{S∈C(X,E)R} λ_{S}S such that

∀U ∈ RC(E, X), ∑_{US=R} λ_{S} = 0 .

(d) Both ω_{E,R} and \overline{ω}_{E,R} are homomorphisms of right k Aut(E, R)-modules.

Proof : (a) This is just an application of Yoneda’s lemma.

(b) The morphism ω_{E,R} is surjective, because f_{R} generates 𝕊_{E,R}. Let u = ∑_{S∈C(X,E)} λ_{S}S be an element of kC(X, E), where λ_{S} ∈ k. Then ω_{E,R,X}(u) = 0 if

and only if the image in 𝕊_{E,R}(X) of the element

∑_{S∈C(X,E)} λ_{S}S ⊗ f_{R} ∈ L_{E,P_{E}f_{R}}(X)

is equal to 0. In other words, this means that ∑_{S∈C(X,E)} λ_{S}S ⊗ f_{R} belongs to

J_{E,P_{E}f_{R}}(X). By definition, this is equivalent to saying that

(4.10)  ∀U ∈ C(E, X), ∑_{S∈C(X,E)} λ_{S}(US) · f_{R} = 0 in P_{E}f_{R} .

Now the action of correspondences on P_{E}f_{R} is described in Proposition 3.10. We obtain (US) · f_{R} = 0, unless there exists τ ∈ Σ_{E} such that

Δ ⊆ Δ_{τ^{−1}}US ⊆ R .

In this case τ is unique, and (US) · f_{R} = Δ_{τ} f_{R}. Thus Condition 4.10 is equivalent to

∀U ⊆ E × X, ∑_{τ∈Σ_{E}} ( ∑_{Δ⊆Δ_{τ^{−1}} US⊆R} λ_{S})Δ_{τ} f_{R} = 0 in P_{E}f_{R} .

Equivalently,

∀U ⊆ E × X, ∀τ ∈ Σ_{E}, ∑_{Δ⊆Δ_{τ^{−1}} US⊆R} λ_{S} = 0 .

Changing U to Δ_{τ} U, this amounts to requiring that

∀U ⊆ E × X, ∑_{Δ⊆US⊆R} λ_{S} = 0 .

(c) We know that Rf_{R} = f_{R}, by Lemma 3.8. Then clearly S−SR ∈ Ker ω_{E,R,X}, for any finite set X and any S ∈ C(X, E). Hence ω_{E,R} factorizes through kC(−, E)R. Replacing S by SR, we can assume that S = SR. Then the condition Δ ⊆ US ⊆ R yields R ⊆ USR ⊆ R^{2} ⊆ R, that is, US = R. The equation ∑_{US=R} λ_{S} = 0 for any

given U is the same as the equation for RU, so we can assume that U ∈ RC(E, X). The description of Ker \overline{ω}_{E,R,X} follows.

(d) The action of k Aut(E, R) on both kC(X, E) and 𝕊_{E,R}(X) is induced by right multiplication of correspondences S ↦ SΔ_{σ}, for every S ∈ C(X, E) and σ ∈ Aut(E, R). We clearly have

ω_{E,R,X}(SΔ_{σ}) = SΔ_{σ}f_{R} = Sf_{R}Δ_{σ} = ω_{E,R,X}(S)Δ_{σ} .

because Δ_{σ} commutes with f_{R} whenever σ ∈ Aut(E, R) (Lemma 3.8).

We end this section with the description of two small examples.

4.11. Example. Let E = ∅ be the empty set and consider the representable functor kC(−, ∅). Then kC(X, ∅) = {∅} is a singleton for any finite set X, so kC(X, ∅) ≅ k. Moreover, any correspondence S ∈ C(Y, X) is sent to the identity map from kC(X, ∅) ≅ k to kC(Y, ∅) ≅ k. This functor deserves to be called the constant functor. We will denote it by \underline{k}.

Assume that k is a field. The algebra kC(∅, ∅) ≅ k has a unique simple module k. It is then easy to check that L_{∅,k} = kC(−, ∅) and J_{∅,k} = {0}. Therefore

kC(−, ∅) = S_{∅,∅,k}

the simple functor indexed by (∅, ∅, k). Here the second ∅ denotes the only relation on the empty set, while k is the only simple module for the group algebra k Aut(E, R) = kΣ_{∅} ≅ k, where Σ_{∅} = {id} is the symmetric group of the empty set. Note that S_{∅,∅,k} is projective, because it is a representable functor.

4.12. Example. Let E = {1} be a set with one element and consider the representable functor C(−, {1}). Then C(X, {1}) is in bijection with the set B(X) of all subsets of X, because X × {1} ≅ X. It is easy to see that a correspondence S ∈ C(Y, X) is sent to the map

B(X) ⟶ B(Y) , A ↦ S_{A} ,

where S_{A} = {y ∈ Y | ∃ a ∈ A such that (y, a) ∈ S}. Thus kB ≅ kC(−, {1}) is a correspondence functor such that kB(X) is a free k-module with basis B(X) and rank 2^{|X|} for every finite set X.

The functor kB has a subfunctor isomorphic to the constant functor S_{∅,∅,k}, because B(X) contains the element ∅ which is mapped to ∅ by any correspondence. We claim that, if k is a field, the quotient kB/S_{∅,∅,k} is a simple functor.

Assume that k is a field. The algebra kC({1}, {1}) has dimension 2, actually isomorphic to k×k with two primitive idempotents ∅ and {(1, 1)}−∅. The essential algebra E_{{1}} is a one-dimensional quotient and its unique simple module W is onedimensional and corresponds to the pair (R, k), where R is the only order relation on {1} and k is the only simple module for the group algebra k Aut(E, R) = kΣ_{{1}} ≅ k, with Σ_{{1}} = {id} the symmetric group of {1}. Thus there is a simple functor S_{{1},W} = S_{{1},R,k}.

The kernel of the quotient map

kB ≅ kC(−, {1}) ⟶ kC(−, {1}) ⊗_{kC({1},{1})} W = L_{{1},W}

is the constant subfunctor S_{∅,∅,k} mentioned above, because ∅ ∈ C(X, {1}) can be written ∅ · ∅, with the second empty set belonging to C({1}, {1}), thus acting by zero on W. Now we know that L_{{1},W} /J_{{1},W} = S_{{1},W} and we are going to show that J_{{1},W} = {0}. It then follows that L_{{1},W} = S_{{1},W} is simple and isomorphic to kB/S_{∅,∅,k}, proving the claim above.

In order to prove that J_{{1},W} = {0}, we let u ∈ L_{{1},W} (X), which can be written

u = ∑_{A∈B(X)} λ_{A}(A × {1}) ⊗ w ,

where w is a generator of W and λ_{A} ∈ k for all A. Since the empty set acts by zero on w, the sum actually runs over nonempty subsets A ∈ B(A). Then u ∈ J_{{1},W} (X) if and only if, for all ({1} × B) ∈ C({1}, X), we have

∑_{A∈B(X)} λ_{A} ({1} × B)(A × {1}) · w = 0 .

Since ∅ acts by zero and {1} acts as the identity, we obtain

(\{1\} × B)(A × \{1\}) · w = \{ \begin{matrix}0 & if B ∩ A = ∅ , \\ w & if B ∩ A ≠ ∅ . \end{matrix}

This yields the condition

∑_{A∩B≠∅} λ_{A} = 0 , for every nonempty B ∈ B(X) .

We prove by induction that λ_{C} = 0 for every nonempty C ∈ B(X). Subtracting the condition for B = X and for B = X − C, we obtain

0 = ∑_{A≠∅} λ_{A} − ∑_{A⊈C} λ_{A} = ∑_{∅≠A⊆C} λ_{A} .

If C = {c} is a singleton, we obtain λ_{{c}} = 0 and this starts the induction. In the general case, we obtain by induction λ_{A} = 0 for ∅ ≠ A ≠ C, so we are left with λ_{C} = 0. Therefore u = 0 and so J_{{1},W} = {0}.

There is a special feature of this small example, namely that the exact sequence

0 ⟶ S_{∅,∅,k} ⟶ kB ⟶ S_{{1},R,k} ⟶ 0

splits. This is because there is a retraction kB → S_{∅,∅,k} defined by

kB(X) ⟶ k , A ↦ 1 ,

which is easily checked to be a homomorphism of functors. Since kB is projective (because it is a representable functor), its direct summand S_{{1},R,k} is projective.

4.13. Remark. In both Example 4.11 and Example 4.12, there is a unique order relation R on E, which is a total order. Actually, these examples are special cases of the general situation of a total order, which is studied in Section 15.

5. Posets and lattices

Finite posets and lattices play an essential role in this paper. Indeed we already know that pairs (E, R), where R is an order relation on E (that is, a finite poset), appear in the parametrization of simple correspondence functors (Theorem 3.12) and in the definition of the fundamental functor 𝕊_{E,R} (Definition 4.7). Moreover, lattices will play a central role in Part 2 (see Section 11).

This section is an interlude, in which we give some definitions, fix some notation, and prove some basic lemmas, which will be used throughout.

5.1. Notation and definitions. Let (E, R) be a finite poset (or more generally a preorder).

(a) We write ≤_{R} for the order relation, so that (a, b) ∈ R if and only if a ≤_{R} b. Moreover a <_{R} b means that a ≤_{R} b and a ≠ b.

(b) If a, b ∈ E with a ≤_{R} b, we define intervals

\begin{matrix}[a, b]_{E} = \{x ∈ E | a ≤_{R} x ≤_{R} b\} , & ]a, b[_{E}= \{x ∈ E | a <_{R} x <_{R} b\} , \\ [a, b[_{E} = \{x ∈ E | a ≤_{R} x <_{R} b\} , & ]a, b]_{E} = \{x ∈ E | A <_{R} x ≤_{R} b\} , \\ [a, ·[_{E} = \{x ∈ E | a ≤_{R} x\} , & ]·, b]_{E} = \{x ∈ E | x ≤_{R} b\} . \end{matrix}

When the context is clear, we write [a, b] instead of [a, b]_{E}.

(c) A subset A of E is a lower R-ideal, or simply a lower ideal, if, whenever a ∈ A and x ≤_{R} a, we have x ∈ A. Similarly, a subset A of E is an upper R-ideal, or simply an upper ideal, if, whenever a ∈ A and a ≤_{R} x, we have x ∈ A.

(d) A principal lower ideal, or simply principal ideal, is a subset of the form ]·, a]_{E}, where a ∈ E. A principal upper ideal is defined similarly.

5.2. Notation and definitions. Let T be a finite lattice.

(a) We write ≤_{T}, or sometimes simply ≤, for the order relation, ∨ for the join (least upper bound), ∧ for the meet (greatest lower bound), ˆ0 for the least element and ˆ1 for the greatest element.

(b) An element e ∈ T is called join-irreducible, or simply irreducible, if, whenever e = ⋁_{a∈A} a for some subset A of T, then e ∈ A. In case A = ∅, the

join is ˆ0 and it follows that ˆ0 is not irreducible. If e ≠ ˆ0 is irreducible and e = s ∨ t with s, t ∈ T, then either e = s or e = t. In other words, if e ≠ ˆ0, then e is irreducible if and only if [ˆ0, e[ has a unique maximal element.

(c) Let E be a subposet of T. We say that E is a full subposet of T if for all e, f ∈ E we have :

e ≤_{R} f ⟺ e ≤_{T} f .

(d) We write Irr(T) for the set of irreducible elements in T, viewed as a full subposet of T.

5.3. Notation. Let (E, R) be a finite poset (or more generally a preorder).

(a) Let I_{↓}(E, R) denote the set of lower R-ideals of E. Then I_{↓}(E, R), ordered by inclusion of subsets, is a lattice : the join operation is union of subsets, and the meet operation is intersection.

(b) Similarly, I^{↑}(E, R) denotes the set of upper R-ideals of E, which is also a lattice. If R^{op} is the relation opposite to R, then clearly I^{↑}(E, R) = I_{↓}(E, R^{op}).

5.4. Lemma. Let R be a preorder on a finite set E. As in Lemma 4.6, let \overline{R} be the corresponding order on the quotient set \overline{E} = E/ ∼. The quotient map E → \overline{E} induces an isomorphism of lattices I_{↓}(E, R) ≅ I_{↓}(\overline{E}, \overline{R}).

Proof : The proof is straightforward and is left to the reader.

Lemma 5.4, as well as Lemma 4.6, imply that it is enough to consider orders rather than preorders. In the rest of this paper, we shall work with orders, without loss of generality.

5.5. Lemma. Let (E, R) be a finite poset.

(a) The irreducible elements in the lattice I_{↓}(E, R) are the principal ideals ]·, e]_{E}, where e ∈ E. Thus the poset E is isomorphic to the poset of all irreducible elements in I_{↓}(E, R) by mapping e ∈ E to the principal ideal ]·, e]_{E}.

(b) I_{↓}(E, R) is a distributive lattice.

(c) If T is a distributive lattice and (E, R) is its subposet of irreducible elements, then T is isomorphic to I_{↓}(E, R).

Proof : This is not difficult and well-known. For details, see Theorem 3.4.1 and Proposition 3.4.2 in [St].

5.6. Convention. In the situation of Lemma 5.5, we shall identify E with its image via the map

E ⟶ I_{↓}(E, R) , e ↦ ]·, e]_{E} .

Thus we view E as a full subposet of I_{↓}(E, R). This abusive convention is a conceptual simplification and has many advantages for the rest of this paper.

Lattices having a given poset (E, R) as poset of irreducible elements will be described in an appendix (Section 21). They are all quotients of I_{↓}(E, R), and I_{↓}(E, R) is the only one which is distributive, up to isomorphism (by part c) of Lemma 5.5).

Note that if (E, R) is the poset of irreducible elements in a finite lattice T, then T is generated by E in the sense that any element x ∈ T is a join of elements of E. To see this, define the height of t ∈ T to be the maximal length of a chain in [ˆ0, t]_{T}. If x is not irreducible and x ≠ ˆ0, then x = t_{1} ∨ t_{2} with t_{1} and t_{2} of smaller height than x. By induction on the height, both t_{1} and t_{2} are joins of elements of E. Therefore x = t_{1} ∨ t_{2} is also a join of elements of E.

Given a poset (E, R), the map

E ⟶ I^{↑}(E, R) , e ↦ [e, ·[_{E}

is order-reversing, so it is in fact (E, R^{op}) which is identified with the poset of irreducible elements in I^{↑}(E, R). Since I^{↑}(E, R) = I_{↓}(E, R^{op}), this is actually just Convention 5.6 applied to R^{op}.

We now introduce a notation which will play an important role in the rest of the paper.

5.7. Notation. Let T be a finite lattice and let (E, R) be the full subposet of its irreducible elements. For any finite set X and any map φ : X → T, we associate the correspondence

Γ_{φ} = {(x, e) ∈ X × E | e ≤_{T} φ(x)} ⊆ X × E .

In the special case where T = I_{↓}(E, R) and in view of Convention 5.6, we obtain

Γ_{φ} = {(x, e) ∈ X × E | e ∈ φ(x)} .

5.8. Lemma. Let T be a finite lattice and let (E, R) be the full subposet of its irreducible elements.

(a) For any map φ : X → T, we have Γ_{φ}R^{op} = Γ_{φ}.

(b) If T = I_{↓}(E, R), then a correspondence S ⊆ X × E has the form S = Γ_{φ}

for some map φ : X → I_{↓}(E, R) if and only if SR^{op} = S.

(c) If T = I^{↑}(E, R), then a correspondence S ⊆ X × E has the form S = Γ_{φ}

for some map φ : X → I^{↑}(E, R) if and only if SR = S.

Proof : (a) Since Δ_{E} ⊆ R^{op}, we always have Γ_{φ} = Γ_{φ}Δ_{E} ⊆ Γ_{φ}R^{op}. Conversely, if (x, f) ∈ Γ_{φ}R^{op}, then there exists e ∈ E such that (x, e) ∈ Γ_{φ} and (e, f) ∈ R^{op}, that is, e ≤_{T} φ(x) and f ≤_{R} e. But f ≤_{R} e if and only if f ≤_{T} e (because (E, R) is a full subposet of T). It follows that f ≤_{T} φ(x), that is, (x, f) ∈ Γ_{φ}. Thus Γ_{φ}R^{op} ⊆ Γ_{φ} and equality follows.

(b) This follows from (c) applied to R^{op}, because of the equality I_{↓}(E, R) = I^{↑}(E, R^{op}).

(c) One direction follows from (a). For the other direction, let S ∈ C(X, E) such that SR = S, or equivalently S ∈ C(X, E)R. Then the set

ψ(x) = {e ∈ E | (x, e) ∈ S}

is an upper R-ideal in E, thus ψ is a function X → I^{↑}(E, R). Clearly Γ_{ψ} = S.

6. Finite generation

In this section, we analyze the property of finite generation for correspondence functors.

6.1. Definition. Let {E_{i} | i ∈ I} be a family of finite sets indexed by a set I and, for every i ∈ I, let m_{i} ∈ M(E_{i}). A correspondence functor M is said to be generated by the set {m_{i} | i ∈ I} if for every finite set X and every element m ∈ M(X), there exists a finite subset J ⊆ I such that

m = ∑_{j∈J} α_{j}m_{j} , for some α_{j} ∈ kC(X, E_{j}) (j ∈ J) .

In the case where I is finite, then M is said to be finitely generated.

We remark that, in the sum above, each α_{j} decomposes as a finite k-linear combination α_{j} = ∑_{S∈C(X,E_{j})} λ_{S} S where λ_{S} ∈ k. Therefore, every m ∈ M(X) decomposes further as a (finite) k-linear combination

m = ∑_{{j∈J \atop S∈C(X,E_{j})}} λ_{S} Sm_{j} .

6.2. Example. If E is a finite set, the representable functor kC(−, E) is finitely generated. It is actually generated by a single element, namely Δ_{E} ∈ kC(E, E).

6.3. Lemma. Let M be a finitely generated correspondence functor over k. Then, for every finite set X, the evaluation M(X) is a finitely generated k-module. In particular, if k is a field, then M(X) is finite-dimensional.

Proof : Let {m_{i} | i = 1, …, n} be a finite set of generators of M, with m_{i} ∈ M(E_{i}). Let B_{X} = {Sm_{i} | S ∈ C(X, E_{i}), i = 1, …, n}. By definition and by the remark above, every element of M(X) is a k-linear combination of elements of B_{X}. But B_{X} is a finite set, so M(X) is finitely generated. If k is a field, this means that M(X) is finite-dimensional.

It follows that, in order to understand finitely generated correspondence functors, we could assume that all their evaluations are finitely generated k-modules. But we do not need this for our next characterizations.

6.4. Proposition. Let M be a correspondence functor over k. The following conditions are equivalent :

(a) M is finitely generated.

(b) M is isomorphic to a quotient of a functor of the form ⨁_{i=1}^{n} kC(−, E_{i}) for

some finite sets E_{i} (i = 1, …, n).

(c) M is isomorphic to a quotient of a functor of the form ⨁_{i∈I} kC(−, E) for

some finite set E and some finite index set I.

(d) There exists a finite set E and a finite subset B of M(E) such that M is generated by B.

Proof : (a) ⇒ (b). Suppose that M is generated by the set {m_{i} | i = 1, …, n}, where m_{i} ∈ M(E_{i}). It follows from Yoneda’s lemma that there is a morphism

ψ_{i} : kC(−, E_{i}) → M

mapping Δ_{E_{i}} ∈ C(E_{i}, E_{i}) to the element m_{i} ∈ M(E_{i}), hence mapping β ∈ C(X, E_{i}) to βm_{i} ∈ M(X). Their sum yields a morphism

ψ : ⨁_{i=1}^{n} kC(−, E_{i}) → M .

For any X and any m ∈ M(X), we have m = ∑_{i=1}^{n} α_{i}m_{i} for some α_{i} ∈ kC(X, E_{i}),

hence m = ψ(α_{1}, …, α_{n}), proving the surjectivity of ψ.

(b) ⇒ (c). Suppose that M is isomorphic to a quotient of a functor of the form

⨁_{i=1}^{n} kC(−, E_{i}). Let F be the largest of the sets E_{i}. By Corollary 4.2, each kC(−, E_{i})

is a direct summand of kC(−, F). Therefore, M is also isomorphic to a quotient of

the functor ⨁_{i=1}^{n} kC(−, F).

(c) ⇒ (d). By Example 6.2, kC(−, E) is generated by Δ_{E} ∈ kC(E, E). Let b_{i} ∈ ⨁_{i∈I} kC(E, E) having zero components everywhere, except the i-th equal to Δ_{E}.

Since M is a quotient of ⨁_{i∈I} kC(−, E), it is generated by the images of the ele-

ments b_{i}. This is a finite set because I is finite by assumption.

(d) ⇒ (a). Since M is generated by B, it is finitely generated.

We apply this to the functors L_{E,V} and S_{E,V} defined in Lemma 2.3 and Notation 2.5.

6.5. Corollary. Let V be a finitely generated R_{E}-module, where E is a finite set and R_{E} is the algebra of relations on E. Then L_{E,V} and S_{E,V} are finitely generated correspondence functors.

Proof : Let {v_{i} | i ∈ I} be a finite set of generators of V as an R_{E}-module. There is a morphism π_{i} : kC(−, E) → L_{E,V} mapping Δ_{E} to v_{i} ∈ L_{E,V} (E) = V. Therefore, we obtain a surjective morphism

∑_{i∈I} π_{i} : ⨁_{i∈I} kC(−, E) ⟶ L_{E,V} ,

showing that L_{E,V} is finitely generated. Now S_{E,V} is a quotient of L_{E,V}, so it is also finitely generated.

6.6. Proposition. Let k be a noetherian ring.

(a) For any finitely generated correspondence functor M over k, the algebra End_{F_{k}}(M) is a finitely generated k-module.

(b) For any two finitely generated correspondence functors M and N, the kmodule Hom_{F_{k}}(M, N) is finitely generated.

(c) If k is a field, the Krull-Remak-Schmidt theorem holds for finitely generated correspondence functors over k.

Proof : (a) Since M is finitely generated, there exists a finite set E and a surjective morphism π : ⨁_{i∈I} kC(−, E) → M for some finite set I (Proposition 6.4).

Denote by A the subalgebra of End_{F_{k}} ( ⨁_{i∈I} kC(−, E)) consisting of endomorphisms

φ such that φ(Ker π) ⊆ Ker π. The algebra A is isomorphic to a k-submodule of End_{F_{k}} ( ⨁_{i∈I} kC(−, E)), which is isomorphic to a matrix algebra of size |I| over the

k-algebra kC(E, E) (because End_{F_{k}} (kC(−, E)) ≅ kC(E, E) by Yoneda’s lemma). This matrix algebra is free of finite rank as a k-module. As k is noetherian, it follows that A is a finitely generated k-module.

Now by definition, any φ ∈ A induces an endomorphism \overline{φ} of M such that \overline{φ}π = πφ. This yields an algebra homomorphism A → End_{F_{k}}(M), which is surjective, since the functor kC(−, E) is projective. It follows that End_{F_{k}}(M) is also a finitely generated k-module.

(b) The functor M ⊕ N is finitely generated, hence V = End_{F_{k}}(M ⊕ N) is a finitely generated k-module, by (a). Since Hom_{F_{k}}(M, N) embeds in V, it is also a finitely generated k-module.

(c) If moreover k is a field, then End_{F_{k}}(M) is a finite dimensional k-vector space, by (a). Finding decompositions of M as a direct sum of subfunctors amounts to splitting the identity of End_{F_{k}}(M) as a sum of orthogonal idempotents. Since End_{F_{k}}(M) is a finite dimensional algebra over the field k, the standard theorems on decomposition of the identity as a sum of primitive idempotents apply. Thus M can be split as a direct sum of indecomposable functors, and such a decomposition is unique up to isomorphism.

After the Krull-Remak-Schmidt theorem, we treat the case of projective covers. Recall (see 2.5.14 in [AF] for categories of modules) that in an abelian category A, a subobject N of an object P is called superfluous if for any subobject X of P, the equality X + N = P implies X = P. Similarly, an epimorphism f : P → M in A is called superfluous if Ker f is superfluous in P, or equivalently, if for any morphism g : L → P in A, the composition f ∘ g is an epimorphism if and only if g is an epimorphism. A projective cover of an object M of A is defined as a pair (P, p), where P is projective and p is a superfluous epimorphism from P to M.

6.7. Proposition. Let M be a finitely generated correspondence functor over a commutative ring k.

(a) Suppose that M is generated by M(E) where E is a finite. If (P, p) is a projective cover of M(E) in kC(E, E)-Mod, then (L_{E,P} , p̃) is a projective cover of M in F_{k}, where p̃: L_{E,P} → M is obtained from p : P → M(E) by the adjunction of Lemma 2.3.

(b) If k is a field, then M admits a projective cover.

(c) In particular, when k is a field, let E be a finite set, let R be an order relation on E, and let V be a simple k Aut(E, R)-module. Let moreover (P, p) be a projective cover of Pf_{R} ⊗_{k Aut(E,R)} V in kC(E, E)-Mod. Then (L_{E,P} , p̃) is a projective cover of the simple correspondence functor S_{E,R,V}.

Proof : (a) (This was already proved in Lemme 2 of [Bo1].) By Lemma 2.3, the functor Q ↦ L_{E,Q} maps projectives to projectives. So the functor L_{E,P} is projective. Since M is generated by M(E), and since the evaluation at E of the morphism p̃: L_{E,P} → M is equal to p : P → M(E), it follows that p̃ is surjective. If N is any subfunctor of L_{E,P} such that p̃(N) = M, then in particular N(E) ⊆ P and p(N(E)) = M(E). Since p is superfluous, it follows that N(E) = P, hence N = L_{E,P} since L_{E,P} is generated by its evaluation P at E.

(b) The algebra kC(E, E) is a finite dimensional algebra over the field k. Hence any finite dimensional kC(E, E)-module admits a projective cover. Therefore (b) follows from (a).

(c) The evaluation of the simple functor S_{E,R,V} at E is the simple kC(E, E)- module Pf_{R} ⊗_{k Aut(E,R)} V. Hence (c) follows from (a) and (b).

7. Bounded type

In this section, we analyze a notion which is more general than finite generation.

7.1. Definition. Let k be a commutative ring and let M be a correspondence functor over k.

(a) We say that M has bounded type if there exists a finite set E such that M is generated by M(E).

(b) We say that M has a bounded presentation if there are projective correspondence functors P and Q of bounded type and an exact sequence of functors

Q → P → M → 0 .

Such a sequence is called a bounded presentation of M.

Suppose that M has bounded type and let E be a finite set such that M is generated by M(E). It is elementary to see that M is finitely generated if and only if M(E) is a finitely generated R_{E}-module (using Example 6.2 and Lemma 6.3). Thus an infinite direct sum of copies of a simple functor S_{E,R,V} has bounded type (because it generated by its evaluation at E) but is not finitely generated. Also, a typical example of a correspondence functor which does not have bounded type is

a direct sum of simple functors ⨁_{n=0}^{∞} S_{E_{n},R_{n},V_{n}}, where |E_{n}| = n for each n. This is

because S_{E_{n},R_{n},V_{n}} cannot be generated by a set of cardinality < n.

7.2. Lemma. Let k be a commutative ring and let M be a correspondence functor over k. Suppose that M has bounded type and let E be a finite set such that M is generated by M(E). For any finite set F with |F | ≥ |E|, the functor M is generated by M(F).

Proof : Let i_{∗} ∈ C(F, E) and i^{∗} ∈ C(E, F) be as in Lemma 4.1, so that i^{∗}i_{∗} = id_{E}. Saying that M is generated by M(E) amounts to saying that M(X) is equal to kC(X, E)M(E), for any finite set X. It follows that

M(X) = kC(X, E)M(E)
= kC(X, E)i^{∗}i_{∗}M(E)
⊆ kC(X, F)i_{∗}M(E)
⊆ kC(X, F)M(F) ⊆ M(X) ,

hence M(X) = kC(X, F)M(F), i.e. M is generated by M(F).

We are going to prove that any correspondence functor having a bounded presentation is isomorphic to some functor L_{E,V}. We first deal with the case of projective functors.

7.3. Lemma. Suppose that a correspondence functor M has bounded type and let E be a finite set such that M is generated by M(E). If M is projective, then for any finite set F with |F | ≥ |E|, the R_{F} -module M(F) is projective, and the counit morphism L_{F,M(F)} → M is an isomorphism.

Proof : By Lemma 7.2, M is generated by M(F). Choosing a set B of generators of M(F) as an R_{F} -module (e.g. B = M(F)), we see that M is also generated by B. As in the beginning of the proof of Proposition 6.4, we can apply Yoneda’s lemma and obtain a surjective morphism ⨁_{b∈B} kC(−, F) → M. Since M is projective, this

morphism splits, and its evaluation at F also splits as a map of R_{F} -modules. Hence M(F) is isomorphic to a direct summand of a free R_{F} -module, that is, a projective R_{F} -module.

By adjunction (Lemma 2.3), there is a morphism θ : L_{F,M(F)} → M which, evaluated at F, gives the identity map of M(F). As M is generated by M(F), it follows that θ is surjective, hence split since M is projective. Let η : M → L_{F,M(F)} be a section of θ. Since, on evaluation at F, we have θ_{F} = id_{M(F)}, the equation θη = id_{M} implies that, on evaluation at F, we get η_{F} = id_{M(F)}. Therefore η_{F} θ_{F} = id_{M(F)}. Now ηθ : L_{F,M(F)} → L_{F,M(F)} corresponds by adjunction to

id_{M(F)} = η_{F} θ_{F} : M(F) ⟶ L_{F,M(F)}(F) = M(F) .

Therefore ηθ must be the identity. It follows that η and θ are mutual inverses. Thus M ≅ L_{F,M(F)}.

We now prove that any functor with a bounded presentation is an L_{E,V}, and conversely. In the case of a noetherian base ring k, this result will be improved in Section 10.

7.4. Theorem.

(a) Suppose that a correspondence functor M has a bounded presentation

Q → P → M → 0 .

Let E be a finite set such that P is generated by P (E) and Q is generated by Q(E). Then for any finite set F with |F | ≥ |E|, the counit morphism

η_{M,F} : L_{F,M(F)} → M

is an isomorphism.

(b) If E is a finite set and V is an R_{E}-module, then the functor L_{E,V} has a bounded presentation. More precisely, if

W_{1} → W_{0} → V → 0

is a projective resolution of V as an R_{E}-module, then

L_{E,W_{1}} → L_{E,W_{0}} → L_{E,V} → 0

is a bounded presentation of L_{E,V}.

Proof : (a) Consider the commutative diagram

\xymatrix{L_{F,Q(F)} \ar[d]^{η_{Q,F}} \ar[r] & L_{F,P (F)} \ar[d]^{η_{P,F}} \ar[r] & L_{F,M(F)} \ar[d]^{η_{M,F}} \ar[r] & 0 \\ Q \ar[r] & P \ar[r] & M \ar[r] & 0}

where the vertical maps are obtained by the adjunction of Lemma 2.3. This lemma also asserts that the first row is exact. By Lemma 7.3, for any finite set F with |F | ≥ |E|, the vertical morphisms η_{Q,F} and η_{P,F} are isomorphisms. Since the rows of this diagram are exact, it follows that η_{M,F} is also an isomorphism.

(b) We use the adjunction of Lemma 2.3. Applying the right exact functor U ↦ L_{E,U} to the exact sequence W_{1} → W_{0} → V → 0 gives the exact sequence

L_{E,W_{1}} → L_{E,W_{0}} → L_{E,V} → 0 .

By Lemma 2.3, L_{E,W_{1}} and L_{E,W_{0}} are projective functors, since W_{1} and W_{0} are projective R_{E}-modules. They all have bounded type since they are generated by their evaluation at E.

Given a finite set E and an R_{E}-module V, we define an induction procedure as follows. For any finite set F, we define the R_{F} -module

V ↑_{E}^{F} := kC(F, E) ⊗_{R_{E}} V .

Notice that, by the definition of L_{E,V}, we have L_{E,V} (F) = V ↑_{E}^{F}. To end this section, we mention the behavior of the functors L_{E,V} under induction.

7.5. Proposition. Let E be a finite set and V be an R_{E}-module. If F is a finite set with |F | ≥ |E|, the equality L_{E,V} (F) = V ↑_{E}^{F} induces an isomorphism of correspondence functors

L_{F,V↑_{E}^{F}} ≅ L_{E,V} .

Proof : Let M = L_{E,V}. Then by Theorem 7.4, there exists a bounded presentation

Q → P → M → 0

where Q = L_{E,W_{1}} is generated by Q(E) and P = L_{E,W_{0}} is generated by P (E). Hence by Theorem 7.4, for any finite set F with |F | ≥ |E|, the counit morphism

η_{M,F} : L_{F,M(F)} → M

is an isomorphism. In other words η_{M,F} : L_{F,V↑_{E}^{F}} → L_{E,V} is an isomorphism.

8. Exponential behavior and finite length

One of our purposes in this section is to show that, if our base ring k is a field, then every finitely generated correspondence functor has finite length. This is based on a lower bound estimate for the dimension of the evaluations of a simple functor S_{E,R,V}, which is proved to behave exponentially. We also prove that the exponential behavior is equivalent to finite generation.

We first need a well-known combinatorial lemma.

8.1. Lemma. Let E be a finite set and let G be a finite set containing E.

(a) For any finite set X, the number s(X, E) of surjective maps φ : X → E is equal to

s(X, E) = ∑_{i=0}^{|E|}(−1)^{|E|−i}\binom{|E|}{i} i^{|X|} ,

or equivalently

s(X, E) = ∑_{j=0}^{|E|}(−1)^{j}\binom{|E|}{j} (|E| − j)^{|X|} .

(b) More generally, for any finite set X, the number ss(X, E, G) of all maps φ : X → G such that E ⊆ φ(X) ⊆ G is equal to

ss(X, E, G) = ∑_{i=0}^{|E|}(−1)^{i}\binom{|E|}{i}(|G| − i)^{|X|} .

Proof : (a) Up to multiplication by |E|!, the number s(X, E) is known as a Stirling number of the second kind. Either by Formula (24a) in Section 1.4 of [St], or by a direct application of Möbius inversion (i.e. inclusion-exclusion principle in the present case), we have

s(X, E) = ∑_{B⊆E}(−1)^{|E−B|}|B|^{|X|} .

Setting |B| = i, the first formula in (a) follows.

(b) Applying (a) to each subset A such that E ⊆ A ⊆ G, we obtain

ss(X, E, G) = ∑_{E⊆A⊆G} s(X, A)
= ∑_{E⊆A⊆G} ∑_{B⊆A}(−1)^{|A−B|}|B|^{|X|}
= ∑_{B⊆G} ( ∑_{E∪B⊆A⊆G}(−1)^{|A−B|})|B|^{|X|}

But the inner sum is zero unless E ∪ B = G. Therefore

ss(X, E, G) = ∑_{{B⊆G \atop E∪B=G}}(−1)^{|G−B|}|B|^{|X|}

= ∑_{C⊆E}(−1)^{|C|}(|G − C|)^{|X|} ,

where the last equality follows by setting C = G − B. This proves part (b).

Now we prove our main lower bound estimate for the dimensions of the evaluations of a simple functor.

8.2. Theorem. Suppose that k is a field and let S_{E,R,V} be a simple correspondence functor, where E is a finite set, R is an order on E, and V is a simple k Aut(E, R)- module. There exists a positive integer N and a positive real number c such that, for any finite set X of cardinality at least N, we have

c|E|^{|X|} ≤ dim (S_{E,R,V} (X)) ≤ (2^{|E|})^{|X|} .

Proof : Let X be a finite set. We first prove the upper bound. This is easy and holds for all X. Since S_{E,R,V} is a quotient of the representable functor kC(−, E), we have

dim (S_{E,R,V} (X)) ≤ dim (kC(X, E)) = 2^{|X×E|} = (2^{|E|})^{|X|} .

Now we want to find first a lower bound for the dimension of the evaluation of the fundamental correspondence functor 𝕊_{E,R}. By Lemma 4.9,

𝕊_{E,R}(X) ≅ kC(X, E)R/ Ker(\overline{ω}_{E,R}) ,

where \overline{ω}_{E,R} : kC(−, E)R ⟶ 𝕊_{E,R} is the natural morphism. We consider the set

A = {Γ_{φ} | φ ∈ Φ} ⊆ C(X, E)R ,

using Notation 5.7, where Φ is the set of all maps φ : X → I^{↑}(E, R) such that Im(φ) = E. We use here Convention 5.6 which identifies (E, R^{op}) with the poset of irreducible elements of I^{↑}(E, R), so that I^{↑}(E, R) contains E as a subset. Note that A is a subset of C(X, E)R because Γ_{φ} belongs to C(X, E)R by Lemma 5.8.

We want to prove that the image of A in kC(X, E)R/ Ker(\overline{ω}_{E,R}) is linearly independent, from which we will deduce that |A| ≤ dim (𝕊_{E,R}(X)). Suppose that ∑_{φ∈Φ} λ_{φ}Γ_{φ} is zero in 𝕊_{E,R}(X), where λ_{φ} ∈ k for every φ ∈ Φ. In other words,

∑_{φ∈Φ} λ_{φ}Γ_{φ} is in the kernel of \overline{ω}_{E,R} and, by Lemma 4.9, we obtain

∀U ∈ RC(E, X), ∑_{UΓ_{φ}=R} λ_{φ} = 0 .

Consider the set Ψ of all maps ψ : X → I_{↓}(E, R) such that Im(ψ) = E and choose U = Γ_{ψ}^{op} ∈ C(E, X). Since Γ_{ψ} = Γ_{ψ}R^{op} by Lemma 5.8, we get Γ_{ψ}^{op} = RΓ_{ψ}^{op}, that is, Γ_{ψ}^{op} ∈ RC(E, X). Therefore we have

∀ψ ∈ Ψ, ∑_{Γ_{ψ}^{op}Γ_{φ}=R} λ_{φ} = 0 .

Define the matrix M indexed by Ψ × Φ by

M_{ψ,φ} = \{ \begin{matrix}1 & if Γ_{ψ}^{op}Γ_{φ} = R , \\ 0 & otherwise . \end{matrix}

Note that M is a square matrix because both Ψ and Φ are in bijection with the set of all surjective maps X → E. The previous relations yield Mλ = 0, where λ is the column vector with entries λ_{φ}, for φ ∈ Φ. We are going to prove that M is invertible.

Suppose M_{ψ,φ} = 1. Then, for any given x ∈ X, we can choose e = ψ(x) and f = φ(x) and we obtain (e, x) ∈ Γ_{ψ}^{op} and (x, f) ∈ Γ_{φ}, hence (e, f) ∈ Γ_{ψ}^{op}Γ_{φ} = R. In other words e ≤_{R} f, that is, ψ(x) ≤_{R} φ(x). This holds for every x ∈ X and we obtain ψ ≤ φ (in the sense that ψ(x) ≤_{R} φ(x) for all x ∈ X). Therefore

M_{ψ,φ} = 1 ⟹ ψ ≤ φ .

This relation ≤ is clearly a partial order relation on the set of all surjective maps X → E and it follows that the matrix M is unitriangular (with an ordering of rows and columns compatible with this partial order). Thus M is invertible (as a square matrix with integer coefficients). Since Mλ = 0, it follows that the column vector λ is zero. In other words, λ_{φ} = 0 for every φ ∈ Φ, proving the linear independence of the image of A in kC(X, E)R/ Ker(\overline{ω}_{E,R}). Therefore |A| ≤ dim (𝕊_{E,R}(X)).

Now we need to estimate |A| and, for simplicity, we write e = |E| and x = |X|. By Lemma 8.1, we have

|A| = ∑_{i=0}^{e}(−1)^{e−i}\binom{e}{i} i^{x} = e^{x} + ∑_{i=0}^{e−1}(−1)^{e−i}\binom{e}{i} i^{x} .

Note that the second sum is negative because the number |A| of surjective maps X → E is smaller than the number e^{x} of all maps X → E. Therefore

|A| = e^{x}(1 + ∑_{i=0}^{e−1}(−1)^{e−i}\binom{e}{i} (\frac{i}{e})^{x}) .

Since \frac{i}{e} ≤ \frac{e−1}{e} < 1, the sum can be made as small as we want, provided x is large enough. Therefore there exists a positive integer N and a positive real number a such that a e^{x} ≤ |A| whenever x ≥ N. In other words, for any finite set X of cardinality at least N, we have

(8.3)  a |E|^{|X|} ≤ |A| ≤ dim (𝕊_{E,R}(X)) ,

giving a lower bound in the case of the fundamental correspondence functor 𝕊_{E,R}.

Now we consider the right action of k Aut(E, R) on each evaluation 𝕊_{E,R}(X), in order to estimate the dimension

dim (S_{E,R,V} (X)) = dim (𝕊_{E,R}(X) ⊗_{k Aut(E,R)} V ) .

By Lemma 5.8, \overline{ω}_{E,R} : kC(X, E)R → 𝕊_{E,R}(X) is a homomorphism of k Aut(E, R)- modules. We claim that the k-subspace kA generated by A = {Γ_{φ} | φ ∈ Φ} is a free right k Aut(E, R)-submodule of kC(X, E)R. Let σ ∈ Aut(E, R). Since

Γ_{φ} = {(x, e) ∈ X × E | e ≤_{R} ϕ(x)}  and  Δ_{σ} = {(σ(f), f) | f ∈ E} ,

we obtain

\begin{aligned}Γ_{φ}Δ_{σ} & = \{(x, f) ∈ X × E | σ(f) ≤_{R} ϕ(x)\} = \{(x, f) ∈ X × E | f ≤_{R} σ^{−1}ϕ(x)\} \\ & = Γ_{σ^{−1}∘φ} , \end{aligned}

using the fact that σ ∈ Aut(E, R) preserves the order relation R. But σ^{−1} ∘ φ ≠ φ if σ ≠ id, because φ is surjective onto E. Therefore the right action of Aut(E, R) on A is free, proving the claim.

If we denote by A′ the image of A in 𝕊_{E,R}(X), we have seen in the first part of the proof that A′ is linearly independent, so that \overline{ω}_{E,R} maps kA isomorphically onto kA′ and kA′ is free. Now kA′ is an injective k Aut(E, R)-module (because the group algebra k Aut(E, R) is symmetric, so that projective and injective modules coincide). Therefore

𝕊_{E,R}(X) = kA′ ⊕ Q

for some right k Aut(E, R)-submodule Q. It follows that

dim (S_{E,R,V} (X)) = dim (𝕊_{E,R}(X) ⊗_{k Aut(E,R)} V )
= dim(kA′ ⊗_{k Aut(E,R)} V) + dim(Q ⊗_{k Aut(E,R)} V) .

But since kA′ is a free k Aut(E, R)-module, we obtain

dim(kA′ ⊗_{k Aut(E,R)} V) = dim(kA′)\frac{dim(V)}{dim(k Aut(E, R))} = |A| b ,

where we put b := \frac{dim(V)}{| Aut(E,R)|} for simplicity. Therefore, from the inequality 8.3, we deduce that

ab |E|^{|X|} ≤ |A| b = dim(kA′ ⊗_{k Aut(E,R)} V) ≤ dim (S_{E,R,V} (X)) .

This produces the required lower bound for dim (S_{E,R,V} (X))

We can now characterize finite generation in terms of exponential behavior.

8.4. Theorem. Let M be a correspondence functor over a field k. The following are equivalent :

(a) M is finitely generated.

(b) There exists positive real numbers a, b, r such that dim(M(X)) ≤ a b^{|X|} for every finite set X with |X| ≥ r.

Proof : (a) ⇒ (b). Let M be a quotient of ⨁_{i∈I} kC(−, E) for some finite set E and

some finite index set I. For every finite set X, we have

dim(M(X)) ≤ |I| dim(kC(X, E)) = |I| 2^{|X×E|} = |I|(2^{|E|})^{|X|} .

(b) ⇒ (a). Let P and Q be subfunctors of M such that Q ⊆ P ⊆ M and P/Q simple, hence P/Q ≅ S_{E,R,V} for some triple (E, R, V). We claim that |E| is bounded above. Indeed, for |X| large enough, we have

c|E|^{|X|} ≤ dim(S_{E,R,V} (X))

for some c > 0, by Theorem 8.2, and

dim(S_{E,R,V} (X)) ≤ dim(M(X)) ≤ a b^{|X|}

by assumption. Therefore, whenever |X| ≥ N for some N, we have

c|E|^{|X|} ≤ a b^{|X|} that is, c ≤ a(\frac{b}{|E|})^{|X|} .

Since c > 0, this forces \frac{b}{|E|} ≥ 1 otherwise a(\frac{b}{|E|})^{|X|} is as small as we want. This

shows the bound |E| ≤ b, proving the claim.

For each set E with |E| ≤ b, we choose a basis {m_{i} | 1 ≤ i ≤ n_{E}} of M(E) and we use Yoneda’s lemma to construct a morphism ψ_{i}^{E} : kC(−, E) → M such that, on evaluation at E, we have ψ_{i,E}^{E}(Δ_{E}) = m_{i}. Starting from the direct sum of n_{E} copies of kC(−, E), we obtain a morphism

ψ^{E} : kC(−, E)^{n_{E}} → M ,

such that, on evaluation at E, the morphism ψ_{E}^{E} : kC(E, E)^{n_{E}} → M(E) is surjective, because the basis of M(E) is in the image. Now the sum of all such morphisms ψ^{E} yields a morphism

ψ : ⨁_{|E|≤b} kC(−, E)^{n_{E}} ⟶ M

which is surjective on evaluation at every set E with |E| ≤ b.

Let N = Im(ψ) and suppose ab absurdo that N ≠ M. Let F be a minimal set such that M(F)/N(F) ≠ {0}. Since ψ is surjective on evaluation at every set E with |E| ≤ b, we must have |F | > b. Now M(F)/N(F) is a module for the finitedimensional algebra R_{F} = kC(F, F) and, by minimality of F, inessential relations act by zero on M(F)/N(F). Let W be a simple submodule of M(F)/N(F) as a module for the essential algebra E_{F}. Associated with W, consider the simple functor S_{F,W}. (Actually, W is parametrized by a pair (R, V) and S_{F,W} = S_{F,R,V} (see Theorem 3.11), but we do not need this.) Now the module W = S_{F,W} (F) is isomorphic to a subquotient of M(F)/N(F). By Proposition 2.7, S_{F,W} is isomorphic to a subquotient of M/N. By the claim proved above, we obtain |F | ≤ b. This contradiction shows that N = M, that is, ψ is surjective. Therefore M is isomorphic, via ψ, to a quotient of ⨁_{|E|≤b} kC(−, E)^{n_{E}}. By Proposition 6.4, M is finitely

generated.

In order to prove that, over a field k, any finitely generated correspondence functor has finite length, we need a lemma.

8.5. Lemma. Let k be a field and let M be a finitely generated correspondence functor over k.

(a) M has a maximal subfunctor.

(b) Any subfunctor of M is finitely generated.

Proof : (a) Since M is finitely generated, M is generated by M(E) for some finite set E (Proposition 6.4). Let N be a maximal submodule of M(E) as a kC(E, E)- module. Note that N exists because M(E) is finite-dimensional by Lemma 6.3. Then M(E)/N is a simple kC(E, E)-module. By Proposition 2.7, there exist two subfunctors F ⊆ G ⊆ M such that G/F is simple, G(E) = M(E), and F (E) = N. Since M is generated by M(E) and G(E) = M(E), we have G = M. Therefore, F is a maximal subfunctor of M.

(b) Let N be a subfunctor of M. Since M is finitely generated, there exist positive numbers a, b such that, for every large enough finite set X, we have

dim(N(X)) ≤ dim(M(X)) ≤ a b^{|X|} ,

by Theorem 8.4. The same theorem then implies that N is finitely generated.

Lemma 8.5 fails for other categories of functors. For instance, in the category of biset functors, the Burnside functor is finitely generated and has a maximal subfunctor which is not finitely generated (see [Bo1] or [Bo2]). Similarly, the following theorem is a specific property of the category of correspondence functors.

8.6. Theorem. Let k be a field and let M be a finitely generated correspondence functor over k. Then M has finite length (that is, M has a finite composition series).

Proof : By Lemma 8.5, M has a maximal subfunctor F_{1} and F_{1} is again finitely generated. Then F_{1} has a maximal subfunctor F_{2} and F_{2} is again finitely generated. We construct in this way a sequence of subfunctors

(8.7)  M = F_{0} ⊃ F_{1} ⊃ F_{2} ⊃ …

such that F_{i}/F_{i+1} is simple whenever F_{i} ≠ 0. We claim that the sequence is finite, that is, F_{m} = 0 for some m.

Let F_{i}/F_{i+1} be one simple subquotient, hence F_{i}/F_{i+1} ≅ S_{E,R,V} for some triple (E, R, V). By Theorem 8.4, since M is finitely generated, there exist positive numbers a, b such that, for every large enough finite set X, we have

dim(M(X)) ≤ a b^{|X|} .

Therefore dim(S_{E,R,V} (X)) ≤ a b^{|X|}. By Theorem 8.2, there exists some constant c > 0 such that c |E|^{|X|} ≤ dim(S_{E,R,V} (X)) for |X| large enough. So we obtain c |E|^{|X|} ≤ a b^{|X|} for |X| large enough, hence |E| < b. This implies that the simple functor F_{i}/F_{i+1} ≅ S_{E,R,V} belongs to a finite set of isomorphism classes of simple functors, because there are finitely many sets |E| with |E| < b and, for any of them, finitely many order relations R on E, and then in turn finitely many k Aut(E, R)- simple modules V (up to isomorphism).

Therefore, if the series (8.7) of subfunctors F_{i} was infinite, then some simple functor S_{E,R,V} would occur infinitely many times (up to isomorphism). But then, on evaluation at E, the simple kC(E, E)-module S_{E,R,V} (E) would occur infinitely many times in M(E). This is impossible because M(E) is finite-dimensional by Lemma 6.3.

Theorem 8.6 was obtained independently by Gitlin [Gi], using a criterion for finite length proved recently by Wiltshire-Gordon [WG].

9. Projective functors and duality

This section is devoted to projective correspondence functors, mainly in the case where k is a field. We use duality, which will again appear in Section 14.

Recall that, by Lemma 7.3, if a projective correspondence functor M is generated by M(E), then M(X) is a projective R_{X}-module, for every set X with |X| ≥ |E|. Recall also that, by Lemma 7.3 again, a projective correspondence functor M is isomorphic to L_{E,M(E)} whenever M is generated by M(E). Thus if we work with functors having bounded type, we can assume that projective functors have the form L_{E,V} for some R_{E}-module V. In such a case, we can also enlarge E because L_{E,V} ≅ L_{F,V↑_{E}^{F}} whenever |F | ≥ |E| (see Proposition 7.5).

9.1. Lemma. Let k be a commutative ring and consider the correspondence functor L_{E,V} for some finite set E and some R_{E}-module V.

(a) L_{E,V} is projective if and only if V is a projective R_{E}-module.

(b) L_{E,V} is finitely generated projective if and only if V is a finitely generated projective R_{E}-module.

(c) L_{E,V} is indecomposable projective if and only if V is an indecomposable projective R_{E}-module.

Proof : (a) If L_{E,V} is projective, then V is projective by Lemma 7.3. Conversely, if V is projective, then L_{E,V} is projective by Lemma 2.3.

(b) If V is a finitely generated R_{E}-module, then L_{E,V} is finitely generated by Corollary 6.5. If L_{E,V} is finitely generated, then its evaluation L_{E,V} (E) = V is finitely generated by Lemma 6.3.

(c) By the adjunction property of Lemma 2.3, End_{F_{k}}(L_{E,V}) ≅ End_{R_{E}}(V), so L_{E,V} is indecomposable if and only if V is indecomposable.

Our main duality result has two aspects, which we both include in the following theorem. The notion of symmetric algebra is standard over a field and can be defined over any commutative ring as in [Br].

9.2. Theorem. Let E be a finite set.

(a) The representable functor kC(−, E) is isomorphic to its dual.

(b) Let R_{E} = kC(E, E) be the k-algebra of relations on E. Then R_{E} is a symmetric algebra. More precisely, let t : R_{E} → k be the k-linear form defined, for all basis elements S ∈ C(X, E), by the formula

t(S) = \{ \begin{matrix}1 & if R ∩ Δ_{E} = ∅ , \\ 0 & otherwise . \end{matrix}

Then t is a symmetrizing form on R_{E}, in the sense that the associated bilinear form (a, b) ↦ t(ab) is symmetric and induces an isomorphism of (R_{E}, R_{E})-bimodules between R_{E} and its dual Hom_{k}(R_{E}, k).

Proof : (a) For every finite set X, consider the symmetric bilinear form

⟨−, −⟩_{X} : kC(X, E) × kC(X, E) ⟶ k

defined, for all basis elements R, S ∈ C(X, E), by the formula

⟨R, S⟩ = \{ \begin{matrix}1 & if R ∩ S = ∅ , \\ 0 & otherwise. \end{matrix}

Then, whenever U ∈ C(Y, X), R ∈ C(Y, E), and S ∈ C(X, E), we have

\begin{aligned}R ∩ US = ∅ & ⟺ ((y, x) ∈ U, (x, e) ∈ S ⇒ (y, e) ∉ R) \\ & ⟺ ((x, y) ∈ U^{op}, (y, e) ∈ R ⇒ (x, e) ∉ S) ⟺ U^{op}R ∩ S = ∅ . \end{aligned}

It follows that ⟨U^{op}R, S⟩_{X} = ⟨R, US⟩_{Y}. In view of the definition of dual functors (Definition 3.5), this implies that the associated family of linear maps

α_{X} : kC(X, E) ⟶ kC(X, E)^{♮}

defines a morphism of correspondence functors α : kC(−, E) ⟶ kC(−, E)^{♮}.

To prove that α is an isomorphism, we fix X and we use the complement cR = (X × E) − R, for any R ∈ C(X, E). Notice that the matrix of α_{X} relative to the canonical basis C(X, E) and its dual is the product of two matrices C and A, where C_{R,S} = 1 if S = ^{c}R and 0 otherwise, while A is the adjacency matrix of the order relation ⊆. This is because R ∩ S = ∅ if and only if R ⊆ ^{c}S. Clearly C is invertible (it has order 2) and A is unitriangular, hence invertible. Therefore α_{X} is an isomorphism.

(b) Let R, S ∈ C(E, E). Then t(RS) is equal to 1 if RS∩Δ_{E} = ∅, and t(RS) = 0 otherwise. Now

RS ∩ Δ_{E} = ∅ ⟺ ∀(e, f) ∈ E × E, (e, f) ∈ R implies (f, e) ∉ S
⟺ R ∩ S^{op} = ∅

where S^{op} is the opposite relation to S. Therefore t(RS) = ⟨R, S^{op}⟩, where ⟨−, −⟩_{E} is the bilinear form on kC(E, E) defined in (a). Since this bilinear form induces an isomorphism with the dual and since the map S ↦ S^{op} is an isomorphism (it has order 2), the bilinear form associated with t induces also an isomorphism with the dual.

Since (R ∩ S^{op})^{op} = S ∩ R^{op} and ∅^{op} = ∅, we have t(RS) = t(SR) for any relations R and S on E, hence the bilinear form (a, b) ↦ t(ab) is symmetric. It is clear that the associated k-linear map R_{E} → Hom_{k}(R_{E}, k) is a morphism of (R_{E}, R_{E})-bimodules.

9.3. Corollary. If k is a field, then the correspondence functor kC(−, E) is both projective and injective.

Proof : Since passing to the dual reverses arrows and since kC(−, E) is projective, its dual is injective. But kC(−, E) is isomorphic to its dual, so it is both projective and injective.

9.4. Remark. Corollary 9.3 holds more generally when k is a self-injective ring (see Corollary 14.11).

9.5. Remark. If R is an order relation on E, then there is a direct sum decomposition

kC(−, E) = kC(−, E)R ⊕ kC(−, E)(1 − R) .

With respect to the bilinear forms defined in the proof of Theorem 9.2, we have

(kC(−, E)R)^{⊥} = kC(−, E)(1 − R^{op})

because

U ∈ (kC(X, E)R)^{⊥} ⟺ ⟨U, V R⟩_{X} = 0 ∀V ∈ kC(X, E)
⟺ ⟨UR^{op}, V ⟩_{X} = 0 ∀V ∈ kC(X, E)
⟺ UR^{op} = 0
⟺ U(1 − R^{op}) = U
⟺ U ∈ kC(X, E)(1 − R^{op}) .

It follows that

kC(−, E)/(kC(−, E)R)^{⊥} = kC(−, E)/kC(−, E)(1 − R^{op}) ≅ kC(−, E)R^{op}

and therefore the bilinear forms ⟨−, −⟩_{X} induce perfect pairings

kC(−, E)R × kC(−, E)R^{op} ⟶ k .

Thus (kC(−, E)R)^{♮} ≅ kC(−, E)R^{op}.

By the Krull-Remak-Schmidt theorem (which holds when k is a field by Proposition 6.6), it is no harm to assume that our functors are indecomposable.

9.6. Theorem. Let k be a field and M be a finitely generated correspondence functor over k. The following conditions are equivalent:

(a) The functor M is projective and indecomposable.

(b) The functor M is projective and admits a unique maximal (proper) subfunctor.

(c) The functor M is projective and admits a unique minimal (nonzero) subfunctor.

(d) The functor M is injective and indecomposable.

(e) The functor M is injective and admits a unique maximal (proper) subfunctor.

(f) The functor M is injective and admits a unique minimal (nonzero) subfunctor.

Proof : (a) ⇒ (b). Suppose first that M is projective and indecomposable. Then M ≅ L_{E,V} for some finite set E and some indecomposable projective kC(E, E)- module V (Lemma 7.3). Since kC(E, E) is a finite dimensional algebra over k, the module V has a unique maximal submodule W. If N is a subfunctor of M, then N(E) is a submodule of V, so there are two cases: either N(E) = V, and then N = M, because M is generated by M(E) = V, or N(E) ⊆ W, and then N(X) ⊆ J_{W} (X) for any finite set X, where

J_{W} (X) = { ∑_{i} φ_{i} ⊗v_{i} ∈ kC(X, E)⊗_{kC(E,E)} V | ∀ψ ∈ C(E, X), ∑_{i}(ψφ_{i})·v_{i} ∈ W } .

One ckecks easily that the assignment X ↦ J_{W} (X) is a subfunctor of L_{E,V}, such that J_{W} (E) = W after the identification L_{E,V} (E) ≅ V. (This subfunctor is similar to the one introduced in Lemma 2.4.) In particular J_{W} is a proper subfunctor of L_{E,V}. It follows that J_{W} is the unique maximal proper subfunctor of L_{E,V}, as it contains any proper subfunctor N of L_{E,V}.

(b) ⇒ (a). Suppose that M admits a unique maximal subfunctor N. If M splits as a direct sum M_{1} ⊕ M_{2} of two nonzero subfunctors M_{1} and M_{2}, then M_{1} and M_{2} are finitely generated. Let N_{1} be a maximal subfunctor of M_{1}, and N_{2} be a maximal subfunctor of M_{2}. Such subfunctors exist by Lemma 8.5. Then M_{1} ⊕N_{2} and N_{1} ⊕ M_{2} are maximal subfunctors of M. This contradiction proves that M is indecomposable.

(a) ⇒ (d). If M is a finitely generated projective functor, then there exists a finite set E such that M is isomorphic to a quotient, hence a direct summand, of ⨁_{i∈I} kC(−, E) for some finite set I (Proposition 6.4). Since k is a field, kC(−, E)

is an injective functor (Corollary 9.3), hence so is the direct sum and its direct summand M.

(d) ⇒ (a). If M is a finitely generated injective functor, then its dual M^{♮} is projective, hence injective, and therefore M ≅ (M^{♮})^{♮} is projective.

(a) ⇒ (c). For a finitely generated functor M, the duality between M and M^{♮} induces an order reversing bijection between the subfunctors of M and the subfunctors of M^{♮}. If M is projective and indecomposable, then so is M^{♮}, that is,

(a) holds for M^{♮}. Thus (b) holds for M^{♮} and the functor M^{♮} has a unique maximal subfunctor. Hence M has a unique minimal subfunctor.

(c) ⇒ (a). If M is projective and admits a unique minimal subfunctor, then M is also injective, and its dual M^{♮} is projective and admits a unique maximal subfunctor. Hence M^{♮} is indecomposable, so M is indecomposable.

It is now clear that (e) and (f) are both equivalent to (a), (b), (c) and (d).

Finally, we prove that the well-known property of indecomposable projective modules over a symmetric algebra also holds for correspondence functors.

9.7. Theorem. Let k be a field.

(a) Let M be a finitely generated projective correspondence functor over k. Then M/ Rad(M) ≅ Soc(M).

(b) Let M and N be finitely generated correspondence functors over k. If M is projective, then dim_{k} Hom_{F_{k}}(M, N) = dim_{k} Hom_{F_{k}}(N, M) < +∞.

Proof : (a) By Proposition 6.6, we can assume that M is indecomposable. In this case, by Theorem 9.6, both M/ Rad(M) and Soc(M) are simple functors. By Proposition 6.4, there is a finite set E such that M is a quotient, hence a direct summand, of F = ⨁_{i∈I} kC(−, E) for some finite set I. Since kC(−, E)^{♮} ≅ kC(−, E),

the dual M^{♮} is a direct summand of F^{♮} ≅ F, and both M and M^{♮} are generated by their evaluations at E. Thus M ≅ L_{E,M(E)} and M^{♮} ≅ L_{E,M^{♮}(E)}, by Lemma 7.3. As M is a direct summand of F and M is indecomposable, M is a direct summand of kC(−, E), by the Krull-Remak-Schmidt Theorem (Proposition 6.6). So there is a primitive idempotent e of kC(E, E) ≅ End_{F_{k}}(kC(−, E)) such that M ≅ kC(−, E)e, and we can assume that M = kC(−, E)e.

If V is a finite dimensional k-vector space, and W is a subspace of V, set

W^{⊥} = {φ ∈ Hom_{k}(V, k) | φ(W) = 0} .

If N is a subfunctor of M, the assignment sending a finite set X to N(X)^{⊥} defines a subfunctor N^{⊥} of M^{♮}, and moreover N ↦ N^{⊥} is an order reversing bijection between the set of subfunctors of M and the set of subfunctors of M^{♮}. In particular Soc(M)^{⊥} = Rad(M^{♮}). Hence Soc(M)^{⊥}(E) = ( Soc(M)(E))^{⊥} = Rad(M^{♮})(E).

Now M^{♮} ≠ Rad(M^{♮}), and M^{♮} is generated by M^{♮}(E). Hence Rad(M^{♮})(E) ≠ M^{♮}(E). It follows that Soc(M)(E) ≠ 0. Then Soc(M)(E) ⊆ kC(E, E)e, and Soc(M)(E) is a left ideal of kC(E, E). It follows that Soc(M)(E) is not contained in the kernel of the map t defined in Theorem 9.2, that is t( Soc(M)(E)) ≠ 0. Hence

0 ≠ t( Soc(M)(E)) = t( Soc(M)(E)e) = t(e Soc(M)(E)) ,

and in particular e Soc(M)(E) ≠ 0. Since

e Soc(M)(E) ≅ Hom_{F_{k}} (kC(−, E)e, Soc(M)) ,

there is a nonzero morphism from M = kC(−, E)e to Soc(M), hence a nonzero morphism from M/ Rad(M) to Soc(M). Since M/ Rad(M) and Soc(M) are simple, it is an isomorphism.

(b) First, by Proposition 6.6, both Hom_{F_{k}}(M, N) and Hom_{F_{k}}(N, M) are finite dimensional k-vectors spaces.

Now we can again assume that M is an indecomposable projective and injective functor. For a finitely generated functor N, set α(N) = dim_{k} Hom_{F_{k}}(M, N) and β(N) = dim_{k} Hom_{F_{k}}(N, M). If 0 → N_{1} → N_{2} → N_{3} → 0 is a short exact sequence of finitely generated functors, then α(N_{2}) = α(N_{1}) + α(N_{3}) because M is projective, and β(N_{2}) = β(N_{1}) + β(N_{3}) because M is injective. So, in order to prove (b), as N has finite length, it is enough to assume that N is simple. In that case α(N) = dim_{k} End_{F_{k}}(N) if M/ Rad(M) ≅ N, and α(N) = 0 otherwise. Similarly β(N) = dim_{k} End_{F_{k}}(N) if Soc(M) ≅ N, and β(N) = 0 otherwise. Hence (b) follows from (a).

10. The noetherian case

In this section, we shall assume that the ground ring k is noetherian, in which case we obtain more results about subfunctors. For instance, we shall prove that any subfunctor of a finitely generated functor is finitely generated. It would be interesting to see if the methods developed recently by Sam and Snowden [SS] for showing noetherian properties of representations of categories can be applied for proving the results of this section.

Our first results hold without any assumption on k.

10.1. Notation. Let k be a commutative ring, let E be a finite set, and let M be a correspondence functor over k. We set

\overline{M}(E) = M(E)/ ∑_{E′⊂E} kC(E, E′)M(E′) ,

where the sum runs over proper subsets E′ of E.

Note that if F is any set of cardinality smaller than |E|, then there exists a bijection σ : E′ → F, where E′ is a proper subset of E. It follows that

kC(E, F)M(F) = kC(E, F)R_{σ}M(E′) ⊆ ∑_{E′⊂E} kC(E, E′)M(E′), where R_{σ} ⊆ F × E′

is the graph of σ.

Note also that \overline{M}(E) is a left module for the essential algebra E_{E}, because the ideal I_{E} = ∑_{|Y |<|E|} kC(E, Y)kC(Y, E) of the algebra R_{E} = kC(E, E) acts by zero

on \overline{M}(E).

10.2. Lemma. Let k be a commutative ring, and let E be a finite set. Let M be a correspondence functor over k. If 𝔭 is a prime ideal of k, denote by M_{𝔭} the localization of M at 𝔭, defined by M_{𝔭}(E) = M(E)_{𝔭} for every finite set E.

(a) M_{𝔭} is a correspondence functor over the localization k_{𝔭}.

(b) If M is finitely generated over k, then M_{𝔭} is finitely generated over k_{𝔭}.

(c) For each finite set E, there is an isomorphism of k_{𝔭}C(E, E)-modules

\overline{M}(E)_{𝔭} ≅ \overline{M_{𝔭}}(E) .

Proof : (a) This is straightforward.

(b) If E is a finite set, then clearly kC(−, E)_{𝔭} ≅ k_{𝔭}C(−, E), because this the localization of a free module (on every evaluation). If M is finitely generated, then there is a finite set F such that M is a quotient of ⨁_{i∈I} kC(−, F) for some finite set I.

Then M_{𝔭} is a quotient of the functor ⨁_{i∈I} kC(−, F)_{𝔭} ≅ ⨁_{i∈I} k_{𝔭}C(−, F), hence it is a

finitely generated functor over k_{𝔭}.

(c) Since localization is an exact functor, the exact sequence of k-modules

⨁_{E′⊂E} kC(E, E′)M(E′) → M(E) → \overline{M}(E) → 0

gives the exact sequence of k_{𝔭}-modules

⨁_{E′⊂E} (kC(E, E′)M(E′))_{𝔭} → M(E)_{𝔭} → \overline{M}(E)_{𝔭} → 0 .

Now clearly (kC(E, E′)M(E′))_{𝔭} = k_{𝔭}C(E, E′)M(E′)_{𝔭} = k_{𝔭}C(E, E′)M_{𝔭}(E′) for each E′ ⊂ E. Hence we get an exact sequence

⨁_{E′⊂E} k_{𝔭}C(E, E′)M_{𝔭}(E′) → M_{𝔭}(E) → \overline{M}(E)_{𝔭} → 0 ,

and it follows that \overline{M}(E)_{𝔭} ≅ \overline{M_{𝔭}}(E).

10.3. Proposition. Let k be a commutative ring, let E be a finite set, and let M be a correspondence functor such that \overline{M}(E) ≠ 0.

(a) There exists a prime ideal 𝔭 of k such that \overline{M_{𝔭}}(E) ≠ 0.

(b) If moreover M(E) is a finitely generated k-module, then there exist subfunctors A and B of M_{𝔭} such that 𝔭M_{𝔭} ⊆ A ⊂ B, and a simple module V for the essential algebra E_{E} of E over k(𝔭) such that B/A ≅ S_{E,V}, where k(𝔭) = k_{𝔭}/𝔭k_{𝔭}.

(c) In this case, there exist positive numbers c and d such that

c |E|^{|X|} ≤ dim_{k(𝔭)}(M_{𝔭}/𝔭M_{𝔭})(X)

whenever X is a finite set such that |X| ≥ d.

Proof : (a) This follows from the well-known fact that the localization map \overline{M}(E) ⟶ ∏_{𝔭∈Spec(k)} \overline{M}(E)_{𝔭} is injective, and from the isomorphism \overline{M}(E)_{𝔭} ≅ \overline{M_{𝔭}}(E)

of Lemma 10.2.

(b) Set N = M_{𝔭}/𝔭M_{𝔭} where 𝔭 is the prime ideal obtained in (a). Then N is a correspondence functor over k(𝔭). Suppose that \overline{N}(E) = 0. Then

M_{𝔭}(E) = 𝔭M_{𝔭}(E) + ∑_{E′⊂E} k_{𝔭}C(E, E′)M_{𝔭}(E′) .

Since M(E) is a finitely generated k-module, M_{𝔭}(E) is a finitely generated k_{𝔭}- module, and Nakayama’s lemma implies that

M_{𝔭}(E) = ∑_{E′⊂E} k_{𝔭}C(E, E′)M_{𝔭}(E′) ,

that is, \overline{M_{𝔭}}(E) = 0. This contradicts (a) and shows that \overline{N}(E) ≠ 0.

Now \overline{N}(E) is a nonzero module for the essential algebra E_{E} of E over k(𝔭), and it is finite dimensional over k(𝔭) (because M_{𝔭}(E) is a finitely generated k_{𝔭}- module). Hence it admits a simple quotient V as E_{E}-module. Then V can be viewed as a simple k(𝔭)R_{E}-module by inflation, and it is also a quotient of N(E). By Proposition 2.7, there exist subfunctors A/𝔭M_{𝔭} ⊂ B/𝔭M_{𝔭} of N such that B/A is isomorphic to the simple functor S_{E,V}, proving (b).

(c) By (b) and Theorem 8.2, there exist positive numbers c and d such that

c |E|^{|X|} ≤ dim_{k(𝔭)}(B/A)(X)

whenever X is a finite set such that |X| ≥ d. Assertion (c) follows.

Now we assume that k is noetherian and we can state the critical result.

10.4. Theorem. Let k be a commutative noetherian ring. Let N be a subfunctor of a correspondence functor M over k. If E and Y are finite sets such that M is generated by M(E) and \overline{N}(Y) ≠ 0, then |Y | ≤ 2^{|E|}.

Proof : Since M is generated by M(E), choosing a set I of generators of M(E) yields a surjection Φ : P = ⨁_{i∈I} kC(−, E) → M. Let L = Φ^{−1}(N). Since Φ induces

a surjection \overline{L}(Y) → \overline{N}(Y), and since P is generated by P (E), we can replace M by P and N by L. Hence we now assume that N is a subfunctor of ⨁_{i∈I} kC(−, E).

Since \overline{N}(Y) ≠ 0, there exists

m ∈ N(Y) − ∑_{Y′⊂Y} kC(Y, Y′)N(Y′) .

Let N′ be the subfunctor of N generated by m. Then clearly \overline{N′}(Y) ≠ 0, because

m ∈ N′(Y) − ∑_{Y′⊂Y} kC(Y, Y′)N′(Y′) .

Moreover N′(Y) = kC(Y, Y)m is a finitely generated k-module, and there is a finite subset S of I such that m ∈ ⨁_{i∈S} kC(Y, E). Therefore N′ ⊆ ⨁_{i∈S} kC(−, E). Replacing

N by N′, we can assume moreover that the set I is finite. In other words, there exists an integer s ∈ ℕ such that N ⊆ kC(−, E)^{⊕s}.

Now by Proposition 10.3, there exists a prime ideal 𝔭 of k such that \overline{N_{𝔭}}(Y) ≠ 0. Moreover N(Y) is a submodule of kC(Y, E)^{⊕s}, which is a finitely generated (free) k-module. Since k is noetherian, it follows that N(Y) is a finitely generated kmodule.

By Proposition 10.3, there exist subfunctors A ⊂ B of N_{𝔭} such that B/A is isomorphic to a simple functor of the form S_{Y,V}, where V is a simple module for the essential algebra of Y over k(𝔭). In particular Y is minimal such that (B/A)(Y) ≠ 0, thus (\overline{B/A})(Y) ≅ (B/A)(Y) ≅ V.

It follows that \overline{B}(Y) ≠ 0, and B is a subfunctor of k_{𝔭}C(−, E)^{⊕s}. In other words, replacing k by k_{𝔭} and N by B, we can assume that k is a noetherian local ring, that 𝔭 is the unique maximal ideal of k, and that N has a subfunctor A such that N/A is isomorphic to S_{Y,V}, where V is a simple module for the essential algebra E_{Y} over k/𝔭.

We claim that there exists an integer n ∈ ℕ such that

N(Y) ≠ A(Y) + (𝔭^{n}C(Y, E)^{⊕s} ∩ N(Y)) .

Indeed N(Y) is a submodule of the finitely generated k-module kC(Y, E)^{⊕s}. By the Artin-Rees lemma (see Theorem 8.5 in [Ma]), there exists an integer l ∈ ℕ such that for any n ≥ l

𝔭^{n}C(Y, E)^{⊕s} ∩ N(Y) = 𝔭^{n−l}(𝔭^{l}C(Y, E)^{⊕s} ∩ N(Y)) .

Let m_{1}, …, m_{r} be generators of N(Y) as a k-module. Suppose that n > l and that N(Y) = A(Y) + (𝔭^{n}C(Y, E)^{⊕s} ∩ N(Y)). Then

N(Y) = A(Y) + 𝔭^{n−l}(𝔭^{l}C(Y, E)^{⊕s} ∩ N(Y)) .

It follows that for each i ∈ {1, …, r}, there exist a_{i} ∈ A(Y) and scalars λ_{i,j} ∈ 𝔭^{n−l}, for 1 ≤ j ≤ r, such that

m_{i} = a_{i} + ∑_{j=1}^{r} λ_{i,j}m_{j} .

In other words the sequence (a_{i})_{i=1,...,r} is the image of the sequence (m_{i})_{i=1,...,r} under the matrix J = id_{r} −Λ, where Λ is the matrix of coefficients λ_{i,j}, and id_{r} is the identity matrix of size r. Since Λ has coefficients in 𝔭^{n−l} ⊆ 𝔭, the determinant of J is congruent to 1 modulo 𝔭, hence J is invertible. It follows that the m_{i}’s are linear combinations of the a_{j}’s with coefficients in k. Hence m_{i} ∈ A(Y) for 1 ≤ i ≤ r, thus N(Y) = A(Y). This is a contradiction since (N/A)(Y) ≅ V ≠ 0. This proves our claim.

We have obtained that N ≠ A + (𝔭^{n}C(−, E)^{⊕s} ∩ N). Since N/A is simple, it follows that 𝔭^{n}C(−, E)^{⊕s} ∩ N = 𝔭^{n}C(−, E)^{⊕s} ∩ A.

Now we reduce modulo 𝔭^{n} and we let respectively Â and N̂ denote the images of A and N in the reduction (k/𝔭^{n})C(−, E)^{⊕s}. Then

N̂/Â = ((N + 𝔭^{n}C(−, E)^{⊕s})/𝔭^{n}C(−, E)^{⊕s})/((A + 𝔭^{n}C(−, E)^{⊕s})/𝔭^{n}C(−, E)^{⊕s})
≅ (N/(𝔭^{n}C(−, E)^{⊕s} ∩ N))/(A/(𝔭^{n}C(−, E)^{⊕s} ∩ A))
≅ N/A ,

and this is isomorphic to the simple functor S_{Y,V} over the field k/𝔭. Hence for any finite set X, the module N̂(X)/Â(X) is a (k/𝔭)-vector space. Moreover, by Proposition 10.3, there exist positive numbers c and d such that the dimension of this vector space is larger than c |Y |^{|X|} whenever |X| ≥ d.

Now for any finite set X, the module (k/𝔭^{n})C(X, E)^{⊕s} is filtered by the submodules Γ_{j} = (𝔭^{j}/𝔭^{n})C(X, E)^{⊕s}, for j = 1, …, n − 1, and the quotient Γ_{j}/Γ_{j+1} is a vector space over k/𝔭, of dimension sd_{j} 2^{|X||E|}, where d_{j} = dim_{k/𝔭}(𝔭^{j}/𝔭^{j+1}). It follows that, for |X| ≥ d,

c |Y |^{|X|} ≤ dim_{k/𝔭} (N̂(X)/Â(X)) ≤ s( ∑_{j=0}^{n−1} d_{j})2^{|X||E|} .

As |X| tends to infinity, this forces |Y | ≤ 2^{|E|}, completing the proof of Theorem 10.4.

10.5. Corollary. Let k be a commutative noetherian ring and let N be a subfunctor of a correspondence functor M over k.

(a) If E is a finite set such that M is generated by M(E) and if F is a finite set with |F | ≥ 2^{|E|}, then N is generated by N(F).

(b) If M has bounded type, then N has bounded type. In particular, over k, any correspondence functor of bounded type has a bounded presentation.

(c) If M is finitely generated, then N is finitely generated. In particular, over k, any finitely generated correspondence functor is finitely presented.

Proof : (a) Let E be a finite set such that M is generated by M(E). If X is a finite set such that \overline{N}(X) ≠ 0, then |X| ≤ 2^{|E|}, by Theorem 10.4. For each integer e ≤ 2^{|E|}, let [e] = {1, …, e} and choose a subset S_{e} of N([e]) which maps to a generating set of \overline{N}([e]) as a k-module. Each i ∈ S_{e} yields a morphism ψ_{i} : kC(−, [e]) → N. Let

Q = ⨁_{e≤2^{|E|}} ⨁_{i∈S_{e}} kC(−, [e])  and  Ψ = ∑_{{e≤2^{|E|} \atop i∈S_{e}}} ψ_{i} : Q → N .

Then by construction the induced map

\overline{Ψ}_{X} : \overline{Q}(X) → \overline{N}(X)

is surjective, for any finite set X, because either \overline{N}(X) = 0 or |X| = e ≤ 2^{|E|}. Suppose that Ψ : Q → N is not surjective and let A be a set of minimal cardinality such that Ψ_{A} : Q(A) → N(A) is not surjective. Let l ∈ N(A) − Im Ψ_{A}. Since the map \overline{Ψ}_{A} is surjective, there is an element q ∈ Q(A) and elements l_{e} ∈ N([e]) and R_{e} ∈ C(A, [e]), for e < |A|, such that

l = Ψ_{A}(q) + ∑_{e<|A|} R_{e}l_{e} .

The minimality of A implies that the map Ψ_{[e]} : Q([e]) → N([e]) is surjective for each e < |A|, so there are elements q_{e} ∈ Q([e]), for e < |A|, such that Ψ_{[e]}(q_{e}) = l_{e}. It follows that l = Ψ_{A}(q + ∑_{e<|A|} R_{e}q_{e}), thus l ∈ Im Ψ_{A}. This contradiction proves

that the morphism Ψ : Q → N is surjective.

Now let F be a set with |F | ≥ 2^{|E|}. For each e ≤ 2^{|E|}, the representable functor kC(−, [e]) is generated by its evaluation at [e], hence also by its evaluation at F, because kC(−, [e]) is a direct summand of kC(−, F) by Corollary 4.2. Therefore Q is generated by Q(F). Since Ψ : Q → N is surjective, it follows that N is generated by N(F).

(b) This follows clearly from (a).

(c) If now M is finitely generated, then the same argument applies, but we can assume moreover that all the set S_{e} appearing in the proof of (a) are finite, since for any finite set X, the module N(X) is finitely generated, being a submodule of the finitely generated module M(X). It follows that the functor Q of the proof of (a) is finitely generated and this proves (c).

It follows from (b) and Theorem 7.4 that, whenever k is noetherian, any correspondence functor of bounded type is isomorphic to L_{F,V} for some F and V. We shall return to this in Theorem 10.9 below.

10.6. Notation. We denote by F_{k}^{b} the full subcategory of F_{k} consisting of correspondence functors having bounded type and by F_{k}^{f} the full subcategory of F_{k} consisting of finitely generated functors.

10.7. Corollary. Let k be a commutative noetherian ring. Then the categories F_{k}^{b} and F_{k}^{f} are abelian full subcategories of F_{k}.

Proof : Any quotient of a functor of bounded type has bounded type and any quotient of a finitely generated functor is finitely generated. When k is noetherian, any subfunctor of a functor of bounded type has bounded type and any subfunctor of a finitely generated functor is finitely generated, by Corollary 10.5.

Recall from Lemma 2.4 that for any finite set E and any R_{E}-module V, we have defined a subfunctor J_{E,V} of L_{E,V} by setting

J_{E,V} (X) = { ∑_{i} R_{i} ⊗_{R_{E}} v_{i} | R_{i} ∈ C(X, E), v_{i} ∈ V, ∀S ∈ C(E, X), ∑_{i}(SR_{i})v_{i} = 0} .

Moreover J_{E,V} (E) = 0 and S_{E,V} = L_{E,V} /J_{E,V}.

We have seen in Proposition 7.5 that L_{F,V↑_{E}^{F}} ≅ L_{E,V} whenever |F | ≥ |E|. The subfunctor J_{F,V↑_{E}^{F}} vanishes at F, hence also at E, so that J_{F,V↑_{E}^{F}} ⊂ J_{E,V}. When k is noetherian, we show that this decreasing sequence reaches zero.

10.8. Theorem. Let k be a commutative noetherian ring, let E be a finite set, and let V be an R_{E}-module. For any finite set F such that |F | ≥ 2^{|E|}, we have J_{F,V↑_{E}^{F}} = 0.

Proof : Let F be a finite set. By Proposition 7.5, there is an isomorphism L_{F,V↑_{E}^{F}} ≅ L_{E,V}. Thus J_{F,V↑_{E}^{F}} is isomorphic to a subfunctor of L_{E,V}. Since J_{F,V↑_{E}^{F}}(F) = 0, it follows from Corollary 4.4 that J_{F,V↑_{E}^{F}}(X) = 0 for any finite set X with |X| ≤ |F |.

We now assume that J_{F,V↑_{E}^{F}} ≠ 0 and we prove that |F | < 2^{|E|}. Let Y be a set of minimal cardinality such that J_{F,V↑_{E}^{F}}(Y) ≠ 0. Then |Y | > |F |. Moreover \overline{J_{F,V↑_{E}^{F}}}(Y) ≅ J_{F,V↑_{E}^{F}}(Y) ≠ 0, hence |Y | ≤ 2^{|E|} by Theorem 10.4, because J_{F,V↑_{E}^{F}} is (isomorphic to) a subfunctor of L_{E,V}, which is generated by L_{E,V} (E) = V. It follows that |F | < |Y | ≤ 2^{|E|}.

We now show that, over a noetherian ring, any correspondence functor of bounded type is isomorphic to L_{F,V} for some F and V, or also isomorphic to S_{G,W} for some G and W (where the symbol S_{G,W} refers to Notation 2.5).

10.9. Theorem. Let k be a commutative noetherian ring. Let M be a correspondence functor over k generated by M(E), for some finite set E.

(a) For any finite set F such that |F | ≥ 2^{|E|}, the counit morphism η_{M,F} : L_{F,M(F)} → M is an isomorphism.

(b) For any finite set G such that |G| ≥ 2^{2^{|E|}}, we have M ≅ L_{G,M(G)} and J_{G,M(G)} = 0, hence M ≅ S_{G,M(G)}.

Proof : (a) If M is generated by M(E), then there is a set I and a surjective morphism P = ⨁_{i∈I} kC(−, E) → M. If F is a finite set with |F | ≥ 2^{|E|}, then by

Corollary 10.5 the kernel K of this morphism is generated by K(F). Then K is in turn covered by a projective functor Q and we have a bounded presentation

Q → P → M → 0

with both Q and P generated by their evaluation at F. By Theorem 7.4, the counit morphism η_{M,F} : L_{F,M(F)} → M is an isomorphism.

(b) For any finite set F such that |F | ≥ 2^{|E|}, we have M ≅ L_{F,M(F)} by (a). For any finite set G such that |G| ≥ 2^{|F |}, that is, |G| ≥ 2^{2^{|E|}}, we obtain J_{G,M(F)↑_{F}^{G}} = 0 by Theorem 10.8. It follows that

M ≅ L_{F,M(F)} ≅ L_{G,M(F)↑_{F}^{G}} = S_{G,M(F)↑_{F}^{G}} .

Finally, notice that, by the definition of L_{F,M(F)}, we have M(G) ≅ L_{F,M(F)}(G) = M(F)↑_{F}^{G}, so we obtain M ≅ L_{G,M(G)} = S_{G,M(G)}.

Other kinds of stabilizations also occur, as the next theorems show.

10.10. Theorem. Let k be a commutative noetherian ring, let M and N be correspondence functors over k, and let E and F be finite sets.

(a) If M is generated by M(E), then for |F | ≥ 2^{|E|}, the evaluation map at F

Hom_{F_{k}}(M, N) → Hom_{R_{F}} (M(F), N(F))

is an isomorphism.

(b) If M has bounded type, then for any integer i ∈ ℕ, there exists an integer n_{i} such that the evaluation map

Ext_{F_{k}}^{i}(M, N) → Ext_{R_{F}}^{i} (M(F), N(F))

is an isomorphism whenever |F | ≥ n_{i}.

Proof : (a) By Theorem 10.9, we have an isomorphism M ≅ L_{F,M(F)} for |F | ≥ 2^{|E|}. Hence

Hom_{F_{k}}(M, N) ≅ Hom_{F_{k}}(L_{F,M(F)}, N) ≅ Hom_{R_{F}} (M(F), N(F)) ,

where the last isomorphism comes from the adjunction of Lemma 2.3, and is given by evaluation at F.

(b) This assertion will follow from (a) by décalage and induction on i. If M is generated by M(E), then there is an exact sequence of correspondence functors

0 → L → P → M → 0

where P is projective and generated by P (E). This gives an exact sequence

0 → Hom_{F_{k}}(M, N) → Hom_{F_{k}}(P, N) → Hom_{F_{k}}(L, N) → Ext_{F_{k}}^{1}(M, N) → 0 ,

and isomorphisms Ext_{F_{k}}^{i}(M, N) ≅ Ext_{F_{k}}^{i−1}(L, N) for i ≥ 2.

Now L has bounded type by Corollary 10.5, and P (F) is a projective R_{F} - module by Lemma 7.3, whenever |F | is large enough. It follows that there is also an exact sequence

0 →Hom_{R_{F}} (M(F), N(F)) → Hom_{R_{F}} (P (F), N(F)) →
Hom_{R_{F}} (L(F), N(F)) →Ext_{R_{F}}^{1} (M(F), N(F)) →0

and isomorphisms Ext_{R_{F}}^{i} (M(F), N(F)) ≅ Ext_{R_{F}}^{i−1}(L(F), N(F)) for i ≥ 2, whenever F is large enough.

Now by (a), the exact sequences

0 → Hom_{F_{k}}(M, N) → Hom_{F_{k}}(P, N) → Hom_{F_{k}}(L, N)

and

0 → Hom_{R_{F}} (M(F), N(F)) → Hom_{R_{F}} (P (F), N(F)) → Hom_{R_{F}} (L(F), N(F))

are isomorphic for F large enough. It follows that

Ext_{F_{k}}^{1}(M, N) ≅ Ext_{R_{F}}^{1} (M(F), N(F)) .

Similarly, for each i ≥ 2, when F is large enough (depending on i), there are isomorphisms Ext_{F_{k}}^{i}(M, N) ≅ Ext_{R_{F}}^{i} (M(F), N(F)).

There is also a stabilization result involving the Tor groups.

10.11. Theorem. Let k be a commutative noetherian ring, and E be a finite set. If F is a finite set with |F | ≥ 2^{2^{|E|}}, then for any finite set X and any left R_{E}-module V, we have

Tor_{1}^{R_{F}} (kC(X, F), V ↑_{E}^{F} ) = 0 .

Proof : Let V be a left R_{E}-module and s : Q → V be a surjective morphism of R_{E}-modules, where Q is projective. Let K denote the kernel of the surjective morphism

L_{E,s} : L_{E,Q} → L_{E,V} .

Since L_{E,Q} is generated by L_{E,Q}(E) ≅ Q, it follows from Corollary 10.5 that K is generated by K(G) whenever G is a finite set with |G| ≥ 2^{|E|}. Now by Theorem 10.9, the counit L_{F,K(F)} → K is an isomorphism whenever F is a finite set with |F | ≥ 2^{|G|}. Hence if |F | ≥ 2^{2^{|E|}}, we have an exact sequence of correspondence functors

(10.12)  0 → L_{F,W_{F}} → L_{E,Q} → L_{E,V} → 0 ,

where W_{F} = K(F), and where the middle term L_{E,Q} is projective. Evaluating this sequence at F, we get the exact sequence of R_{F} -modules

0 → W_{F} → kC(F, E) ⊗_{R_{E}} Q → kC(F, E) ⊗_{R_{E}} V → 0 ,

where the middle term is projective.

Let X be a finite set. Applying the functor kC(X, F) ⊗_{R_{F}} (−) to this sequence yields the exact sequence

0 → Tor_{1}^{R_{F}} (kC(X, F), V ↑_{E}^{F} ) →

kC(X, F) ⊗_{R_{F}} W_{F} → kC(X, E) ⊗_{R_{E}} Q → kC(X, E) ⊗_{R_{E}} V → 0 ,

because kC(X, F) ⊗_{R_{F}} kC(F, E) ≅ kC(X, E) by Corollary 4.5, as |F | ≥ |E|. On the other hand, evaluating at X the exact sequence 10.12 gives the exact sequence

0 → kC(X, F) ⊗_{R_{F}} W_{F} → kC(X, E) ⊗_{R_{E}} Q → kC(X, E) ⊗_{R_{E}} V → 0 .

In both latter exact sequences, the maps are exactly the same. It follows that

Tor_{1}^{R_{F}} (kC(X, F), V ↑_{E}^{F} ) = 0 ,

as was to be shown.

As a final approach to stabilization, we introduce the following definition.

10.13. Definition. Let G_{k} denote the following category:

• The objects of G_{k} are pairs (E, U) consisting of a finite set E and a left R_{E}-module U.

• A morphism φ : (E, U) → (F, V) in G_{k} is a morphism of R_{E}-modules U → kC(E, F) ⊗_{R_{F}} V.

• The composition of morphisms φ : (E, U) → (F, V) and ψ : (F, V) → (G, W) is the morphism obtained by composition

U →^{φ} kC(E, F) ⊗_{R_{F}} V ⟶^{id ⊗ψ} kC(E, F) ⊗_{R_{F}} kC(F, G) ⊗_{R_{G}} W ⟶^{μ⊗id} kC(E, G) ⊗_{R_{G}} W ,

where μ : kC(E, F) ⊗_{R_{F}} kC(F, G) → kC(E, G) is the composition in the category kC.

• The identity morphism of (E, U) is the canonical isomorphism

U → kC(E, E) ⊗_{R_{E}} U

resulting from the definition R_{E} = kC(E, E).

One can check easily that G_{k} is a k-linear category.

10.14. Theorem. Let k be a commutative ring. Let 𝕃 : G_{k} → F_{k}^{b} be the assignment sending (E, U) to L_{E,U}, and φ : (E, U) → (F, V) to the morphism L_{E,U} → L_{F,V} associated by adjunction to φ : U → L_{F,V} (E).

(a) 𝕃 is a fully faithful k-linear functor.

(b) 𝕃 is an equivalence of categories if k is noetherian.

(c) G_{k} is an abelian category if k is noetherian.

Proof : It is straightforward to check that 𝕃 is a k-linear functor. It is moreover fully faithful, since

Hom_{F_{k}^{b}}(L_{E,U}, L_{F,V}) ≅ Hom_{R_{E}}(U, kC(E, F) ⊗_{F} V ) ≅ Hom_{G_{k}}((E, U), (F, V)) .

Finally, if k is noetherian, then any correspondence functor M of bounded type is isomorphic to a functor of the form L_{E,U}, by Theorem 10.9, for E large enough and U = M(E). Hence 𝕃 is essentially surjective, so it is an equivalence of categories. In particular, G_{k} is abelian by Corollary 10.7.

Part 2

FUNCTORS AND LATTICES

11. Functors associated to lattices

A fundamental construction associates a correspondence functor F_{T} to any finite lattice T. This is one of our main tools for the analysis of correspondence functors. Throughout this section, k is an arbitrary commutative ring.

11.1. Definition. Let T be a finite lattice. For a finite set X, we define F_{T} (X) to be the free k-module with basis the set T^{X} of all functions from X to T :

F_{T} (X) = k(T^{X}) .

For two finite sets X and Y and a correspondence R ⊆ Y × X, we define a map F_{T} (R) : F_{T} (X) → F_{T} (Y) as follows : to a function φ : X → T, we associate the function F_{T} (R)(φ) : Y → T, also simply denoted by Rφ, defined by

(Rφ)(y) = ⋁_{(y,x)∈R} φ(x) ,

with the usual rule that a join over the empty set is equal to ˆ0. The map

F_{T} (R) : F_{T} (X) → F_{T} (Y)

is the unique k-linear extension of this construction. More generally, if α = ∑_{R∈C(Y,X)} α_{R}R is any element of kC(Y, X), where α_{R} ∈ k, we set

F_{T} (α) = ∑_{R∈C(Y,X)} α_{R}F_{T} (R) .

11.2. Proposition. The assignment sending a finite set X to F_{T} (X) and α ∈ kC(Y, X) to F_{T} (α) : F_{T} (X) → F_{T} (Y) is a correspondence functor.

Proof : First it is clear that if X is a finite set and Δ_{X} ∈ C(X, X) is the identity correspondence, then for any φ : X → T and any y ∈ X

(Δ_{X}φ)(y) = ⋁_{(y,x)∈Δ_{X}} φ(x) = φ(y) ,

hence Δ_{X}φ = φ and F_{T} (Δ_{X}) is the identity map of F_{T} (X).

Now if X, Y, and Z are finite sets, if R ∈ C(Y, X) and S ∈ C(Z, Y), then for any φ : X → T and any z ∈ Z

(S(Rφ))(z) = ⋁_{(z,y)∈S}(Rφ)(y)
= ⋁_{(z,y)∈S} ⋁_{(y,x)∈R} φ(x)
= ⋁_{(z,x)∈SR} φ(x)
= (SRφ)(z) .

By linearity, it follows that F_{T} (β) ∘ F_{T} (α) = F_{T} (β α), for any β ∈ kC(Z, Y) and any α ∈ kC(Y, X).

We now establish the link between the action of correspondences on functions φ : X → T (as in Definition 11.1 above) and the correspondences Γ_{φ} defined in Notation 5.7.

11.3. Lemma. Let T be a finite lattice, let E = Irr(T), and assume that T is distributive. Then, for any finite sets X, Y, any correspondence S ∈ C(Y, X), and any function φ : X → T, we have Γ_{Sφ} = SΓ_{φ}, where Γ_{φ} is defined in Notation 5.7.

Proof : Let y ∈ Y and e ∈ E. Then

\begin{aligned}(y, e) ∈ Γ_{Sφ} & ⟺ e ≤_{T} (Sφ)(y) ⟺ e ≤_{T} ⋁_{(y,x)∈S} φ(x) \\ & ⟺ e = e ∧ ( ⋁_{(y,x)∈S} φ(x)) . \end{aligned}

But, since T is distributive, the latter equality is equivalent to e = ⋁_{(y,x)∈S} (e∧φ(x)).

Now, since e is irreducible, this is in turn equivalent to

∃x ∈ X , (y, x) ∈ S and e ∧ φ(x) = e ⟺ ∃x ∈ X , (y, x) ∈ S and e ≤_{T} φ(x) ⟺ ∃x ∈ X , (y, x) ∈ S and (x, e) ∈ Γ_{φ} ⟺ (y, e) ∈ SΓ_{φ} .

This completes the proof.

11.4. Proposition. Let (E, R) be a finite poset.

(a) For any finite set X

{Γ_{φ} | φ : X → I^{↑}(E, R)} = {S ∈ C(X, E) | SR = S} = C(X, E)R .

(b) The correspondence functor F_{I^{↑}(E,R)} is isomorphic to kC(−, E)R. In particular F_{I^{↑}(E,R)} is a projective object of F_{k}.

(c) The correspondence functor F_{I_{↓}(E,R)} is isomorphic to kC(−, E)R^{op}. In particular F_{I_{↓}(E,R)} is a projective object of F_{k}.

Proof : (a) This is a restatement of Lemma 5.8.

(b) The map

F_{I^{↑}(E,R)}(X) ⟶ kC(X, E)R , φ ↦ Γ_{φ}

is an isomorphism of correspondence functors, by (a) and Lemma 11.3.

(c) follows from (b) and the obvious equality I_{↓}(E, R) = I^{↑}(E, R^{op}).

We now introduce a suitable category L of lattices, as well as its k-linearization kL (see Definition 2.1), such that the assignment T ↦ F_{T} becomes a k-linear functor from kL to F_{k}.

11.5. Definition. Let L denote the following category :

• The objects of L are the finite lattices.

• For any two lattices T and T′, the set L(T′, T) is the set of all maps f : T → T′ which commute with joins, i.e. such that

f( ⋁_{a∈A} a) = ⋁_{a∈A} f(a) ,

for any subset A of T.

• The composition of morphisms in L is the composition of maps.

11.6. Remark. Morphisms in L preserve the order relation, but they are generally not morphisms of lattices in the sense that they need not commute with the meet operation. On the other hand, the case A = ∅ in Definition 11.5 shows that a morphism f : T → T′ in L always maps ˆ0 ∈ T to ˆ0 ∈ T′.

Conversely, if f : T → T′ satisfies f(ˆ0) = ˆ0 and f(a ∨ b) = f(a) ∨ f(b) for all a, b ∈ T, then f is a morphism in L.

Let f : T → T′ be a morphism in the category L. For a finite set X, let F_{f,X} : F_{T} (X) → F_{T′}(X) be the k-linear map sending the function φ : X → T to the function f ∘ φ : X → T′.

11.7. Theorem.

(a) Let f : T → T′ be a morphism in the category L. Then the collection of maps F_{f,X} : F_{T} (X) → F_{T′}(X), for all finite sets X, yields a natural transformation F_{f} : F_{T} → F_{T′} of correspondence functors.

(b) The assignment sending a lattice T to F_{T}, and a morphism f : T → T′ in L to F_{f} : F_{T} → F_{T′}, yields a functor L → F_{k}. This functor extends uniquely to a k-linear functor

F_{?} : kL → F_{k} .

(c) The functor F_{?} is fully faithful.

Proof : (a) Let X and Y be finite sets, let φ : X → T be a function, and let R ∈ C(Y, X) be a correspondence. Then F_{T′}(R)(F_{f,X}(φ)) = F_{T′}(R)(f ∘ φ) = R(f ∘ φ) and F_{f,Y} (F_{T} (R)(φ)) = F_{f,Y} (Rφ) = f ∘ Rφ. We show that they are equal by evaluating at any y ∈ Y :

R(f ∘ φ)(y) = ⋁_{(y,x)∈R}(f ∘ φ)(x)
= ⋁_{(y,x)∈R} f(φ(x))
= f( ⋁_{(y,x)∈R} φ(x))
= (f ∘ Rφ)(y) ,

hence R(f ∘ φ) = f ∘ Rφ, which proves (a).

(b) It follows that the assignment T ↦ F_{T} is a functor L → F_{k}. Since kL is the k-linearization of L, this functor extends uniquely to a k-linear functor F_{?} : kL → F_{k}.

(c) Let S and T be finite lattices, and Φ : F_{S} → F_{T} be a morphism of functors. Thus, for any finite set X, we have a morphism of k-modules Φ_{X} : F_{S}(X) → F_{T} (X) such that for any finite set Y and any correspondence R ⊆ (Y × X), the diagram

\xymatrix{F_{S}(X) \ar[r]^{Φ_{X}} \ar[d]_{F_{S}(R)} & F_{T} (X) \ar[d]^{F_{T} (R)} \\ F_{S}(Y) \ar[r]^{Φ_{Y}} & F_{T} (Y)}

is commutative. In other words, for any function α : X → S

(11.8)  RΦ_{X}(α) = Φ_{Y} (Rα) .

Taking X = S and α = id_{S} in this relation, and setting

φ = Φ_{S}(id_{S}) = ∑_{λ:S→T} u_{λ}λ ,

where u_{λ} ∈ k, this gives

Rφ = Φ_{Y} (R id_{S}) ,

for any Y and any R ⊆ (Y × S).

Given a function β : Y → S and taking R = Ω_{β} := {(y, β(y)) | y ∈ Y }, one can check easily that Ω_{β} id_{S} = β. It follows that

(11.9)  Φ_{Y} (β) = Ω_{β} φ .

Hence Φ is entirely determined by φ. Now Condition 11.8 is fulfilled if and only if, for any finite sets X and Y, any correspondence R ⊆ (Y × X), and any function α : X → S, we have

RΩ_{α}(φ) = Ω_{Rα}(φ) .

In other words

∑_{λ} u_{λ}RΩ_{α}(λ) = ∑_{λ} u_{λ}Ω_{Rα}(λ) .

Hence Condition 11.8 is satisfied if and only if, for any finite sets X and Y, any correspondence R ⊆ (Y ×X), any function α : X → S, and any function ψ : Y → T, we have

(11.10)  ∑_{RΩ_{α}(λ)=ψ} u_{λ} = ∑_{Ω_{Rα}(λ)=ψ} u_{λ} .

But for y ∈ Y

RΩ_{α}(λ)(y) = ⋁_{(y,s)∈RΩ_{α}} λ(s)
= ⋁_{(y,x)∈R} λα(x) .

On the other hand

Ω_{Rα}(λ)(y) = ⋁_{(y,s)∈Ω_{Rα}} λ(s)
= λ(Rα(y))
= λ( ⋁_{(y,x)∈R} α(x)) .

Now take X = S and α = id_{S} in 11.10. Then let Y = B(S) be the set of subsets of S and let R ⊆ (Y × S) be the set of pairs (A, s), where A ⊆ S and s ∈ A.

Then for a given map λ : S → T, let us define ψ : Y → T by ψ = Ω_{R id_{S}}(λ), in other words

∀A ⊆ S, ψ(A) = λ( ⋁_{s∈A} s) .

Suppose that there exists λ′ : S → T such that Ω_{R id_{S}}(λ′) = ψ. Then for A ⊆ S

ψ(A) = λ′( ⋁_{s∈A} s) .

Taking A = {s}, it follows that λ′ = λ. Hence in 11.10 with our specific choices, the right hand side is simply equal to u_{λ}.

On the other hand the left hand side is equal to the sum of u_{λ′}, for all λ′ such that Rλ′ = ψ, that is, satisfying

∀A ⊆ S, ψ(A) = ⋁_{s∈A} λ′(s) .

Again, taking A = {s}, it follows that λ′ = λ. With our specific choices, the left hand side of 11.10 is then equal to u_{λ} if and only if Rλ = Ω_{R id_{S}}(λ), that is, for any A ⊆ S,

⋁_{s∈A} λ(s) = λ( ⋁_{s∈A} s) .

If this condition is not satisfied, then the left hand side of 11.10 is zero (empty sum). In other words u_{λ} = 0 if λ is not a morphism in the category L. It follows that φ = ∑_{λ} u_{λ}λ is a morphism in kL, from S to T. We claim that the image of

this morphism via the functor F_{?} is equal to Φ and this will prove that the functor F_{?} : kL → F_{k} is full. To prove the claim, notice that, for any function β : Y → S, we have

F_{φ,Y} (β) = ∑_{λ} u_{λ}F_{λ,Y} (β) = ∑_{λ} u_{λ}(λ ∘ β) = ∑_{λ} u_{λ}Ω_{β} λ = Ω_{β} φ = Φ_{Y} (β) ,

using the equation 11.9. This proves the claim and completes the proof that F_{?} is full.

It remains to show that the functor F_{?} is faithful. So let φ and ψ be two linear combinations of morphisms S → T in L, which induce the same morphism θ = F_{φ} = F_{ψ} : F_{S} → F_{T}. Evaluating this morphism at the set S gives a map θ_{S} : F_{S}(S) → F_{T} (S), and moreover

θ_{S}(id_{S}) = F_{φ,S}(id_{S}) = φ ∘ id_{S} = φ ∈ F_{T} (S) = k(T^{S}) .

For the same reason, θ_{S}(id_{S}) = F_{ψ,S}(id_{S}) = ψ, hence φ = ψ. This completes the proof of Theorem 11.7.

The connection between finite lattices and correspondence functors also has the following rather remarkable feature.

11.11. Theorem. Let T be a finite lattice. The functor F_{T} is projective in F_{k} if and only if T is distributive.

Proof : Let B(T) be the lattice of subsets of T. Let υ : B(T) → T be the morphism in the category L defined by

∀A ⊆ T, υ(A) = ⋁_{t∈A} t .

This morphism induces a morphism of functors F_{υ} : F_{B(T )} → F_{T}, and F_{υ} is surjective : indeed, if X is a finite set and α : X → T is a function, and if we define α̂ : X → B(T) by

∀x ∈ X, α̂(x) = {α(x)} ,

then, for any x ∈ X

F_{υ}(α̂)(x) = (υ ∘ α̂)(x) = ⋁_{t∈α̂(x)} t = α(x) ,

thus F_{υ}(α̂) = α, so F_{υ} is surjective.

Now if F_{T} is projective, then the morphism F_{υ} splits and there exists a morphism Φ : F_{T} → F_{B(T)} such that F_{υ} ∘ Φ is the identity morphism of F_{T}. It follows from Theorem 11.7 that Φ is of the form ∑_{σ∈M} u_{σ}F_{σ}, where M is a finite set of

morphisms σ : T → B(T) in L, and u_{σ} ∈ k. Moreover F_{υ} ∘ Φ is then equal to ∑_{σ∈M} u_{σ}F_{υ∘σ}, hence there exists at least one such σ ∈ M such that υ ∘ σ is equal to

the identity of T. This means that

∀t ∈ T, t = ⋁_{x∈σ(t)} x .

In particular σ(t) ⊆ [ˆ0, t]_{T} for any t ∈ T. Then for r, s ∈ T

[ˆ0, r ∧ s]_{T} = [ˆ0, r]_{T} ∩ [ˆ0, s]_{T} ⊇ σ(r) ∩ σ(s) ⊇ σ(r ∧ s) ,

because σ is order-preserving. It follows that

r ∧ s ≥ ⋁_{x∈σ(r)∩σ(s)} x ≥ ⋁_{x∈σ(r∧s)} x = r ∧ s ,

hence

r ∧ s = ⋁_{x∈σ(r)∩σ(s)} x .

Now, since σ preserves joins, we obtain, for all r, s, t ∈ T,

t ∧ (r ∨ s) = ⋁_{x∈σ(t)∩σ(r∨s)} x
= ⋁ ; x∈σ(t)∩(σ(r)∪σ(s)) x

= ⋁ ; x∈(σ(t)∩σ(r))∪(σ(t)∩σ(s)) x

= ( ⋁_{x∈σ(t)∩σ(r)} x) ∨ ( ⋁_{x∈σ(t)∩σ(s)} x)
= (t ∧ r) ∨ (t ∧ s) .

In other words the lattice T is distributive.

Conversely, by Theorem 3.4.1 in [St], any finite distributive lattice T is isomorphic to the lattice I_{↓}(E, R) of lower ideals of a finite poset (E, R). By Proposition 11.4, the associated functor F_{T} is projective in F_{k}. This completes the proof of Theorem 11.11.

12. Quotients of functors associated to lattices

We now introduce, for any finite lattice T, a subfunctor of F_{T} naturally associated with the set of irreducible elements of T.

12.1. Notation. Let T be a finite lattice and let (E, R) be the full subposet of its irreducible elements. For a finite set X, let H_{T} (X) denote the k-submodule of F_{T} (X) = k(T^{X}) generated by all functions φ : X → T such that E ⊈ φ(X).

12.2. Proposition. Let T be a finite lattice and let (E, R) be the full subposet of its irreducible elements.

(a) The assignment sending a finite set X to H_{T} (X) ⊆ F_{T} (X) is a subfunctor H_{T} of F_{T}.

(b) The evaluation (F_{T} /H_{T})(X) has a k-basis consisting of (the classes of) all functions φ : X → T such that E ⊆ φ(X).

Proof : (a) Let X and Y be finite sets, let Q ∈ C(Y, X) be a correspondence, and let φ : X → T be a function. Then

(12.3)  (Qφ)(Y) ∩ E ⊆ φ(X) ∩ E .

Indeed, if e ∈ E and e = (Qφ)(y), for y ∈ Y, then

e = ⋁_{(y,x)∈Q} φ(x) .

As e is irreducible in T, there exists x ∈ X such that (y, x) ∈ Q and e = φ(x), and 12.3 follows.

In particular, if φ(X)∩E is a proper subset of E, then (Qφ)(Y)∩E is a proper subset of E. Hence H_{T} is a subfunctor of F_{T}.

(b) This follows from the definitions of F_{T} and H_{T}.

The quotient functor F_{T} /H_{T} plays a important role in the rest of this paper, in particular for the description of the fundamental functors 𝕊_{E,R} in Section 17. Note that the module (F_{T} /H_{T})(X) = F_{T} (X)/H_{T} (X) has a k-basis consisting of (the classes of) all maps φ : X → T such that φ(X) ⊇ E. We now give another characterization of H_{T} (X).

12.4. Proposition. Let T = I^{↑}(E, R) for a finite poset (E, R) and let X be a finite set.

(a) Under the isomorphism F_{T} → kC(−, E)R of Proposition 11.4, H_{T} (X) is isomorphic to the k-submodule of kC(X, E)R generated by the correspondences S which have no retraction, that is, for which there is no U ∈ C(E, X) such that US = R.

(b) Under the isomorphism F_{T} → kC(−, E)R of Proposition 11.4, the image of F_{T} (X)/H_{T} (X) is a free k-module with basis consisting of all the correspondences S ∈ C(X, E)R which have a retraction U ∈ C(E, X).

Proof : By Proposition 11.4, the functor F_{T} is isomorphic to the functor kC(−, E)R by sending, for a finite set X, a function φ : X → I^{↑}(E, R) to the correspondence Γ_{φ} = {(x, e) ∈ X × E | e ∈ φ(x)}.

(a) The set E^{↑} of irreducible elements of the lattice I^{↑}(E, R) is the set of principal upper ideals

[e, ·[_{R}= {f ∈ E | (e, f) ∈ R} ,

for e ∈ E. Let φ : X → I^{↑}(E, R) be such that φ ∉ H_{T} (X), that is, φ(X) ⊇ E^{↑}. Then, for each e ∈ E, there exists x_{e} ∈ X such that φ(x_{e}) = [e, ·[_{R}. Let U ∈ C(E, X) be defined by

U = {(e, x_{e}) | e ∈ E} ⊆ E × X .

Then for any e ∈ E

(Uφ)(e) = ⋃_{(e,x)∈U} φ(x) = φ(x_{e}) = [e, ·[_{R} .

By Lemma 11.3, it follows that

UΓ_{φ} = Γ_{Uφ} = {(e, f) ∈ E × E | f ∈ [e, ·[_{R}} = R ,

so Γ_{φ} has a retraction.

Conversely, let S ∈ C(X, E)R be a correspondence such that there exists a correspondence U ∈ C(X, E) with US = R. Then S = Γ_{φ}, where φ : X → I^{↑}(E, R) is the function defined by φ(x) = {e ∈ E | (x, e) ∈ S}, for any x ∈ X. It follows that US = Γ_{Uφ} = R, or in other words

∀e, f ∈ E, (e, f) ∈ R ⟺ ∃x ∈ X, (e, x) ∈ U, (x, e) ∈ S .

As Δ_{E} ⊆ R, for any e ∈ E, there exists x_{e} ∈ X such that (e, x_{e}) ∈ U and (x_{e}, e) ∈ S. Moreover if (x_{e}, f) ∈ S, then (e, f) ∈ R, and conversely, if (e, f) ∈ R, then (x_{e}, f) ∈ SR = S. In other words, f ∈ φ(x_{e}) if and only if (e, f) ∈ R. It follows that φ(x_{e}) = [e, ·[_{R}, hence φ(X) ⊇ E^{↑}. This proves that φ ∉ H_{T} (X).

(b) This follows from (a).

12.5. Remark. In the special case when R = Δ_{E} is the equality relation, then C(X, E)R = C(X, E) and a retraction of S ∈ C(X, E) is a correspondence U ∈ C(E, X) such that US = id_{E} (a retraction in the usual sense). Moreover, if S ∈ C(X, E) has a retraction, then S is a monomorphism in the category C. It can be shown conversely that any monomorphism in the category C has a retraction. Thus in this case, the evaluation F_{T} (X)/H_{T} (X) of the quotient functor F_{T} /H_{T} has a k-basis consisting of all the monomorphisms in C(X, E).

In order to deal with quotients of the functor F_{T}, we need information on morphisms starting from F_{T}. But we first need a lemma.

12.6. Lemma. Let T be a finite lattice, let (E, R) be the full subposet of its irreducible elements, and let ι : E → T denote the inclusion map. If φ : X → T is a function, then Γ_{φ} ι = φ and Γ_{φ}R^{op} = Γ_{φ}, where Γ_{φ} is defined in Notation 5.7.

Proof : By definition, the map Γ_{φ} ι : X → T satisfies

∀x ∈ X, (Γ_{φ} ι)(x) = ⋁_{(x,e)∈Γ_{φ}} ι(e) = ⋁_{e≤_{T} φ(x)} e = φ(x) ,

as any element t of T is equal to the join of the irreducible elements of T smaller than t. Thus we have Γ_{φ} ι = φ.

The equality Γ_{φ}R^{op} = Γ_{φ} was proved in Lemma 5.8.

12.7. Proposition. Let T be a finite lattice, let (E, R) be the full subposet of its irreducible elements, and let ι : E → T denote the inclusion map. Let M be a correspondence functor.

(a) The k-linear map

Hom_{F_{k}}(F_{T} , M) ⟶ M(E) , Φ ↦ Φ_{E}(ι)

is injective. Its image is contained in the k-submodule

R^{op}M(E) = {m ∈ M(E) | R^{op}m = m} .

(b) If T is distributive, then the image of the above map is equal to R^{op}M(E), so that Hom_{F_{k}}(F_{T} , M) ≅ R^{op}M(E) as k-modules.

Proof : By Lemma 12.6, for any Φ ∈ Hom_{F_{k}}(F_{T} , M) and any map φ : X → T, we have

Φ_{X}(φ) = Φ_{X}(Γ_{φ} ι) = Γ_{φ} Φ_{E}(ι) .

This shows that Φ is entirely determined by Φ_{E}(ι), proving the injectivity of the map Φ ↦ Φ_{E}(ι).

In the special case where φ = ι, we have

Γ_{ι} = {(x, e) ∈ E × E | e ≤_{T} ι(x)} = {(x, e) ∈ E × E | e ≤_{R} x} = R^{op} .

Moreover R^{op}ι = Γ_{ι} ι = ι, so its image Φ_{E}(ι) must also be invariant by R^{op}, proving that the image of the map Φ ↦ Φ_{E}(ι) is contained in R^{op}M(E).

(b) Since T is distributive, we have

Γ_{Qφ} = QΓ_{φ}

by Lemma 11.3. Now given m ∈ R^{op}M(E), we can define Φ : F_{T} → M by setting

Φ_{X}(φ) = Γ_{φ}m , ∀φ : X → T .

This is indeed a natural transformation of functors since

Φ_{Y} (Qφ) = Γ_{Qφ}m = QΓ_{φ}m = QΦ_{X}(φ)

for any correspondence Q ⊆ Y × X. Moreover,

Φ_{E}(ι) = Γ_{ι}m = R^{op}m = m ,

because m ∈ R^{op}M(E) by assumption and R^{op} is idempotent. Thus m is indeed in the image of the map Φ ↦ Φ_{E}(ι).

When k is a field, we can now give some information on simple functors S_{F,Q,V} appearing as quotients of F_{T}. As usual, we prove a more general result over an arbitrary commutative ring k, involving the not necessarily simple functors S_{F,Q,V}.

12.8. Theorem. Let T be a finite lattice and let (E, R) be the full subposet of its irreducible elements. Let (F, Q) be a poset and let V be a k Aut(F, Q)-module generated by a single element (e.g. a simple module).

(a) If S_{F,Q,V} is isomorphic to a quotient of F_{T} , then |F | ≤ |E|.

(b) Assume that F = E. If S_{E,Q,V} is isomorphic to a quotient of F_{T}, then there exists a permutation σ ∈ Σ_{E} such that R^{op} ⊆ ^{σ}Q.

(c) Assume that F = E and that T is distributive. Then S_{E,Q,V} is isomorphic to a quotient of F_{T} if and only if there exists a permutation σ ∈ Σ_{E} such that R^{op} ⊆ ^{σ}Q.

Proof : (a) If S_{F,Q,V} is isomorphic to a quotient of F_{T}, then Hom_{F_{k}}(F_{T} , S_{F,Q,V}) ≠ {0}, so we have S_{F,Q,V} (E) ≠ {0} by Proposition 12.7. But we know that F is a minimal set for S_{F,Q,V}, so |F | ≤ |E|.

(b) If S_{E,Q,V} is isomorphic to a quotient of F_{T}, there exists a nonzero morphism Φ : F_{T} → S_{E,Q,V}. By Proposition 12.7, Φ_{E}(ι) = m ≠ 0 ∈ R^{op}S_{E,Q,V} (E). Now

S_{E,Q,V} (E) = P_{E}f_{Q} ⊗_{k Aut(E,Q)} V

and P_{E}f_{Q} is a free k-module with basis {Δ_{σ}f_{Q} | σ ∈ Σ_{E}}, by Proposition 3.10. Thus we can write

m = ∑_{σ∈Σ_{E}} λ_{σ}Δ_{σ}f_{Q} ⊗ v (λ_{σ} ∈ k) .

Since m ∈ R^{op}S_{E,Q,V} (E), we have R^{op}m = m and so there exists σ ∈ Σ_{E} such that R^{op}Δ_{σ}f_{Q} ≠ 0. Hence R^{op}Δ_{σ}f_{Q}Δ_{σ}^{−1} ≠ 0, that is, R^{op} ⊆ ^{σ}Q, by Lemma 3.8.

(c) One implication follows from (b). Assume now that there exists a permutation σ ∈ Σ_{E} such that R^{op} ⊆ ^{σ}Q. We first note that S_{E,Q,V} is generated by f_{Q} ⊗ v ∈ S_{E,Q,V} (E) = P_{E}f_{Q} ⊗_{k Aut(E,Q)} V, where v is a generator of V. This follows from the definition of S_{E,Q,V} as a quotient of L_{E,P_{E}f_{Q}⊗V} and the fact that any functor L_{E,W} is generated by L_{E,W} (E) = W by definition.

Also S_{E,Q,V} ≅ S_{E,^{σ}Q,^{σ}V} for any σ ∈ Σ_{E}, by construction. Since R^{op} is contained in a conjugate of Q, we can assume that R^{op} ⊆ Q. This is equivalent to R^{op}f_{Q} = f_{Q}, by Lemma 3.8.

Thus m = f_{Q} ⊗v ∈ S_{E,Q,V} (E) is invariant under left multiplication by R^{op}. By Proposition 12.7 and the assumption that T is distributive, there exists a morphism Φ : F_{T} → S_{E,Q,V} such that Φ_{E}(ι) = f_{Q} ⊗ v. Since this is nonzero and generates S_{E,Q,V}, this functor is isomorphic to a quotient of F_{T}.

13. The fundamental functor associated to a poset

The fundamental functor 𝕊_{E,R} associated to a poset (E, R) was introduced in Definition 4.7 and some description of its evaluations is given in Lemma 4.9. One of our important goals is to give a more precise description of its evaluations, but this will be fully achieved only in Section 17. We prepare the ground by proving several main results on 𝕊_{E,R}. By Lemma 3.14, we already know that E is a minimal set for 𝕊_{E,R} and that 𝕊_{E,R}(E) is the fundamental module P_{E}f_{R}, which is described in Proposition 3.10.

The following theorem establishes the link between the quotient functor F_{T} /H_{T} and the fundamental correspondence functors.

13.1. Theorem. Let T be a finite lattice, let (E, R) be the full subposet of its irreducible elements, and let ι : E → T denote the inclusion map.

(a) There is a unique homomorphism of kC(E, E)-module

θ : (F_{T} /H_{T})(E) → P_{E}f_{R^{op}}

sending ι + H_{T} (E) to f_{R^{op}}. Moreover θ is an isomorphism.

(b) There exists a unique surjective morphism of correspondence functors

Θ_{T} : F_{T} /H_{T} → 𝕊_{E,R^{op}}

such that θ is equal to the composition of Θ_{T,E} : (F_{T} /H_{T})(E) → 𝕊_{E,R^{op}}(E) with the canonical isomorphism 𝕊_{E,R^{op}}(E) →^{≅} P_{E}f_{R^{op}}.

(c) The functor F_{T} is generated by ι ∈ F_{T} (E) and 𝕊_{E,R^{op}} is generated by Θ_{T,E}(ι) ∈ 𝕊_{E,R^{op}}(E).

Proof : (a) The kC(E, E)-module (F_{T} /H_{T})(E) has a k-basis consisting of the classes of maps φ : E → T such that φ(E) ⊇ E, in other words the classes of maps d_{σ} = ισ^{−1} : E → T, where σ ∈ Σ_{E} (the reason for inverting σ here will become clear below). By a small abuse, we view the elements d_{σ} as basis elements of (F_{T} /H_{T})(E).

By Proposition 3.10, the kC(E, E)-module P_{E}f_{R^{op}} has a k-basis {Δ_{σ}f_{R^{op}} | σ ∈ Σ_{E}}. We define

θ : (F_{T} /H_{T})(E) → P_{E}f_{R^{op}} , θ(d_{σ}) = Δ_{σ}f_{R^{op}} ,

extended k-linearly. We only have to check that θ is a homomorphism of kC(E, E)- modules.

Let Q ∈ C(E, E). Then the map Qd_{σ} is defined by

∀e ∈ E, (Qd_{σ})(e) = ⋁_{(e,f)∈Q} d_{σ}(f) = ⋁_{(e,f)∈Q} σ^{−1}(f) .

The image of Qd_{σ} in (F_{T} /H_{T})(E) is nonzero if and only if there exists a permutation ρ ∈ Σ_{E} such that (Qd_{σ})(e) = ρ^{−1}(e), for any e ∈ E, i.e.

∀e ∈ E, ⋁_{(e,f)∈Q} σ^{−1}(f) = ρ^{−1}(e) .

Since ρ^{−1}(e) is irreducible in T, this is equivalent to the following two conditions :

\{ \begin{matrix}∀e ∈ E, ∃f_{e} ∈ E such that (e, f_{e}) ∈ Q and σ^{−1}(f_{e}) = ρ^{−1}(e) , \\ (e, f) ∈ Q ⟹ σ^{−1}(f) ≤_{T} ρ^{−1}(e) . \end{matrix}

The first of these conditions gives f_{e} = σρ^{−1}(e), and (e, σρ^{−1}(e)) ∈ Q for any e ∈ E. In other words Δ_{(σρ^{−1})^{−1}} ⊆ Q, i.e. Δ ⊆ Δ_{σρ^{−1}}Q. For the second condition, note that we have equivalences

σ^{−1}(f) ≤_{T} ρ^{−1}(e) ⟺ σ^{−1}(f) ≤_{R} ρ^{−1}(e) ⟺ (ρ^{−1}(e), σ^{−1}(f)) ∈ R^{op} .

So the second condition is equivalent to Q ⊆ Δ_{ρ}R^{op}Δ_{σ^{−1}}, that is, Q ⊆ Δ_{ρσ^{−1}} ^{σ}R^{op}, or equivalently Δ_{σρ^{−1}}Q ⊆ ^{σ}R^{op}.

Finally, the class of Qd_{σ} in (F_{T} /H_{T})(E) is zero, unless there exists a permutation τ = ρσ^{−1} ∈ Σ_{E} such that

Δ ⊆ Δ_{τ^{−1}}Q ⊆ ^{σ}R^{op} ,

and in that case Qd_{σ} = d_{ρ} = d_{τσ}, hence θ(Qd_{σ}) = θ(d_{ρ}) = Δ_{ρ}f_{R^{op}}. On the other hand, by Proposition 3.10, Qθ(d_{σ}) = QΔ_{σ}f_{R^{op}} is zero, unless there exists a permutation τ ∈ Σ_{E} such that Δ ⊆ Δ_{τ^{−1}}Q ⊆ ^{σ}R^{op}, in which case

Qθ(d_{σ}) = QΔ_{σ}f_{R^{op}} = Δ_{τσ}f_{R^{op}} = Δ_{ρ}f_{R^{op}} = θ(Qd_{σ}) .

It follows that the map θ : (F_{T} /H_{T})(E) → P_{E}f_{R^{op}} is a homomorphism of kC(E, E)- modules. It sends ι = d_{id} to Δ_{id}f_{R^{op}} = f_{R^{op}}. Such a homomorphism is unique, since (F_{T} /H_{T})(E) is generated by ι as a kC(E, E)-module. Moreover θ is clearly an isomorphism, which completes the proof of (a).

(b) Observe first that the functor F_{T} is generated by ι ∈ F_{T} (E) : indeed, if X is a finite set and φ : X → T is any map, recall that Γ_{φ} ι = φ, where Γ_{φ} is the correspondence defined in Lemma 12.6.

By Assertion (a), the map θ^{−1} : P_{E}f_{R^{op}} → (F_{T} /H_{T})(E) is an isomorphism of kC(E, E)-modules, mapping f_{R^{op}} to ι. By Lemma 2.3, there is a unique morphism of correspondence functors π : L_{E,P_{E}f_{R^{op}}} → F_{T} /H_{T} such that π_{E} = θ^{−1}. Since π_{E} is surjective, and since F_{T} /H_{T} is generated by the image of ι in (F_{T} /H_{T})(E), it follows that π is surjective. Moreover, since L_{E,P_{E}f_{R^{op}}} (E) ≅ P_{E}f_{R^{op}}, there exists a unique morphism of correspondence functors λ : kC(−, E) → L_{E,P_{E}f_{R^{op}}} such that λ_{E}(Δ_{E}) = f_{R^{op}} in P_{E}f_{R^{op}}. The morphism λ is surjective, because L_{E,P_{E}f_{R^{op}}} is generated by f_{R^{op}} ∈ L_{E,P_{E}f_{R^{op}}} (E). So we have a diagram

\xymatrix{kC(−, E) \ar@{->>}[d]^{λ} \ar@{->>}[dr]^{πλ} \\ L_{E,P_{E}f_{R^{op}}} \ar@{->>}[r]^{π} \ar@{->>}[dr]_{Θ̃_{T}} & F_{T} /H_{T} \\  & 𝕊_{E,R^{op}}}

where Θ̃_{T} is the canonical surjection L_{E,P_{E}f_{R^{op}}} → S_{E,R^{op},P_{E}f_{R^{op}}} = 𝕊_{E,R^{op}}. Moreover, the composition Θ̃_{T} λ is equal to the morphism

ω_{E,R^{op}} : kC(−E) ⟶ 𝕊_{E,R}

given by Lemma 4.9. All we have to do is to show that Θ̃_{T} factors through π, i.e. that Ker π ⊆ Ker Θ̃_{T}. Equivalently, we have to show that Ker πλ ⊆ Ker ω_{E,R^{op}}.

To see this, let X be a finite set, and let u = ∑_{S∈C(X,E)} u_{S}S be an element of

kC(X, E), where u_{S} ∈ k. Since π_{X}λ_{X}(S) = π_{X}(S ⊗ f_{R^{op}}) = Sπ_{E}(f_{R^{op}}) = Sι, we obtain that u ∈ Ker π_{X}λ_{X} if and only if

(13.2)  ∀φ : X → T, φ(X) ⊇ E ⟹ ∑_{{S∈C(X,E) \atop Sι=φ}} u_{S} = 0 .

On the other hand, by Lemma 4.9, the element u belongs to Ker ω_{E,R^{op},X} if and only if for any V ∈ C(E, X),

∑_{Δ⊆V S⊆R^{op}} u_{S} = 0 .

13.3. Claim. Let Q ∈ C(E, E). Then Δ ⊆ Q ⊆ R^{op} if and only if Qι = ι. Postponing the proof of this claim, we have u ∈ Ker ω_{E,R^{op},X} if and only if

∀V ∈ C(E, X), ∑_{{S∈C(X,E) \atop V Sι=ι}} u_{S} = 0 .

Equivalently

∀V ∈ C(E, X), ∑_{{φ:X→T \atop V φ=ι}} ∑_{{S∈C(X,E) \atop Sι=φ}} u_{S} = 0 .

Now if V φ = ι, then φ ∉ H_{T} (X), otherwise we would have ι ∈ H_{T} (E). Hence φ(X) ⊇ E. Now Condition 13.2 shows that Ker π_{X}λ_{X} ⊆ Ker ω_{E,R^{op},X}.

It remains to prove Claim 13.3. Suppose first that Δ ⊆ Q ⊆ R^{op}. Then for any e ∈ E,

(Qι)(e) = ⋁_{(e,f)∈Q} ι(f) = ⋁_{(e,f)∈Q} f .

Since (e, e) ∈ Q, it follows that (Qι)(e) ≥_{T} e. On the other hand since Q ⊆ R^{op}, if (e, f) ∈ Q, then f ≤_{R} e, hence f ≤_{T} e. Thus (Qι)(e) ≤_{T} e. It follows that (Qι)(e) = e for any e ∈ E, i.e. Qι = ι.

Conversely, if Qι = ι, then

∀e ∈ E, ⋁_{(e,f)∈Q} f = e .

As e is irreducible, it follows that (e, e) ∈ Q for any e ∈ E, i.e. Δ ⊆ Q. Moreover if (e, f) ∈ Q, then f ≤_{T} e, hence f ≤_{R} e and so Q ⊆ R^{op}. This proves the claim and completes the proof of (b).

(c) At the beginning of the proof of (b), we have already noticed that F_{T} is generated by ι ∈ F_{T} (E). Since Θ_{T} : F_{T} /H_{T} → 𝕊_{E,R^{op}} is surjective, 𝕊_{E,R^{op}} is generated by Θ_{T,E}(ι). This completes the proof of Theorem 13.1.

Whenever we need it, we shall view Θ_{T,X} as a map F_{T} (X) → 𝕊_{E,R^{op}}(X) having H_{T} (X) in its kernel. Thus we can evaluate Θ_{T,X} on any element of F_{T} (X), that is, on any map φ : X → T. The following result will be our main tool for analyzing evaluations of fundamental functors and simple functors (see Section 17).

13.4. Theorem. Let T be a finite lattice, let (E, R) be the full subposet of its irreducible elements, and let X be a finite set. The kernel of the map

Θ_{T,X} : F_{T} (X) → 𝕊_{E,R^{op}}(X)

is equal to the set of linear combinations ∑_{φ:X→T} λ_{φ}φ, where λ_{φ} ∈ k, such that for

any map ψ : X → I^{↑}(E, R)

∑_{{φ \atop Γ_{ψ}^{op}Γ_{φ}=R^{op}}} λ_{φ} = 0 .

Here Γ_{φ} = {(x, e) ∈ X × E | e ∈ φ(x)} ⊆ C(X, E), as in Notation 5.7.

Proof : As in the proof of Theorem 13.1 above, there is a sequence of surjective morphisms of correspondence functors

\xymatrix{kC(−, E) \ar@{->>}[r]^{λ} & L_{E,P_{E}f_{R^{op}}} \ar@{->>}[r]^{π} & F_{T} /H_{T} \ar@{->>}[r]^{Θ_{T}} & 𝕊_{E,R^{op}}}

The generator ι ∈ (F_{T} /H_{T})(E) is the image under π_{E} of the element f_{R^{op}} ∈ P_{E}f_{R^{op}} = L_{E,P_{E}f_{R^{op}}} (E). Therefore, any map φ : X → T, viewed as an element of (F_{T} /H_{T})(X), is the image under π_{X} of the element Γ_{φ} ⊗ f_{R^{op}} ∈ L_{E,P_{E}f_{R^{op}}} (X), because

π_{X}(Γ_{φ} ⊗ f_{R^{op}}) = Γ_{φ}π_{E}(f_{R^{op}}) = Γ_{φ} ι = φ ,

by Lemma 12.6. Then Γ_{φ} ⊗ f_{R^{op}} is in turn the image under λ_{X} of Γ_{φ} ∈ kC(X, E).

Recall that the composition Θ_{T} ∘ π ∘ λ is equal to the morphism ω_{E,R^{op}} of Lemma 4.9. Therefore the image of φ : X → T under the map Θ_{T,X} is equal to ω_{E,R^{op}}(Γ_{φ}). It follows that the linear combination u = ∑_{φ:X→T} λ_{φ}φ lies in Ker Θ_{T,X}

if and only if ∑_{φ:X→T} λ_{φ}Γ_{φ} belongs to Ker ω_{E,R^{op}}. By Lemma 4.9

(13.5)  ∀S ∈ C(E, X), ∑_{Δ⊆SΓ_{φ}⊆R^{op}} λ_{φ} = 0 .

But Γ_{φ}R^{op} = Γ_{φ} by Lemma 5.8. Multiplying on the right by R^{op} the inclusions Δ ⊆ SΓ_{φ} ⊆ R^{op}, we see that the sum runs over all φ satisfying SΓ_{φ} = R^{op}. It follows that Condition 13.5 is equivalent to

(13.6)  ∀S ∈ C(E, X), ∑_{SΓ_{φ}=R^{op}} λ_{φ} = 0 .

This holds for S if and only if it holds for R^{op}S, so we can assume that R^{op}S = S. Now R^{op}S = S if and only if S^{op}R = S^{op}. By Lemma 5.8, there exists a map ψ : X → I^{↑}(E, R) such that S^{op} = Γ_{ψ}, that is,

S = Γ_{ψ}^{op} = {(e, x) | e ∈ ψ(x)} .

Thus the condition SΓ_{φ} = R^{op} becomes Γ_{ψ}^{op}Γ_{φ} = R^{op}.

13.7. Corollary. Let X be a finite set. The kernel of the map

\overline{ω}_{E,R^{op},X} : kC(X, E)R^{op} → 𝕊_{E,R^{op}}(X)

is equal to the set of linear combinations ∑_{φ:X→T} λ_{φ}Γ_{φ}, where λ_{φ} ∈ k, such that for

any map ψ : X → I^{↑}(E, R)

∑_{{φ \atop Γ_{ψ}^{op}Γ_{φ}=R^{op}}} λ_{φ} = 0 .

Proof : For the lattice T, we choose T = I_{↓}(E, R), so that F_{T} ≅ kC(−, E)R^{op} by Proposition 11.4. Since R^{op}f_{R^{op}} = f_{R^{op}}, the morphism λ : kC(−, E) → L_{E,P_{E}f_{R^{op}}} factorizes through \overline{λ} : kC(−, E)R^{op} → L_{E,P_{E}f_{R^{op}}}. Moreover the composite

\xymatrix{kC(−, E)R^{op} \ar@{->>}[r]^{\overline{λ}} & L_{E,P_{E}f_{R^{op}}} \ar@{->>}[r]^{π} & F_{T} /H_{T} \ar@{->>}[r]^{Θ_{T}} & 𝕊_{E,R^{op}}}

is equal to the morphism \overline{ω}_{E,R^{op}} of Lemma 4.9, and π ∘\overline{λ} maps Γ_{φ} to φ for any map φ : X → T (because again Γ_{φ} ι = φ). It follows that ∑_{φ:X→T} λ_{φ}φ is in the kernel of

Θ_{T,X} if and only if ∑_{φ:X→T} λ_{φ}Γ_{φ} is in the kernel of \overline{ω}_{E,R^{op},X}. But by Lemma 5.8,

any element V ∈ C(X, E)R^{op} has the form V = Γ_{φ} for some map φ : X → T. It follows that any element in kC(X, E)R^{op} can be written ∑_{φ:X→T} λ_{φ}Γ_{φ} and the result

follows from Theorem 13.4.

Now we want to characterize the condition Γ_{ψ}^{op}Γ_{φ} = R^{op} which appears in both Theorem 13.4 and Corollary 13.7. We use the following notation.

13.8. Notation. Let T be a finite lattice, let (E, R) be the full subposet of its irreducible elements, let ψ : X → I^{↑}(E, R) be any map, and let φ : X → T be any map. We define the function ∧ψ : X → T by

∀x ∈ X, ∧ψ(x) = ⋀_{e∈ψ(x)} e ,

where ⋀ is the meet in the lattice T.

Moreover, the notation φ ≤ ∧ψ means that φ(x) ≤_{T} ∧ψ(x) for all x ∈ X.

13.9. Lemma. Let T be a finite lattice, let (E, R) be the full subposet of its irreducible elements, and let X be a finite set. Let φ : X → T be a map and let Γ_{φ} = {(x, e) ∈ X × E | e ≤_{T} φ(x)} be the associated correspondence. Let ψ : X → I^{↑}(E, R) be a map and let Γ_{ψ}^{op} = {(e, x) ∈ E × X | e ∈ ψ(x)} be the associated correspondence. The following conditions are equivalent.

(a) Γ_{ψ}^{op}φ = ι.
(b) Γ_{ψ}^{op}Γ_{φ}ι = ι.
(c) Δ ⊆ Γ_{ψ}^{op}Γ_{φ} ⊆ R^{op}.
(d) Γ_{ψ}^{op}Γ_{φ} = R^{op}.
(e) φ ≤ ∧ψ and ∀e ∈ E, ∃x ∈ X such that φ(x) = e and ψ(x) = [e, ·[_{E}.
(f) ∀t ∈ T, ψ(φ^{−1}(t)) ⊆ [t, ·[_{T} ∩E and ∀e ∈ E, ψ(φ^{−1}(e)) = [e, ·[_{E}.

Proof : (a) ⇔ (b). By Lemma 12.6, we have φ = Γ_{φ}ι.

(b) ⇔ (c). This follows from Claim 13.3.

(c) ⇔ (d). If (c) holds, multiply on the right by R^{op} and use the equality Γ_{φ}R^{op} = Γ_{φ} of Lemma 5.8 to obtain (d). On the other hand, it is clear that (d) implies (c).

(d) ⇒ (e). Suppose that Γ_{ψ}^{op}Γ_{φ} = R^{op} and let x ∈ X. Then for all f ≤_{T} φ(x) and for all e ∈ ψ(x), we have (e, x) ∈ Γ_{ψ}^{op} and (x, f) ∈ Γ_{φ}, hence (e, f) ∈ R^{op}, that is, f ≤_{R} e, hence f ≤_{T} e. Therefore φ(x) = ⋁_{f≤_{T} φ(x)} f ≤_{T} e, whenever e ∈ ψ(x).

Thus

∀x ∈ X, φ(x) ≤_{T} ⋀_{e∈ψ(x)} e = ∧ψ(x) ,

that is, φ ≤ ∧ψ. This shows that the first property in (e) holds.

Since (e, e) ∈ R^{op}, there exists x_{e} ∈ X such that e ≤_{T} φ(x_{e}) and e ∈ ψ(x_{e}). Then for all f ≤_{T} φ(x_{e}), we have (e, x_{e}) ∈ Γ_{ψ}^{op} and (x_{e}, f) ∈ Γ_{φ}, hence (e, f) ∈ R^{op}, that is, f ≤_{R} e, or in other words f ≤_{T} e. Thus again φ(x_{e}) = ⋁_{f≤_{T} φ(x_{e})} f ≤_{T} e,

hence φ(x_{e}) = e. Moreover, if g ∈ E with g ∈ ψ(x_{e}), then (g, x_{e}) ∈ Γ_{ψ}^{op} and (x_{e}, e) ∈ Γ_{φ}, hence (g, e) ∈ R^{op}, that is, e ≤_{T} g. Therefore ψ(x_{e}) ⊆ [e, ·[_{E}. But we also have [e, ·[_{E}⊆ ψ(x_{e}), as e ∈ ψ(x_{e}) and ψ(x_{e}) is an upper ideal of E. Thus ψ(x_{e}) = [e, ·[_{E}. This shows that the second property in (e) holds.

(e) ⇒ (d). For any e ∈ E, there exists x_{e} ∈ X such that φ(x_{e}) = e and ψ(x_{e}) = [e, ·[_{E}. If now (f, e) ∈ R^{op}, then e ≤_{R} f, hence f ∈ ψ(x_{e}). Since we also have e ≤_{R} φ(x_{e}), we obtain (f, x_{e}) ∈ Γ_{ψ}^{op} and (x_{e}, e) ∈ Γ_{φ}. Thus R^{op} ⊆ Γ_{ψ}^{op}Γ_{φ}.

Moreover if (f, e) ∈ Γ_{ψ}^{op}Γ_{φ}, then there exists x ∈ X such that f ∈ ψ(x) and e ≤_{T} φ(x). Since φ ≤ ∧ψ, we have φ(x) ≤_{T} ⋀_{f∈ψ(x)} f. It follows that e ≤_{T} f,

hence e ≤_{R} f, that is, (f, e) ∈ R^{op}. Thus Γ_{ψ}^{op}Γ_{φ} ⊆ R^{op}. Therefore we obtain Γ_{ψ}^{op}Γ_{φ} = R^{op}.

(e) ⇔ (f). We are going to slightly abuse notation by setting, for any subset Y of X, ψ(Y) = ⋃_{x∈Y} ψ(x). Taking t = φ(x), the first condition in (e) is equivalent to

∀t ∈ T, e ∈ ψ(φ^{−1}(t)) ⟹ t ≤_{T} e ,

which in turn is equivalent to

∀t ∈ T, ψ(φ^{−1}(t)) ⊆ [t, ·[_{T} ∩E .

In particular ψ(φ^{−1}(e)) ⊆ [e, ·[_{E} for all e ∈ E because [e, ·[_{T} ∩E = [e, ·[_{E}. But the second condition in (e) says that e must belong to ψ(φ^{−1}(e)), so we get ψ(φ^{−1}(e)) = [e, ·[_{E}. This shows that the second condition in (e) is equivalent to

∀e ∈ E, ψ(φ^{−1}(e)) = [e, ·[_{E} .

This completes the proof of Lemma 13.9.

Condition (d) will play an important role in the proof of Theorem 14.16, condition (f) will be essential in the proof of Theorem 17.5, and condition (e) will be a main tool used in the proof of Theorem 17.10.

14. Duality and opposite lattices

By Theorem 13.1, any fundamental functor 𝕊_{E,R} is isomorphic to a quotient of some functor associated to a lattice. One of our main purposes in this section is to use duality to realize 𝕊_{E,R} as a subfunctor of some functor associated to a lattice. This requires to define a duality between F_{T} and F_{T^{op}}.

Let F be a correspondence functor over k. Recall from Definition 3.5, that the dual F^{♮} of F is the correspondence functor defined on a finite set X by

F^{♮}(X) = Hom_{k} (F (X), k) .

If Y is a finite set and R ⊆ Y ×X, then the map F^{♮}(R) : F^{♮}(X) → F^{♮}(Y) is defined by

∀α ∈ F^{♮}(X), F^{♮}(R)(α) = α ∘ F (R^{op}) .

Recall that L denotes the category of finite lattices (Definition 11.5), and kL its k-linearization (Definition 2.1). For any finite lattice T = (T, ∨, ∧), denote by T^{op} = (T, ∧, ∨) the opposite lattice, i.e. the set T ordered with the opposite partial order. For simplicity throughout this section, we write ≤ for ≤_{T} and ≤^{op} for ≤_{T^{op}}.

14.1. Lemma. The assignment T ↦ T^{op} extends to an isomorphism L → L^{op}, and to a k-linear isomorphism kL → kL^{op}.

Proof : Let f : T_{1} → T_{2} be a morphism in the category L. For any t ∈ T_{2}, let f^{op}(t) denote the join in T_{1} of all the elements x such that f(x) ≤ t, i.e.

(14.2)  f^{op}(t) = ⋁_{f(x)≤t} x .

Then f(f^{op}(t)) = ⋁_{f(x)≤t} f(x) ≤ t, so f^{op}(t) is actually the greatest element of

f^{−1}([ˆ0, t]_{T_{2}}), i.e. f^{−1}([ˆ0, t]_{T_{2}}) = [ˆ0, f^{op}(t)]_{T_{1}}. In other words,

(14.3)  ∀t_{1} ∈ T_{1}, ∀t_{2} ∈ T_{2}, f(t_{1}) ≤ t_{2} ⟺ t_{1} ≤ f^{op}(t_{2}) ,

that is, the pair (f, f^{op}) is an adjoint pair of functors between the posets T_{1} and T_{2}, viewed as categories. In those terms, saying that f is a morphism in L is equivalent to saying that f commutes with colimits in T_{1} and T_{2}. Hence f^{op} commutes with limits, that is, f^{op} commutes with the meet operation, i.e. it is a morphism of lattices T_{2}^{op} → T_{1}^{op}.

In more elementary terms, for any subset A ⊆ T_{2},

[ˆ0, f^{op}( ⋀_{t∈A} t)]_{T_{1}} = f^{−1}([ˆ0, ⋀_{t∈A} t]_{T_{2}})
= f^{−1}( ⋂_{t∈A}[ˆ0, t]_{T_{2}})
= ⋂_{t∈A} f^{−1}([ˆ0, t]_{T_{2}})
= ⋂_{t∈A}[ˆ0, f^{op}(t)]_{T_{1}}
= [ˆ0, ⋀_{t∈A} f^{op}(t)]_{T_{1}} .

It follows that f^{op}( ⋀_{t∈A} t) = ⋀_{t∈A} f^{op}(t), i.e. f^{op} is a morphism T_{2}^{op} → T_{1}^{op} in L.

Now denoting by ≤^{op} the opposite order relations on both T_{1} and T_{2}, Equation 14.3 reads

∀t_{2} ∈ T_{2}, ∀t_{1} ∈ T_{1}, f^{op}(t_{2}) ≤^{op} t_{1} ⟺ t_{2} ≤^{op} f(t_{1}) ,

which shows that the same construction applied to the morphism f^{op} : T_{2}^{op} → T_{1}^{op} yields (f^{op})^{op} = f. This proves that the map f ↦ f^{op} is a bijection from L(T_{2}, T_{1}) to L(T_{1}^{op}, T_{2}^{op}) (recall our notational convention of Section 2 on morphisms in a category).

Now if f : T_{1} → T_{2} and g : T_{2} → T_{3} are morphisms in L, the adjunction 14.3 easily implies that (gf)^{op} = f^{op}g^{op}. It is clear moreover that (id_{T})^{op} = id_{T^{op}} for any finite lattice T. Hence the assignment T ↦ T^{op} and f ↦ f^{op} is an isomorphism L → L^{op}, which extends linearly to an isomorphism kL → kL^{op}.

14.4. Definition. Let T be a finite lattice and let X be a finite set. For two functions φ : X → T and ψ : X → T^{op}, set

(φ, ψ)_{X} = \{ \begin{matrix}1 & if φ ≤ ψ, i.e. if φ(x) ≤_{T} ψ(x), ∀x ∈ X, \\ 0 & otherwise. \end{matrix}

This definition extends uniquely to a k-bilinear form

(−, −)_{X} : F_{T} (X) × F_{T^{op}}(X) → k .

This bilinear form induces a k-linear map Ψ_{T,X} : F_{T^{op}}(X) → (F_{T})^{♮}(X) defined by Ψ_{T,X}(ψ)(φ) = (φ, ψ)_{X}.

We need some notation.

14.5. Notation. Let T be a finite lattice, X and Y finite sets, Q ⊆ Y × X a correspondence, and ψ : X → T^{op} a map. We denote by Q ⋆ ψ the action of the correspondence Q on ψ. In other words, Q ⋆ ψ is the map F_{T^{op}}(Q)(ψ) : Y → T^{op}. Recall that it is defined by

∀y ∈ Y, (Q ⋆ ψ)(y) = ⋀_{(y,x)∈Q} ψ(x) ,

because the join in T^{op} is the meet in T.

14.6. Lemma.

(a) With the notation 14.5, the family of bilinear forms in Definition 14.4 satisfy

(φ, Q ⋆ ψ)_{Y} = (Q^{op}φ, ψ)_{X} .

(b) The family of maps Ψ_{T,X} : F_{T^{op}}(X) → (F_{T})^{♮}(X) form a morphism of correspondence functors Ψ_{T} : F_{T^{op}} → (F_{T})^{♮}.

Proof : (a) We have

φ ≤ Q ⋆ ψ ⟺ ∀y ∈ Y, φ(y) ≤_{T} Q ⋆ ψ(y)
⟺ ∀y ∈ Y, φ(y) ≤_{T} ⋀_{(y,x)∈Q} ψ(x)
⟺ ∀(y, x) ∈ Q, φ(y) ≤_{T} ψ(x)
⟺ ∀x ∈ X, ⋁_{(x,y)∈Q^{op}} φ(y) ≤_{T} ψ(x)
⟺ Q^{op}φ ≤ ψ .

(b) The equation in part (a) also reads

Ψ_{T,X}(ψ)(Q^{op}φ) = Ψ_{T,Y} (Q ⋆ ψ)(φ) ,

that is QΨ_{T,X}(ψ) = Ψ_{T,Y} (Q ⋆ ψ).

14.7. Remark. Let T = I_{↓}(E, R^{op}) be the lattice corresponding to a poset (E, R^{op}). Then T^{op} = I_{↓}(E, R^{op})^{op} is isomorphic, via complementation, to the lattice I_{↓}(E, R). Using the isomorphisms of Proposition 11.4

F_{T} = F_{I_{↓}(E,R^{op})} ≅ kC(−, E)R , F_{T^{op}} = F_{I_{↓}(E,R)} ≅ kC(−, E)R^{op} ,

we can transport the bilinear forms (−, −)_{X} defined in 14.4 and obtain a pairing

kC(−, E)R × kC(−, E)R^{op} ⟶ k .

It is easy to check, using complementation, that this pairing coincides with the one obtained in Remark 9.5.

14.8. Notation. Let T be a finite lattice, X a finite set, and φ : X → T a map. We denote by φ^{⋆} the element of F_{T^{op}}(X) defined by

φ^{⋆} = ∑_{{ρ:X→T \atop ρ≤φ}} μ(ρ, φ)ρ^{∘} ,

where ρ^{∘} is the function ρ, viewed as a map X → T^{op}, and where μ(ρ, φ) is the Möbius function of the poset of maps from X to T, in which ρ ≤ φ if and only if ρ(x) ≤ φ(x) in T for any x ∈ X. Recall that μ(ρ, φ) can be computed as follows :

μ(ρ, φ) = ∏_{x∈X} μ_{T} (ρ(x), φ(x)) ,

where μ_{T} is the Möbius function of the poset T.

14.9. Theorem. Let T be a finite lattice.

(a) Let X be a finite set. The bilinear form (14.4) is nondegenerate, in the strong sense, namely it induces an isomorphism

Ψ_{T,X} : F_{T^{op}}(X) → (F_{T})^{♮}(X) .

More precisely, {φ^{⋆} | φ : X → T } is the dual basis, in F_{T^{op}}(X), of the k-basis of functions X → T, in F_{T} (X).

(b) Ψ_{T} : F_{T^{op}} → (F_{T})^{♮} is an isomorphism of correspondence functors.

(c) The functor T ↦ F_{T^{op}} and the functor T ↦ (F_{T})^{♮} are naturally isomorphic functors from kL to F_{k}^{op}. More precisely, the family of isomorphisms Ψ_{T}, for finite lattices T, form a natural transformation Ψ between the functor T ↦ F_{T^{op}} and the functor T ↦ (F_{T})^{♮}.

Proof : (a) The set {ρ^{∘} | ρ^{∘} : X → T^{op}} is a k-basis of the free k-module F_{T^{op}}(X). It follows that {φ^{⋆} | φ : X → T } is also a k-basis of F_{T^{op}}(X), because the integral matrix of Möbius coefficients μ(ρ, ϕ) is unitriangular, hence invertible over ℤ. Actually its inverse is the adjacency matrix of the order relation ρ ≤ φ on the set of maps X → T.

Now, for any two functions φ, λ : X → T,

(14.10)  (λ, φ^{⋆})_{X} = ∑_{{ρ:X→T \atop ρ≤φ}} μ(ρ, φ)(λ, ρ^{∘})_{X} = ∑_{{ρ:X→T \atop λ≤ρ≤φ}} μ(ρ, φ) = δ_{λ,φ} ,

where δ_{λ,φ} is the Kronecker symbol (the last equality coming from the definition of the Möbius function). This shows that {φ^{⋆} | φ : X → T } is the dual basis, in F_{T^{op}}(X), of the k-basis of functions X → T, in F_{T} (X).

(b) This follows immediately from (a). Another way of seeing this is to build an explicit inverse Φ_{T} of Ψ_{T}. For each finite set X, we define a linear map Φ_{T,X} : (F_{T})^{♮}(X) → F_{T^{op}}(X) by setting

∀α ∈ (F_{T})^{♮}(X), Φ_{T,X}(α) = ∑_{φ:X→T} α(φ)φ^{⋆} .

Then, for any function λ : X → T,

(Ψ_{T,X}Φ_{T,X}(α))(λ) = (λ, Φ_{T,X}(α))_{X} = ∑_{φ:X→T} α(φ)(λ, φ^{⋆})_{X} = α(λ) ,

so Ψ_{T,X}Φ_{T,X} is the identity map of (F_{T})^{♮}(X).

On the other hand, Ψ_{T,X} is injective, because if Ψ_{T,X}(β) = 0, then we write β = ∑_{φ:X→T} a_{φ}φ^{⋆}, where a_{φ} ∈ k, and then for all λ : X → T, we get

0 = Ψ_{T,X}( ∑_{φ:X→T} a_{φ}φ^{⋆})(λ) = (λ, ∑_{φ:X→T} a_{φ}φ^{⋆})_{X} = ∑_{φ:X→T} a_{φ}(λ, φ^{⋆})_{X} = a_{λ} ,

so that β = 0. Therefore Ψ_{T,X} is an isomorphism and Φ_{T,X} is its inverse.

(c) Let T′ be another finite lattice, and let Ψ_{T′} : F_{T′}^{op} → (F_{T′})^{♮} be the corresponding morphism. Let moreover f : T → T′ be a morphism in L. We claim that for any finite set X, the square

\xymatrix{F_{T^{op}}(X) \ar[r]^{Ψ_{T,X}} & (F_{T})^{♮}(X) \\ F_{T′}^{op}(X) \ar[r]_{Ψ_{T′,X}} \ar[u]^{F_{f^{op}}} & (F_{T′})^{♮}(X) \ar[u]_{(F_{f})^{♮}}}

is commutative: indeed, for any functions ψ : X → T′^{op} and φ : X → T,

((F_{f})^{♮}Ψ_{T′,X}(ψ))(φ) = Ψ_{T′,X}(ψ)(f ∘ φ) = (f ∘ φ, ψ)_{X} ,

whereas

(Ψ_{T,X}F_{f^{op}}(ψ))(φ) = (φ, F_{f^{op}}(ψ))_{X} = (φ, f^{op} ∘ ψ)_{X} .

Now by 14.3, we have that

f ∘ φ ≤ ψ ⟺ ∀x ∈ X, f(φ(x)) ≤ ψ(x)
⟺ ∀x ∈ X, φ(x) ≤ f^{op}(ψ(x))
⟺ φ ≤ f^{op} ∘ ψ ,

which proves our claim. This shows that the isomorphisms Ψ_{T}, for finite lattices T, form a natural transformation Ψ of the functor T ↦ F_{T^{op}} to the functor T ↦ (F_{T})^{♮} from kL to F_{k}^{op}. This completes the proof of Theorem 14.9.

14.11. Corollary. Let k be a self-injective ring. Then for any distributive lattice T, the functor F_{T} is projective and injective in F_{k}.

Proof : Since T is distributive, the functor F_{T} is projective by Theorem 11.11, without further assumption on k.

If k is self-injective, the functor sending a k-module A to its k-dual Hom_{k}(A, k) is exact. It follows that the functor M ↦ M^{♮} is an exact contravariant endofunctor of the category F_{k} (where M^{♮} denotes the dual correspondence functor defined in 3.5). Let α : M → N be an injective morphism in F_{k}, and let λ : M → F_{T} be any morphism. Then α^{♮} : N^{♮} → M^{♮} is surjective, and we have the following diagram with exact row in F_{k}

\xymatrix{ & (F_{T})^{♮} \ar[d]^{λ^{♮}} \\ N^{♮} \ar[r]^{α^{♮}} & M^{♮} \ar[r] & 0}

Now (F_{T})^{♮} ≅ F_{T^{op}} by Theorem 14.9, and T^{op} is distributive. Hence F_{T^{op}} is projective in F_{k}, and there exists a morphism β : (F_{T})^{♮} → N^{♮} such that α^{♮} ∘ β = λ^{♮}.

Dualizing once again the previous diagram yields the commutative diagram

\xymatrix{ & (F_{T})^{♮}^{♮} & F_{T} \ar[l]_{η_{F_{T}}} \\ N^{♮}^{♮} \ar[ur]^{β^{♮}} & M^{♮}^{♮} \ar[u]_{λ^{♮}^{♮}} \ar[l]^{α^{♮}^{♮}} \\ N \ar[u]^{η_{N}} & M \ar[u]_{η_{M}} \ar[l]^{α} \ar[uur]_{λ}}

where for any functor M, we denote by η_{M} the canonical morphism from M to M^{♮}^{♮}. Now η_{F_{T}} is an isomorphism, because for any finite set X, the module F_{T} (X) is a finitely generated free k-module. Let ε : N → F_{T} be defined by ε = η_{F_{T}}^{−1} ∘ β^{♮} ∘ η_{N}. Then

ε ∘ α = η_{F_{T}}^{−1} ∘ β^{♮} ∘ η_{N} ∘ α = η_{F_{T}}^{−1} ∘ λ^{♮}^{♮} ∘ η_{M} = η_{F_{T}}^{−1} ∘ η_{F_{T}} ∘ λ = λ .

Thus for any injective morphism α : M → N and any morphism λ : M → F_{T}, there exists a morphism ε : N → F_{T} such that ε ∘ α = λ. Hence F_{T} is injective in F_{k}.

We now want to study the subfunctor generated by a specific element of F_{T^{op}}(E) which will be defined below. We need some more notation. 14.12. Notation. Let T be a finite lattice. If t ∈ T, let r(t) denote the join of all the elements of T strictly smaller than t, i.e.

r(t) = ⋁_{s<t} s

Thus r(t) = t if and only if t is not irreducible. If t is irreducible, then r(t) is the unique maximal element of [ˆ0, t[. 14.13. Notation. Let T be a finite lattice and let (E, R) be the full subposet of its irreducible elements. If A ⊆ E, let η_{A} : E → T be the map defined by

∀e ∈ E, η_{A}(e) = \{ \begin{matrix}r(e) & if e ∈ A \\ e & if e ∉ A \end{matrix} .

Moreover, let γ_{T} denote the element of F_{T^{op}}(E) defined by

γ_{T} = ∑_{A⊆E}(−1)^{|A|}η_{A}^{∘} ,

where η_{A}^{∘} denotes the function η_{A}, viewed as a map E → T^{op}.

We now show that this element γ_{T} has another characterization. Recall that we use a star ⋆, as in Notation 14.5, for the action of a correspondence on evaluations of F_{T^{op}}. 14.14. Lemma. Let T be a finite lattice, let (E, R) be the full subposet of its irreducible elements, and let ι : E → T be the inclusion map.

(a) The element γ_{T} is equal to ι^{∗} (using the notation defined in 14.8).

(b) R ⋆ γ_{T} = γ_{T}.

Proof : (a) By definition, ι^{⋆} = ∑_{ρ≤ι} μ(ρ, ι)ρ^{∘}, where ρ^{∘} is the function ρ, viewed as

a map X → T^{op}, and where μ is the Möbius function of the poset of functions from X to T (see Notation 14.8). Furthermore

μ(ρ, ι) = ∏_{e∈E} μ_{T} (ρ(e), ι(e)) ,

where μ_{T} is the Möbius function of the poset T. Now μ_{T} (ρ(e), ι(e)) = μ_{T} (ρ(e), e) is equal to 0 if ρ(e) < r(e), because in that case the interval ]ρ(e), e[_{T} has a greatest element r(e). Moreover μ_{T} (ρ(e), e) is equal to -1 if ρ(e) = r(e), and to +1 if ρ(e) = e. It follows that the only maps ρ appearing in the sum above are of the form ρ = η_{A} for some subset A ⊆ E and μ(η_{A}, ι) = (−1)^{|A|}. Therefore

ι^{⋆} = ∑_{A⊆E}(−1)^{|A|}η_{A}^{∘} = γ_{T} .

(b) For any A ⊆ E and any e ∈ E,

(R ∗ η_{A})(e) = ⋀_{(e,e′)∈R} η_{A}(e′) = ⋀_{e≤e′} η_{A}(e′) = η_{A}(e) ,

since e < e′ implies η_{A}(e) ≤ e ≤ r(e′) ≤ η_{A}(e′). Therefore R ⋆ γ_{T} = γ_{T}.

Our aim is to show that the subfunctor ⟨γ_{T} ⟩ of F_{T^{op}} generated by γ_{T} is isomorphic to the fundamental correspondence functor 𝕊_{E,R}. We first show that ⟨γ_{T} ⟩ is independent of the choice of T.

14.15. Lemma. Let f : T → T′ be a morphism in L, let E = Irr(T), and let E′ = Irr(T′). Suppose that the restriction of f to E is an isomorphism of posets

f_{|} : E ⟶^{≅} E′.

(a) The map f is surjective and fr(e) = rf(e) for any e ∈ E.

(b) The map f^{op} : T′^{op} → T^{op} restricts to a bijection f_{|}^{op} : E′ ⟶^{≅} E, which is inverse to f_{|}. Moreover f^{op}r(e′) = rf^{op}(e′) for any e′ ∈ E′.

(c) f^{op} : T′^{op} → T^{op} induces an injective morphism F_{f^{op}} : F_{T′}^{op} → F_{T^{op}} and an isomorphism ⟨γ_{T′}⟩ ≅ ⟨γ_{T} ⟩.

Proof : (a) Since any element of T′ is a join of irreducible elements, which are in the image of f, and since f commutes with joins, the map f is surjective.

Let e ∈ E. By assumption f(e) ∈ E′. The condition r(e) < e implies f(r(e)) ≤ f(e). Moreover r(e) = ⋁_{{e_{1}∈E \atop e_{1}<e}} e_{1}, hence fr(e) = ⋁_{{e_{1}∈E \atop e_{1}<e}} f(e_{1}). Thus if fr(e) = f(e),

then there exists e_{1} < e such that f(e_{1}) = f(e), contradicting the assumption on f. It follows that fr(e) ≤ rf(e).

Now rf(e) = ⋁_{{e′∈E′ \atop e′<f(e)}} e′, and each e′ ∈ E′ with e′ < f(e) can be written e′ =

f(e_{1}), for e_{1} ∈ E with e_{1} < e. It follows that rf(e) ≤ ⋁_{{e_{1}∈E \atop e_{1}<e}} f(e_{1}) = fr(e). Thus

rf(e) = fr(e), as was to be shown.

(b) Recall from Equation 14.2 that f^{op} is defined by f^{op}(t′) = ⋁_{f(t)≤t′} t. Let

e′ ∈ E′. Then there exists e ∈ E such that f(e) = e′. Let t ∈ T be such that f(t) ≤ e′ and write t = ⋁_{{e_{1}∈E \atop e_{1}≤t}} e_{1}. For each e_{1} ∈ E with e_{1} ≤ t, we have f(e_{1}) ≤

f(t) ≤ e′ = f(e), hence e_{1} ≤ e, and t ≤ e. It follows that f^{op}(e′) = ⋁_{f(t)≤e′} t = e, so

f_{|}^{op} is a bijection E′ → E, inverse to f_{|}. This proves the first statement in (b).

Now let e ∈ E, and set e′ = f(e) ∈ E′. First we have r(e′) ≤ e′, thus

f^{op}r(e′) ≤ f^{op}(e′) = e. If f^{op}r(e′) = e, then ⋁_{f(t)≤r(e′)} t = e, hence

f(e) ≤ r(e′) < e′ = f(e) ,

a contradiction. Thus f^{op}r(e′) ≤ r(e) = rf^{op}(e′). But we also have

rf^{op}(e′) = r(e) = ⋁_{{e_{1}∈E \atop e_{1}<e}} e_{1} = ⋁_{{e_{1}∈E \atop f(e_{1})<f(e)}} e_{1} ≤ ⋁_{{t∈T \atop f(t)≤rf(e)}} t = f^{op}rf(e) = f^{op}r(e′) ,

so f^{op}r(e′) = rf^{op}(e′), which proves the second statement in (b).

(c) Since f is surjective by (a), so is the morphism F_{f} : F_{T} → F_{T′}. By duality and Theorem 14.9, the morphism F_{f^{op}} : F_{T′}^{op} → F_{T^{op}} can be identified with the dual of F_{f} and is therefore injective. This proves the first statement in (c).

Now for any B ⊆ E′, consider the map η_{B}^{∘} : E′ → T′^{op}. Then for any e′ ∈ E′

f^{op}η_{B}^{∘}(e′) = \{ \begin{matrix}f^{op}(e′) & if e′ ∉ B, \\ f^{op}r(e′) = rf^{op}(e′) & if e′ ∈ B. \end{matrix}

Hence f^{op} ∘ η_{B}^{∘} = η_{f_{|}^{op}(B)}^{∘} ∘ f_{|}^{op}, and therefore f^{op} ∘ γ_{T′} = γ_{T} ∘ f_{|}^{op}. It follows that

F_{f^{op}}(γ_{T′} ∘ f_{|}) = f^{op} ∘ γ_{T′} ∘ f_{|} = γ_{T} ∘ f_{|}^{op} ∘ f_{|} = γ_{T} .

Therefore the injective morphism F_{f^{op}} maps the subfunctor ⟨ γ_{T′} ∘ f_{|} ⟩ isomorphically to the subfunctor ⟨γ_{T} ⟩. But since f_{|} : E → E′ is a bijection, the subfunctor ⟨ γ_{T′} ∘f_{|} ⟩ of F_{T′}^{op} is equal to the subfunctor ⟨γ_{T′}⟩. This proves the second statement in (c).

Recall that we use a star ⋆, as in Notation 14.5, for the action of a correspondence on evaluations of F_{T^{op}}. We now come to our main result.

14.16. Theorem. Let T be a finite lattice and let (E, R) be the full subposet of its irreducible elements.

(a) The subfunctor ⟨γ_{T} ⟩ of F_{T^{op}} generated by γ_{T} is isomorphic to 𝕊_{E,R}.

(b) In other words, for any finite set X, the module 𝕊_{E,R}(X) is isomorphic to the k-submodule of F_{T^{op}}(X) generated by the elements S ⋆ γ_{T}, for S ⊆ X × E.

Proof : In view of Lemma 14.15, it suffices to prove the result in the case when T is the lattice I_{↓}(E, R): indeed, for any other lattice T′ with the same poset (E, R) of irreducible elements, the inclusion E ⊆ T′ extends to a unique map of lattices f : T = I_{↓}(E, R) → T′ which induces the identity E = Irr(T) ⟶^{=} E = Irr(T′). This result is not difficult and will be proved later in an appendix (Proposition 21.4). Then ⟨γ_{T′}⟩ is isomorphic to ⟨γ_{T} ⟩ by Lemma 14.15, so we now assume that T = I_{↓}(E, R).

By Yoneda’s lemma, there is a unique morphism of correspondence functors ξ : kC(−, E) → F_{T^{op}} such that ξ_{E}(Δ_{E}) = γ_{T}, and the image of ξ is the subfunctor ⟨γ_{T} ⟩ of F_{T^{op}} generated by γ_{T}.

Then for any finite set X, the surjection ξ_{X} : kC(X, E) → ⟨γ_{T} ⟩(X) maps the correspondence S ∈ C(X, E) to S ⋆ γ_{T} ∈ ⟨γ_{T} ⟩(X). Since R ⋆ γ_{T} = γ_{T} by Lemma 14.14, the morphism ξ factorizes as

\xymatrix{kC(−, E) \ar@{->>}[r] & kC(−, E)R \ar@{->>}[r]^{\overline{ξ}} & ⟨γ_{T} ⟩ ,}

where the left hand side morphism is right multiplication by R.

By Proposition 11.4, for a finite set X, the set C(X, E)R is the set of correspondences of the form

Γ_{ψ} = {(x, e) ∈ X × E | e ∈ ψ(x)} ,

where ψ is a map from X to I^{↑}(E, R).

We want to prove that, for any finite set X, the kernel of the surjection

\overline{ξ}_{X} : kC(X, E)R → ⟨γ_{T} ⟩(X)

is equal to the kernel of the surjection

\overline{ω}_{E,R,X} : kC(X, E)R → 𝕊_{E,R}(X) .

The kernel of the surjection \overline{ξ}_{X} is the set of linear combinations ∑_{ψ:X→I^{↑}(E,R)} λ_{ψ}Γ_{ψ},

with λ_{ψ} ∈ k, such that

∑_{ψ:X→I^{↑}(E,R)} λ_{ψ}Γ_{ψ} ⋆ γ_{T} = 0 .

Equivalently,

∀φ : X → T, (φ, ∑_{ψ} λ_{ψ}Γ_{ψ} ⋆ γ_{T})_{X} = ∑_{ψ} λ_{ψ}(Γ_{ψ}^{op}φ, γ_{T})_{X} = 0 ,

where (−, −)_{X} is the bilinear form of 14.4, using also Equation 14.6.

Now γ_{T} = ι^{⋆} by Lemma 14.14 and we use ι^{⋆} instead. By Equation 14.10, we have

(Γ_{ψ}^{op}φ, γ_{T})_{X} = (Γ_{ψ}^{op}φ, ι^{⋆}) = δ_{Γ_{ψ}^{op}φ,ι}

and therefore we obtain the condition

∀φ : X → T, ∑_{{ψ:X→I^{↑}(E,R) \atop Γ_{ψ}^{op}φ=ι}} λ_{ψ} = 0 .

For φ : X → T and ψ : X → I^{↑}(E, R), we know from Lemma 13.9 that the property Γ_{ψ}^{op}φ = ι is equivalent to Γ_{ψ}^{op}Γ_{φ} = R^{op}.

It follows that u = ∑_{ψ:X→I^{↑}(E,R)} λ_{ψ}Γ_{ψ} is in the kernel of \overline{ξ}_{X} if and only if

∀φ : X → I_{↓}(E, R), ∑_{{ψ \atop Γ_{ψ}^{op}Γ_{φ}=R^{op}}} λ_{ψ} = 0 .

But the condition Γ_{ψ}^{op}Γ_{φ} = R^{op} is in turn is equivalent to Γ_{φ}^{op}Γ_{ψ} = R, and moreover I_{↓}(E, R) = I^{↑}(E, R^{op}) and I^{↑}(E, R) = I_{↓}(E, R^{op}). This is where we use that the lattice T is equal to I_{↓}(E, R).

Therefore u = ∑_{ψ:X→I_{↓}(E,R^{op})} λ_{ψ}Γ_{ψ} is in the kernel of \overline{ξ}_{X} if and only if

∀φ : X → I^{↑}(E, R^{op}), ∑_{{ψ \atop Γ_{φ}^{op}Γ_{ψ}=R}} λ_{ψ} = 0 .

By Corollary 13.7, this is equivalent to requiring that u ∈ Ker \overline{ω}_{E,R,X}. It follows that Ker \overline{ξ}_{X} = Ker \overline{ω}_{E,R,X}. Consequently, the images of \overline{ξ}_{X} and \overline{ω}_{E,R,X} are isomorphic, that is, ⟨γ_{T} ⟩ ≅ 𝕊_{E,R}. This completes the proof of Theorem 14.16.

Since we now know that the subfunctor ⟨γ_{T} ⟩ of F_{T^{op}} is isomorphic to 𝕊_{E,R}, we use again duality to obtain more.

14.17. Theorem. Let T be a finite lattice and let (E, R) be the full subposet of its irreducible elements. We consider orthogonal k-submodules with respect to the family of k-bilinear forms (−, −)_{X} defined in 14.4.

(a) ⟨γ_{T} ⟩^{⊥} = Ker Θ_{T} (as subfunctors of F_{T}).

(b) F_{T} /⟨γ_{T} ⟩^{⊥} ≅ 𝕊_{E,R^{op}}.

(c) ⟨γ_{T} ⟩^{⊥⊥} ≅ 𝕊E,R^{op}^{♮}.

(d) There is a canonical injective morphism α_{E,R} : 𝕊_{E,R} ⟶ 𝕊E,R^{op}^{♮}.

Proof : (a) Let ∑_{φ:X→T} λ_{φ}φ ∈ F_{T} (X), where X is a finite set. Then

∑_{φ} λ_{φ}φ ∈ ⟨γ_{T} ⟩(X)^{⊥} ⟺ ( ∑_{φ} λ_{φ}φ, Q ⋆ γ_{T} )_{X} = 0 ∀Q ∈ C(X, E)

⟺ ( ∑_{φ} λ_{φ}φ, Q ⋆ γ_{T} )_{X} = 0 ∀Q ∈ C(X, E)R (because R ⋆ γ_{T} = γ_{T})

⟺ ∑_{φ} λ_{φ}(Q^{op}φ, γ_{T} )_{E} = 0 ∀Q ∈ C(X, E)R (by 14.6)

⟺ ∑_{{φ \atop Q^{op}φ=ι}} λ_{φ} = 0 ∀Q ∈ C(X, E)R (by 14.10 and Lemma 14.14)

⟺ ∑_{{φ \atop Γ_{ψ}^{op}φ=ι}} λ_{φ} = 0 ∀ψ : X → I^{↑}(E, R) (by Proposition 11.4)

⟺ ∑_{{φ \atop Γ_{ψ}^{op}Γ_{φ}=R^{op}}} λ_{φ} = 0 ∀ψ : X → I^{↑}(E, R)  (by Lemma 13.9)

⟺ ∑_{φ:X→T} λ_{φ}φ ∈ Ker Θ_{T,X} (by Theorem 13.4)

Therefore ⟨γ_{T} ⟩(X)^{⊥} = Ker Θ_{T,X}.

(b) This follows immediately from (a) and Theorem 13.1.

(c) This follows immediately from (b) and duality.

(d) There is an obvious inclusion ⟨γ_{T} ⟩ ⊆ ⟨γ_{T} ⟩^{⊥⊥}. Now we have ⟨γ_{T} ⟩ ≅ 𝕊_{E,R} by Theorem 14.16 and ⟨γ_{T} ⟩^{⊥⊥} ≅ 𝕊E,R^{op}^{♮} by (c). Thus we obtain a canonical injective

morphism 𝕊_{E,R} ⟶ 𝕊E,R^{op}^{♮}.

14.18. Remark. We will prove later that α_{E,R} : 𝕊_{E,R} ⟶ 𝕊E,R^{op}^{♮} is an isomorphism (see Theorem 18.1). This is easy to prove if k is a field, because the inclusion ⟨γ_{T} ⟩ ⊆ ⟨γ_{T} ⟩^{⊥⊥} must be an equality since the pairing 14.4 is nondegenerate, by Theorem 14.9.

We end this section with a description of the dual of a simple functor. We assume that k is a field and we let S_{E,R,V} be a simple correspondence functor over k. Part (e) of Theorem 14.17 suggests that the index R must become R^{op} after applying duality. We now show that this is indeed the case.

14.19. Theorem. Let k be a field. The dual SE,R,V^{♮} of the simple functor S_{E,R,V} is isomorphic to S_{E,R^{op},V^{♮}}, where V^{♮} denotes the ordinary dual of the k Aut(E, R)- module V.

Proof : We use Section 3 and consider the P_{E}-simple module

T_{R,V} = P_{E}f_{R} ⊗_{k Aut(E,R)} V ,

which is also a simple R_{E}-module (see Theorem 3.11). Recall that P_{E} is a quotient algebra of the algebra R_{E} = kC(E, E). Then we have S_{E,R,V} = S_{E,T_{R,V}}. Clearly the dual SE,T_{R,V}^{♮} is again a simple functor and its minimal set is E again. Moreover, by evaluation at E, we find that

SE,T_{R,V}^{♮} (E) ≅ S_{E,T_{R,V}} (E)^{♮} ≅ TR,V^{♮} .

Here the action of a relation Q ∈ R_{E} on a R_{E}-module W^{♮} is defined by

(Q · α)(w) = α(Q^{op} · w) , ∀α ∈ W^{♮} , ∀w ∈ W .

We are going to define a nondegenerate pairing

⟨−, −⟩ : T_{R,V} × T_{R^{op},V^{♮}} ⟶ k ,

satisfying ⟨Q · x , y⟩ = ⟨x , Q^{op} · y⟩ for all x ∈ T_{R,V}, y ∈ T_{R^{op},V^{♮}}, and Q ∈ R_{E}. This will induce an isomorphism of R_{E}-modules

TR,V^{♮} ≅ T_{R^{op},V^{♮}} .

It will then follow that

SE,T_{R,V}^{♮} ≅ S_{E,T_{R^{op},V ^{♮}}} ,  that is ,  SE,R,V^{♮} ≅ S_{E,R^{op},V^{♮}} ,

as required.

By Proposition 3.10, P_{E}f_{R} has a k-basis {Δ_{σ}f_{R} | σ ∈ Σ_{E}}, where Σ_{E} is the group of all permutations of E. Moreover, it is a free right k Aut(E, R)-module and it follows that we can write

T_{R,V} = P_{E}f_{R} ⊗_{k Aut(E,R)} V = ⨁_{σ∈[Σ_{E}/ Aut(E,R)]} Δ_{σ}f_{R} ⊗ V ,

where [Σ_{E}/ Aut(E, R)] denotes a set of representatives of the left cosets of Aut(E, R) in Σ_{E}. Noticing that Aut(E, R^{op}) = Aut(E, R), we have a similar decomposition

T_{R^{op},V^{♮}} = P_{E}f_{R^{op}} ⊗_{k Aut(E,R)} V^{♮} = ⨁_{τ∈[Σ_{E}/ Aut(E,R)]} Δ_{τ} f_{R^{op}} ⊗ V^{♮} .

We define the pairing

⟨−, −⟩ : T_{R,V} × T_{R^{op},V^{♮}} ⟶ k , ⟨Δ_{σ}f_{R} ⊗ v , Δ_{τ} f_{R^{op}} ⊗ α⟩ = δ_{σ,τ} α(v) ,

where σ, τ ∈ [Σ_{E}/ Aut(E, R)], v ∈ V, α ∈ V^{♮}.

By choosing dual bases of V and V^{♮} respectively, we easily find dual bases of T_{R,V} and T_{R^{op},V^{♮}} respectively, and it follows that this pairing is nondegenerate. We are left with the proof of the required property of this pairing. We consider the action of a relation Q ∈ R_{E}. For its action on P_{E}f_{R}, it suffices to consider its image in the quotient algebra P_{E}, so we can assume that Q is a permuted order, and even an order relation, as the case where Q is a permutation is clear.

By Lemma 3.8, any order relation Q ∈ P_{E} acts on P_{E}f_{R} via

Q · Δ_{σ}f_{R} = Q · ^{σ}f_{R} Δ_{σ} = Q · f_{^{σ}R} Δ_{σ} = \{ \begin{matrix}Δ_{σ}f_{R} & if Q ⊆ ^{σ}R , \\ 0 & otherwise . \end{matrix}

It follows that

⟨Q · Δ_{σ}f_{R} ⊗ v , Δ_{τ} f_{R^{op}} ⊗ α⟩ = \{ \begin{matrix}α(v) & if Q ⊆ ^{σ}R and σ = τ , \\ 0 & otherwise . \end{matrix}

On the other hand

⟨Δ_{σ}f_{R} ⊗ v , Q^{op} · Δ_{τ} f_{R^{op}} ⊗ α⟩ = \{ \begin{matrix}α(v) & if Q^{op} ⊆ ^{τ}R^{op} and σ = τ , \\ 0 & otherwise . \end{matrix}

When σ = τ, the conditions Q ⊆ ^{σ}R and Q^{op} ⊆ ^{τ}R^{op} are equivalent. Therefore, we obtain the required equality

⟨Q · Δ_{σ}f_{R} ⊗ v , Δ_{τ} f_{R^{op}} ⊗ α⟩ = ⟨Δ_{σ}f_{R} ⊗ v , Q^{op} · Δ_{τ} f_{R^{op}} ⊗ α⟩ ,

from which it follows that we have an isomorphism of R_{E}-modules TR,V^{♮} ≅ T_{R^{op},V^{♮}}. This completes the proof.

15. The case of a total order

In this section, we consider in full detail the case of a totally ordered lattice. For a (non negative) integer n ∈ ℕ, we denote by \underline{n} the set \{0, 1, …, n\}, linearly ordered by 0 < 1 < … < n. Then \underline{n} is a lattice, with least element 0 and greatest element n. Moreover x ∨ y = sup(x, y) and x ∧ y = inf(x, y), for any x, y ∈ \underline{n}. We denote by [n] = \{1, …, n\} the set of irreducible elements of \underline{n}, viewed as a full subposet of \underline{n}. If n = 0 (in which case [n] = ∅) and if n = 1 (in which case [n] = {1}), we recover the cases already considered in Examples 4.11 and 4.12. The purpose of this section is to treat the general case.

Let n ∈ ℕ and let T be a finite lattice. By Definition 11.5 and Remark 11.6, a map f : \underline{n} → T is a morphism in L if and only if f(0) = ˆ0 and f is order-preserving, i.e. if f(i) ≤ f(j) for any (i, j) ∈ \underline{n} with i ≤ j.

15.1. Notation. Let n ∈ ℕ, let A ⊆ [n], and let l = |A|. We denote by s_{A} : \underline{n} → \underline{l} the map defined by

∀j ∈ \underline{n}, s_{A}(j) = |]0, j] ∩ A| .

If moreover C ⊆ [l], and A = \{a_{1}, a_{2}, …, a_{l}\} in increasing order, let i_{A,C} : \underline{l} → \underline{n} denote the map defined by

∀j ∈ \underline{l}, i_{A,C}(j) = \{ \begin{matrix}0 & if j = 0 , \\ a_{j} & if j ∉ C , \\ a_{j} − 1 & if j ∈ C . \end{matrix}

Clearly all the functions s_{A} and i_{A,C} are order-preserving and map 0 to 0, so they are morphisms in the category L. Set finally

i_{A} = ∑_{C⊆[l]}(−1)^{|C|}i_{A,C} ∈ kL(\underline{n}, \underline{l}) .

In view of Section 11, i_{A} induces, for every finite set X, a k-linear map

F_{i_{A}} : F\underline{l}(X) ⟶ F\underline{n}(X) ,

where F\underline{l} and F\underline{n} are the correspondence functors associated to the lattices \underline{l} and \underline{n}.

15.2. Lemma. Let n ∈ ℕ, let A ⊆ [n], and let l = |A|. Let X be a finite set, and φ : X → \underline{l} be any function. Then F_{i_{A}}(φ) = 0 unless φ(X) ⊇ [l].

Proof : We have

F_{i_{A}}(φ) = ∑_{C⊆[l]}(−1)^{|C|}i_{A,C} ∘ φ

= ∑_{ψ:X→\underline{n}} ( ∑_{{C⊆[l] \atop i_{A,C}∘φ=ψ}} (−1)^{|C|})ψ .

Moreover

∀x ∈ X, i_{A,C} ∘ φ(x) = \{ \begin{matrix}0 & if φ(x) = 0 \\ a_{φ(x)} & if φ(x) ∉ C \\ a_{φ(x)} − 1 & if φ(x) ∈ C \end{matrix} .

For a given function ψ : X → \underline{n}, setting

Z_{ψ} = ψ^{−1}(0) ,
U_{ψ} = {x ∈ X − Z_{ψ} | ψ(x) = a_{φ(x)} − 1} ,
V_{ψ} = {x ∈ X − Z_{ψ} | ψ(x) = a_{φ(x)}} ,

we have i_{A,C} ∘ φ = ψ if and only if the following three conditions hold :

X = Z_{ψ}⊔U_{ψ}⊔V_{ψ} , (x ∈ U_{ψ} ⟺ φ(x) ∈ C) , and (x ∈ V_{ψ} ⟺ φ(x) ∉ {0}⊔C) .

In other words

φ(U_{ψ}) ⊆ C ⊆ {1, …, l} − φ(V_{ψ}) .

So if ψ : X → \underline{n} appears in F_{i_{A}}(φ) with a nonzero coefficient, then

(15.3)  φ(U_{ψ}) ⊆ [l] − φ(V_{ψ}) ,

i.e. φ(U_{ψ}) ⊆ [l] and φ(U_{ψ}) ∩ φ(V_{ψ}) = ∅. But if the inclusion in (15.3) is proper, then the coefficient of ψ in F_{i_{A}}(φ) is equal to

∑_{φ(U_{ψ})⊆C⊆[l]−φ(V_{ψ})}(−1)^{|C|} = 0 .

Hence the coefficient of ψ in F_{i_{A}}(φ) is nonzero if and only if

(15.4)  [l] = φ(U_{ψ}) ⊔ (φ(V_{ψ}) ∩ [l]) .

In particular, if [l] ⊈ φ(X), there is no such ψ and F_{i_{A}}(φ) = 0.

15.5. Notation. Let n ∈ ℕ and let A and B be subsets of [n] such that |A| = |B|. We write f_{A,B} = i_{A}s_{B} ∈ End_{kL}(\underline{n}).

15.6. Lemma. Let n ∈ ℕ, let A, B, C, D be subsets of [n] with l = |A| = |B| and m = |C| = |D|. Then :

(a) If C ⊈ B, then s_{B}i_{C} = 0 in kL(\underline{l}, \underline{m}).
(b) In kL(\underline{n}, \underline{m})

i_{A}s_{B}i_{C} = \{ \begin{matrix}0 & if B ≠ C , \\ i_{A} & if B = C . \end{matrix}

(c) In End_{kL}(\underline{n})

f_{A,B}f_{C,D} = \{ \begin{matrix}0 & if B ≠ C , \\ f_{A,D} & if B = C . \end{matrix}

Proof : (a) By definition

s_{B}i_{C} = ∑_{G⊆[m]}(−1)^{|G|}s_{B}i_{C,G}

= ∑_{ψ:\underline{m}→\underline{l}} ( ∑_{{G⊆[m] \atop s_{B}i_{C,G}=ψ}} (−1)^{|G|})ψ .

Write C = \{c_{1}, …, c_{m}\} in increasing order. For j ∈ \underline{l}

s_{B}i_{C,G}(j) = \{ \begin{matrix}0 & if j = 0 , \\ |]0, c_{j}] ∩ B| & if j ∉ G , \\ |]0, c_{j} − 1] ∩ B| & if j ∈ G . \end{matrix}

Set

U_{ψ} = {j ∈ [m] | ψ(j) = |]0, c_{j} −1]∩B|} and V_{ψ} = {j ∈ [m] | ψ(j) = |]0, c_{j}]∩B|} .

Then s_{B}i_{C,G} = ψ if and only if G ⊆ U_{ψ} and [m] − G ⊆ V_{ψ}, i.e.

(15.7)  [m] − V_{ψ} ⊆ G ⊆ U_{ψ}

Thus the coefficient of ψ in s_{B}i_{C} is equal to

∑_{[m]−V_{ψ}⊆G⊆U_{ψ}}(−1)^{|G|} .

This is equal to zero if [m]−V_{ψ} ⊈ U_{ψ} or if [m]−V_{ψ} is a proper subset of U_{ψ}. Hence if ψ appears in s_{B}i_{C}, then [m] = U_{ψ} ⊔ V_{ψ}, i.e.

[m] = {j ∈ [m] | ψ(j) = |]0, c_{j} − 1] ∩ B|} ⊔ {j ∈ [m] | ψ(j) = |]0, c_{j}] ∩ B|} ,

and in this case, the coefficient of ψ in s_{B}i_{C} is equal to (−1)^{|U_{ψ}|}. In particular, for any j ∈ [m], we have |]0, c_{j} − 1] ∩ B| ≠ |]0, c_{j}] ∩ B|, hence c_{j} ∈ B, and C ⊆ B. Thus s_{B}i_{C} = 0 if C ⊈ B.

(b) In particular, if s_{B}i_{C} ≠ 0, then C ⊆ B, and m ≤ l. Now s_{B}i_{C} is a k-linear combination of morphisms ψ ∈ Hom_{L}(\underline{l}, \underline{m}). By Lemma 15.2, the product i_{A}ψ (which is F_{i_{A}}(ψ) by definition) is equal to zero, unless the image of ψ contains [l]. In the latter case, ψ is surjective because ψ(0) = 0. Hence m ≥ l, thus l = m and ψ is bijective. As ψ is order-preserving, it is the identity map.

Moreover B = C because C ⊆ B and l = m. Then, with the notation of part (a), we have

|]0, c_{j}] ∩ B| = |]0, c_{j}] ∩ C| = j = ψ(j) ,

for any j ∈ [m]. Therefore U_{ψ} = ∅, hence (−1)^{|U_{ψ}|} = 1 and the coefficient of ψ = id in s_{B}i_{C} is equal to 1. It follows that i_{A}s_{B}i_{C} = i_{A} id = i_{A} in this case.

(c) This follows from (a) and (b).

Our next result describes completely the structure of the k-algebra End_{kL}(\underline{n}) of all endomorphisms of the lattice \underline{n}.

15.8. Theorem. Let n ∈ ℕ. For l ∈ \underline{n}, let M\binom{n}{l}(k) denote the algebra of square matrices indexed by the set of subsets of [n] of cardinality l, with coefficients in k. For subsets A and B of [n] of cardinality l, let m_{A,B} denote the matrix with coefficient 1 in position (A, B), and 0 elsewhere.

(a) There is an isomorphism of k-algebras

I : ⨁_{l=0}^{n} M\binom{n}{l}(k) ⟶ End_{kL}(\underline{n})

sending m_{A,B} ∈ M\binom{n}{l}(k) to f_{A,B}, for any l ∈ \underline{n} and any pair (A, B) of subsets of [n] with |A| = |B| = l.

(b) In particular, if k is a field, then End_{kL}(\underline{n}) is semi-simple.

Proof : Throughout this proof, X_{n} denotes the set of pairs (A, B) of subsets of [n] such that |A| = |B|. The matrices m_{A,B}, for (A, B) ∈ X_{n}, form a k-basis of

M = ⨁_{l=0}^{n} M\binom{n}{l}(k). They satisfy the relations

m_{A,B} m_{C,D} = \{ \begin{matrix}0 & if B ≠ C \\ m_{A,D} & if B = C \end{matrix} .

Indeed, if |B| ≠ |C|, then m_{A,B} and m_{C,D} are not in the same block of M, so their product is 0. And if |B| = |C| = l, then the corresponding relations are the standard relations within the matrix algebra M\binom{n}{l}(k).

By Lemma 15.6, it follows that I is an algebra homomorphism. Proving that I is an isomorphism is equivalent to proving that the elements f_{A,B}, for (A, B) ∈ X_{n}, form a k-basis of End_{kL}(\underline{n}). Let B_{n} denote the standard k-basis of the latter algebra, i.e. the set of order-preserving functions \underline{n} → \underline{n} which map 0 to 0. For such a function φ, let

A_{φ} = φ(\underline{n}) ∩ [n] and B_{φ} = \{j ∈ [n] | φ(j) ≠ φ(j − 1)\} .

Clearly |A_{φ}| = |φ(\underline{n})| − 1 = |B_{φ}|, thus (A_{φ}, B_{φ}) ∈ X_{n}. Conversely, if (A, B) ∈ X_{n}, let φ_{A,B} denote the element i_{A,∅} s_{B} of End_{L}(\underline{n}). We claim that the maps

σ : φ ↦ (A_{φ}, B_{φ}) and τ : (A, B) ↦ φ_{A,B}

are inverse to each other, hence bijections between B_{n} and X_{n}.

To see this, let (A, B) ∈ X_{n}, and set φ = φ_{A,B}. Let A = {a_{1}, …, a_{r}} in increasing order, and set moreover a_{0} = 0. Then by definition

∀j ∈ \underline{n}, φ(j) = a_{|]0,j]∩B|} .

In particular, for j ∈ [n], we have φ(j) ≠ φ(j − 1) if and only if |]0, j] ∩ B| ≠ |]0, j − 1] ∩ B|, i.e. if and only if j ∈ B. In other words B_{φ} = B. Moreover φ(\underline{n}) = \{a_{0}, a_{1}, …, a_{r}\}, hence A_{φ} = φ(\underline{n}) ∩ [n] = \{a_{1}, …, a_{r}\} = A. This proves that σ ∘ τ is the identity map of X_{n}. In particular τ is injective, and σ is surjective.

Conversely, let φ ∈ B_{n}, and set A = A_{φ} and B = B_{φ}. Let A = {a_{1}, …, a_{r}} in increasing order, and set moreover a_{0} = 0. Similarly, let B = {b_{1}, …, b_{r}} in increasing order, and set moreover b_{0} = 0 and b_{r+1} = n + 1. Then the intervals [b_{i}, …, b_{i+1} − 1], for i ∈ \{0, …, r\} form a partition of \underline{n}, and φ is constant on each of these intervals, and takes two distinct values on any two of these distinct intervals, by definition of B. More precisely φ([b_{i}, …, b_{i+1} − 1]) = {a_{i}}, for any i ∈ \underline{n}. It follows that the map φ can be recovered from the knowledge of the pair (A, B). Therefore σ is injective, hence bijective. Since σ ∘ τ ∘ σ = σ, it follows that τ ∘ σ is the identity map of B_{n}, which completes the proof of the claim.

Now for (A, B) ∈ X_{n}

f_{A,B} = ∑_{C⊆[l]}(−1)^{|C|} i_{A,C} s_{B} ,

and it is clear form the definition of i_{A,C} that i_{A,C} s_{B}(j) ≤ i_{A,∅}s_{B}(j), for all j ∈ \underline{n}. It follows that f_{A,B} is a linear combination of functions all smaller than φ_{A,B}, for the standard ordering of functions \underline{n} → \underline{n}. Moreover, the coefficient of φ_{A,B} in the expansion of f_{A,B} in the basis B_{n} is equal to 1. It follows that the transition matrix from B_{n} to the elements f_{A,B} is triangular, with ones on the diagonal. Hence the elements f_{A,B}, for (A, B) ∈ X_{n}, form a basis of End_{kL}(\underline{n}), and this completes the proof Theorem 15.8.

Theorem 15.8 is similar to the result proved in [FHH] about the planar rook algebra. Over the field ℂ of complex numbers, this algebra is actually isomorphic to End_{ℂL}(\underline{n}). However, the planar rook monoid is not isomorphic to the monoid of endomorphisms of \underline{n} in L. Only the corresponding monoid algebras become isomorphic (over ℂ).

Now we want to use the functor F_{?} : kL → F_{k} to deduce information on the correspondence functor F\underline{n}. By Theorem 11.11, we already know that F\underline{n} is projective, because the total order \underline{n} is a distributive lattice.

15.9. Notation. For n ∈ ℕ, denote by ε_{n} the element i_{[n]} of End_{kL}(\underline{n}). In other words

ε_{n} = ∑_{C⊆[n]}(−1)^{|C|}i_{[n],C} ,

where, as before, i_{[n],C} ∈ End_{L}(\underline{n}) is defined by i_{[n],C}(j) = j − 1 if j ∈ C and i_{[n],C}(j) = j otherwise, for any j ∈ \underline{n}.

15.10. Proposition.

(a) The element ε_{n} is a central idempotent of End_{kL}(\underline{n}). Moreover

ε_{n} End_{kL}(\underline{n}) = kε_{n} .

(b) The kernel of the idempotent endomorphism F_{ε_{n}} of F\underline{n} is equal to the subfunctor H\underline{n}, defined in Notation 12.1.

Proof : (a) If A = [n], then the map s_{A} is the identity map of \underline{n}. Thus ε_{n} = i_{[n]} = i_{[n]}s_{[n]} = f_{[n],[n]}, which is an idempotent by Lemma 15.6. Moreover the inverse image of ε_{n} under the algebra isomorphism

M = ⨁_{l=0}^{n} M\binom{n}{l}(k) ⟶ End_{kL}(\underline{n})

of Theorem 15.8 is the matrix e_{n} = m_{[n],[n]} of the component M\binom{n}{n}(k) = k of M. But e_{n} is central in M and e_{n}M = ke_{n}.

(b) Let φ : X → \underline{n} be any map. Then F_{ε_{n}}(φ) = i_{[n]}φ = 0 by Lemma 15.2, unless φ(X) contains [n]. It follows that H\underline{n} ⊆ Ker F_{ε_{n}}.

To show that Ker F_{ε_{n}} ⊆ H\underline{n}, let X be a finite set and denote by Z_{n}(X) the set of functions φ : X → \underline{n} such that φ(X) ⊇ [n]. Then by definition

F\underline{n}(X) = H\underline{n}(X) ⊕ kZ_{n}(X) .

Let π = π_{X} denote the idempotent endomorphism of the k-module F\underline{n}(X) with image kZ_{n}(X) and kernel H\underline{n}(X). For any φ : X → \underline{n}, we have that

πF_{ε_{n}}(φ) = π( ∑_{C⊆[n]}(−1)^{|C|}(i_{[n],C} ∘ φ)) .

But π(i_{[n],C} ∘ φ) = 0, unless the image of i_{[n],C} ∘ φ contains [n]. In particular i_{[n],C}(\underline{n}) ⊇ [n], hence C = ∅ and i_{[n],C} = id\underline{n}. It follows that πF_{ε_{n}} = π. Hence Ker F_{ε_{n}} ⊆ Ker π = H\underline{n}(X), and Ker F_{ε_{n}} = H\underline{n}. This completes the proof of Proposition 15.10.

15.11. Lemma. Let n ∈ ℕ, let A ⊆ [n], and let l = |A|. Then s_{A}i_{A} = ε_{l} in End_{kL}(\underline{l}).

Proof : By definition

s_{A}i_{A} = ∑_{C⊆[l]}(−1)^{|C|}s_{A}i_{A,C} .

Moreover, if A = \{a_{1}, …, a_{l}\} in increasing order, then for j ∈ \underline{l}

s_{A}i_{A,C}(j) = \{ \begin{matrix}0 & if j = 0 \\ |]0, a_{j}] ∩ A| & if j ∉ C \\ |]0, a_{j} − 1] ∩ A| & if j ∈ C \end{matrix} .

But ]0, a_{j}] ∩ A = {a_{1}, …, a_{j}} and ]0, a_{j} − 1] ∩ A = {a_{1}, …, a_{j−1}}. It follows that |]0, a_{j}] ∩ A| = j, for any j ∈ \underline{l}, and |]0, a_{j} − 1] ∩ A| = j − 1. Hence s_{A}i_{A,C} = i_{[l],C}. Now Lemma 15.11 follows from Notation 15.9.

15.12. Theorem. Let n ∈ ℕ and let 𝕊_{n} denote the image of the idempotent endomorphism F_{ε_{n}} of F\underline{n}.

(a) For an integer l with 0 ≤ l ≤ n, set

β_{l} = ∑_{{A⊆[n] \atop |A|=l}} f_{A,A} .

Then the elements β_{l}, for 0 ≤ l ≤ n, are orthogonal central idempotents of End_{kL}(\underline{n}), and their sum is equal to the identity.

(b) There are isomorphisms of correspondence functors

F_{β_{l}}F\underline{n} ≅ 𝕊_{l}^{⊕\binom{n}{l}} , for 0 ≤ l ≤ n ,
F\underline{n} ≅ ⨁_{A⊆[n]} 𝕊_{|A|} .

Proof : (a) For A ≤ [n] with |A| = l, the inverse image of f_{A,A} under the algebra isomorphism

I : M = ⨁_{l=0}^{n} M\binom{n}{l}(k) ⟶ End_{kL}(\underline{n})

of Theorem 15.8 is the matrix m_{A,A} of the component indexed by l of M. The coefficients of this matrix are all 0, except one coefficient equal to 1 on the diagonal, at place indexed by the subset A of [n]. It follows that the inverse image of β_{l} under I is the identity element of the component indexed by l of M. Assertion (a) follows.

(b) By Theorem 11.7, the functor F_{?} induces an isomorphism of k-algebras

End_{kL}(\underline{n}) ≅ End_{F_{k}}(F\underline{n}) .

Now the idempotents f_{A,A} of End_{kL}(\underline{n}), for A ⊆ [n], are orthogonal, and their sum is equal to the identity. It follows that the endomorphisms F_{f_{A,A}} of F\underline{n} are orthogonal idempotents, and their sum is the identity. Hence we obtain a decomposition of correspondence functors

F\underline{n} = ⨁_{A⊆[n]} F_{f_{A,A}}(F\underline{n}) .

Moreover for any A ⊆ [n] with |A| = l, we have s_{A}f_{A,A} = s_{A}i_{A}s_{A} = ε_{l}s_{A} by Lemma 15.11. It follows that the morphism F_{s_{A}} : F\underline{n} → F\underline{l} restricts to a morphism π_{A} : F_{f_{A,A}}(F\underline{n}) → F_{ε_{l}}(F\underline{l}). Conversely, since f_{A,A}i_{A} = i_{A}s_{A}i_{A} = i_{A} by Lemma 15.6, the morphism F_{i_{A}} : F\underline{l} → F\underline{n} restricts to ρ_{A} : F_{ε_{l}}(F\underline{l}) → F_{f_{A,A}}(F\underline{n}). Since moreover i_{A}s_{A}f_{A,A} = f_{A,A}^{2} = f_{A,A} and s_{A}i_{A} = ε_{l}, the morphisms π_{A} and ρ_{A} are inverse to each other. They are therefore isomorphisms between F_{f_{A,A}}(F\underline{n})

and F_{ε_{l}}(F\underline{l}) ≅ 𝕊_{l}. It follows in particular that F_{β_{l}}F\underline{n} ≅ 𝕊_{l}^{⊕\binom{n}{l}}, which completes the proof.

15.13. Corollary. Let m, n ∈ ℕ. Then

Hom_{F_{k}}(𝕊_{n}, 𝕊_{m}) = \{ \begin{matrix}0 & if n ≠ m , \\ k · id_{𝕊_{n}} & if n = m . \end{matrix}

Proof : Since 𝕊_{n} = F_{ε_{n}}F\underline{n}, the case n = m follows from Assertion (a) of Proposition 15.10. Now for integers l, m ∈ {0, …, n}, we have that

Hom_{F_{k}}(F_{β_{l}}F\underline{n}, F_{β_{m}}F\underline{n}) ≅ Hom_{F_{k}}(𝕊_{l}, 𝕊_{m})^{⊕\binom{n}{l}\binom{n}{m}} .

Since F_{β_{l}} and F_{β_{m}} are central idempotents of End_{F_{k}}(F\underline{n}), and since they are orthogonal if l ≠ m. It follows that Hom_{F_{k}}(F_{β_{l}}F\underline{n}, F_{β_{m}}F\underline{n}) = 0 if l ≠ m, hence Hom_{F_{k}}(𝕊_{l}, 𝕊_{m}) = 0.

Now we prove that the functor 𝕊_{n} is actually isomorphic to a fundamental functor and we compute the dimensions of all its evaluations.

15.14. Theorem. Let 𝕊_{n} denote the image of the idempotent endomorphism F_{ε_{n}} of F\underline{n}.

(a) 𝕊_{n} ≅ F\underline{n}/H\underline{n}.

(b) 𝕊_{n} is isomorphic to the fundamental functor 𝕊_{[n],tot}, where tot denotes the total order on [n].

(c) For any finite set X, the k-module 𝕊_{n}(X) is free of rank

rank(𝕊_{n}(X)) = ∑_{i=0}^{n}(−1)^{n−i}\binom{n}{i}(i + 1)^{|X|} .

Proof : (a) This follows from the second statement in Proposition 15.10.

(b) We are going to use the results of Section 14 applied to the lattice T = \underline{n}^{op}. The set of its irreducible elements is

E = {0, 1, …, n−1} ,

with a total order R being the opposite of the usual order. Now we have

F_{T^{op}} = F_{(\underline{n}^{op})^{op}} = F\underline{n}

and its evaluation at E contains an element

γ_{T} = γ\underline{n}^{op} = ∑_{A⊆E}(−1)^{|A|}η_{A}^{∘} .

Recall from Notation 14.13 that η_{A}^{∘} : E → T^{op} = \underline{n} denotes the same map as η : E → T = \underline{n}^{op} and that η is defined by

∀e ∈ E, η_{A}(e) = \{ \begin{matrix}r(e) & if e ∈ A , \\ e & if e ∉ A , \end{matrix} that is, η_{A}(e) = \{ \begin{matrix}e + 1 & if e ∈ A , \\ e & if e ∉ A , \end{matrix}

because r(e) = e + 1 in the lattice \underline{n}^{op}.

Now we define ω : E → \underline{n} by ω(e) = e + 1. Then ω ∈ F\underline{n}(E) and when we apply the idempotent F_{ε_{n}} we claim that we obtain

(15.15)  F_{ε_{n}}(ω) = (−1)^{n}γ\underline{n}^{op} .

The definition of ε_{n} yields

F_{ε_{n}}(ω) = ∑_{C⊆[n]}(−1)^{|C|}i_{[n],C} ∘ ω

and we have

(i_{[n],C} ∘ ω)(e) = i_{[n],C}(e + 1) = \{ \begin{matrix}e + 1 & if e + 1 ∉ C \\ e & if e + 1 ∈ C \end{matrix}

For each C ⊆ [n] = {1, …, n}, define A to be the complement of C − 1 in E = {0, …, n − 1}, so that C is the complement of A + 1 in [n]. In other words

j ∉ A ⟺ j + 1 ∉ A + 1 ⟺ j + 1 ∈ C .

Then we see that i_{[n],C} ∘ ω = η_{A}^{∘} and it follows that

F_{ε_{n}}(ω) = ∑_{C⊆[n]}(−1)^{|C|}i_{[n],C} ∘ ω
= ∑_{A⊆E}(−1)^{n−|A|}η_{A}^{∘}
= (−1)^{n} ∑_{A⊆E}(−1)^{|A|}η_{A}^{∘}
= (−1)^{n}γ\underline{n}^{op} .

This proves Claim 15.15 above.

Now F\underline{n} is generated by ω ∈ F\underline{n}(E), because it is generated by ι ∈ F\underline{n}([n]) (where ι : [n] → \underline{n} is the inclusion), hence also by any injection from a set E of cardinality n to \underline{n} (by composing with a bijection between E and [n]). Since F_{ε_{n}} is an idempotent endomorphism of the correspondence functor F\underline{n}, we see that F_{ε_{n}}F\underline{n} is generated by F_{ε_{n}}(ω). In other words, in view of Claim 15.15 above, 𝕊_{n} is generated by γ\underline{n}^{op} ∈ F\underline{n}(E). Now Theorem 14.16 asserts that the subfunctor generated by γ_{T} = γ\underline{n}^{op} is isomorphic to 𝕊_{E,R}. But (E, R) ≅ ([n], tot) via the map e ↦ n − e, so we obtain

𝕊_{n} = F_{ε_{n}}F\underline{n} = ⟨γ\underline{n}^{op}⟩ ≅ 𝕊_{E,R} ≅ 𝕊_{[n],tot} .

(c) The k-module 𝕊_{n}(X) is isomorphic to kZ_{n}(X), where Z_{n}(X) is as in the proof of Proposition 15.10 above, namely the set of all maps φ : X → \underline{n} such that [n] ⊆ φ(X) ⊆ \underline{n}. Therefore 𝕊_{n}(X) is free of rank |Z_{n}(X)|. Lemma 8.1 shows that this rank is equal to

|Z_{n}(X)| = ∑_{i=0}^{n}(−1)^{i}\binom{n}{i}(n + 1 − i)^{|X|} = ∑_{j=0}^{n}(−1)^{n−j}\binom{n}{j}(j + 1)^{|X|}

as required.

15.16. Remark. We shall see in Section 17 that a similar formula holds for the rank of the evaluation of any fundamental functor, but the proof in the general case is much more elaborate.

Also, Corollary 15.13 holds more generally for fundamental functors and the general case will be proved in Section 18.

15.17. Corollary. Let k be a field.

(a) The functor 𝕊_{n} is simple, isomorphic to S_{[n],tot,k}, where k is the trivial module for the trivial group Aut([n], tot) = {id}.

(b) 𝕊_{n} is simple, projective, and injective in F_{k}.

Proof : (a) It is clear that Aut([n], tot) is the trivial group, with a single simple module k. Since 𝕊_{n} ≅ 𝕊_{[n],tot} by Theorem 15.14, we obtain (using Remark 4.8) the simple functor

S_{[n],tot,k} = 𝕊_{[n],tot} ⊗_{k} k ≅ 𝕊_{[n],tot} ≅ 𝕊_{n} .

(b) Since \underline{n} is a distributive lattice, F\underline{n} is projective and injective by Corollary 14.11. Therefore so is its direct summand 𝕊_{n}. Hence 𝕊_{n} is simple, projective, and injective.

Our last purpose in this section is to find, for any finite lattice T, direct summands of F_{T} isomorphic to 𝕊_{n}. We start with a lemma, which is probably well known.

15.18. Lemma. Let n ∈ ℕ, and T be a finite lattice. Let U_{n} denote the set of non decreasing sequences u_{0} ≤ u_{1} ≤ … ≤ u_{n−1} ≤ u_{n} = ˆ1 of elements of T, and let V_{n} be the subset of U_{n} consisting of (strictly) increasing sequences.

For u ∈ U_{n}, let ǔ : T → \underline{n} be the map defined by

∀t ∈ T, ǔ(t) = Min\{j ∈ \underline{n} | t ≤ u_{j}\} .

Conversely, if φ is a map from T to \underline{n}, let φ̂ be the sequence of elements of T defined by

∀j ∈ \underline{n}, φ̂_{j} = ⋁_{{t∈T \atop φ(t)≤j}} t .

(a) The assigments u ↦ ǔ and φ ↦ φ̂ are inverse bijections between U_{n} and Hom_{L}(T, \underline{n}).

(b) Let u ∈ U_{n}. The following assertions are equivalent:

(i) The map ǔ is surjective.

(ii) ε_{n}ǔ ≠ 0. (iii) u ∈ V_{n}.

Proof : (a) Let u ∈ U_{n}. Then ǔ ∈ Hom_{L}(T, \underline{n}): indeed for A ⊆ T

ǔ( ⋁_{t∈A} t) = Min\{j ∈ \underline{n} | ⋁_{t∈A} t ≤ u_{j}\}
= Min( ⋂_{t∈A}\{j ∈ \underline{n} | t ≤ u_{j}\})
= Max_{t∈A} (Min\{j ∈ \underline{n} | t ≤ u_{j}\})
= Max_{t∈A} ǔ(t) = ⋁_{t∈A} ǔ(t) .

Conversely, if φ ∈ Hom_{L}(T, \underline{n}), then the sequence φ̂_{j} = ⋁_{{t∈T \atop φ(t)≤j}} t, for j ∈ \underline{n}, is a

nondecreasing sequence of elements of T, and φ̂_{n} = ˆ1. Hence φ̂ ∈ U_{n}.

Now let u ∈ U_{n}, and set φ = ǔ ∈ Hom_{L}(T, \underline{n}). Then for j ∈ \underline{n} and any t ∈ T

φ̂_{j} = ⋁_{{t∈T \atop φ(t)≤j}} t = ⋁_{{t∈T \atop Min\{l∈\underline{n}|t≤u_{l}\}≤j}}t = ⋁_{{t∈T \atop t≤u_{j}}} t = u_{j} ,

thus φ̂ = u.

Conversely, let φ ∈ Hom_{L}(T, \underline{n}), and set u = φ̂. Then for t ∈ T

ǔ(t) = Min\{j ∈ \underline{n} | t ≤ u_{j}\} = Min\{j ∈ \underline{n} | t ≤ ⋁_{{t′∈T \atop φ(t′)≤j}} t′\}

But if t ≤ ⋁_{{t′∈T \atop φ(t′)≤j}} t′, then

φ(t) ≤ φ( ⋁_{{t′∈T \atop φ(t′)≤j}} t′) = Max_{t′∈T, φ(t′)≤j} φ(t′) ≤ j ,

so we obtain

ǔ(t) = Min\{j ∈ \underline{n} | t ≤ ⋁_{{t′∈T \atop φ(t′)≤j}} t′\} ≥ Min\{j ∈ \underline{n} | φ(t) ≤ j\} = φ(t) .

Now if l = φ(t), then in particular φ(t) ≤ l, thus

u_{l} = φ_{l} = ⋁_{{t′∈T \atop φ(t′)≤l}} t′ ≥ t .

Hence φ(t) = l ≥ Min\{j ∈ \underline{n} | t ≤ u_{j}\} = ǔ(t). It follows that φ(t) = ǔ(t), i.e. φ = ǔ, as was to be shown.

(b) Let u ∈ U_{n}. Clearly, if u_{j} = u_{j−1} for some j ∈ [n], then j ∉ Im(ǔ), so ǔ is not surjective. Conversely, if u is strictly increasing, then ǔ(u_{j}) = j for any j ∈ \underline{n}, so ǔ is surjective. This proves that (i) is equivalent to (iii).

Now it follows from Proposition 15.10 that ε_{n}ǔ ≠ 0 if and only if ǔ ∉ H_{n}(T), i.e. the image of ǔ contains [n]. But this holds if and only if ǔ is surjective, since ǔ(ˆ0) = 0. Hence (ii) is equivalent to (i).

15.19. Theorem. Let T be a finite lattice. For n ∈ ℕ, and u ∈ V_{n}, let τ_{u} denote the composition of morphisms

\xymatrix{F_{T} \ar[r]^{Fǔ} & F\underline{n} \ar[r]^{F_{εn}} & 𝕊_{n} .}

(a) The maps τ_{u}, for u ∈ V_{n}, form a k-basis of Hom_{F_{k}}(F_{T} , 𝕊_{n}).

(b) The map

τ = ⨁_{{n∈ℕ \atop u∈V_{n}}} τ_{u} : F_{T} → ⨁_{{n∈ℕ \atop u∈V_{n}}} 𝕊_{n}

is split surjective.

Proof : (a) Since the functor kL → F_{k} defined by T ↦ F_{T} is fully faithful by Theorem 11.7, Hom_{F_{k}}(F_{T} , F\underline{n}) is isomorphic to Hom_{kL}(T, \underline{n}). By Lemma 15.18, this is a free k-module with basis the set of maps ǔ : T → \underline{n}, for u ∈ U_{n}. Therefore Hom_{F_{k}}(F_{T} , F\underline{n}) is a free k-module with basis the set of maps Fǔ : F_{T} → F\underline{n}, for u ∈ U_{n}. Its submodule Hom_{F_{k}}(F_{T} , 𝕊_{n}) is the image under composition with the idempotent F_{ε_{n}}. If u ∈ U_{n} − V_{n}, then ε_{n}ǔ = 0 by Lemma 15.18. Therefore τ_{u} = F_{ε_{n}ǔ} = 0 if u ∈ U_{n} − V_{n} and it follows that the elements τ_{u}, for u ∈ V_{n}, generate Hom_{F_{k}}(F_{T} , 𝕊_{n}). These elements are also linearly independent, since by Proposition 15.10, the kernel of F_{ε_{n}} is equal to H\underline{n}, and since (the images of) the maps ǔ, for u ∈ V_{n}, form part of the k-basis of (F\underline{n}/H\underline{n})(T).

(b) It suffices to show that the map τ is surjective. It will be split because the functors 𝕊_{n} are projective in F_{k}.

We first claim that, if M and N are objects in an abelian category A, if M is projective, and if Hom_{A}(M, N) = 0, then any subobject of M ⊕ N which maps onto both M and N is equal to M ⊕ N. Indeed for such an object L, there is an isomorphism θ : M/(L ∩ M) →^{≅} N/(L ∩ N) sending the class m + (L ∩ M) for m ∈ M, to the class n + (L ∩ N), where n ∈ N is such that m + n ∈ L. (This is a special case of the Goursat lemma on subgroups in a direct product of groups). As M is projective, this isomorphism can be lifted to a morphism ψ : M → N making the following diagram commutative

\xymatrix{M \ar@{->>}[d] \ar[r] \ar[r]^{ψ} & N \ar@{->>}[d] \\ M/(L ∩ M) \ar[r]^{θ}_{≅} & N/(L ∩ N) ,}

where the vertical arrows are the canonical surjections. But ψ = 0 by assumption, hence θ = 0, thus M = L ∩ M and N = L ∩ N, i.e. L = M ⊕ N.

In order to prove the surjectivity of τ, it is enough to show that for any n ∈ ℕ, the map

τ_{n} = ⨁_{u∈V_{n}} τ_{u} : F_{T} → ⨁_{u∈V_{n}} 𝕊_{n}

is surjective. This reduction follows from the claim above, using the fact that, for n ≠ m, we have Hom_{F_{k}}(𝕊_{n}, 𝕊_{m}) = 0 by Corollary 15.13.

We now fix an n ∈ ℕ and show that the map τ_{n} is surjective. We consider the special case where the ground ring is ℤ, and we use the notation F_{T}^{ℤ}, 𝕊_{n}^{ℤ}, τ_{n}^{ℤ} to denote F_{T}, 𝕊_{n}, and τ_{n} in this case. Let C be the cokernel of τ_{n}^{ℤ}. We have an exact sequence

\xymatrix{F_{T}^{ℤ} \ar[r]^{τ_{n}^{ℤ}} & ⨁_{u∈V_{n}} 𝕊_{n}^{ℤ} \ar[r] & C \ar[r] & 0} .

Tensoring this sequence with an arbitrary commutative ring k gives the top row of the following commutative diagram with exact rows

[figure: k ⊗_{ℤ} F_{T}^{ℤ}; ≅; k⊗_{ℤ}τ_{n}^{ℤ}; ⨁_{u∈V_{n}} k ⊗_{ℤ} 𝕊_{n}^{ℤ}; ≅; k ⊗_{ℤ} C; 0; F_{T}; τ_{n}; ⨁_{u∈V_{n}} 𝕊_{n}; k ⊗_{ℤ} C; 0]

Hence if we show that τ_{n} is surjective when k is a field, we get that k ⊗_{ℤ} C = 0 for any field k. In particular, 𝔽_{p} ⊗_{ℤ} C = 0 for any prime p, hence C = 0, and then τ_{n} is surjective for any commutative ring k.

Thus we can assume that k is a field. In this case 𝕊_{n} is a simple functor by Corollary 15.17 and End_{F_{k}}(𝕊_{n}) ≅ k by Corollary 15.13. By (a), we know that the maps τ_{u}, for u ∈ V_{n}, form a k-basis of Hom_{F_{k}}(F_{T} , 𝕊_{n}). By a standard argument of linear algebra, which we sketch below, it follows that the map τ_{n} is surjective, which completes the proof of Theorem 15.19.

The linear algebra argument goes as follows. Let M = ⋂_{u∈V_{n}} Ker τ_{u} ⊆ F_{T}. Then

F_{T} /M embeds via τ_{n} into the semi-simple functor ⨁_{u∈V_{n}} 𝕊_{n}, hence it is semi-simple,

a direct sum of m copies of 𝕊_{n} for some m. Moreover m ≤ v, where v = |V_{n}|. The matrix of the map τ_{n} : F_{T} /M → ⨁_{u∈V_{n}} 𝕊_{n} has coefficients in End_{F_{k}}(𝕊_{n}) ≅ k and has

v rows and m columns. Since the maps τ_{u}, for u ∈ V_{n}, are linearly independent by (a), the rows of this matrix are linearly independent. Therefore the rank of the matrix is v and this forces m ≥ v. Hence m = v and consequently the v columns of the matrix generate the image. This means that τ_{n} is surjective.

16. On the structure of lattices

The purpose of this section is to define, in any finite lattice, two operations r^{∞} and s^{∞}, as well as some special elements which we call bulbs, lying at the bottom of a totally ordered subset with strong properties. Such bulbs play a crucial role in the description of the evaluation of fundamental functors and simple functors, which will be carried out in Section 17.

Let T be a finite lattice and let (E, R) be the full subposet of irreducible elements in T. For simplicity, we write ≤ for ≤_{T} and [t_{1}, t_{2}] for [t_{1}, t_{2}]_{T}. Recall that T is generated by E in the sense that any element x ∈ T is a join of elements of E (see Section 5).

If t ∈ T, recall from Notation 14.12 that r(t) denotes the join of all elements in [ˆ0, t[. If t ∉ E, then t can be written as the join of two smaller elements, so r(t) = t. If e ∈ E, then r(e) is the unique maximal element of [ˆ0, e[. We put r^{k}(t) = r(r^{k−1}(t)) and r^{∞}(t) = r^{n}(t) if n is such that r^{n}(t) = r^{n+1}(t). Note that the map r : T → T is order-preserving.

16.1. Lemma. Let t ∈ T.

(a) r^{∞}(t) ∉ E.

(b) r^{∞}(t) = t if and only if t ∈ T − E.

(c) If e ∈ E, r^{∞}(e) is the unique greatest element of T − E smaller than e.

(d) If t′ ∈ T with t ≤ t′, then r^{∞}(t) ≤ r^{∞}(t′).

Proof : The proof is a straightforward consequence of the definitions.

16.2. Lemma. Let e ∈ E.

(a) [ˆ0, e] = [ˆ0, r^{∞}(e)] ⊔ ]r^{∞}(e), e]

(b) ]r^{∞}(e), e] is contained in E.

(c) [r^{∞}(e), e] is totally ordered.

(d) More precisely, [r^{∞}(e), e] = {r^{n}(e), …, r^{1}(e), e} if r^{∞}(e) = r^{n}(e).

(e) r^{∞}(r^{i}(e)) = r^{∞}(e) for all 0 ≤ i ≤ n − 1.

Proof : (b) This is an immediate consequence of the definition of r^{∞}(e).

(a) Let f ∈ [ˆ0, e]. Then f ∨ r^{∞}(e) ∈ [r^{∞}(e), e]. If f ∨ r^{∞}(e) = r^{∞}(e), then f ∈ [ˆ0, r^{∞}(e)]. Otherwise, f ∨ r^{∞}(e) ∈ ]r^{∞}(e), e], hence f ∨ r^{∞}(e) ∈ E by (b), that is, f ∨ r^{∞}(e) is irreducible. It follows that f ∨ r^{∞}(e) = f or f ∨ r^{∞}(e) = r^{∞}(e). But the second case is impossible since r^{∞}(e) ∉ E. Therefore f ∨ r^{∞}(e) = f, that is, f ∈ ]r^{∞}(e), e].

(c) Let f, g ∈ [r^{∞}(e), e]. We may suppose that they are not both equal to r^{∞}(e). Then f ∨ g ∈ ]r^{∞}(e), e] and again f ∨ g is irreducible by (b). Thus f ∨ g is equal to either f or g. Consequently g ≤ f or f ≤ g.

(d) This is an immediate consequence of (b) and (c).

(e) This follows from the definition of r^{∞}(e).

We write ∧E for the subset of T consisting of all meets of elements of E, that is, elements of the form ⋀_{i∈I} e_{i} where I is a finite set of indices and e_{i} ∈ E for every

i ∈ I. Note that we include the possibility that I be the empty set, in which case one gets the unique greatest element ˆ1.

16.3. Notation. If t ∈ T, define s(t) to be the meet of all the irreducible elements which are strictly larger than t.

Notice that this definition of s is ‘dual’ to the definition of r, because r(t) is the join of all the irreducible elements which are strictly smaller than t. It is clear that the map s : T → T is order-preserving.

In order to describe the effect of s, note first that s(t) = t if t ∈ ∧E − E and s(t) > t if t ∉ ∧E. Now if e ∈ E, there are 3 cases:

(1) If ]e, ˆ1] ∩ E has at least two minimal elements, then s(e) = e.

(2) If ]e, ˆ1] ∩ E has a unique minimal element e^{+}, then s(e) = e^{+}.

(3) If ]e, ˆ1] ∩ E is empty (that is, e is maximal in E), then s(e) = ˆ1.

Note that the equality s(e) = e also occurs in the third case for e = ˆ1, provided ˆ1 is irreducible. We define inductively s^{k}(t) = s(s^{k−1}(t)) and s^{∞}(t) = s^{n}(t) where n is such that s^{n}(t) = s^{n+1}(t).

16.4. Definition. An element t ∈ T is called a bulb if t ∉ ∧E and if there exists e ∈ E such that t = r^{∞}(e) and s(e) = e.

Since the poset ]r^{∞}(e), e] is totally ordered and consists of elements of E, it is clear that r^{∞}(f) = r^{∞}(e) for every f ∈ ]r^{∞}(e), e] (see Lemma 16.2). Since it may happen that s(f) = f for certain elements f ∈ ]r^{∞}(e), e], the element e in the definition above is not necessarily unique. The following lemma shows that e becomes unique if it is chosen minimal among all elements f ∈ E such that r^{∞}(f) = t and s(f) = f.

16.5. Lemma. Let t ∈ T be a bulb and let e ∈ E be a minimal element such that t = r^{∞}(e) and s(e) = e. Let k ≥ 1 be the smallest positive integer such that r^{k}(e) = r^{∞}(e), so that [t, e] = {r^{k}(e), r^{k−1}(e), …, r^{1}(e), e}.

(a) s^{i}(t) = r^{k−i}(e) for 1 ≤ i ≤ k and s^{k}(t) = s^{∞}(t) = e.
(b) [t, s^{∞}(t)] = {r^{k}(e), …, e} = {t, s(t), …, s^{k}(t)}.
(c) e = s^{∞}(t), in other words e is unique. Moreover, s^{∞}(t) ∈ E.
(d) r^{∞}(s^{i}(t)) = t for 1 ≤ i ≤ k.

Proof : By Lemma 16.2, r^{∞}(r^{i}(e)) = r^{∞}(e) = t for all 0 ≤ i ≤ k − 1. If 1 ≤ i ≤ k − 1, then r^{i}(e) ∈ E, r^{∞}(r^{i}(e)) = t and r^{i}(e) < e. By minimality of e, it follows that s(r^{i}(e)) ≠ r^{i}(e), hence s(r^{i}(e)) > r^{i}(e). Moreover, s(r^{i}(e)) ≤ r^{i−1}(e) by definition of s(r^{i}(e)) and the fact that r^{i−1}(e) ∈ E. Since [t, e] is totally ordered, this forces the equality s(r^{i}(e)) = r^{i−1}(e). This equality also holds if i = k because r^{k}(e) = t and t ∉ ∧E, so s(t) > t, and again s(t) ≤ r^{k−1}(e) so that s(t) = r^{k−1}(e).

Then one obtains s^{i}(t) = r^{k−i}(e) for 1 ≤ i ≤ k and in particular s^{k}(t) = s^{∞}(t) = r^{0}(e) = e. The first three statements follow. The fourth is a consequence of Lemma 16.2.

16.6. Example. If T = \underline{n} = \{0, 1, …, n\} is totally ordered, then E = Irr(T) = {1, …, n} and 0 is a bulb in T.

16.7. Notation. We let G_{1} = G_{1}(T) be the set of all bulbs in T and we define

G = ∧E ⊔ G_{1} .

We first show that G has another characterization.

16.8. Lemma. Let G be as above. Then

G = E ⊔ G^{♯} ,

where G^{♯} = {a ∈ T | a = r^{∞}s^{∞}(a)}.

Proof : First observe that E ⊔ G^{♯} is a disjoint union because an element of the form a = r^{∞}s^{∞}(a) satisfies r(a) = a, so it cannot belong to E.

In order to prove that G ⊆ E ⊔ G^{♯}, let a ∈ G. If a ∈ E, then obviously a ∈ E ⊔ G^{♯}. If a ∈ ∧E − E, then a = s(a), hence a = s^{∞}(a). Moreover a = r^{∞}(a) since a ∉ E. Hence a = r^{∞}s^{∞}(a), that is a ∈ G^{♯}. Finally if a ∈ G_{1}, then a ∉ E and a = r^{∞}s^{∞}(a), by definition of a bulb. This proves that G ⊆ E ⊔ G^{♯}.

For the reverse inclusion, first note that E ⊆ G because E ⊆ ∧E. Now let a ∈ G^{♯} and set b = s^{∞}(a). If b ∉ E, then b = r^{∞}(b), hence b = a and a = s(a). It follows that a ∈ ∧E, hence a ∈ G. If now b ∈ E, there are two cases. Either a ∈ ∧E, hence a ∈ G and we are done, or a ∉ ∧E. But then we have b = s(b) and a = r^{∞}(b), so a is a bulb by definition, so that a ∈ G. This proves the inclusion E ⊔ G^{♯} ⊆ G.

The following two propositions will be crucial for our results on evaluations of fundamental functors in Section 17.

16.9. Proposition. Let a ∈ T − ∧E and suppose that s^{∞}(a) ∈ E. Write s^{∞}(a) = s^{m}(a) where m is such that s^{m−1}(a) < s^{m}(a). Let b := r^{∞}(s^{∞}(a)).

(a) There exists 0 ≤ r ≤ m − 1 such that s^{r}(a) < b < s^{r+1}(a).
(b) b ∈ G.
(c) ]s^{r}(a), ˆ1] ∩ E = [b, ˆ1] ∩ E = [s^{r+1}(a), ˆ1] ∩ E.

Proof : (a) Define e_{i} = s^{i}(a) for all 0 ≤ i ≤ m. Note that e_{1}, …, e_{m−1} all belong to E because they belong to ∧E (since they are in the image of the operator s) and moreover s(e_{i}) > e_{i}. Also e_{m} = s^{∞}(a) ∈ E by assumption.

We have a ≤ r^{∞}(s^{∞}(a)) < s^{∞}(a), because s^{∞}(a) ∈ E by assumption. Therefore, there is an integer r ≤ m − 1 such that b ≤ e_{r+1} but b ≰ e_{r}. The inequality b < e_{r+1} is strict because b ∉ E while e_{r+1} ∈ E. (The case r = 0 occurs when a ≤ b < e_{1}.)

In particular b ≤ r^{∞}(e_{r+1}) ≤ r^{∞}(s^{∞}(a)) = b, hence b = r^{∞}(e_{r+1}). We want to show that the element e_{r}∨b cannot be irreducible. Otherwise e_{r}∨b = b or e_{r}∨b = e_{r}. The first case is impossible because b is not irreducible since b = r^{∞}(e_{r+1}) ∉ E. The second case is impossible because it would imply b ≤ e_{r}, contrary the the definition of r. Therefore e_{r} ∨ b ∉ E and e_{r} ∨ b ≤ e_{r+1}. By definition of r^{∞}(e_{r+1}), we obtain e_{r} ∨ b ≤ r^{∞}(e_{r+1}) = b. It follows that e_{r} < b < e_{r+1}, as required.

(b) To prove that b ∈ G, we first note that if b ∈ ∧E, then b ∈ G. Otherwise, b ∉ ∧E, b = r^{∞}(s^{∞}(a)) and s(s^{∞}(a)) = s^{∞}(a), proving that b is a bulb, that is, b ∈ G_{1}.

(c) By the definition of s(e_{r}), there is a unique minimal element in ]e_{r}, ˆ1] ∩ E, namely s(e_{r}) = e_{r+1}. Therefore, ]e_{r}, ˆ1]∩E = [e_{r+1}, ˆ1]∩E. In particular ]e_{r}, ˆ1]∩E = [b, ˆ1] ∩ E.

16.10. Notation. Let σ : G → I^{↑}(E, R) be the map defined by

σ(t) = \{ \begin{matrix}[t, ˆ1] ∩ E & if t ∈ E , \\ ]s^{∞}(t), ˆ1] ∩ E & if t ∉ E . \end{matrix}

For any B ∈ I^{↑}(E, R), define ∧B = ∧_{e∈B} e. By definition of s^{∞}(t), we obtain

∧σ(t) = \{ \begin{matrix}t & if t ∈ E , \\ s^{∞}(t) & if t ∉ E . \end{matrix}

16.11. Proposition. Let t ∈ G and t′ ∈ T such that t′ ≤ ∧σ(t).

(a) s^{∞}(t′) ≤ s^{∞}(t) and r^{∞}(t′) ≤ r^{∞}(t).

(b) t′ ≤ t, except possibly if t is a bulb (i.e. t ∈ G_{1}).

(c) If t′ ≰ t, then t ∈ G_{1} and t = r^{∞}(t′).

Proof : We have G = (∧E − E) ⊔ E ⊔ G_{1} and one considers the three cases for t successively.

If t ∈ ∧E − E, then ∧σ(t) = s^{∞}(t) = t, hence t′ ≤ t and consequently s^{∞}(t′) ≤ s^{∞}(t) and r^{∞}(t′) ≤ r^{∞}(t).

If t ∈ E, then ∧σ(t) = t, hence t′ ≤ t and consequently s^{∞}(t′) ≤ s^{∞}(t) and r^{∞}(t′) ≤ r^{∞}(t).

Finally, if t ∈ G_{1}, then ∧σ(t) = s^{∞}(t), thus s^{∞}(t′) ≤ s^{∞}(t). Moreover, using part (d) of Lemma 16.1 and part (d) of Lemma 16.5, we obtain

r^{∞}(t′) ≤ r^{∞}(∧σ(t)) = r^{∞}(s^{∞}(t)) = t = r^{∞}(t) .

This proves (a), and also (b) because the relation t′ ≰ t can appear only if t ∈ G_{1}. In that case case, by Lemma 16.5, we can write t = r^{∞}(e) where e = s^{∞}(t). Since t′ ≤ s^{∞}(t), we get t′ ∈ [ˆ0, t] ⊔]t, s^{∞}(t)] by Lemma 16.2, hence t′ ∈ ]t, s^{∞}(t)] because t′ ≰ t. In other words, t′ ∈ {s(t), …, s^{k}(t)} by Lemma 16.5. Therefore t′ = s^{i}(t) for some i ≥ 1 and so r^{∞}(t′) = t by Lemma 16.5. This proves (c) and completes the proof.

17. Evaluations of fundamental functors and simple functors

As usual, E denotes a fixed finite set and R an order relation on E. Our purpose is to prove that, for an arbitrary commutative ring k and for any finite set X, the evaluation 𝕊_{E,R}(X) of the fundamental correspondence functor 𝕊_{E,R} is a free k-module, by finding an explicit k-basis. In case k is a field, we will then find the dimension of S_{E,R,V} (X), where S_{E,R,V} is a simple correspondence functor.

Let T be any lattice such that (E, R) is the full subposet of irreducible elements in T. Note that, by Theorem 21.16 in the first appendix, I_{↓}(E, R) is the largest such lattice and all the others are described as quotients of I_{↓}(E, R). By Theorem 13.1, the fundamental functor 𝕊_{E,R^{op}} is isomorphic to a quotient of F_{T} via a morphism Θ_{T} : F_{T} → 𝕊_{E,R^{op}}. For this reason, we work with 𝕊_{E,R^{op}} rather than 𝕊_{E,R}.

17.1. Notation. Let G = G(T) be the subset defined in Notation 16.7 and let X be a finite set. We define B_{X} to be the set of all maps φ : X → T such that E ⊆ φ(X) ⊆ G.

Our main purpose is to prove that the set Θ_{T,X}(B_{X}) is a k-basis of 𝕊_{E,R^{op}}(X). We shall first prove that Θ_{T,X}(B_{X}) generates k-linearly 𝕊_{E,R^{op}}(X). Then, in the second part of this section, we shall show that Θ_{T,X}(B_{X}) is k-linearly independent.

By Lemma 16.8, we have G = E ⊔ G^{♯} and we denote by Γ the complement of G in T, namely

Γ = {a ∈ T | a ∉ E, a < r^{∞}s^{∞}(a)} ,

where r^{∞} and s^{∞} are defined as in Section 16.

17.2. Lemma. Let a ∈ Γ, and let b = r^{∞}s^{∞}(a). There exists an integer r ≥ 0 such that

a < s(a) < … < s^{r}(a) < b = r^{∞}s^{∞}(a) ≤ s^{r+1}(a) ,

and s^{j}(a) ∈ E for j ∈ {1, …, r}. Moreover b = r^{∞}s^{∞}(s^{j}(a)) for j ∈ {1, …, r}, and b ∈ G^{♯}.

Proof : We know that a = s^{0}(a) ∉ E because a ∈ Γ. Suppose first that there exists an integer r ≥ 0 such that c = s^{r+1}(a) ∉ E. In this case, we choose r minimal with this property, so that s^{j}(a) ∈ E for 1 ≤ j ≤ r. We have c = s(s^{r}(a)) ∈ ∧E − E, hence c = s(c) = s^{∞}(c) = s^{∞}(a). Moreover b = r^{∞}(c) = c, because c ∉ E, and b = r^{∞}s^{∞}(b), so b ∈ G^{♯}.

Suppose now that s^{r}(a) ∈ E for all r ∈ ℤ_{>0}. Then Proposition 16.9 applies and there exists an integer r ≥ 0 such that s^{r}(a) < b < s^{r+1}(a). Moreover b ∈ G and b ∉ E, because b = r^{∞}(b), so b ∈ G^{♯}.

17.3. Definition. For a ∈ Γ, the sequence a < s(a) < … < s^{r}(a) < b defined in Lemma 17.2 is called the reduction sequence associated to a.

17.4. Notation. Let n ≥ 1 and let \underline{a} = (a_{0}, a_{1}, …, a_{n}) be a sequence of distinct elements of T. We denote by [a_{0}, …, a_{n}] : T → T the map defined by

∀t ∈ T, [a_{0}, …, a_{n}](t) = \{ \begin{matrix}a_{j+1} & if t = a_{j}, j ∈ \{0, …, n − 1\} \\ t & otherwise. \end{matrix}

We denote by κ\underline{a} the element of k(T^{T}) = F_{T} (T) defined by

κ\underline{a} = [a_{0}, a_{1}] − [a_{0}, a_{1}, a_{2}] + … + (−1)^{n−1}[a_{0}, a_{1}, …, a_{n}] .

For a ∈ Γ, we set u_{a} = κ\underline{a}, where \underline{a} is the reduction sequence associated to a.

17.5. Theorem. Let T be a finite lattice, let (E, R) be the full subposet of its irreducible elements, let Γ = {a ∈ T | a ∉ E, a < r^{∞}s^{∞}(a)}, and let X be a finite set. Then

∀a ∈ Γ, ∀φ : X → T, φ − u_{a} ∘ φ ∈ Ker Θ_{T,X} ,

where u_{a} ∘φ is defined by bilinearity from the composition of maps T^{T} ×T^{X} → T^{X}.

Proof : The kernel of the map Θ_{T,X} : F_{T} (X) → 𝕊_{E,R^{op}}(X) was given in Theo-

rem 13.4. Let ∑_{φ:T →X} λ_{φ}φ ∈ F_{T} (X), where λ_{φ} ∈ k. Then ∑_{φ:T →X} λ_{φ}φ ∈ Ker Θ_{T,X} if

and only if the coefficients λ_{φ} satisfy a system of linear equations indexed by maps ψ : X → I^{↑}(E, R). The equation (E_{ψ}) indexed by such a map ψ is the following :

(E_{ψ}) :  ∑_{φ ⊢_{E,R} ψ} λ_{φ} = 0 ,

where φ ⊢_{E,R} ψ means that φ : X → T and ψ : X → I^{↑}(E, R) satisfy the equivalent

conditions of Lemma 13.9. We shall use condition (f) of Lemma 13.9, namely

φ ⊢_{E,R} ψ ⟺ \{ \begin{matrix}∀t ∈ T, ψ(φ^{−1}(t)) ⊆ [t, ·[_{T} ∩E , \\ ∀e ∈ E, ψ(φ^{−1}(e)) = [e, ·[_{E} . \end{matrix}

Let a ∈ Γ, and let (a, e_{1}, e_{2}, …, e_{r}, b) be the associated reduction sequence. For i ∈ {1, …, r}, set φ_{i} = [a, e_{1}, …, e_{i}] ∘ φ. Also define φ_{r+1} = [a, e_{1}, …, e_{r}, b] ∘ φ. In particular, for any i ∈ {1, …, r + 1},

if φ(x) ∈ T − {a, e_{1}, …, e_{r}} , then φ_{i}(x) = φ(x) .

The other values of the maps φ_{i} are given in the following table:

(17.6)  \begin{matrix}x ∈ & φ^{−1}(a) & φ^{−1}(e_{1}) & φ^{−1}(e_{2}) & … & φ^{−1}(e_{r−1}) & φ^{−1}(e_{r}) \\ φ(x) & a & e_{1} & e_{2} & … & e_{r−1} & e_{r} \\ φ_{1}(x) & e_{1} & e_{1} & e_{2} & … & e_{r−1} & e_{r} \\ φ_{2}(x) & e_{1} & e_{2} & e_{2} & … & e_{r−1} & e_{r} \\ φ_{3}(x) & e_{1} & e_{2} & e_{3} & … & e_{r−1} & e_{r} \\ … & … & … & … & … & … & … \\ φ_{r}(x) & e_{1} & e_{2} & e_{3} & … & e_{r} & e_{r} \\ φ_{r+1}(x) & e_{1} & e_{2} & e_{3} & … & e_{r} & b \end{matrix}

We want to prove that the element

φ − u_{a} ∘ φ = φ − φ_{1} + φ_{2} − … + (−1)^{r−1}φ_{r+1}

belongs to Ker Θ_{T,X}. We must prove that it satisfies the equation (E_{ψ}) for every ψ, so we must find which of the functions φ, φ_{1}, φ_{2}, …, φ_{r+1} are linked with ψ under the relation ⊢_{E,R}. We are going to prove that only two consecutive functions can be

linked with a given ψ, from which it follows that the corresponding equation (E_{ψ}) reduces to either 1 − 1 = 0, or −1 + 1 = 0. It follows from this that φ − u_{a} ∘ φ satisfies all equations (E_{ψ}), hence belongs to Ker Θ_{T,X}, as required.

The linking with ψ is controlled by the following conditions:

φ ⊢_{E,R} ψ ⟺ \{ \begin{matrix}∀t ∈ U, ψ(φ^{−1}(t)) ⊆ [t, ·[_{T} ∩E \\ ∀e ∈ V, ψ(φ^{−1}(e)) = [e, ·[_{E} \\ ψ(φ^{−1}(a)) ⊆ [a, ·[_{T} ∩E = [e_{1}, ·[_{E} \\ ψ(φ^{−1}(e_{i})) = [e_{i}, ·[_{E} ∀i ∈ \{1, …, r\} . \end{matrix}

Similarly, since φ_{1}^{−1}(e_{1}) = φ^{−1}(a) ⊔ φ^{−1}(e_{1}),

φ_{1} ⊢_{E,R} ψ ⟺ \{ \begin{matrix}∀t ∈ U, ψ(φ^{−1}(t)) ⊆ [t, ·[_{T} ∩E \\ ∀e ∈ V, ψ(φ^{−1}(e)) = [e, ·[_{E} \\ ψ(φ^{−1}(a) ⊔ φ^{−1}(e_{1})) = [e_{1}, ·[_{E} \\ ψ(φ^{−1}(e_{i})) = [e_{i}, ·[_{E} ∀i ∈ \{2, …, r\} . \end{matrix}

Similarly,

φ_{2} ⊢_{E,R} ψ ⟺ \{ \begin{matrix}∀t ∈ U, ψ(φ^{−1}(t)) ⊆ [t, ·[_{T} ∩E \\ ∀e ∈ V, ψ(φ^{−1}(e)) = [e, ·[_{E} \\ ψ(φ^{−1}(a)) = [e_{1}, ·[_{E} \\ ψ(φ^{−1}(e_{1}) ⊔ φ^{−1}(e_{2})) = [e_{2}, ·[_{E} \\ ψ(φ^{−1}(e_{i})) = [e_{i}, ·[_{E} ∀i ∈ \{3, …, r\} . \end{matrix}

φ_{3} ⊢_{E,R} ψ ⟺ \{ \begin{matrix}∀t ∈ U, ψ(φ^{−1}(t)) ⊆ [t, ·[_{T} ∩E \\ ∀e ∈ V, ψ(φ^{−1}(e)) = [e, ·[_{E} \\ ψ(φ^{−1}(a)) = [e_{1}, ·[_{E} \\ ψ(φ^{−1}(e_{1})) = [e_{2}, ·[_{E} \\ ψ(φ^{−1}(e_{2}) ⊔ φ^{−1}(e_{3})) = [e_{3}, ·[_{E} \\ ψ(φ^{−1}(e_{i})) = [e_{i}, ·[_{E} ∀i ∈ \{4, …, r\} . \end{matrix}

…

φ_{r−1} ⊢_{E,R} ψ ⟺ \{ \begin{matrix}∀t ∈ U, ψ(φ^{−1}(t)) ⊆ [t, ·[_{T} ∩E \\ ∀e ∈ V, ψ(φ^{−1}(e)) = [e, ·[_{E} \\ ψ(φ^{−1}(a)) = [e_{1}, ·[_{E} \\ ψ(φ^{−1}(e_{i})) = [e_{i+1}, ·[_{E} ∀i ∈ \{1, … r − 3\} \\ ψ(φ^{−1}(e_{r−2}) ⊔ φ^{−1}(e_{r−1})) = [e_{r−1}, ·[_{E} \\ ψ(φ^{−1}(e_{r})) = [e_{r}, ·[_{E} . \end{matrix}

φ_{r} ⊢_{E,R} ψ ⟺ \{ \begin{matrix}∀t ∈ U, ψ(φ^{−1}(t)) ⊆ [t, ·[_{T} ∩E \\ ∀e ∈ V, ψ(φ^{−1}(e)) = [e, ·[_{E} \\ ψ(φ^{−1}(a)) = [e_{1}, ·[_{E} \\ ψ(φ^{−1}(e_{i})) = [e_{i+1}, ·[_{E} ∀i ∈ \{1, …, r − 2\} \\ ψ(φ^{−1}(e_{r−1}) ⊔ φ^{−1}(e_{r})) = [e_{r}, ·[_{E} . \end{matrix}

φ_{r+1} ⊢_{E,R} ψ ⟺ \{ \begin{matrix}∀t ∈ U, ψ(φ^{−1}(t)) ⊆ [t, ·[_{T} ∩E \\ ∀e ∈ V, ψ(φ^{−1}(e)) = [e, ·[_{E} \\ ψ(φ^{−1}(a)) = [e_{1}, ·[_{E} \\ ψ(φ^{−1}(e_{i})) = [e_{i+1}, ·[_{E} ∀i ∈ \{1, …, r − 1\} \\ ψ(φ^{−1}(e_{r})) ⊆ [b, ·[_{T} ∩E . \end{matrix}

In particular φ ⊢_{E,R} ψ implies φ_{1} ⊢_{E,R} ψ, and also φ_{i} ⊬_{E,R} ψ for i ≥ 2, since φ ⊢_{E,R} ψ

implies ψ(φ^{−1}(e_{1})) = [e_{1}, ·[_{E}, but φ_{i} ⊢_{E,R} ψ implies ψ(φ^{−1}(e_{1})) ⊆ [e_{2}, ·[_{E} when i ≥ 2.

Therefore only φ and φ_{1} are involved in this case.

Now if φ_{1} ⊢_{E,R} ψ but φ ⊬_{E,R} ψ, then ψ(φ^{−1}(e_{1})) ⊆]e_{1}, ·[_{T} ∩E = [e_{2}, ·[_{E} (as e_{2} =

s(e_{1})) and ψ(φ^{−1}(a)) = [e_{1}, ·[_{E}, hence in particular φ_{2} ⊢_{E,R} ψ, since φ_{1} ⊢_{E,R} ψ also implies ψ(φ^{−1}(e_{i})) = [e_{i}, ·[_{E} for i ∈ {2, …, n}. On the other hand φ_{i} ⊬_{E,R} ψ, for

i ≥ 3, since φ_{1} ⊢_{E,R} ψ implies ψ(φ^{−1}(e_{2})) = [e_{2}, ·[_{E}, but if i ≥ 3, then φ_{i} ⊢_{E,R} ψ

implies ψ(φ^{−1}(e_{1})) ⊆ [e_{3}, ·[_{E}. Therefore only φ_{1} and φ_{2} are involved in this case.

By induction, the same argument shows, for i ∈ {1, …, r−1}, that φ_{i} ⊢_{E,R} ψ but φ_{i−1} ⊬_{E,R} ψ, then φ_{i+1} ⊢_{E,R} ψ. Therefore only φ_{i} and φ_{i+1} are involved in this case.

If now φ_{r} ⊢_{E,R} ψ but φ_{r−1} ⊬_{E,R} ψ, then ψ(φ^{−1}(e_{r−1}) ⊔ φ^{−1}(e_{r})) = [e_{r}, ·[_{E} but

ψ(φ^{−1}(e_{r})) ≠ [e_{r}, ·[_{E}. Hence ψ(φ^{−1}(e_{r})) ⊆]e_{r}, ·[_{E}⊆ [b, ·[_{T} ∩E, since s(e_{r}) ≥ b. Moreover e_{r} ∈ ψ(φ^{−1}(e_{r−1}) ⊔ φ^{−1}(e_{r})) and e_{r} ∉ ψ(φ^{−1}(e_{r})). It follows that

e_{r} ∈ ψ(φ^{−1}(e_{r−1})), hence ψ(φ^{−1}(e_{r−1})) = [e_{r}, ·[_{E}, and φ_{r+1} ⊢_{E,R} ψ. Therefore only

φ_{r} and φ_{r+1} are involved in this case.

Finally, if φ_{r+1} ⊢_{E,R} ψ, then ψ(φ^{−1}(e_{r−1})) = [e_{r}, ·[_{E} and ψ(φ^{−1}(e_{r})) ⊆ [b, ·[_{T} ∩E ⊆

[e_{r}, ·[_{E}. Thus ψ(φ^{−1}(e_{r−1}) ⊔ φ^{−1}(e_{r})) = [e_{r}, ·[_{E}, and φ_{r} ⊢_{E,R} ψ. Therefore we are

again in the case when only φ_{r} and φ_{r+1} are involved.

Of course, if none of φ, φ_{1}, φ_{2}, …, φ_{r+1} is linked with ψ, then the corresponding equation (E_{ψ}) is just 0 = 0. It follows that φ − u_{a} ∘ φ satisfies the equation (E_{ψ}) for every ψ, as was to be shown.

We have now paved the way for finding generators of 𝕊_{E,R^{op}}(X).

17.7. Theorem. Let T be a finite lattice, let (E, R) be the full subposet of its irreducible elements, and let Γ = {a ∈ T | a ∉ E, a < r^{∞}s^{∞}(a)}. For a ∈ Γ, let u_{a} be the element of k(T^{T}) introduced in Notation 17.4, and let u_{T} denote the composition of all the elements u_{a}, for a ∈ Γ, in any order (they actually commute, see Theorem 17.9 below).

(a) Let X be a finite set. Then for any φ : X → T, the element u_{T} ∘ φ is a k-linear combination of functions f : X → T such that f(X) ⊆ G.

(b) Let B_{X} be the set of all maps φ : X → T such that E ⊆ φ(X) ⊆ G. Then the set Θ_{T,X}(B_{X}) generates 𝕊_{E,R^{op}}(X) as a k-module.

Proof : (a) It follows from Table 17.6 that for any a ∈ Γ and any φ : X → T, the element u_{a} ∘ φ = φ_{1} − φ_{2} + … + (−1)^{r}φ_{r+1} is a k-linear combination of functions φ_{i} such that φ_{i}(X) ∩ Γ ⊆ (φ(X) ∩ Γ) − {a}. We now remove successively all such elements a by applying successively all u_{a} for a ∈ Γ. It follows that u_{T} ∘ φ is a k-linear combination of functions f : X → T such that f(X) ∩ Γ = ∅, that is, f(X) ⊆ G.

(b) Since Θ_{T,X} : F_{T} (X) → 𝕊_{E,R^{op}}(X) is surjective, 𝕊_{E,R^{op}}(X) is generated as a k-module by the images of all maps φ : X → T. For any a ∈ Γ, u_{a} ∘ φ has the same image as φ under Θ_{T,X}, by Theorem 17.5. Therefore u_{T} ∘ φ has the same image as φ under Θ_{T,X}. Moreover u_{T} ∘ φ is a k-linear combination of functions f : X → T such that f(X) ⊆ G, by (a). Finally, if E ⊈ f(X), then f ∈ H_{T} (X) ⊆ Ker Θ_{T,X}. So we are left with all maps φ : X → T such that E ⊆ φ(X) ⊆ G.

We now mention that much more can be said about the elements u_{a} appearing in Theorem 17.7.

17.8. Definition. Let T be a finite lattice. Recall that Γ denotes the complement of G in T. We define an oriented graph structure G(T) on T in the following way: for x, y ∈ T, there is an edge [figure: x; y] from x to y in G(T) if there exists a ∈ Γ such that (x, y) is a pair of consecutive elements in the reduction sequence associated to a.

17.9. Theorem. Let T be a finite lattice, let (E, R) the full subposet of its irreducible elements, let Γ = {a ∈ T | a ∉ E, a < r^{∞}s^{∞}(a)}, and let G(T) be the graph structure on T introduced in Definition 17.8. For a ∈ Γ, let u_{a} be the element of k(T^{T}) introduced in Notation 17.4, and let u_{T} denote the composition of all the elements u_{a}, for a ∈ Γ.

(a) The graph G(T) has no (oriented or unoriented) cycles, and each vertex has at most one outgoing edge. Hence it is a forest.

(b) For a ∈ Γ, the element u_{a} is an idempotent of k(T^{T}).

(c) u_{a} ∘ u_{b} = u_{b} ∘ u_{a} for any a, b ∈ Γ.

(d) The element u_{T} is an idempotent of k(T^{T}).

Theorem 17.9 has apparently no direct implication on the structure of correspondence functors. Therefore, it will be proved in an appendix (Section 22), as a special case of a general results on forests.

We now move to our second part.

17.10. Theorem. Let T be a finite lattice, let (E, R) be the full subposet of its irreducible elements, and let X be a finite set. The set Θ_{T,X}(B_{X}) is a k-linearly independent subset of 𝕊_{E,R^{op}}(X).

Proof : We consider again the kernel of the map Θ_{T,X} : F_{T} (X) → 𝕊_{E,R^{op}}(X), which was described in Theorem 13.4 by a system of linear equations. This can be reformulated by introducing the k-linear map

[figure: η_{E,R,X} : F_{T} (X); F_{I^{↑}(E,R)}(X); φ; ∑_{{ψ:X→I^{↑}(E,R) \atop φ ⊢_{E,R} ψ}} ψ]

where the notation φ ⊢_{E,R} ψ means, as before, that φ : X → T and ψ : X → I^{↑}(E, R)

satisfy the equivalent conditions of Lemma 13.9. Theorem 13.4 asserts that

Ker(Θ_{T,X}) = Ker(η_{E,R,X}) .

For handling the condition φ ⊢_{E,R} ψ, we shall use part (e) of Lemma 13.9, namely

(17.11)

φ ⊢_{E,R} ψ ⟺ \{ \begin{matrix}φ ≤ ∧ψ , \\ ∀e ∈ E, ∃x ∈ X such that φ(x) = e and ψ(x) = [e, ·[_{E} . \end{matrix}

Let N = N_{E,R,X} be the matrix of η_{E,R,X} with respect to the standard basis of F_{T} (X), consisting of maps φ : X → T, and the standard basis of F_{I^{↑}(E,R)}(X), consisting of maps ψ : X → I^{↑}(E, R). Explicitly,

(17.12)  N_{ψ,φ} = \{ \begin{matrix}1 & if φ ⊢_{E,R} ψ , \\ 0 & otherwise. \end{matrix}

Note that N is a square matrix in the special case when T = I_{↓}(E, R), because complementation yields a bijection between I_{↓}(E, R) and I^{↑}(E, R). However, if T is a proper quotient of I_{↓}(E, R), then N has less columns.

In order to prove that the elements Θ_{T,X}(φ), for φ ∈ B_{X}, are k-linearly independent, we shall prove that the elements η_{E,R,X}(φ), for φ ∈ B_{X}, are k-linearly independent. In other words, we have to show that the columns of N indexed by φ ∈ B_{X} are k-linearly independent. Now we consider only the lines indexed by elements of the form ψ = σ ∘ φ′, where φ′ : X → G is any function, and σ : G → I^{↑}(E, R) is the map defined in (16.10). We then define the square matrix M, indexed by B_{X} × B_{X}, by

∀φ, φ′ ∈ B_{X}, M_{φ′,φ} = N_{σ∘φ′,φ} .

We are going to prove that M is invertible and this will prove the required linear independence.

The invertibility of M implies in particular that the map σ must be injective, otherwise two lines of M would be equal. Therefore M turns out to be a submatrix of N, but this cannot be seen directly from its definition (unless an independent proof of the injectivity of σ is provided).

The characterization of the condition φ ⊢_{E,R} ψ given in (17.11) implies that

M_{φ′,φ} = \{ \begin{matrix}1 & if φ ≤ ∧σφ′ and ∀e ∈ E, ∃x ∈ X, φ(x) = e = φ′(x) , \\ 0 & otherwise , \end{matrix}

because the equality σφ′(x) = [e, ·[_{E} is equivalent to φ′(x) = e, by definition of σ (see Notation 16.10).

By Proposition 16.11, if t, t′ ∈ G are such that t ≤ ∧σ(t′), then r^{∞}(t) ≤ r^{∞}(t′) and s^{∞}(t) ≤ s^{∞}(t′). Let ⪯ be the preorder on G defined by these conditions, i.e. for all t, t′ ∈ G,

t ⪯ t′ ⟺ r^{∞}(t) ≤ r^{∞}(t′) and s^{∞}(t) ≤ s^{∞}(t′) .

We extend this preorder to B_{X} by setting, for all φ′, φ ∈ B_{X},

φ ⪯ φ′ ⟺ ∀x ∈ X, φ(x) ⪯ φ′(x) ,

which makes sense because φ(x), φ′(x) ∈ G by definition of B_{X}. We denote by ⪯⪰ the equivalence relation defined by this preorder.

Clearly the condition M_{φ′,φ} ≠ 0 implies φ ≤ ∧σφ′, hence φ ⪯ φ′ by Proposition 16.11 quoted above. In other words the matrix M is block triangular, the blocks being indexed by the equivalence classes of the preorder ⪯ on B_{X}. Showing that M is invertible is equivalent to showing that all its diagonal blocks are invertible. In other words, we must prove that, for each equivalence class C of B_{X} for the relation ⪯⪰, the matrix M_{C} = (M_{φ′,φ})_{φ′,φ∈C} is invertible. Let C be such a fixed equivalence class.

Recall that G = ∧E ⊔ G_{1}, where G_{1} is the set of bulbs. If t ∈ G_{1}, then s^{∞}(t) ∈ E by Lemma 16.5. By Lemma 16.2, all elements of [ t, s^{∞}(t) ]_{T} belong to E except t itself. Moreover, if x ∈ [ t, s^{∞}(t) ]_{T} then r^{∞}(x) = t by Lemma 16.5. It follows that the sets G_{t} = [ t, s^{∞}(t) ]_{T}, for t ∈ G_{1}, are disjoint, and contained in G. Let

G_{∗} = G − ⨆_{t∈G_{1}} G_{t} ,

so that we get a partition

G = ⨆_{t∈{∗}⊔G_{1}} G_{t} .

17.13. Lemma. Let φ′, φ ∈ B_{X}. If φ′⪯⪰φ, then for all t ∈ {∗} ⊔ G_{1},

φ^{′−1}(G_{t}) = φ^{−1}(G_{t}) .

Proof : Let t ∈ G_{1} and x ∈ φ^{−1}([ t, s^{∞}(t) ]_{T} ). Then φ(x) ∈ [ t, s^{∞}(t) ]_{T}, hence r^{∞}φ(x) = t and s^{∞}φ(x) = s^{∞}(t), by Lemma 16.5. But the relation φ′⪯⪰φ implies that r^{∞}φ′(x) = r^{∞}φ(x) and s^{∞}φ′(x) = s^{∞}φ(x). Therefore r^{∞}φ′(x) = t and s^{∞}φ′(x) = s^{∞}(t), from which it follows that φ′(x) ∈ [ t, s^{∞}(t) ]_{T}, that is, x ∈ φ^{′−1}([ t, s^{∞}(t) ]_{T} ). This shows that φ^{−1}(G_{t}) ⊆ φ^{′−1}(G_{t}). By exchanging the roles of φ and φ′, we obtain φ^{′−1}(G_{t}) = φ^{−1}(G_{t}).

Now G_{∗} is the complement of ⨆_{t∈G_{1}} G_{t} in G and the functions φ′, φ have their

values in G (by definition of B_{X}). So we must have also φ^{′−1}(G_{∗}) = φ^{−1}(G_{∗}).

For every t ∈ {∗} ⊔ G_{1}, we define
X_{t} = φ_{0}^{−1}(G_{t})

where φ_{0} is an arbitrary element of C. It follows from Lemma 17.13 that this definition does not depend on the choice of φ_{0}. Therefore, the equivalence class C yields a partition

X = ⨆_{t∈{∗}⊔G_{1}} X_{t} ,

and every function φ ∈ C decomposes as the disjoint union of the functions φ_{t}, where φ_{t} : X_{t} → G_{t} is the restriction of φ to X_{t}.

For t ∈ G_{1}, define

E_{t} =] t, s^{∞}(t) ]_{T} .

By Lemma 16.2, this consists of elements of E, so E_{t} = E ∩ G_{t}. Then we define E_{∗} = E − ⨆_{t∈G_{1}} E_{t} = E ∩ G_{∗}, so that we get a partition

E = ⨆_{t∈{∗}⊔G_{1}} E_{t} .

For every t ∈ {∗} ⊔ G_{1} and for every φ ∈ C, the function φ_{t} satisfies the condition E_{t} ⊆ φ_{t}(X_{t}) ⊆ G_{t}, by definition of B_{X}. Moreover, if φ′, φ ∈ C, then

M_{φ′,φ} = 1 ⟺ ∀t ∈ \{∗\} ⊔ G_{1}, \{ \begin{matrix}∀x ∈ X_{t}, φ′_{t}(x) ≤ ∧σφ_{t}(x) \\ ∀e ∈ E_{t}, ∃x ∈ X_{t}, φ′(x) = e = φ(x) . \end{matrix}

It follows that the matrix M_{C} is a tensor product of square matrices M_{C,t}, for t ∈ {∗} ⊔ G_{1}. The matrix M_{C,t} is indexed by the functions φ_{t} : X_{t} → G_{t} whose image contains E_{t} and satisfies

(M_{C,t})_{φ′,φ} = 1 ⟺ \{ \begin{matrix}∀x ∈ X_{t}, φ′_{t}(x) ≤ ∧σφ_{t}(x) \\ ∀e ∈ E_{t}, ∃x ∈ X_{t}, φ′(x) = e = φ(x) . \end{matrix}

In order to show that M_{C} is invertible, we shall prove that each matrix M_{C,t} is invertible.

If x ∈ X_{∗}, then φ′(x) is not a bulb, by construction of X_{∗}. Therefore, the condition φ′_{∗}(x) ≤ ∧σφ_{∗}(x) implies φ′_{∗}(x) ≤ φ_{∗}(x) by Proposition 16.11. It follows that the matrix M_{C,∗} is unitriangular, hence invertible, as required.

We are left with the matrices M_{C,t} for t ∈ G_{1} and we now discuss the special role played by the set G_{1} of all bulbs. If t is a bulb, then G_{t} = [t, s^{∞}(t)]_{T} is isomorphic to the totally ordered lattice \underline{n}, for some n ≥ 0, and the set of irreducible elements E_{t} =]t, s^{∞}(t)]_{T} is isomorphic to [n] = {1, …, n} (or ∅ if n = 0). Composing the maps φ_{t} : X_{t} → G_{t} with this isomorphism, we obtain maps X_{t} → \underline{n}. Changing notation for simplicity, we write X for X_{t} and φ for φ_{t} and we consider all such maps φ : X → \underline{n} lying in B_{X}, i.e. satisfying the condition [n] ⊆ φ(X) ⊆ \underline{n}. The matrix M_{C,t}, which we write M for simplicity, is now indexed by all such maps and we have

M_{φ′,φ} = 1 ⟺ \{ \begin{matrix}∀x ∈ X, φ′(x) ≤ ∧σφ(x) \\ ∀e ∈ [n], ∃x ∈ X, φ′(x) = e = φ(x) . \end{matrix}

(Actually, the map σ : \underline{n} → I^{↑}([n], tot), defined in (16.10), is easily seen to be a bijection (mapping 0 to ∅) and the matrix M_{C,t} is equal to a square submatrix of the matrix N_{[n],tot,X}, analogous to the matrix N_{E,R,X} of 17.12. Hence we are actually dealing with a reduction to the case of a total order.)

For every a ∈ \underline{n} we have r^{∞}(a) = 0 and s^{∞}(a) = n. Thus for φ ∈ B_{X} and x ∈ X, we have

∧φ(x) = \{ \begin{matrix}φ(x) & if φ(x) ≠ 0 , \\ n & if φ(x) = 0 . \end{matrix}

Hence for φ, φ′ ∈ B_{X}, we have φ′ ≤ ∧σφ if and only if φ′(x) ≤ φ(x) for any x ∈ X − φ^{−1}(0).

Now the set B_{X} consists of those functions φ : X → \underline{n} such that φ(X) ⊇ [n]. For such a map φ, the sets A_{i} = φ^{−1}(i), for i ∈ [n], are nonempty disjoint subsets of X. Conversely, if A = (A_{1}, …, A_{n}) is a sequence of nonempty disjoint subsets of X, setting A_{0} = X − ⊔_{i=1}^{n} A_{i}, there is a unique map φ_{A} : X → \underline{n} defined for x ∈ X

by φ_{A}(x) = i if x ∈ A_{i}. This gives a bijective correspondence

P_{n}(X) ⟶ B_{X} , A ↦ φ_{A}

where P_{n}(X) denotes the set of all sequences (A_{1}, …, A_{n}) of nonempty disjoint subsets of X.

For A, A′ ∈ P_{n}(X), we have M_{φ_{A′} ,φ_{A}} = 1 if and only if the following two conditions hold:

(1) If x ∈ X is such that φ′(x) = i and φ(x) ≠ 0, then φ(x) ≥ i. In other words A′_{i} ⊆ A_{0} ⊔ A_{i} ⊔ A_{i+1} ⊔ … ⊔ A_{n}, for any i ∈ [n]. Equivalently A′_{i} ∩ A_{j} = ∅ for 1 ≤ j < i ≤ n.

(2) For any i ∈ [n], there exists x ∈ X such that φ_{A}(x) = i = φ_{A′}(x). In otherwords A_{i} ∩ A′_{i} ≠ ∅ for any i ∈ [n].

We can now view M as a square matrix indexed by P_{n}(X). For all A, A′ ∈ P_{n}(X), we have

M_{A′,A} = 1 ⟺ \{ \begin{matrix}∀1 ≤ j < i ≤ n, A′_{i} ∩ A_{j} = ∅ , \\ ∀i ∈ [n], A′_{i} ∩ A_{i} ≠ ∅ . \end{matrix}

In order to show that the matrix M is invertible, we introduce a square matrix U indexed by P_{n}(X), defined as follows. For all A, B ∈ P_{n}(X),

U_{A,B} = \{ \begin{matrix}(−1)^{∑_{i=1}^{n} |A_{i}−B_{i}|} & if ∀i ∈ [n], A_{i} ⊇ B_{i} , \\ 0 & otherwise . \end{matrix}

We now compute the product V = U · M ·^{t}U. For A, A′ ∈ P_{n}(X), we have

V_{A′,A} = ∑_{B′} ∑_{B} U_{A′,B′} M_{B′,B} U_{A,B} = ∑_{(B′,B)}(−1)^{∑_{i=1}^{n} |A′_{i}−B′_{i}|+|A_{i}−B_{i}|} ,

where (B′, B) runs through the set of pairs of elements of P_{n}(X) such that

∀1 ≤ i ≤ n , B′_{i} ⊆ A′_{i} and B_{i} ⊆ A_{i} ,
∀1 ≤ j < i ≤ n , B′_{i} ∩ B_{j} = ∅ ,
∀1 ≤ i ≤ n , B_{i} ∩ B′_{i} ≠ ∅ .

We compute the above sum giving V_{A′,A} by first choosing B ∈ P_{n}(X) such that B_{i} ⊆ A_{i} for all i ∈ [n], and then computing

(−1)^{∑_{i=1}^{n}(|A_{i}−B_{i}|+|A′_{i}|)} ∑_{B′}(−1)^{∑_{i=1}^{n} |B′_{i}|} ,

where B′ runs through the set of elements of P_{n}(X) such that

(17.14)  \{ \begin{matrix}∀1 ≤ i ≤ n , & B′_{i} ⊆ A′_{i} , \\ ∀1 ≤ j < i ≤ n , & B′_{i} ∩ B_{j} = ∅ , \\ ∀1 ≤ i ≤ n , & B′_{i} ∩ B_{i} ≠ ∅ . \end{matrix}

For i ∈ [n], set D_{i} = A′_{i} − ⊔_{j=1}^{i−1} B_{j} and E_{i} = A′_{i} ∩ B_{i}, and observe that E_{i} ⊆ D_{i}.

Conditions 17.14 hold if and only if, for every i ∈ [n],

(17.15)  B′_{i} ⊆ D_{i} and B′_{i} ∩ E_{i} ≠ ∅ ,

and we have to compute

∑_{{B′∈P_{n}(X) \atop s.t.(17.14)}}(−1)^{∑_{i=1}^{n} |B′_{i}|} = ∏_{i=1}^{n} ( ∑_{{B′_{i}⊆X \atop s.t.(17.15)}}(−1)^{|B′_{i}|}) .

By Condition (17.15), B′_{i} = E′_{i} ⊔ D′_{i}, where E′_{i} is a nonempty subset of E_{i} and D′_{i} is a subset of D_{i} − E_{i}, and we have to sum over all such E′_{i} and D′_{i}. For a given i ∈ [n],

∑_{{B′_{i}⊆X \atop s.t.(17.15)}}(−1)^{|B′_{i}|} = ∑_{∅≠E′_{i}⊆E_{i}} ∑_{D′_{i}⊆D_{i}−E_{i}}(−1)^{|E′_{i}|+|D′_{i}|} ,

and the inner sum is zero unless D_{i} − E_{i} = ∅, or equivalently A′_{i} − ⊔_{j=1}^{i−1} B_{j} ⊆ B_{i}. In

this case E_{i} = D_{i} = A′_{i} − ⊔_{j=1}^{i−1} B_{j} and D′_{i} = ∅ is the only choice. So we are left with

the sum

∑_{∅≠E′_{i}⊆D_{i}}(−1)^{|E′_{i}|} ,

which is equal to −1 if D_{i} ≠ ∅, and to 0 if D_{i} = ∅.

It follows that ∑_{B′}(−1)^{∑_{i=1}^{n} |B′_{i}|} is equal to 0, unless B satisfies

(17.16)  ∀i ∈ [n], ∅ ≠ A′_{i} − ⊔_{j=1}^{i−1} B_{j} ⊆ B_{i} ⊆ A_{i} ,

in which case we obtain

∑_{B′}(−1)^{∑_{i=1}^{n} |B′_{i}|} = ∏_{i=1}^{n} ( ∑_{B′_{i}}(−1)^{|B′_{i}|}) = (−1)^{n} .

Condition 17.16 imply that A′_{i} ⊆ ⊔_{j=1}^{i} B_{j} ⊆ ⊔_{j=1}^{i} A_{j}, for any i ∈ [n], or equivalently

⊔_{j=1}^{i} A′_{j} ⊆ ⊔_{j=1}^{i} A_{j}, for every i ∈ [n]. Let ⪯ be the partial order relation on P_{n}(X)

defined, for all A′, A ∈ P_{n}(X), by the requirement

A′ ⪯ A ⟺ ∀i ∈ [n], ⊔_{j=1}^{i} A′_{j} ⊆ ⊔_{j=1}^{i} A_{j} .

The previous discussion shows that V_{A′,A} = 0 unless A′ ⪯ A. Moreover if A′ = A, we obtain

V_{A,A} = ∑_{{B∈P_{n}(X) \atop B_{i}⊆A_{i} ∀i}}(−1)^{∑_{i=1}^{n}(|A_{i}−B_{i}|+|A_{i}|)} ∑_{{B′∈P_{n}(X) \atop s.t.(17.14)}}(−1)^{∑_{i=1}^{n} |B′_{i}|}

= ∑_{{B∈P_{n}(X) \atop s.t.(17.16)}}(−1)^{∑_{i=1}^{n} |B_{i}|}(−1)^{n}

= (−1)^{n+ ∑_{i=1}^{n} |A_{i}|} ,

because, when A′ = A, Condition 17.16 is equivalent to B_{i} = A_{i} for every i ∈ [n], so the sum over B reduces to a single term.

It follows now that for a total ordering of P_{n}(X) finer than ⪯, the matrix V = U ·M ·^{t}U is triangular with diagonal coefficients equal to ±1, hence invertible. Moreover U_{A,B} ≠ 0 implies that A_{i} ⊇ B_{i} for all i ∈ [n], and this is also a partial order on P_{n}(X). As U_{A,A} = 1 for any A ∈ P_{n}(X), the matrix U is unitriangular for a suitable total order on P_{n}(X), hence it is also invertible. Therefore the matrix

M = U^{−1} · V ·^{t}U^{−1}

is invertible, as was to be shown. This completes the proof of Theorem 17.10.

We can finally prove one of our main results about fundamental correspondence functors, which generalizes the formula obtained in Theorem 15.14 in the case of a total order.

17.17. Corollary. Let (E, R) be a finite poset and let T be any lattice such that (E, R) is the full subposet of irreducible elements in T.

(a) Θ_{T,X}(B_{X}) is a k-basis of 𝕊_{E,R^{op}}(X).
(b) The k-module 𝕊_{E,R^{op}}(X) is free of rank

rk_{k}(𝕊_{E,R^{op}}(X)) = ∑_{i=0}^{|E|}(−1)^{i}\binom{|E|}{i}(|G| − i)^{|X|} .

In particular, |G| only depends on (E, R), and not on the choice of T.

Proof : (a) follows from Theorem 17.7 and Theorem 17.10.

(b) The formula follows immediately from (a) and Lemma 8.1. Moreover, it implies that

rk_{k}(𝕊_{E,R^{op}}(X)) ∼ |G|^{|X|} as |X| → ∞ .

Since 𝕊_{E,R^{op}} only depends on (E, R), it follows that |G| only depends on (E, R).

We shall now determine the dimension of any evaluation of a simple correspondence functor. We first need a lemma.

17.18. Lemma. Let (E, R) be a poset.

(a) For any lattice T such that Irr(T) ≅ (E, R), restriction induces an injective group homomorphism Aut(T) → Aut(E, R).

(b) There exists a lattice T such that Irr(T) ≅ (E, R) and such that the restriction homomorphism Aut(T) → Aut(E, R) is an isomorphism.

Proof : (a) For any lattice T such that Irr(T) ≅ (E, R), any lattice automorphism of T induces an automorphism of the poset (E, R). This gives a group homomorphism Aut(T) → Aut(E, R), which is injective since any element t of T is equal to the join of the irreducible elements e ≤_{T} t.

(b) Requiring that Aut(T) ≅ Aut(E, R) amounts to requiring that any automorphism of (E, R) can be extended to an automorphism of T. This is clearly possible if we choose T = I_{↓}(E, R).

We can finally prove one of our main results about simple correspondence functors.

17.19. Theorem. Let k be a field. Let (E, R) be a poset and let V be a simple left k Aut(E, R)-module. Let T be a lattice such that Irr(T) ≅ (E, R) and such that the restriction homomorphism Aut(T) → Aut(E, R) is an isomorphism. Let G = E ⊔ {t ∈ T | t = r^{∞}s^{∞}(t)} ⊆ T. Then for any finite set X, the dimension of S_{E,R,V} (X) is given by

dim_{k} S_{E,R,V} (X) = \frac{dim_{k} V}{| Aut(E, R)|} ∑_{i=0}^{|E|}(−1)^{i}\binom{|E|}{i}(|G| − i)^{|X|}

Proof : Observe first that for any poset (E, R), the subgroups Aut(E, R) and Aut(E, R^{op}) of the group of permutations of E are actually equal. For any finite set X, let B_{X} be the set of all functions φ : X → T such that E ⊆ φ(X) ⊆ G. By Corollary 17.17, the image of B_{X} under Θ_{T,X} : F_{T} (X) → 𝕊_{E,R^{op}}(X) is a basis of the k-module 𝕊_{E,R^{op}}(X). The group Aut(T) ≅ Aut(E, R) = Aut(E, R^{op}) acts on the right on B_{X} via φ · α = α^{−1} ∘ φ, for φ ∈ B_{X} and α ∈ Aut(E, R). This action is moreover free: indeed, if φ · α = φ, then α^{−1}(φ(x)) = φ(x) for any x ∈ X, thus α(e) = e for any e ∈ E, since E ⊆ φ(X), hence α = id.

It follows that 𝕊_{E,R^{op}}(X) is isomorphic to a direct sum of copies of the free right module k Aut(E, R). Applying this to R^{op} instead of R, we obtain that 𝕊_{E,R}(X) is isomorphic to the direct sum of n_{X} copies of the free right module k Aut(E, R), for some n_{X} ∈ ℕ. In particular

dim_{k} 𝕊_{E,R}(X) = n_{X} | Aut(E, R)| .

Moreover, by Remark 4.8, the simple functor S_{E,R,V} satisfies

S_{E,R,V} ≅ 𝕊_{E,R} ⊗_{k Aut(E,R)} V ,

so that

S_{E,R,V} (X) = 𝕊_{E,R}(X) ⊗_{k Aut(E,R)} V ≅ n_{X}(k Aut(E, R)) ⊗_{k Aut(E,R)} V ≅ n_{X}V .

Hence dim_{k} 𝕊_{E,R,V} (X) = n_{X} dim_{k} V. Therefore

dim_{k} 𝕊_{E,R,V} (X) = \frac{dim_{k} V}{| Aut(E, R)|} dim_{k} 𝕊_{E,R}(X) ,

and Theorem 17.19 now follows from Corollary 17.17.

18. Properties of fundamental functors

By using the results of the previous section, we can obtain more information on fundamental functors, in particular about the dual of a fundamental functor, homomorphisms between fundamental functors, and the description of the action of correspondences on a fundamental functor.

At the beginning of the proof of Theorem 17.10, we considered the k-linear map

[figure: η_{E,R,X} : F_{T} (X); F_{I^{↑}(E,R)}(X); φ; ∑_{{ψ:X→I^{↑}(E,R) \atop φ ⊢_{E,R} ψ}} ψ]

where the notation φ ⊢_{E,R} ψ means, as before, that φ : X → T and ψ : X → I^{↑}(E, R)

satisfy the equivalent conditions of Lemma 13.9. Since η_{E,R,X} has the same kernel as the surjective map Θ_{T,X} : F_{T} (X) → 𝕊_{E,R^{op}}(X), we see that 𝕊_{E,R^{op}}(X) is isomorphic, as a k-module, to the image of η_{E,R,X}.

18.1. Proposition. Let N be the matrix of the k-linear map

η_{E,R,X} : F_{T} (X) ⟶ F_{I^{↑}(E,R)}(X)

with respect to the canonical bases (see 17.12).

(a) Let k = ℤ and let

d = rk_{k}(𝕊_{E,R^{op}}(X)) = ∑_{i=0}^{|E|}(−1)^{i}\binom{|E|}{i}(|G| − i)^{|X|} .

Then the first d elementary divisors of the matrix N are equal to 1, and the next ones are equal to 0.

(b) The image of the k-linear map η_{E,R,X} is a direct summand of F_{I^{↑}(E,R)}(X) as a k-module.

Proof : (a) For any commutative ring k, let I_{k} be the image of η_{E,R,X}. It is isomorphic to 𝕊_{E,R}(X), hence it is a free k-module of rank d, independently of k, by Corollary 17.17. Now for k = ℤ, we have two exact sequences of abelian groups

\xymatrix{0 \ar[r] & K_{ℤ} \ar[r] & F_{T} (X) \ar[r]^{η_{E,R,X}} & I_{ℤ} \ar[r] & 0 \\ 0 \ar[r] & I_{ℤ} \ar[r] & F_{I^{↑}(E,R)}(X) \ar[r] & C_{ℤ} \ar[r] & 0 ,}

where K_{ℤ} and C_{ℤ} are the kernel and cokernel of η_{E,R,X}, respectively. The first exact sequence is actually split, because I_{ℤ} is a free ℤ-module (either because it is a subgroup of the finitely generated free ℤ-module F_{I^{↑}(E,R)}(X), or by Corollary 17.17). Hence tensoring these exact sequences with k gives the exact sequences

\xymatrix{ & 0 \ar[r] & k ⊗_{ℤ} K_{ℤ} \ar[r] & k ⊗_{ℤ} F_{T} (X) \ar[r]^{k⊗_{ℤ}η_{E,R,X}} & k ⊗_{ℤ} I_{ℤ} \ar[r] & 0 \\ 0 \ar[r] & Tor_{1}^{ℤ}(k, C_{ℤ}) \ar[r] & k ⊗_{ℤ} I_{ℤ} \ar[r] & k ⊗_{ℤ} F_{I^{↑}(E,R)}(X) \ar[r] & k ⊗_{ℤ} C_{ℤ} \ar[r] & 0 .}

The image of the middle map

k ⊗_{ℤ} I_{ℤ} ⟶ k ⊗_{ℤ} F_{I^{↑}(E,R)}(X) = k(I^{↑}(E, R)^{X})

is precisely I_{k}. It follows that I_{k} ≅ (k⊗_{ℤ}I_{ℤ})/Tor_{1}^{ℤ}(k, C_{ℤ}), i.e. we have the following exact sequence

\xymatrix{0 \ar[r] & Tor_{1}^{ℤ}(k, C_{ℤ}) \ar[r] & k ⊗_{ℤ} I_{ℤ} \ar[r] & I_{k} \ar[r] & 0 .}

But k ⊗_{ℤ} I_{ℤ} ≅ I_{k} as k-modules, because both are free k-modules of rank d. Therefore, if k is a field, then Tor_{1}^{ℤ}(k, C_{ℤ}) = 0. Taking k = 𝔽_{p} for some prime number p, it follows that Tor_{1}^{ℤ}(𝔽_{p}, C_{ℤ}) = 0, and since p is an arbitrary prime, the group C_{ℤ} is torsion free, hence it is a free ℤ-module. Thus I_{ℤ} is a pure submodule of F_{I^{↑}(E,R)}(X), that is, the nonzero elementary divisors of the matrix N are all equal to 1.

(b) The proof of part (a) shows that

\xymatrix{0 \ar[r] & I_{ℤ} \ar[r] & F_{I^{↑}(E,R)}(X) \ar[r] & C_{ℤ} \ar[r] & 0}

is an exact sequence of free ℤ-modules, hence split, and that k⊗_{ℤ}C_{ℤ} is isomorphic to the cokernel of the k-linear map η_{E,R,X}. Therefore this cokernel is a free k-module and so the image I_{k} is a direct summand of F_{I^{↑}(E,R)}(X) as a k-module.

We can now settle the question left open in Theorem 14.17 concerning the structure of the dual of a fundamental functor.

18.2. Theorem. Let (E, R) be a poset and let T = I_{↓}(E, R). Consider the subfunctor ⟨γ_{T} ⟩ of F_{T^{op}} generated by γ_{T} (isomorphic to 𝕊_{E,R} by Theorem 14.16).

(a) For every finite set X, the evaluation ⟨γ_{T} ⟩(X) is a direct summand of F_{T^{op}}(X) as a k-module.

(b) The injective morphism α_{E,R} : 𝕊_{E,R} ⟶ 𝕊E,R^{op}^{♮} (see Theorem 14.17) is an isomorphism. Thus 𝕊E,R^{♮} ≅ 𝕊_{E,R^{op}}.

Proof : (a) By Example 12.4, F_{I_{↓}(E,R^{op})} ≅ kC(−, E)R. The proof of Theorem 14.16 shows that there is a surjective morphism

\overline{ξ} : F_{I_{↓}(E,R^{op})} ≅ kC(−, E)R ⟶ ⟨γ_{T} ⟩ ⊆ F_{T^{op}} ≅ FT^{♮} ,

where T = I_{↓}(E, R). Moreover \overline{ξ} induces the isomorphism 𝕊_{E,R} ≅ ⟨γ_{T} ⟩. Consider the canonical k-basis {ψ : X → I_{↓}(E, R^{op})} of F_{I_{↓}(E,R^{op})}(X) and the k-basis

{φ^{⋆} | φ : X → I_{↓}(E, R)}

of F_{I_{↓}(E,R)^{op}}(X), where φ^{⋆} is defined as in Notation 14.8. The matrix M of the k-linear map

\overline{ξ}_{X} : F_{I_{↓}(E,R^{op})}(X) ⟶ F_{I_{↓}(E,R)^{op}}(X)

with respect to these bases is easy to compute using the arguments of the proof of Theorem 14.16. It is given by

M_{φ^{⋆},ψ} = \{ \begin{matrix}1 & if Γ_{ψ}^{op}Γ_{φ} = R^{op} , \\ 0 & otherwise , \end{matrix} φ : X → I_{↓}(E, R) , ψ : X → I_{↓}(E, R^{op}) .

Since the condition Γ_{ψ}^{op}Γ_{φ} = R^{op} is the same as the condition φ ⊢_{E,R} ψ (see (17.11)

and Lemma 13.9) and since I^{↑}(E, R) = I_{↓}(E, R^{op}), we obtain

M_{φ^{⋆},ψ} = \{ \begin{matrix}1 & if φ ⊢_{E,R} ψ , \\ 0 & otherwise , \end{matrix} φ : X → I_{↓}(E, R) , ψ : X → I^{↑}(E, R) .

But this is the transpose of the matrix N of the k-linear map

η_{E,R,X} : F_{T} (X) ⟶ F_{I^{↑}(E,R)}(X) .

By Proposition 18.1, the nonzero elementary divisors of N are all equal to 1. Therefore, the same holds for M (because this property does not change by transposition). Equivalently, the image of the map \overline{ξ}_{X} : F_{I_{↓}(E,R^{op})}(X) ⟶ F_{I_{↓}(E,R)^{op}}(X) is a direct summand of F_{I_{↓}(E,R)^{op}}(X). In other words, ⟨γ_{T} ⟩(X) is a direct summand of F_{T^{op}}(X) as a k-module.

(b) In the proof of Theorem 14.17, we obtained the injective morphism α_{E,R} : 𝕊_{E,R} ⟶ 𝕊E,R^{op}^{♮} from the inclusion ⟨γ_{T} ⟩ ⊆ ⟨γ_{T} ⟩^{⊥⊥}. Now by part (a), ⟨γ_{T} ⟩(X) is a direct summand of F_{T^{op}}(X) as a k-module for every finite set X. Therefore ⟨γ_{T} ⟩^{⊥⊥}(X) = ⟨γ_{T} ⟩(X) (as both are free k-modules with the same rank), that is, the inclusion above is an equality. It follows that α_{E,R} is an isomorphism.

Our next result on fundamental functors is similar to Schur’s lemma.

18.3. Theorem. Let (E, R) and (E′, R′) be two finite posets.

(a) If (E, R) ≅ (E′, R′), then Hom_{F_{k}}(𝕊_{E,R}, 𝕊_{E′,R′}) = {0}.

(b) End_{F_{k}}(𝕊_{E,R}) ≅ k Aut(E, R) as k-algebras.

Proof : (a) Let α : 𝕊_{E,R} → 𝕊_{E′,R′} be a nonzero morphism of functors. Since 𝕊_{E,R} is generated by its evaluation 𝕊_{E,R}(E) (by Theorem 13.1), α_{E} cannot be zero, hence 𝕊_{E′,R′}(E) ≠ {0}. Since E′ is a minimal set for 𝕊_{E′,R′} by Lemma 3.14, we have |E′| ≤ |E|. Passing to the dual, we have a nonzero morphism α^{♮} : 𝕊E′,R′^{♮} → 𝕊E,R^{♮}, hence a nonzero morphism 𝕊_{E′,R′^{op}} → 𝕊_{E,R^{op}} by Theorem 18.2. The same argument yields |E| ≤ |E′|, hence |E′| = |E|.

We can now assume that E′ = E. By evaluation at E, we have a nonzero map

α_{E} : 𝕊_{E,R}(E) = P_{E}f_{R} ⟶ 𝕊_{E,R′}(E) = P_{E}f_{R′} ,

hence a nonzero element α_{E}(f_{R}) ∈ f_{R}P_{E}f_{R′} (the image of f_{R}). Since P_{E}f_{R′} has k-basis {Δ_{σ}f_{R′} | σ ∈ Σ_{E}} by Proposition 3.10, we must have f_{R}Δ_{σ}f_{R′} ≠ 0 for some σ ∈ Σ_{E}. Therefore f_{R}f_{^{σ}R}′ = f_{R}Δ_{σ}f_{R′}Δ_{σ}^{−1} ≠ 0 and this implies that R = ^{σ}R′ because otherwise we would have orthogonal idempotents. It follows that conjugation by σ induces an isomorphism of posets (E, R) ≅ (E′, R′).

(b) By the argument of part (a), any α ∈ End_{F_{k}}(𝕊_{E,R}) yields, by evaluation at E, an element of End_{P_{E}}(P_{E}f_{R}) ≅ f_{R}P_{E}f_{R}. If σ ∈ Σ_{E} − Aut(E, R), then f_{R} ≠ f_{^{σ}R}, hence f_{R}Δ_{σ}f_{R} = 0 (orthogonal idempotents). If σ ∈ Aut(E, R), then f_{R}Δ_{σ}f_{R} = Δ_{σ}f_{R}. Therefore f_{R}P_{E}f_{R} has a k-basis {Δ_{σ}f_{R} | σ ∈ Aut(E, R)}. It follows that End_{P_{E}}(P_{E}f_{R}) ≅ k Aut(E, R), by mapping a ∈ k Aut(E, R) to left multiplication by a in P_{E}f_{R}. Therefore, evaluation at E induces a k-algebra map

End_{F_{k}}(𝕊_{E,R}) ⟶ End_{P_{E}}(P_{E}f_{R}) ≅ k Aut(E, R) .

This map is injective because 𝕊_{E,R} is generated by its evaluation at E (Theorem 13.1).

In order to show that this map is surjective, we construct an inverse. Let T = I_{↓}(E, R^{op}) = I^{↑}(E, R), so that Irr(T) = (E, R^{op}). Any element τ ∈ Aut(E, R) induces an automorphism of T, hence an automorphism of F_{T}, namely, for any finite set X,

F_{τ} : F_{T} (X) ⟶ F_{T} (X) , φ ↦ τφ .

The surjection Θ_{T} : F_{T} → 𝕊_{E,R} is given by Θ_{T,X}(φ) = Γ_{φ} ι, where Γ_{φ} is defined as in Notation 5.7 and ι : E → T is the inclusion.

We claim that Γ_{τφ} = Γ_{φ}Δ_{τ^{−1}}. Indeed

\begin{aligned}Γ_{τφ} & = \{(x, e) | e ≤_{T} τφ(x)\} = \{(x, e) | τ^{−1}(e) ≤_{T} φ(x)\} \\ & = \{(x, f) | f ≤_{T} φ(x)\} · \{(τ^{−1}(e), e) | e ∈ E\} = Γ_{φ}Δ_{τ^{−1}} . \end{aligned}

By Theorem 13.4, ∑_{φ:X→T} λ_{φ}φ belongs to Ker(Θ_{T,X}) if and only if

∀ψ : X → I_{↓}(E, R) , ∑_{{φ \atop Γ_{ψ}^{op}Γ_{φ}=R}} λ_{φ} = 0 .

Since Δ_{τ} RΔ_{τ^{−1}} = R, we have equivalences

Γ_{ψ}^{op}Γ_{φ} = R ⟺ Δ_{τ} Γ_{ψ}^{op}Γ_{φ}Δ_{τ^{−1}} = R
⟺ Γ_{τψ}^{op}Γ_{τφ} = R ,

by the claim above and the fact that Δ_{τ} Γ_{ψ}^{op} = Δ_{τ^{−1}}^{op}Γ_{ψ}^{op} = (Γ_{ψ}Δ_{τ^{−1}})^{op} = Γ_{τψ}^{op}. It follows that ∑_{φ:X→T} λ_{φ}φ satisfies the condition for a given ψ : X → I_{↓}(E, R) if and

only if ∑_{φ:X→T} λ_{φ}τφ satisfies the condition for τψ. Therefore Ker(Θ_{T,X}) is invariant

under the automorphism F_{τ} and consequently F_{τ} induces an automorphism

\overline{F_{τ}} : 𝕊_{E,R}(X) ⟶ 𝕊_{E,R}(X) .

Then the map τ ↦ \overline{F_{τ}} induces a k-algebra homomorphism

k Aut(E, R) ⟶ End_{F_{k}}(𝕊_{E,R}) .

It is straightforward to check that this provides the required inverse.

Our next result provides a description of the action of correspondences on a fundamental functor.

18.4. Theorem. Let k be a commutative ring. Let (E, R) be a poset, let T be a lattice such that Irr(T) ≅ (E, R), and let Θ_{T} : F_{T} → 𝕊_{E,R^{op}} be the canonical surjection of Theorem 13.1. Let G = G(T) be the subset defined in Notation 16.7. For a finite set X, let 𝕊(X) denote the free k-submodule of F_{T} (X) = k(T^{X}) with basis the set B_{X} of maps φ : X → T such that E ⊆ φ(X) ⊆ G. Let u_{T} be the element of k(T^{T}) introduced in Theorem 17.7. Let π_{T,X} be the k-linear idempotent endomorphism of k(T^{X}) defined by

∀φ : X → T, π_{T,X}(φ) = \{ \begin{matrix}φ & if φ(X) ⊇ E , \\ 0 & otherwise . \end{matrix}

(a) k(T^{X}) = 𝕊(X) ⊕ Ker Θ_{T,X}.

(b) Let φ : X → T. Then:

(i) φ − u_{T} ∘ φ ∈ Ker Θ_{T,X}.

(ii) if φ(X) ⊉ E, then u_{T} ∘ φ ∈ H_{T} (X).

(iii) if φ(X) ⊆ G, then φ − u_{T} ∘ φ ∈ H_{T} (X). Moreover π_{T,X}(φ) ∈ 𝕊(X). (iv) π_{T,X}(u_{T} ∘ φ) ∈ 𝕊(X).

(c) Ker Θ_{T,X} = {φ ∈ F_{T} (X) | u_{T} ∘ φ ∈ H_{T} (X)}.

(d) The elements φ − u_{T} ∘ φ, where φ runs through the set of maps from X to T such that E ⊆ φ(X) ⊈ G form a k-basis of a complement of H_{T} (X) in Ker Θ_{T,X}.

(e) For a finite set Y, and for U ⊆ Y × X, let 𝕊(U) : 𝕊(X) → 𝕊(Y) be the k-linear map defined by

∀φ : X → T, E ⊆ φ(X) ⊆ G, 𝕊(U)(φ) = π_{T,Y} (u_{T} ∘ Uφ) .

With this definition, the assignment X ↦ 𝕊(X) becomes a correspondence functor, isomorphic to the fundamental functor 𝕊_{E,R^{op}}.

18.5. Remark. In general, 𝕊 is not a subfunctor of F_{T}, since the action of a correspondence U on φ is not Uφ, but π_{T,Y} (u_{T} ∘ Uφ). Nevertheless, (d) provides a formula for the action of a correspondence U on a fundamental functor.

Proof : (a) This is a straightforward consequence of Corollary 17.17.

(b) (i) By Theorem 17.5, for any φ : X → T, the difference φ − u_{a} ∘ φ lies in Ker Θ_{T,X}, where a ∈ T is such that a ∉ E and a < r^{∞}s^{∞}(a). In other words u_{a} ∘ φ is equal to φ modulo Ker Θ_{T,X}. As u_{T} is the composition of all maps u_{a}, for a ∉ E and a < r^{∞}s^{∞}(a), the same congruence holds with u_{a} replaced by u_{T}, that is φ − u_{T} ∘ φ ∈ H_{T} (X).

(ii) If φ(X) ⊉ E, then φ ∈ H_{T} (X) ⊆ Ker Θ_{T,X}, so u_{T} ∘ φ ∈ Ker Θ_{T,X} by (i), hence π_{T,X}(u_{T} ∘ φ) ∈ Ker Θ_{T,X}, since

Im(id −π_{T,X}) = Ker π_{T,X} = H_{T} (X) ⊆ Ker Θ_{T,X} .

But π_{T,X}(u_{T} ∘ φ) is a linear combination of elements of B_{X}, thus π_{T,X}(u_{T} ∘ φ) = 0 by Theorem 17.10, that is u_{T} ∘ φ ∈ H_{T} (X).

(iii) If φ(X) ⊆ G, then φ − u_{T} ∘ φ is a linear combination of maps from X to T with image contained in G, by Theorem 17.7. Moreover φ − u_{T} ∘ φ lies in the kernel of Θ_{T,X}, by (i). Then π_{T,X}(φ − u_{T} ∘ φ) is a linear combination of elements of B_{X}, which is also in Ker Θ_{T,X}, again because Im(id −π_{T,X}) ⊆ Ker Θ_{T,X}. Hence π_{T,X}(φ − u_{T} ∘ φ) = 0 by Theorem 17.10, that is φ − u_{T} ∘ φ ∈ H_{T} (X). Moreover, if φ(X) ⊆ G, then π_{T,X}(φ) = φ if φ(X) ⊇ E, i.e. φ ∈ B_{X}, and π_{T,X}(φ) = 0 otherwise. Hence π_{T,X}(φ) ∈ 𝕊(X).

(iv) follows from (iii), since u_{T} ∘ φ is a linear combination of maps from X to T whose image is contained in G, by Theorem 17.7.

(c) By (b)(i), φ − u_{T} ∘ φ ∈ Ker Θ_{T,X} for φ ∈ F_{T} (X). Hence if u_{T} ∘ φ ∈ H_{T} (X), then φ ∈ Ker Θ_{T,X}, as H_{T} (X) ⊆ Ker Θ_{T,X}. Conversely, if φ ∈ Ker Θ_{T,X}, then u_{T} ∘ φ ∈ Ker Θ_{T,X} also, hence π_{T,X}(u_{T} ∘ φ) ∈ Ker Θ_{T,X}, again because Im(id −π_{T,X}) ⊆ Ker Θ_{T,X}. But π_{T,X}(u_{T} ∘φ) ∈ 𝕊(X) by (b)(iv), so π_{T,X}(u_{T} ∘φ) = 0 by (a). This means that u_{T} ∘ φ ∈ H_{T} (X).

(d) Since Im(id −π_{T,X}) = Ker π_{T,X} = H_{T} (X), proving (d) is equivalent to proving that the elements π_{T,X}(φ − u_{T} ∘ φ), for E ⊆ φ(X) ⊈ G, form a basis of a complement of H_{T} (X) in Ker Θ_{T,X}.

For this, we observe that k(T^{X}) = Ker π_{T,X} ⊕ Im π_{T,X} = H_{T} (X) ⊕ Im π_{T,X}. Since H_{T} (X) is contained in Ker Θ_{T,X}, it follows that

Ker Θ_{T,X} = H_{T} (X) ⊕ ( Im π_{T,X} ∩ Ker Θ_{T,X}) = H_{T} (X) ⊕ π_{T,X}(Ker Θ_{T,X}) ,

as π_{T,X} is a projector. We will prove that the elements π_{T,X}(φ − u_{T} ∘ φ), for E ⊆ φ(X) ⊈ G, form a basis of π_{T,X}(Ker Θ_{T,X}).

Let w = ∑_{φ:X→T} λ_{φ}φ be an element of Ker Θ_{T,X}, where λ_{φ} ∈ k for φ : X → T.

Then

w = ∑_{φ:X→T} λ_{φ}(φ − u_{T} ∘ φ) + ∑_{φ:X→T} λ_{φ}u_{T} ∘ φ .

By (b)(i), the sum ∑_{φ:X→T} λ_{φ}u_{T} ∘ φ is in Ker Θ_{T,X}, so π_{T,X}( ∑_{φ:X→T} λ_{φ}u_{T} ∘ φ) also.

But this is a linear combination of elements of B_{X}, hence it is equal to 0 by Theorem 17.10. In other words

π_{T,X}(w) = π_{T,X}( ∑_{φ:X→T} λ_{φ}(φ − u_{T} ∘ φ)) = ∑_{φ:X→T} λ_{φ}π_{T,X}(φ − u_{T} ∘ φ) .

Now by (b)(ii) and (b)(iii), we have π_{T,X}(φ−u_{T} ∘φ) = 0 if φ(X) ⊉ E or φ(X) ⊆ G. It follows that

π_{T,X}(w) = ∑_{{φ:X→T \atop E⊆φ(X)⊈G}} λ_{φ}π_{T,X}(φ − u_{T} ∘ φ) .

Hence π_{T,X}(Ker Θ_{T,X}) is generated by the elements π_{T,X}(φ−u_{T} ∘φ), for φ : X → T such that E ⊆ φ(X) ⊈ G. Moreover, these elements are linearly independent: indeed, if

∑_{{φ:X→T \atop E⊆φ(X)⊈G}} λ_{φ}π_{T,X}(φ − u_{T} ∘ φ) = 0 ,

then as π_{T,X}(φ) = φ if φ(X) ⊇ E, we have

∑_{{φ:X→T \atop E⊆φ(X)⊈G}} λ_{φ}φ = ∑_{{φ:X→T \atop E⊆φ(X)⊈G}} λ_{φ}π_{T,X}(u_{T} ∘ φ) .

On the right hand side, we have an element of 𝕊(X), by (b)(iv), but none of the maps appearing in the left hand side lies in B_{X}. It follows that both sides are equal to 0, and all the coefficients λ_{φ} are equal to 0. This completes the proof of (d).

(e) It follows from (a) that the restriction of Θ_{T,X} induces an isomorphism of k-modules

𝕊(X) → 𝕊_{E,R^{op}}(X) .

Now if U ⊆ Y × X, we have a diagram

\xymatrix{𝕊(X) \ar@{^{(}->}[r]^{i_{X}} \ar[d]^{𝕊(U)} & F_{T} (X) \ar@{->>}[r]^{Θ_{T,X}} \ar[d]^{F_{T} (U)} & 𝕊_{E,R^{op}}(X) \ar[d]^{𝕊_{E,R^{op}} (U)} \\ 𝕊(Y) \ar@{^{(}->}[r]^{i_{Y}} & F_{T} (Y) \ar@{->>}[r]^{Θ_{T,Y}} & 𝕊_{E,R^{op}}(Y)}

(where i_{X} is the inclusion map). This diagram is not commutative because the left hand square is not commutative (but the right hand square is). But for φ ∈ B_{X}, we have

Θ_{T,Y} ∘ i_{Y} ∘ 𝕊(U)(φ) = Θ_{T,Y} (π_{T,Y} (u_{T} ∘ Uφ))
= Θ_{T,Y} (u_{T} ∘ Uφ)  [as Im(id −π_{T,Y}) ⊆ Ker Θ_{T,Y}]
= Θ_{T,Y} (Uφ)  [by (b)(i)]
= Θ_{T,Y} (F_{T} (U)(φ)) [by definition of F_{T} (U)(φ)]
= 𝕊_{E,R^{op}}(U)(Θ_{T,X}(φ))
= 𝕊_{E,R^{op}}(U) ∘ Θ_{T,X} ∘ i_{X}(φ) .

It follows that the outer rectangle is commutative. In other words, we have a commutative diagram

\xymatrix{𝕊(X) \ar[r]^{Θ_{T,X}i_{X}}_{≅} \ar[d]^{𝕊(U)} & 𝕊_{E,R^{op}}(X) \ar[d]^{𝕊_{E,R^{op}} (U)} \\ 𝕊(Y) \ar[r]^{Θ_{T,Y} i_{Y}}_{≅} & 𝕊_{E,R^{op}}(Y)}

in which the horizontal maps are isomorphisms. This shows that (e) holds and the proof of Theorem 18.4 is complete.

Our final goal in this section is to shed some light on a special case, namely when T = ∧E, where T is a finite lattice and (E, R) is the subposet of irreducible elements in T. By definition of the set G in Notation 16.7, we have G = T (and there are no bulbs). This allows us to obtain a much finer result in that case. 18.6. Theorem. Let (E, R) be a finite poset and let T be any lattice such that (E, R) is the full subposet of irreducible elements in T. Assume that T = ∧E.

(a) F_{T} /H_{T} ≅ 𝕊_{E,R}.

(b) If k is a field and Aut(E, R) is the trivial group, then F_{T} /H_{T} is simple, namely F_{T} /H_{T} ≅ S_{E,R,k}.

Proof : (a) The surjection Θ_{T} : F_{T} → 𝕊_{E,R^{op}} factorizes as the composition of the morphisms Π : F_{T} ⟶ F_{T} /H_{T} and \overline{Θ}_{T} : F_{T} /H_{T} ⟶ 𝕊_{E,R^{op}}. For any finite set X, let

B_{X} = {φ : X → T | E ⊆ φ(X)} .

Then Π_{X}(B_{X}) is a k-basis of (F_{T} /H_{T})(X). Our assumption T = ∧E implies that G = T, where G is, as usual, the subset defined in Notation 16.7. Therefore the set B_{X} coincides with the set defined in Notation 17.1 and used throughout Section 17. Hence Θ_{T,X}(B_{X}) is a k-basis of 𝕊_{E,R^{op}}(X) by Corollary 17.17. It follows that \overline{Θ}_{T,X} maps a k-basis to a k-basis, hence it is an isomorphism.

(b) Our assumption on Aut(E, R) implies that

S_{E,R,k} = 𝕊_{E,R} ⊗_{k Aut(E,R)} k ≅ 𝕊_{E,R} ⊗_{k} k ≅ 𝕊_{E,R} .

By (a), it follows that S_{E,R,k} ≅ F_{T} /H_{T}.

There are many examples of the situation described in Theorem 18.6.

18.7. Example. Suppose that (E, R) is a forest, namely a disjoint union of trees with all edges oriented from the leaves to the root, that is, in such a way that the roots of the trees are the minimal elements of E. Consider the lattice T = E ⊔ {ˆ0, ˆ1}, with ˆ0 minimal and ˆ1 maximal. We assume that E has at least two maximal elements (i.e. E is not totally ordered), so that ˆ1 is not irreducible. Then E is the set of irreducible elements of the lattice T. We claim that G = T, where G is, as usual, the subset defined in Notation 16.7. If E has at least two minimal elements (i.e. at least two connected components), then this follows from the observation that ∧E = T. If E has a single minimal element (i.e. E is a tree), then ∧E = E ⊔ {ˆ1} and ˆ0 is a bulb, so G = ∧E ⊔ {ˆ0} = T.

In all such cases, F_{T} /H_{T} is a fundamental functor. Moreover, there are many cases when Aut(E, R) is the trivial group (take branches of different length). In such a case, F_{T} /H_{T} is simple, provided k is a field.

19. Tensor product and internal hom

In this section, we introduce two constructions in the category F_{k}: when M and M′ are correspondence functors, we define their tensor product M ⊗ M′ and their internal hom H(M, M′), which are both correspondence functors. These functors are associated to the symmetric monoidal structure on C given by the disjoint union of finite sets: if X, X′, Y, Y′ are finite sets, and if U ∈ C(X′, X) and V ∈ C(Y′, Y), then U ⊔ V can be viewed as a subset of (X′ ⊔ Y′) × (X ⊔ Y). We will represent this correspondence in a matrix form, as follows:

( \begin{matrix}U & ∅ \\ ∅ & V \end{matrix} ) .

This yields a bifunctor C × C → C, still denoted by a disjoint union symbol.

19.1. Definition. Let M and M′ be correspondence functors over a commutative ring k.

(a) The tensor product of M and M′ is the correspondence M ⊗ M′ defined for a finite set X by

(M ⊗ M′)(X) = M(X) ⊗_{k} M′(X) .

If Y is a finite set and U ∈ C(Y, X), the map

(M ⊗ M′)(U) : (M ⊗ M′)(X) → (M ⊗ M′)(Y)

is the k-linear map defined by

∀m ∈ M(X), ∀m′ ∈ M′(X), (M ⊗ M′)(U)(m ⊗ m′) = Um ⊗ Um′ ,

where as usual Um = M(U)(m).

(b) Let E be a finite set. Let M_{E} be the correspondence functor obtained from M by precomposition with the endofunctor t_{E} : X ↦ X ⊔ E of C. When E′ is a finite set and V ∈ C(E′, E), let M_{V} : M_{E} → M_{E′} be the morphism obtained by precomposition with the natural transformation id ⊔V : t_{E} → t_{E′}.

(c) Let H(M, M′) be the correspondence functor defined on a finite set E by

H(M, M′)(E) = Hom_{F_{k}}(M, M′_{E}) ,

and for V ∈ C(E′, E), by composition with M′_{V} : M′_{E} → M′_{E′}.

19.2. Theorem. Let k be a commutative ring, and let M, M′ and M′′ be correspondence functors over k.

(a) The assignment (M, M′) ↦ M ⊗M′ is a k-linear bifunctor F_{k} ×F_{k} → F_{k}, right exact in M and M′. The assignment (M, M′) ↦ H(M, M′) is a klinear bifunctor F_{k}^{op} × F_{k} → F_{k}, left exact in M and M′.

(b) There are natural isomorphisms of correspondence functors

M ⊗ (M′ ⊗ M′′) ≅ (M ⊗ M′) ⊗ M′′
M ⊗ M′ ≅ M′ ⊗ M
\underline{k} ⊗ M ≅ M ,

where \underline{k} is the constant functor introduced in Example 4.11.

(c) The k-module Hom_{F_{k}}(M′ ⊗ M, M′′) is isomorphic to the k-module of bilinear pairings M′ × M → M′′, i.e. the k-module of natural transformations of the bifunctor (X, Y) ↦ M′(X) × M(Y) to the bifunctor (X, Y) ↦ M′′(X ⊔ Y) from C × C to k-Mod.

(d) There are isomorphisms of k-modules

Hom_{F_{k}}(M′ ⊗ M, M′′) ≅ Hom_{F_{k}} (M, H(M′, M′′))

natural in M, M′, M′′. In particular, for any correspondence functor M′ over k, the endofunctor M ↦ M′ ⊗ M of F_{k} is left adjoint to the endofunctor M ↦ H(M′, M)

Proof : (a) The first assertion is clear, and the second assertion follows from the fact that the assignment (M, E) ↦ M_{E} is a functor F_{k} × C → F_{k}, exact in M.

(b) For any finite set X, the standard k-linear isomorphisms

M(X) ⊗_{k} (M′(X) ⊗_{k} M′′(X)) ≅ (M(X) ⊗_{k} M′(X)) ⊗_{k} M′′(X)
M(X) ⊗_{k} M′(X) ≅ M′(X) ⊗_{k} M(X)
\underline{k}(X) ⊗_{k} M(X) = k ⊗_{k} M(X) ≅ M(X) ,

are compatible with the action of correspondences.

(c) Let ψ : M′ ⊗ M → M′′ be a morphism of correspondence functors. Equivalently, for any finite set X, let ψ_{X} : M′(X) ⊗_{k} M(X) → M′′(X) be a linear map with the property that for any finite set Z and any U ∈ C(Z, X), the diagram

\xymatrix{M′(X) ⊗_{k} M(X) \ar[r]^{ψ_{X}} \ar[d]_{M′(U)⊗_{k}M(U)} & M′′(X) \ar[d]^{M′′(U)} \\ M′(Z) ⊗_{k} M(Z) \ar[r]^{ψ_{Z}} & M′′(Z)}

is commutative. If X and Y are finite sets, let \binom{Δ_{X}}{∅} ∈ C(X ⊔ Y, X) be the graph of the inclusion i_{X} of X in X ⊔ Y, and let \binom{∅}{Δ_{Y}} ∈ C(X ⊔ Y, Y) be the graph of the inclusion of Y in X ⊔ Y. We define a map

ψ̂_{X,Y} : M′(X) ⊗_{k} M(Y) → M′′(X ⊔ Y)

as the following composition

\xymatrix{M′(X) ⊗_{k} M(Y) \ar[r]^{M′\binom{Δ_{X}}{∅}⊗M\binom{∅}{Δ_{Y}}} & M′(X ⊔ Y) ⊗_{k} M(X ⊔ Y) \ar[r]^{ψ_{X⊔Y}} & M′′(X ⊔ Y)}

i.e. ψ̂_{X,Y} = ψ_{X⊔Y} ∘ (\binom{Δ_{X}}{∅} ⊗ \binom{∅}{Δ_{Y}}), with the usual abuse of notation identifying a correspondence and its action on a correspondence functor. If X′ and Y′ are finite sets, if U ∈ C(X′, X) and V ∈ C(Y′, Y), then we have the following commutative diagram

\xymatrix{M′(X) ⊗_{k} M(Y) \ar[d]_{U}^{V} \ar[r]^{\binom{ΔX}{∅}⊗\binom{∅}{ΔY}} & M′(X ⊔ Y) ⊗_{k} M(X ⊔ Y) \ar[d]^{_{( \begin{matrix}U & ∅ \\ ∅ & V \end{matrix} )}}_{_{( \begin{matrix}U & ∅ \\ ∅ & V \end{matrix} )}} \ar[r]^{ψ_{X⊔Y}} & M′′(X ⊔ Y) \ar[d]_{_{( \begin{matrix}U & ∅ \\ ∅ & V \end{matrix} )}} \\ M′(X′) ⊗_{k} M(Y′) \ar[r]_{\binom{ΔX′}{∅}⊗\binom{∅}{ΔY ′}} & M′(X′ ⊔ Y′) ⊗_{k} M(X′ ⊔ Y′) \ar[r]_{ψ_{X′⊔Y ′}} & M′′(X′ ⊔ Y′) .}

The left hand side square is commutative because ( \begin{matrix}U & ∅ \\ ∅ & V \end{matrix} )\binom{Δ_{X}}{∅} = \binom{U}{∅} = \binom{Δ_{X′}}{∅}U,

and similarly ( \begin{matrix}U & ∅ \\ ∅ & V \end{matrix} )\binom{∅}{Δ_{Y}} = \binom{∅}{V} = \binom{∅}{Δ_{Y ′}}V. The right hand square is commu-

tative by the defining property of the morphism ψ : M′ ⊗ M → M′′. It follows that ( \begin{matrix}U & ∅ \\ ∅ & V \end{matrix} ) ∘ ψ̂_{X,Y} = ψ̂_{X′,Y′} ∘ (U ⊗_{k} V), so the maps ψ̂_{X,Y} define a natu-

ral transformation of the bifunctor (X, Y) ↦ M′(X) ⊗_{k} M(Y) to the bifunctor (X, Y) ↦ M′′(X ⊔ Y), in other words a bilinear pairing ψ̂: M′ × M → M′′.

Conversely, let η : M′ × M → M′′ be a bilinear pairing. This means that for any finite sets X, Y, there is a map η_{X,Y} : M′(X) ⊗_{k} M(Y) → M′′(X ⊔ Y), such that for any finite set X′ and any correspondences U ∈ C(X′, X) and V ∈ C(Y′, Y), the diagram

(19.3)  \xymatrix{M′(X) ⊗_{k} M(Y) \ar[d]_{U}^{V} \ar[r]^{η_{X,Y}} & M′′(X ⊔ Y) \ar[d]^{_{( \begin{matrix}U & ∅ \\ ∅ & V \end{matrix} )}} \\ M′(X′) ⊗_{k} M(Y′) \ar[r]^{η_{X′,Y ′}} & M′′(X′ ⊔ Y′)}

is commutative.

In particular, for X = Y, we have a map η_{X,X} : M′(X)⊗_{k}M(X) → M′′(X⊔X), which we can compose with the map M′′(X ⊔ X) → M′′(X) given by the action of the “codiagonal” correspondence (Δ_{X}, Δ_{X}) ∈ C(X, X ⊔ X), to get a map

η̃_{X} = (Δ_{X}, Δ_{X}) ∘ η_{X,X} : M′(X) ⊗ M(X) → M′′(X) .

If Z is a finite set and U ∈ C(Z, X), the diagram

\xymatrix{M′(X) ⊗ M(X) \ar[d]_{U}^{U} \ar[r]^{η_{X,X}} & M′′(X ⊔ X) \ar[d]^{_{( \begin{matrix}U & ∅ \\ ∅ & U \end{matrix} )}} \ar[r]^{(Δ_{X},Δ_{X})} & M′′(X) \ar[d]^{U} \\ M′(Z) ⊗ M(Z) \ar[r]_{η_{Z,Z}} & M′′(Z ⊔ Z) \ar[r]_{(Δ_{Z},Δ_{Z})} & M′′(Z)}

is commutative: the left hand square is commutative because η is a bilinear pairing, and the right hand square is commutative because

U(Δ_{X}, Δ_{X}) = (U, U) = (Δ_{Z}, Δ_{Z})( \begin{matrix}U & ∅ \\ ∅ & U \end{matrix} ) .

Hence the maps η̃_{X} define a morphism of correspondence functors η̃ from M′ ⊗ M to M′′.

The constructions ψ ↦ ψ̂ and η ↦ η̃ are k-linear and inverse to each other: indeed, for any finite set X,

˜ψ̂_{X} = (Δ_{X},Δ_{X})ψ̂_{X,X} = (Δ_{X},Δ_{X})ψ_{X⊔X}(\binom{Δ_{X}}{∅}⊗\binom{∅}{Δ_{X}})
= ψ_{X}((Δ_{X},Δ_{X}) ⊗_{k} (Δ_{X},Δ_{X}))(\binom{Δ_{X}}{∅}⊗\binom{∅}{Δ_{X}})
= ψ_{X} (Δ_{X}⊗Δ_{X}) = ψ_{X} ,

since (Δ_{X},Δ_{X})\binom{Δ_{X}}{∅} = Δ_{X} and (Δ_{X},Δ_{X})\binom{∅}{Δ_{X}} = Δ_{X}. Similarly, for finite sets X and Y,

ˆη̃_{X,Y} = η̃_{X⊔Y,X⊔Y} (\binom{Δ_{X}}{∅}⊗_{k}\binom{∅}{Δ_{Y}}) = (Δ_{X⊔Y} ,Δ_{X⊔Y} )η_{X⊔Y,X⊔Y} (\binom{Δ_{X}}{∅}⊗_{k}\binom{∅}{Δ_{Y}})
= (Δ_{X⊔Y} ,Δ_{X⊔Y} ) ( \begin{matrix}\binom{Δ_{X}}{∅} & ∅ \\ ∅ & \binom{∅}{Δ_{Y}} \end{matrix} ) η_{X,Y}
= ( \begin{matrix}Δ_{X} & ∅ \\ ∅ & Δ_{Y} \end{matrix} ) η_{X,Y} = η_{X,Y} .

(d) Let ψ : M′ × M → M′′ be a morphism of correspondence functors. By (c), for any finite sets X and Y, we get a linear map ψ̂_{X,Y} : M′(X) ⊗_{k} M(Y) → M′′(X ⊔ Y), or equivalently, a linear map

\overline{ψ}_{Y,X} : M(Y) → Hom_{k} (M′(X), M′′(X ⊔ Y))

defined by \overline{ψ}_{Y,X}(m)(m′) = ψ̂_{X,Y} (m′ ⊗ m), for m ∈ M(Y) and m′ ∈ M′(X). Now M′′(X ⊔ Y) = M′′_{Y}(X). Moreover, for any finite set X′ and any U ∈ C(X′, X), the commutative diagram 19.3, for Y′ = Y and V = Δ_{Y}, becomes

\xymatrix{M′(X) ⊗_{k} M(Y) \ar[d]_{U}^{Δ_{Y}} \ar[r]^{ψ̂_{X,Y}} & M′′(X ⊔ Y) \ar[d]^{_{( \begin{matrix}U & ∅ \\ ∅ & Δ_{Y} \end{matrix} )}} \\ M′(X′) ⊗_{k} M(Y) \ar[r]^{ψ̂_{X′,Y}} & M′′(X′ ⊔ Y) .}

In other words \overline{ψ}_{Y,X′}(Um′) = U\overline{ψ}_{Y,X}(m′) for any m′ ∈ M′(X). Hence for a given set Y, the maps \overline{ψ}_{Y,X} define a morphism of correspondence functors M′ → M′′_{Y}, and this gives a map \overline{ψ}_{Y} : M(Y) → H(M′, M′′)(Y).

Now if Y′ is a finite set and V ∈ C(Y′, Y), the commutative diagram 19.3, for X′ = X and U = Δ_{X}, becomes

\xymatrix{M′(X) ⊗_{k} M(Y) \ar[d]_{Δ_{X}}^{V} \ar[r]^{ψ̂_{X,Y}} & M′′(X ⊔ Y) \ar[d]^{_{( \begin{matrix}Δ_{X} & ∅ \\ ∅ & V \end{matrix} )}} \\ M′(X) ⊗_{k} M(Y′) \ar[r]^{ψ̂_{X,Y ′}} & M′′(X ⊔ Y′) ,}

and it follows that the maps \overline{ψ}_{Y} define a morphism of correspondence functors \overline{ψ} : M → H(M′, M′′).

Conversely, a morphism of correspondence functors ξ : M → H(M′, M′′) is determined by maps ξ_{Y} : M(Y) → H(M′, M′′)(Y) = Hom_{F_{k}}(M′, M′′_{Y}), for all finite sets Y. Furthermore, for m ∈ M(Y), the map ξ_{Y} (m) is in turn determined by maps ξ_{Y} (m)_{X} : M′(X) → M′′(X ⊔ Y), for all finite sets X. It is easy to see that ξ is a morphism of correspondence functors from M to H(M′, M′′) if and only if the maps

[figure: ξ̊_{X,Y} : M′(X) ⊗_{k} M(Y); M′′(X ⊔ Y); m′ ⊗ m; ξ_{Y} (m)(m′)]

define a bilinear pairing ξ̊: M′ × M → M′′, i.e. if for any finite sets X, Y, X′, Y′, and any correspondences U ∈ C(X′, X) and V ∈ C(Y′, Y), the diagram

\xymatrix{M′(X) ⊗_{k} M(Y) \ar[d]_{U}^{V} \ar[r]^{ξ̊_{X,Y}} & M′′(X ⊔ Y) \ar[d]^{_{( \begin{matrix}U & ∅ \\ ∅ & V \end{matrix} )}} \\ M′(X′) ⊗_{k} M(Y′) \ar[r]^{ξ̊_{X′,Y ′}} & M′′(X′ ⊔ Y′)}

is commutative. By (c), we get a morphism of correspondence functors ξ̌ = ˜ ξ̊from M′ ⊗ M to M′′.

Now it is straightforward to check that the maps ψ ↦ \overline{ψ} and ξ ↦ ξ̌ are inverse isomorphisms between Hom_{F_{k}}(M′ ⊗ M, M′′) and Hom_{F_{k}} (M, H(M′, M′′)). Assertion (d) follows.

Now we want to apply the tensor product construction to functors of the form F_{T}, where T is a finite lattice. Thus we again consider the category kL of lattices, introduced in Definition 11.5.

19.4. Theorem. Let k be a commutative ring.

(a) There is a well defined k-linear bifunctor kL × kL → kL sending

• the pair (T, T′) of lattices to the product lattice T × T′,

• and the pair of morphisms (∑_{i=1}^{n} λ_{i}f_{i}, ∑_{j=1}^{m} μ_{j}g_{j}), where λ_{i}, μ_{j} ∈ k and

f_{i} : T → T_{1} and g_{j} : T′ → T′_{1} are morphisms in L, to the morphism

∑_{{1≤i≤n \atop 1≤j≤m}} λ_{i}μ_{j}(f_{i} × g_{j}) : T × T′ → T_{1} × T′_{1}.

(b) The bifunctors (T, T′) ↦ F_{T} ⊗ F_{T′} and (T, T′) ↦ F_{T ×T′} from kL × kL to F_{k} are isomorphic.

(c) In particular, if E and F are finite sets, then

kC(−, E) ⊗ kC(−, F) ≅ kC(−, E ⊔ F) .

Proof : (a) If f : T → T_{1} and g : T′ → T′_{1} are morphisms in L, then f × g is a morphism in L from T × T′ to T_{1} × T′_{1}: indeed, if A ⊆ T × T′, let B (resp. B′) denote the projection of A on T (resp. T′). Then

(f × g)( ⋁_{(t,t′)∈A}(t, t′)) = (f × g)( ⋁_{(t,t′)∈A} ((t, ˆ0) ∨ (ˆ0, t′)))

= (f × g)( ⋁_{t∈B}(t, ˆ0) ∨ ⋁_{t′∈B′}(ˆ0, t′))
= (f × g)( ⋁_{t∈B} t, ⋁_{t′∈B′} t′)
= ( ⋁_{t∈B} f(t), ⋁_{t′∈B′} g(t′))

On the other hand

⋁_{(t,t′)∈A} (f(t), g(t′)) = ⋁_{t∈B} (f(t), ˆ0) ∨ ⋁_{t′∈B′} (ˆ0, g(t′))

= ( ⋁_{t∈B} f(t), ˆ0) ∨ (ˆ0, ⋁_{t′∈B′} g(t′))
= ( ⋁_{t∈B} f(t), ⋁_{t′∈B′} g(t′))

It follows that (f ×g)( ⋁_{(t,t′)∈A}(t, t′)) = ⋁_{(t,t′)∈A}(f ×g)(t, t′), as claimed. Assertion (a)

follows by bilinearity.

(b) Let T and T′ be finite lattices. Then for any finite set X, there is a unique isomorphism of k-modules

τ_{X} : (F_{T} ⊗ F_{T′})(X) = k(T^{X}) ⊗_{k} k(T^{′X}) ⟶ k(T × T′)^{X} ,

sending φ ⊗ φ′, for φ : X → T and φ′ : X → T′, to the map φ × φ′ : X → T × T′. If Y is a finite set and U ∈ C(Y, X), then for any y ∈ Y,

U(φ × φ′)(y) = ⋁_{(y,x)∈U} (φ(x), φ′(x))
= ( ⋁_{(y,x)∈U} φ(x), ⋁_{(y,x′)∈U} φ′(x′))
= (Uφ(y), Uφ′(y)) .

Thus Uτ_{X}(φ ⊗_{k} φ′) = Uφ × Uφ′ = τ_{Y} (Uφ ⊗_{k} Uφ′), hence τ : F_{T} ⊗ F_{T′} → F_{T ×T′} is an isomorphism of correspondence functors.

If f : T → T_{1} and f′ : T′ → T′_{1} are morphisms in L, then by (a) the map f × f′ : T × T′ → T_{1} × T′_{1} is a morphism in L. Moreover, the diagram

\xymatrix{F_{T} ⊗ F_{T′} \ar[d]_{F_{f} ⊗F_{f′}} \ar[r]^{τ} & F_{T ×T′} \ar[d]^{F_{f×f′}} \\ F_{T_{1}} ⊗ F_{T′_{1}} \ar[r]^{τ_{1}} & F_{T_{1}×T′_{1}}}

where τ_{1} : F_{T_{1}} ⊗ F_{T′_{1}} → F_{T_{1}×T′_{1}} is the corresponding isomorphism for the lattices T_{1} and T′_{1}, is commutative, since for any finite set X, any φ : X → T and any φ′ : X′ → T′

F_{f×f′}τ_{X}(φ ⊗_{k} φ′) = F_{f×f′}(φ × φ′) = (f ∘ φ) × (f′ ∘ φ′) = τ_{1,X}(F_{f}(φ) ⊗ F_{f′}(φ′)) .

Hence τ is an isomorphism of bifunctors.

(c) This follows from (b), applied to the lattice T of subsets of E and the lattice T′ of subsets of F. Then F_{T} ≅ kC(−, E) and T′ ≅ kC(−, F). Moreover T × T′ is isomorphic to the lattice of subsets of E ⊔ F.

19.5. Proposition. Let T be a finite lattice.

(a) The map

υ : T × T → T , υ(a, b) = a ∨ b ,

is a morphism in the category L.

(b) The map υ induces a morphism μ : F_{T} ⊗ F_{T} → F_{T}, defined for a finite set X and maps φ, ψ : X → T by μ(φ ⊗ ψ) = φ ∨ ψ, where (φ ∨ ψ)(x) = φ(x) ∨ ψ(x), for x ∈ X.

(c) For finite sets X and Y, let μ̂ : F_{T} (X) × F_{T} (Y) → F_{T} (X ⊔ Y) be the bilinear pairing associated to μ, in the sense of part (c) of Theorem 19.2. Then for φ : X → T and ψ : Y → T, the element μ̂(φ, ψ) ∈ F_{T} (X ⊔ Y) is the function from X ⊔ Y to T equal to φ on X and to ψ on Y.

(d) The triple (F_{T} , μ, ϵ) is a commutative algebra in the tensor category F_{k}, where ϵ : \underline{k} → F_{T} is the morphism sending 1 ∈ \underline{k}(X) = k to the function X → T mapping any element of X to ˆ0, for any finite set X.

Proof : (a) Clearly for a, b, c, d ∈ T, the join (a, b)∨(c, d) = (a∨c, b∨d) is mapped by υ to (a ∨ c) ∨ (b ∨ d) = (a ∨ b) ∨ (c ∨ d), so υ((a, b) ∨ (c, d)) = υ(a, b) ∨ υ(c, d). Moreover the zero element of T × T is (ˆ0, ˆ0), and υ(ˆ0, ˆ0) = ˆ0 ∨ ˆ0 = ˆ0.

(b) Follows from (a), by Theorem 11.7 and Theorem 19.4.

(c) One checks easily that \binom{Δ_{X}}{∅}φ is the function from X ⊔ Y to T equal to φ on X, and to ˆ0 on Y. Similarly \binom{∅}{Δ_{Y}}ψ is the map equal to ˆ0 on X and to ψ on Y.

Thus (\binom{Δ_{X}}{∅}φ) ∨ (\binom{∅}{Δ_{Y}}ψ) is the map equal to φ on X and to ψ on Y.

(d) Clearly μ is associative, commutative, and ϵ is an identity element.

19.6. Theorem. Let k be a commutative ring. Let M and N be correspondence functors over k.

(a) Let E and F be finite sets, and let G = E ⊔ F. Let V be an R_{E}-module and W be an R_{F} -module. Then there is an isomorphism of correspondence functors

L_{E,V} ⊗ L_{F,W} ≅ L_{G,V ↑_{E}^{G}⊗_{k}W ↑_{F}^{G}} ,

where the R_{G}-module structure on V ↑_{E}^{G} ⊗_{k}W ↑_{F}^{G} is induced by the comultiplication ν : R_{G} → R_{G} ⊗ R_{G} defined by ν(U) = U ⊗ U for U ∈ C(G, G).

(b) If M and N are projective, then M ⊗ N is projective.

(c) If E and F are finite sets such that M = ⟨M(E)⟩ and N = ⟨N(F)⟩, then M ⊗ N = ⟨(M ⊗ N)(E ⊔ F)⟩.

(d) If M and N have bounded type, so has M ⊗ N.

(e) If M and N are finitely generated, so is M ⊗ N.

Proof : (a) Since (L_{E,V} ⊗ L_{F,W})(G) = V ↑_{E}^{G} ⊗_{k} W ↑_{F}^{G}, by adjunction, we get a morphism

Φ : L_{G,V↑_{E}^{G}⊗_{k}W↑_{F}^{G}} → L_{E,V} ⊗ L_{F,W}

which can be described as follows: for a finite set X,

L_{G,V ↑_{E}^{G}⊗_{k}W ↑_{F}^{G}}(X) = kC(X, G) ⊗_{R_{G}} ((kC(G, E) ⊗_{R_{E}} V ) ⊗_{k} (kC(G, F) ⊗_{R_{F}} W ))

and

(L_{E,V} ⊗ L_{F,W})(X) = (kC(X, E) ⊗_{R_{E}} V ) ⊗_{k} (kC(X, F) ⊗_{R_{F}} W ) .

The morphism Φ_{X} sends the element

C ⊗_{R_{G}} ((A ⊗_{R_{E}} v) ⊗_{k} (B ⊗_{R_{F}} w))

of L_{G,V ↑_{E}^{G}⊗_{k}W ↑_{F}^{G}}(X), where C ∈ C(X, G), A ∈ C(G, E), v ∈ V, B ∈ C(G, F), and w ∈ W, to the element

(CA ⊗_{R_{E}} v) ⊗_{k} (CB ⊗_{R_{F}} w)

of (L_{E,V} ⊗ L_{F,W})(X).

Conversely, there is a morphism Ψ_{X} : (L_{E,V} ⊗ L_{F,W})(X) → L_{G,V ↑_{E}^{G}⊗_{k}W ↑_{F}^{G}}(X) sending the element

(P ⊗_{R_{E}} v) ⊗ (Q ⊗_{R_{F}} w)

of (L_{E,V} ⊗ L_{F,W})(X), where P ∈ C(X, E), v ∈ V, Q ∈ C(X, F), and w ∈ W, to the element

(P, Q) ⊗_{R_{G}} ((\binom{Δ_{E}}{∅} ⊗_{R_{E}} v) ⊗_{k} (\binom{∅}{Δ_{F}} ⊗_{R_{F}} w))

of L_{G,V ↑_{E}^{G}⊗_{k}W ↑_{F}^{G}}(X), where (P, Q) ∈ C(X, E ⊔ F), and where \binom{Δ_{E}}{∅} ∈ C(E ⊔ F, E) and \binom{∅}{Δ_{F}} ∈ C(E ⊔ F, F).

The map Ψ_{X} is well defined, for if R ∈ C(E, E) and S ∈ C(F, F)

Ψ_{X}((P ⊗_{R_{E}} Rv) ⊗_{k} (Q ⊗_{R_{F}} Sw))

is equal to

(P, Q) ⊗_{R_{G}} ((\binom{Δ_{E}}{∅} ⊗_{R_{E}} Rv) ⊗_{k} (\binom{∅}{Δ_{F}} ⊗_{R_{F}} Sw)) =
(P, Q) ⊗_{R_{G}} ((\binom{R}{∅} ⊗_{R_{E}} v) ⊗_{k} (\binom{∅}{S} ⊗_{R_{F}} w)) =
(P, Q) ⊗_{R_{G}} ( \begin{matrix}R & ∅ \\ ∅ & S \end{matrix} ) ((\binom{Δ_{E}}{∅} ⊗_{R_{E}} v) ⊗_{k} (\binom{∅}{Δ_{F}} ⊗_{R_{F}} w)) =

(P R, QS)((\binom{Δ_{E}}{∅} ⊗_{R_{E}} v) ⊗_{k} (\binom{∅}{Δ_{F}} ⊗_{R_{F}} w)) ,

which is equal to Ψ_{X}((P R ⊗_{R_{E}} v) ⊗_{k} (QS ⊗_{R_{F}} w)).

Moreover, if Y is a finite set and U ∈ C(Y, X), then

Ψ_{X}(U((P ⊗_{R_{E}} Rv) ⊗_{k} (Q ⊗_{R_{F}} Sw))) =
Ψ_{X}((UP ⊗_{R_{E}} Rv) ⊗_{k} (UQ ⊗_{R_{F}} Sw)) =
(UP, UQ) ⊗_{R_{G}} ((\binom{Δ_{E}}{∅} ⊗_{R_{E}} v) ⊗_{k} (\binom{∅}{Δ_{F}} ⊗_{R_{F}} w)) =
UΨ_{X}((P ⊗_{R_{E}} Rv) ⊗_{k} (Q ⊗_{R_{F}} Sw)) .

It follows that the maps Ψ_{X} define a morphism of correspondence functors

Ψ : L_{E,V} ⊗ L_{F,W} → L_{G,V ↑_{E}^{G}⊗_{k}W ↑_{F}^{G}} .

Moreover, setting u = (P ⊗_{R_{E}} v) ⊗ (Q ⊗_{R_{F}} w)

Φ_{X}Ψ_{X}(u) = Φ_{X} (((P, Q) ⊗_{R_{G}} ((\binom{Δ_{E}}{∅} ⊗_{R_{E}} v) ⊗_{k} (\binom{∅}{Δ_{F}} ⊗_{R_{F}} w)))
= ((P, Q)\binom{Δ_{E}}{∅} ⊗_{R_{E}} v) ⊗_{k} ((P, Q)\binom{∅}{Δ_{F}} ⊗_{R_{E}} w)
= (P ⊗_{R_{E}} v) ⊗ (Q ⊗_{R_{F}} w) = u ,

so ΦΨ is equal to the identity morphism.

Similarly, setting s = C ⊗_{R_{G}} ((A ⊗_{R_{E}} v) ⊗_{k} (B ⊗_{R_{F}} w)),
Ψ_{X}Φ_{X}(s) = Ψ_{X}((CA ⊗_{R_{E}} v) ⊗_{k} (CB ⊗_{R_{F}} w))
= (CA, CB) ⊗_{R_{G}} ((\binom{Δ_{E}}{∅} ⊗_{R_{E}} v) ⊗_{k} (\binom{∅}{Δ_{F}} ⊗_{R_{F}} w))
= C(A, B) ⊗_{R_{G}} ((\binom{Δ_{E}}{∅} ⊗_{R_{E}} v) ⊗_{k} (\binom{∅}{Δ_{F}} ⊗_{R_{F}} w))
= C ⊗_{R_{G}} (A, B)((\binom{Δ_{E}}{∅} ⊗_{R_{E}} v) ⊗_{k} (\binom{∅}{Δ_{F}} ⊗_{R_{F}} w))
= C ⊗_{R_{G}} ((A ⊗_{R_{E}} v) ⊗_{k} (B ⊗_{R_{F}} w)) = s ,

so ΨΦ is also equal to the identity morphism.

(b) It suffices to assume that M = ⨁_{i∈I} kC(−, E_{i}) and N = ⨁_{j∈J} kC(−, F_{j}), where

I and J are sets, and E_{i}, for i ∈ I, and F_{j}, for j ∈ J, are finite sets. In this case

M ⊗ N ≅ ⨁_{i∈I, j∈J} kC(−, E_{i}) ⊗ kC(−, F_{j}) ≅ ⨁_{i∈I, j∈J} kC(−, E_{i} ⊔ F_{j}) ,

so M ⊗ N is projective.

(c) Suppose that M = ⟨M(E)⟩ and N = ⟨N(F)⟩. Equivalently the counit morphisms L_{E,M(E)} → M and L_{F,N(F)} → N are surjective. Then M ⊗ N is a quotient of

L_{E,M(E)} ⊗ L_{F,N(F)} ≅ L_{E⊔F,M(E)↑_{E}^{E⊔F} ⊗_{k}N(F)↑_{F}^{E⊔F}} .

Since L_{E⊔F,M(E)↑_{E}^{E⊔F} ⊗_{k}N(F)↑_{F}^{E⊔F}} is generated by its evaluation at E ⊔ F, so does M ⊗ N.

(d) This follows from (c).

(e) With the assuption of (c), if moreover M(E) and N(F) are finitely generated k-modules, so is M(E) ↑_{E}^{E⊔F} ⊗_{k}N(F) ↑_{F}^{E⊔F}. Therefore, if M and N are finitely generated, so is M ⊗ N.

19.7. Theorem. Let k be a commutative ring. Let M and N be correspondence functors over k.

(a) Let E and F be finite sets. If M = ⟨M(E)⟩, then M_{F} = ⟨M_{F} (E)⟩. Therefore, if M has bounded type, so has M_{F}. If M is finitely generated, so is M_{F}.

(b) Let F be a finite set. There is an isomorphism of correspondence functors H(kC(−, F), N) ≅ N_{F}.

(c) Assume that k is noetherian. If M is finitely generated and if N has bounded type (resp. is finitely generated), then H(M, N) has bounded type (resp. is finitely generated).

Proof : (a) Saying that M = ⟨M(E)⟩ is equivalent to saying that

M(X) = kC(X, E)M(E)

for each finite set X. Replacing X by X ⊔ F gives

M(X ⊔ F) = kC(X ⊔ F, E)M(E) .

Now any U ∈ C(X ⊔F, E) can be written in matrix form \binom{V}{W}, where V ∈ C(X, E)

and W ∈ C(F, E). Moreover for m ∈ M(E)

\binom{V}{W}(m) = ( \begin{matrix}V & ∅ \\ ∅ & Δ_{F} \end{matrix} ) \binom{Δ_{E}}{W}(m)

and ( \begin{matrix}V & ∅ \\ ∅ & Δ_{F} \end{matrix} ) \binom{Δ_{E}}{W}(m) is the image of \binom{Δ_{E}}{W}(m) ∈ M_{F} (E) by the corre-

spondence V, in the functor M_{F}. It follows that \binom{V}{W}(m) ∈ kC(X, E)M_{F} (E),

for any m ∈ M(E). Thus UM(E) ⊆ kC(X, E)M_{F} (E), for any U ∈ C(X ⊔ F, E). Hence M(X ⊔ F) = M_{F} (X) = kC(X, E)M_{F} (E), as was to be shown. The last two assertions are then clear.

(b) Let X be a finite set. Then

H(kC(−, E), N)(X) = Hom_{F_{k}} (kC(−, E), N_{X})
≅ N_{X}(E) = N(E ⊔ X) ≅ N(X ⊔ E) = N_{E}(X) ,

and it is straightforward to check that the resulting isomorphism

H(kC(−, E), N)(X) ≅ N_{E}(X)

is compatible with correspondences, i.e. that it yields an isomorphism of correspondence functors H(kC(−, E), N) → N_{E}.

(c) If M is finitely generated, then M is a quotient of a finite direct sum

⨁_{i=1}^{n} kC(−, E_{i}) of representable functors. Then

\xymatrix{H(M, N) \ar@{^{(}->}[r] & ⨁_{i=1}^{n} H(kC(−, E_{i}), N) ≅ ⨁_{i=1}^{n} N_{E_{i}} .}

If N has bounded type, then each N_{E_{i}} has bounded type also, by (a), so the finite

direct sum ⨁_{i=1}^{n} N_{E_{i}} also has bounded type. Then H(M, N) is a subfunctor of a

functor of bounded type. If k is noetherian, this implies that H(M, N) has bounded type, by Corollary 10.5. The same argument with “bounded type” replaced by “finitely generated” goes through, completing the proof.

20. Examples of decompositions

We state here without complete proofs a list of examples of decomposition of functors F_{T} associated to some particular lattices T. We assume that k is a field, as this allows for much simpler arguments. We emphasize however that the decompositions we obtain as direct sums of fundamental functors actually hold over an arbitrary commutative ring k.

20.1. Example. Let T = ♢ be the lozenge, in other words the lattice of subsets of a set of cardinality 2:

♢ = [figure: •; ∘; ∘; •]

Then we know from Theorem 15.19 that F_{♢} splits as a direct sum

F_{♢} ≅ 𝕊_{0} ⊕ 3𝕊_{1} ⊕ 2𝕊_{2} ⊕ L

for some subfunctor L, and we also know that the direct sum

𝕊_{0} ⊕ 3𝕊_{1} ⊕ 2𝕊_{2}

corresponding to totally ordered subsets of ♢ lies in the subfunctor H_{♢}, as no such subset contains the two irreducible elements of ♢ (figured with an empty circle in the above picture).

We can evaluate this at a set X of cardinality x, and take dimensions over k. Using Corollary 17.17, this gives

4^{x} = 1^{x} + 3(2^{x} − 1^{x}) + 2(3^{x} − 2 · 2^{x} + 1^{x}) + dim_{k} L(X) .

It follows that

dim_{k} L(X) = 4^{x} − 2 · 3^{x} + 2^{x} .

Moreover we know that F_{♢}/H_{♢} maps surjectively on the fundamental functor 𝕊_{∘∘} associated to the (opposite) poset of irreducible elements of ♢, that is, a set of cardinality 2 ordered by the equality relation. The set G is the whole of T in this case, so dim_{k} 𝕊_{∘∘}(X) = 4^{x} − 2 · 3^{x} + 2^{x}.

It follows that L maps surjectively onto 𝕊_{∘∘}, and since dim_{k} L(X) = dim_{k} 𝕊_{∘∘}(X) for any finite set X, this surjection is an isomorphism. Hence

F_{♢} ≅ 𝕊_{0} ⊕ 3𝕊_{1} ⊕ 2𝕊_{2} ⊕ 𝕊_{∘∘} .

Again, the argument given here works only over a field k, but one can show that this direct sum decomposition of F_{♢} actually holds over any commutative ring k. In particular, as ♢ is distributive, F_{♢} is projective by Theorem 11.11. It follows that 𝕊_{∘∘} is a projective object in F_{k}. When k is self-injective, 𝕊_{∘∘} is also injective in F_{k}.

20.2. Example. Let T be the following lattice:

T = [figure: •; ∘; ∘; ∘; •]

We know from Theorem 15.19 that F_{T} admits a direct summand isomorphic to

𝕊_{0} ⊕ 4𝕊_{1} ⊕ 3𝕊_{2} .

Moreover, there are three obvious sublattices of T isomorphic to ♢, which provide three direct summands of F_{T} isomorphic to 𝕊_{∘∘}. Thus we have a decomposition

F_{T} ≅ 𝕊_{0} ⊕ 4𝕊_{1} ⊕ 3𝕊_{2} ⊕ 3𝕊_{∘∘} ⊕ M

for some subfunctor M of F_{T}. Taking the dimensions of the evaluations at a set X of cardinality x gives

5^{x} = 1^{x} + 4(2^{x} − 1^{x}) + 3(3^{x} − 2 · 2^{x} + 1^{x}) + 3(4^{x} − 2 · 3^{x} + 2^{x}) + dim_{k} M(X) .

It follows that

dim_{k} M(X) = 5^{x} − 3 · 4^{x} + 3 · 3^{x} − 2^{x} .

We also know that F_{T} /H_{T} maps surjectively onto the fundamental functor 𝕊_{∘∘∘} associated to the (opposite) poset of its irreducible elements, that is, a set of cardinality 3 ordered by the equality relation. The set G is the whole of T in this case, so

dim_{k} 𝕊_{∘∘∘}(X) = 5^{x} − 3 · 4^{x} + 3 · 3^{x} − 2^{x} .

Moreover, all the direct summands of F_{T} different from M in the above decomposition are contained in H_{T}, since none of the corresponding sublattices contains all the irreducible elements of T. Thus M maps onto 𝕊_{∘∘∘}, and since dim_{k} M(X) = dim_{k} 𝕊_{∘∘∘}(X), this surjection is an isomorphism. Hence we get

F_{T} ≅ 𝕊_{0} ⊕ 4𝕊_{1} ⊕ 3𝕊_{2} ⊕ 3𝕊_{∘∘} ⊕ 𝕊_{∘∘∘} .

As in the the previous example, our argument works only when k is a field, but one can show that the result still holds over an arbitrary commutative ring k. It should be noted that all the summands in this decomposition of F_{T}, except possibly 𝕊_{∘∘∘}, are projective functors. Since the lattice T is not distributive, the functor F_{T} is not projective, thus 𝕊_{∘∘∘} is actually not projective either.

20.3. Example. Let D be the following lattice:

D = [figure: •; ∘; ∘; ∘; •]

Again, we know from Theorem 15.19 that F_{T} admits a direct summand isomorphic to 𝕊_{0} ⊕ 4𝕊_{1} ⊕ 4𝕊_{2} ⊕ 𝕊_{3}. Moreover, there are two inclusions

[figure: •; •; ∘; ∘; ∘; ∘; ∘; •; •; and; •; •; ∘; ∘; ∘; ∘; ∘; •; •]

of the lattice ♢ into D, which yield two direct summands of F_{D} isomorphic to 𝕊_{∘∘}. So there is a decomposition

F_{D} ≅ 𝕊_{0} ⊕ 4𝕊_{1} ⊕ 4𝕊_{2} ⊕ 𝕊_{3} ⊕ 2𝕊_{∘∘} ⊕ N

for a suitable subfunctor N of F_{D}. As in the previous examples, the subfunctor N maps surjectively onto the fundamental functor 𝕊 \begin{matrix}∘ \\ ∘ \end{matrix} _{∘} associated to the (opposite)

poset \begin{matrix}∘ \\ ∘ \end{matrix} ∘ of irreducible elements of D. The set G is the whole of T in this case also. Taking dimensions of the evaluations of these functors at a set X of cardinality x gives

5^{x} = 1^{x} + 4(2^{x} − 1^{x}) + 4(3^{x} − 2 · 2^{x} + 1^{x})

+ (4^{x} − 3 · 3^{x} + 3 · 2^{x} − 1^{x}) + 2(4^{x} − 2 · 3^{x} + 2^{x}) + dim_{k} N(X) ,

hence

dim_{k} N(X) = 5^{x} − 3 · 4^{x} + 3 · 3^{x} − 2^{x} .

This is also equal to dim_{k} 𝕊 {∘ \atop ∘} _{∘}(X), so N ≅ 𝕊 \begin{matrix}∘ \\ ∘ \end{matrix} _{∘}, and we get

F_{D} ≅ 𝕊_{0} ⊕ 4𝕊_{1} ⊕ 4𝕊_{2} ⊕ 𝕊_{3} ⊕ 2𝕊_{∘∘} ⊕ 𝕊 \begin{matrix}∘ \\ ∘ \end{matrix} _{∘} .

Again this decomposition holds over any commutative ring k. As D is not distributive (the lattice D and the lattice T of the previous example are the smallest non-distributive lattices), the functor 𝕊 \begin{matrix}∘ \\ ∘ \end{matrix} _{∘} is not projective either.

20.4. Example. Let C be the following lattice:

C = [figure: •; ∘; ∘; ∘; •]

We know from Theorem 15.19 that F_{C} admits a direct summand isomorphic to 𝕊_{0} ⊕ 4𝕊_{1} ⊕ 5𝕊_{2} ⊕ 2𝕊_{3} and this summand is contained in H_{C}. The inclusion

[figure: •; •; ∘; ∘; ∘; ∘; ∘; •; •]

of ♢ in C yields a direct summand of F_{C} isomorphic to 𝕊_{∘∘}, also contained in H_{C}. So there is a decomposition

F_{C} ≅ 𝕊_{0} ⊕ 4𝕊_{1} ⊕ 5𝕊_{2} ⊕ 2𝕊_{3} ⊕ 𝕊_{∘∘} ⊕ Q

for some direct summand Q of F_{C}. Evaluating at a set X of cardinality x and taking dimensions gives

5^{x} = 1^{x} + 4(2^{x} − 1^{x}) + 5(3^{x} − 2 · 2^{x} + 1^{x})

+ 2(4^{x} − 3 · 3^{x} + 3 · 2^{x} − 1^{x}) + (4^{x} − 2 · 3^{x} + 2^{x}) + dim_{k} Q(X) .

This gives

dim_{k} Q(X) = 5^{x} − 3 · 4^{x} + 3 · 3^{x} − 2^{x} .

Now F_{C} maps surjectively onto the fundamental functor 𝕊_{∘ ; {/\ \atop ∘ ∘}} associated to the

opposite poset of its irreducible elements, and the corresponding subset G of C is the whole of C. It follows that

dim_{k} 𝕊_{∘ ; {/\ \atop ∘ ∘}}(X) = 5^{x} − 3 · 4^{x} + 3 · 3^{x} − 2^{x} ,

thus Q ≅ 𝕊_{∘ ; {/\ \atop ∘ ∘}}. Finally, we get a decomposition

F_{C} ≅ 𝕊_{0} ⊕ 4𝕊_{1} ⊕ 5𝕊_{2} ⊕ 2𝕊_{3} ⊕ 𝕊_{∘∘} ⊕ 𝕊_{∘ ; {/\ \atop ∘ ∘}} .

Again, this decomposition actually holds for an arbitrary commutative ring k. Since C is distributive, F_{C} is projective and we conclude that 𝕊_{∘ ; {/\ \atop ∘ ∘}} is projective. Taking

dual functors, we get a decomposition

F_{C^{op}} ≅ 𝕊_{0} ⊕ 4𝕊_{1} ⊕ 5𝕊_{2} ⊕ 2𝕊_{3} ⊕ 𝕊_{∘∘} ⊕ 𝕊_{{∘ ∘ \atop \/} ; ∘} ,

so 𝕊_{{∘ ∘ \atop \/} ; ∘} is also projective.

20.5. Example. Let P be the following lattice:

P = [figure: •; ∘; •; ∘; ∘; •]

that is, the direct product of a totally ordered lattice of cardinality 3 with a totally ordered lattice of cardinality 2.

We know from Theorem 15.19 that F_{P} admits a direct summand isomorphic to 𝕊_{0} ⊕ 5𝕊_{1} ⊕ 7𝕊_{2} ⊕ 3𝕊_{3} and this summand is contained in H_{P}. The inclusions

[figure: •; •; ∘; •; ∘; ∘; ∘; ∘; •; •; •; •; ∘; •; ∘; ∘; ∘; ∘; •; •; •; •; ∘; •; ∘; ∘; ∘; ∘; •; •]

of ♢ in P yield 3 direct summands of F_{P} isomorphic to 𝕊_{∘∘}, and contained in H_{P}. Moreover, the inclusions

[figure: •; •; ∘; ∘; ∘; •; ∘; ∘; ∘; •; •; and; ∘; •; •; ∘; •; ∘; ∘; ∘; ∘; •; •]

of C and C^{op} in P yield direct summands 𝕊_{{∘ ∘ \atop \/} ; ∘} and 𝕊_{∘ ; {/\ \atop ∘ ∘}} of F_{P}, also contained in H_{P}.

Hence there is a direct summand U of F_{P} such that

F_{P} ≅ 𝕊_{0} ⊕ 5𝕊_{1} ⊕ 7𝕊_{2} ⊕ 3𝕊_{3} ⊕ 3𝕊_{∘∘} ⊕ 𝕊_{{∘ ∘ \atop \/} ; ∘} ⊕ 𝕊_{∘ ; {/\ \atop ∘ ∘}} ⊕ U .

Since the lattice P is distributive, the functor F_{P} is projective, hence U is projective. Now F_{P} maps surjectively onto the fundamental functor 𝕊 \begin{matrix}∘ \\ ∘ \end{matrix} _{∘}, and H_{P} is contained

in the kernel of this surjection. It follows that U maps surjectively onto 𝕊 \begin{matrix}∘ \\ ∘ \end{matrix} _{∘}, which

is a simple functor, as k is a field and the poset \begin{matrix}∘ \\ ∘ \end{matrix} ∘ has no nontrivial automorphisms.

Using the fact that dim Hom_{F_{k}}(M, N) = dim Hom_{F_{k}}(N, M) if M is projective (by Theorem 9.7), a tedious analysis shows that U is indecomposable, so U is a projective cover of the simple functor 𝕊 \begin{matrix}∘ \\ ∘ \end{matrix} _{∘}. Moreover, using Theorem 18.4, one can

show that the functor U is uniserial, with a filtration

0  _{𝕊 _{\begin{matrix}∘ \\ ∘ \end{matrix} _{∘}}} ⊂  W  _{𝕊 ∘ ∘ ; /\/ ; ∘ ∘} ⊂  V  _{𝕊 _{\begin{matrix}∘ \\ ∘ \end{matrix} _{∘}}} ⊂  U ,

where W ≅ U/V ≅ 𝕊 \begin{matrix}∘ \\ ∘ \end{matrix} _{∘}, and V/W is isomorphic to the simple functor 𝕊_{∘ ∘ ; /\/ ; ∘ ∘}

associated to the poset [figure: ∘; ∘; ∘; ∘] of cardinality 4. An easy consequence of this is

that (when k is a field)

Ext_{F_{k}}^{1}(𝕊 \begin{matrix}∘ \\ ∘ \end{matrix} _{∘}, 𝕊 /\/_{∘ ∘}^{∘ ∘}) ≅ Ext_{F_{k}}^{1}(𝕊 /\/_{∘ ∘}^{∘ ∘}, 𝕊 \begin{matrix}∘ \\ ∘ \end{matrix} _{∘}) ≅ k .

Part 3

APPENDICES

21. Lattices associated to posets

Posets occur in the parametrization of simple functors. In order to be able to use our results on lattices, we needed to associate a lattice T to any given finite poset (E, R). The easiest choice is the lattice I_{↓}(E, R). The purpose of this section is to describe the various possibilities for T.

We let (E, R) be a finite poset. Throughout this section, we write the order relation ≤_{R}. Also, we use a subscript E for intervals in E.

If A ⊆ E, we let Ub A be the set of all upper bounds of A and Lb A the set of all lower bounds of A, that is,

Ub A = {e ∈ E | a ≤_{R} e, ∀a ∈ A} and Lb A = {e ∈ E | e ≤_{R} a, ∀a ∈ A} .

Clearly Ub A is an upper ideal in E while Lb A is a lower ideal in E.

Let I_{↓}(E, R) be the lattice of all lower ideals in E, with respect to unions and intersections. The irreducible elements in I_{↓}(E, R) are the principal ideals ]·, e]_{E}, where e ∈ E. Thus the poset E is isomorphic to the poset of all irreducible elements in I_{↓}(E, R) by mapping e ∈ E to the principal ideal ]·, e]_{E}. Note that I_{↓}(E, R) is actually a distributive lattice. For more details, see Theorem 3.4.1 and Proposition 3.4.2 in [St].

21.1. Convention. Recall our Convention 5.6. We identify E with its image in I_{↓}(E, R) via the map e ↦ ]·, e]_{E}. Thus we view E as a subposet of I_{↓}(E, R).

21.2. Definition. A closure operation on I_{↓}(E, R) is an order-preserving map

I_{↓}(E, R) → I_{↓}(E, R) , A ↦ Ã ,

such that A ⊆ Ã and ˜ Ã = Ã for all A ∈ I_{↓}(E, R). We say that A ∈ I_{↓}(E, R) is closed if Ã = A.

We shall require that every principal ideal ]·, e]_{E} is closed. In other words, the closure operation is the identity on E (identified with a subposet of I_{↓}(E, R)).

As before, we say that T is generated by E if any element of T is a join of elements of E. For instance, if we view E as a subposet of I_{↓}(E, R) as above, then clearly E is a full subposet and I_{↓}(E, R) is generated by E since any lower ideal is a union of principal ideals.

21.3. Lemma. Let (E, R) be a finite poset. Let A ↦ Ã be a closure operation on I_{↓}(E, R) which is the identity on E. Let T be the subset of I_{↓}(E, R) consisting of all closed elements of I_{↓}(E, R), viewed as a poset via inclusion.

(a) T is a lattice. Explicitly, if A, B ∈ T, then

A ∧ B = A ∩ B  and  A ∨ B = { \atop A ∪ B} .

Moreover, E is a full subposet of T.

(b) The map π = π_{T} : I_{↓}(E, R) → T defined by π_{T} (A) = Ã is a surjective map of posets which preserves joins and which is the identity on E (identified with a full subposet of I_{↓}(E, R) and T). In other words, π_{T} is a morphism in the category L (see Definition 11.5).

(c) T is generated by E (identified with a subposet of T).

Proof : (a) Since closure is order-preserving, we have { \atop A ∩ B} ⊆ Ã = A and similarly { \atop A ∩ B} ⊆ B. Therefore { \atop A ∩ B} ⊆ A ∩ B, forcing equality { \atop A ∩ B} = A ∩ B.

Now if C ∈ T satisfies A ⊆ C and B ⊆ C, then A ∪ B ⊆ C, hence { \atop A ∪ B} ⊆ C̃ = C. This shows that { \atop A ∪ B} is the least upper bound of A and B.

Finally, E is a full subposet of T because ]·, e]_{E} ⊆]·, f]_{E} if and only if e ≤_{R} f.

(b) Clearly π is a map of posets because the closure operation is order-preserving. It is the identity on E because so is the closure operation. If A, B ∈ I_{↓}(E, R), then

π(A ∪ B) = { \atop A ∪ B} = { \atop Ã ∪ B̃} = Ã ∨ B̃ = π(A) ∨ π(B) ,

so π preserves joins.

(c) Since I_{↓}(E, R) is generated by E, so is its image T, because π preserves joins and is the identity on E.

We now prove the converse.

21.4. Proposition. Let (E, R) be a finite poset. Let T be a lattice with order relation ≤_{T}, containing E as a full subposet, and suppose that T is generated by E.

(a) There is a unique surjective map of posets

π = π_{T} : I_{↓}(E, R) ⟶ T

which preserves joins and which is the identity on E (viewed as a subposet of I_{↓}(E, R)).

(b) For every t ∈ T, the inverse image π^{−1}(t) has a greatest element.

(c) If A ∈ I_{↓}(E, R) with π(A) = t, let Ã be the greatest element of π^{−1}(t).

Then

Ã = {e ∈ E | e ≤_{T} t} .

(d) The map A ↦ Ã is a closure operation on I_{↓}(E, R) which is the identity on E.

(e) The poset of closed elements in I_{↓}(E, R) is isomorphic to T via π.

Proof : (a) Let A ∈ I_{↓}(E, R) and write A = {e_{1}, …, e_{k}}. Then we have A = ]·, e_{1}]_{E} ∪ … ∪ ]·, e_{k}]_{E} and we define

π(A) = e_{1} ∨ … ∨ e_{k} ∈ T .

It follows that π(A ∪ B) = π(A) ∨ π(B), for all A, B ∈ I_{↓}(E, R). In the special case when A ⊆ B, we get π(B) = π(A) ∨ π(B), that is, π(A) ≤_{T} π(B). Therefore π is a map of posets. It is surjective because T is generated by E. To prove that it is the identity on E, let A = ]·, e_{k}]_{E} = {e_{1}, …, e_{k}}. Then e_{i} ≤_{R} e_{k} for all i, hence e_{i} ≤_{T} e_{k} because E is a subposet of T, and therefore π(A) = e_{1} ∨ … ∨ e_{k} = e_{k}. The uniqueness of π follows from the fact that it is the identity on E and that it preserves joins.

(b) If A, B ∈ π^{−1}(t), then π(A ∪ B) = π(A) ∨ π(B) = t ∨ t = t. Therefore A ∪ B ∈ π^{−1}(t). It follows that the union of all A ∈ π^{−1}(t) is in π^{−1}(t) and is its greatest element.

(c) Let A ∈ I_{↓}(E, R) with π(A) = t and write Ă = {e ∈ E | e ≤_{T} t}. We want to show that Ă = Ã. If e ∈ Ã, then ]·, e]_{E} ⊆ Ã (because the elements of I_{↓}(E, R) are lower ideals). Since π is a map of posets and is the identity on E, we obtain e ≤_{T} π(Ã) = t, hence e ∈ Ă. Thus Ã ⊆ Ă. On the other hand, π(Ă) = ⋁_{e≤_{T} t} e = t

because T is generated by E. Therefore Ă ∈ π^{−1}(t). Since Ã is the greatest element of π^{−1}(t), we obtain Ă ⊆ Ã.

(d) Since Ã is the greatest element of π^{−1}(π(A)), it is clear that A ⊆ Ã and

˜Ã = Ã. If A ⊆ B with π(A) = t and π(B) = v, then t ≤_{T} v by part (a). The characterization of Ã given in (c) implies that

Ã = {e ∈ E | e ≤_{T} t} ⊆ {e ∈ E | e ≤_{T} v} = B̃ ,

so the closure operation is order-preserving. To show that it is the identity on E, let x ∈ E, which is identified with the principal ideal A = ]·, x]_{E} ∈ I_{↓}(E, R). Then π(A) = x because π is the identity on E. Then Ã = {e ∈ E | e ≤_{T} x} by (c) and since E is a full subposet of T, we obtain

Ã = {e ∈ E | e ≤_{T} x} = {e ∈ E | e ≤_{R} x} = ]·, x]_{E} = A ,

showing that the closure is indeed the identity on E.

(e) Since π is surjective and each fibre π^{−1}(t) has a unique closed element (its greatest element), the restriction of π to closed elements is a bijection. Now π is order-preserving and we have to prove that its inverse is also order-preserving. So suppose that A and B are closed, with π(A) ≤_{T} π(B). Again, the characterization of Ã given in (c) implies that Ã ⊆ B̃, that is, A ⊆ B.

21.5. Corollary. For any given finite poset E, there is a bijection between the set of isomorphism classes of lattices generated by the full subposet E and the set of all closure operations on I_{↓}(E, R).

Proof : This follows immediately from Lemma 21.3 and Proposition 21.4.

Now we construct a lattice L(E, R) which will turn out to be the smallest lattice generated by E. It will be associated with the closure operation Lb Ub on I_{↓}(E, R). We first check this.

21.6. Lemma. Let E be a finite poset.

(a) Ub and Lb are order-reversing.

(b) Ub Lb Ub = Ub and Lb Ub Lb = Lb.

(c) The map A ↦ Lb Ub A is a closure operation on I_{↓}(E, R) which is the identity on E.

Proof : (a) follows from the definitions.

(b) Let A be a lower ideal and B an upper ideal of E. It is clear that B ⊆ Ub Lb B and, applying this to B = Ub A, we get Ub A ⊆ Ub Lb Ub A. Similarly A ⊆ Lb Ub A and, applying Ub, we get Ub A ⊇ Ub Lb Ub A. It follows that Ub A = Ub Lb Ub A. The equality Lb Ub Lb B = Lb B is proved in the same manner.

(c) The inclusion A ⊆ Lb Ub A has already been observed and the equality Lb Ub Lb Ub A = Lb Ub A follows from (b). Now if ]·, x]_{E} is a principal ideal, then Ub ]·, x]_{E} = [x, ·[_{E} and Lb Ub ]·, x]_{E} = ]·, x]_{E}, so Lb Ub is the identity on E.

We define L(E, R) to be the lattice associated with the closure operation Lb Ub by the procedure of Lemma 21.3. For any A ∈ I_{↓}(E, R), we write \overline{A} = Lb Ub A. Thus L(E, R) consists of all lower ideals which are closed, that is, of the form \overline{A} for some A ∈ I_{↓}(E, R).

21.7. Lemma. Let L(E, R) be the lattice associated with the closure operation Lb Ub in I_{↓}(E, R).

(a) L(E, R) is the subset of I_{↓}(E, R) consisting of all lower ideals of the form Lb B for some upper ideal B of E.

(b) L(E, R) is the subset of I_{↓}(E, R) consisting of all intersections of principal ideals (with the usual convention that an empty intersection yields the whole set E).

(c) The join operation in L(E, R) is described as follows. If B_{1} and B_{2} are two upper ideals of E such that Ub Lb B_{1} = B_{1} and Ub Lb B_{2} = B_{2}, then Lb B_{1} ∨ Lb B_{2} = Lb(B_{1} ∩ B_{2}).

Proof : (a) We have Lb B = Lb Ub Lb B by part (b) of Lemma 21.6.

(b) Lb B = ⋂_{b∈B} ]·, b]_{E}.

(c) Note that B_{1} ∩ B_{2} = Ub Lb B_{1} ∩ Ub Lb B_{2} = Ub(Lb B_{1} ∪ Lb B_{2}). Therefore

Lb(B_{1} ∩ B_{2}) = Lb Ub(Lb B_{1} ∪ Lb B_{2}) = \overline{Lb B_{1} ∪ Lb B_{2}} = Lb B_{1} ∨ Lb B_{2}

using Lemma 21.3.

Now we prove that any lattice generated by E is sandwiched between I_{↓}(E, R) and L(E, R).

21.8. Theorem. Let (E, R) be a finite poset, let T be a lattice containing E as a full subposet, and suppose that T is generated by E.

(a) There is a unique surjective map of posets

π_{T} : I_{↓}(E, R) ⟶ T

which preserves joins and which is the identity on E (viewed as a subposet of I_{↓}(E, R)).

(b) There is a unique surjective map of posets

ϕ_{T} : T ⟶ L(E, R)

which preserves joins and which is the identity on E (viewed as a subposet of L(E, R)).

(c) The composite ϕ_{T} ∘ π_{T} is equal to the map π_{L(E,R)}.

Proof : As before, the order relation in T will be written ≤_{T} in order to avoid confusion. Similarly, we write intervals in T with a subscript T in order to emphasize that they are considered in the lattice T.

(a) follows from Proposition 21.4.

(b) We define

ϕ_{T} : T → L(E, R) , ϕ_{T} (t) = Lb ([t, ˆ1]_{T} ∩ E) ,

where the operator Lb is considered within the set of all subsets of E. Since T is generated by E, we can write any t ∈ T as t = f_{1} ∨ … ∨ f_{r} where f_{1}, …, f_{r} ∈ E. For any e ∈ E, there are equivalences

e ∈ [t, ˆ1]_{T} ∩ E ⟺ t ≤_{T} e ⟺ f_{i} ≤_{T} e , ∀i ,

using the fact that t is the join t = f_{1} ∨ … ∨ f_{r}. Since E is a full subposet of T, we get further equivalences

f_{i} ≤_{T} e , ∀i ⟺ f_{i} ≤_{R} e , ∀i ⟺ e ∈ ⋂_{i=1}^{r} [f_{i}, ·[_{E} .

Therefore [t, ˆ1]_{T} ∩ E = ⋂_{i=1}^{r} [f_{i}, ·[_{E} and ϕ_{T} (t) = Lb ( ⋂_{i=1}^{r} [f_{i}, ·[_{E}).

Now part (c) of Lemma 21.7 applies, because Ub Lb [f_{i}, ·[_{E}= Ub ]·, f_{i}]_{E} = [f_{i}, ·[_{E}. It follows that

ϕ_{T} (t) = Lb ( ⋂_{i=1}^{r} [f_{i}, ·[_{E}) = ⋁_{i=1}^{r} Lb [f_{i}, ·[_{E}= ⋁_{i=1}^{r} ]·, f_{i}]_{E} .

It follows easily from this that ϕ_{T} preserves joins, because if s = e_{1} ∨ … ∨ e_{q} and t = f_{1} ∨…∨f_{r} with e_{1}, …, e_{q}, f_{1}, …, f_{r} ∈ E, then s∨t = e_{1} ∨…∨e_{q} ∨f_{1} ∨…∨f_{r} and

ϕ_{T} (s ∨ t) = ( ⋁_{j=1}^{q} ]·, e_{j}]_{E}) ⋁ ( ⋁_{i=1}^{r} ]·, f_{i}]_{E}) = ϕ_{T} (s) ∨ ϕ_{T} (t) .

In particular, ϕ_{T} is a map of posets. It is the identity on E because if e ∈ E, then

ϕ_{T} (e) = Lb ([e, ˆ1]_{T} ∩ E) = [ˆ0, e]_{T} ∩ E = ]·, e]_{E} ,

the last equality using the assumption that E is a full subposet of T. The surjectivity of ϕ_{T} follows from (c) and the surjectivity of π_{L(E,R)}.

(c) By Proposition 21.4, there is a unique surjective map

π_{L(E,R)} : I_{↓}(E, R) → L(E, R)

which preserves joins and is the identity on E. Therefore ϕ_{T} ∘ π_{T} = π_{L(E,R)}.

21.9. Remark. Note that π_{T} and ϕ_{T} may not preserve meets. As in Remark 11.6, we see that the relevant maps between lattices are join-preserving, but not necessarily meet-preserving.

Theorem 21.8 shows that the lattice L(E, R) is the unique smallest lattice generated by the poset E. Its subset of irreducible elements is contained in E but may not be equal to E. Our purpose now is to consider lattices whose subset of irreducible elements is the whole of E.

21.10. Definition. An element e of E is called reducible in E if it is the least upper bound of some subset B of ]·, e[_{E}. In other words, [e, ·[_{E}= Ub B for some B ⊆ ]·, e[_{E}. Note that the possibility B = ∅ occurs when E has a least element e_{0}, in which case Ub ∅ = E = [e_{0}, ·[, so that e_{0} is reducible in E.

21.11. Lemma. Let (E, R) be a finite poset and e ∈ E.

(a) e is reducible in E if and only if there exists a nonprincipal ideal A ∈ I_{↓}(E, R) such that \overline{A} = ]·, e]_{E}.

(b) e is irreducible in L(E, R) if and only if e is not reducible in E.

Proof : (a) If e is reducible in E, then [e, ·[_{E}= Ub(B) for some B ⊆ ]·, e[_{E}. Let A be the lower ideal generated by B, that is, A = {a ∈ E | a ≤_{R} b for some b ∈ B}. Note that A ⊆ ]·, e[_{E} again (with A = ∅ in case B = ∅). Then Ub A = Ub B = [e, ·[_{E}, hence

\overline{A} = Lb Ub A = Lb [e, ·[_{E}= ]·, e]_{E} .

We have to show that A is not principal. If we had A = ]·, f]_{E}, then we would have Lb Ub A = Lb [f, ·[_{E}= ]·, f]_{E} = A and [e, ·[_{E}= Ub A = [f, ·[_{E}, so that e = f. But then A = ]·, e]_{E}, contrary to the fact that A ⊆ ]·, e[_{E}.

Suppose conversely that \overline{A} = ]·, e]_{E} for some nonprincipal A ∈ I_{↓}(E, R). Then

Ub A = Ub Lb Ub A = Ub \overline{A} = Ub ]·, e]_{E} = [e, ·[_{E} ,

showing that e is the least upper bound of the subset A. Since A is nonprincipal and contained in ]·, e]_{E}, we have A ⊆ ]·, e[_{E}. This completes the proof that e is reducible in E.

(b) If e is reducible in E, then again [e, ·[_{E}= Ub(A) for some lower ideal A ∈ I_{↓}(E, R) such that A ⊆ ]·, e[_{E}. Then

]·, e]_{E} = \overline{A} = \overline{⋃_{a∈A}]·, a]_{E}} = ⋁_{a∈A}]·, a]_{E}

and this shows that ]·, e]_{E} is not irreducible in the lattice L(E, R). (Note that this includes the case when E has a least element e_{0} and A = ∅, because in that case L(E, R) has a least element \{e_{0}\} = \overline{∅}, which is not irreducible by definition.)

Conversely, if ]·, e]_{E} is not irreducible in the lattice L(E, R), then ]·, e]_{E} = ⋁_{a∈A}]·, a]_{E} where A = ]·, e[_{E}. Therefore

]·, e]_{E} = \overline{⋃_{a∈A}]·, a]_{E}} = \overline{A} .

Moreover, A is not principal, otherwise A = \overline{A}, hence A = ]·, e]_{E}, contrary to the fact that A = ]·, e[_{E}. By part (a), e is reducible in E.

We are going to construct a lattice K(E, R) having E as its subset of irreducible elements and minimal with this property. In order to define K(E, R), we need to define a new closure operation on I_{↓}(E, R), which we write A ↦ Â and which is defined as follows, with two cases :

(K1) If A is not principal and \overline{A} = ]·, a]_{E} , then Â = ]·, a[_{E}.

(K2) Otherwise Â = \overline{A}.

In the first case, a is reducible in E, by Lemma 21.11. Thus we have kept the closure operation defining L(E, R), except on subsets A generating a reducible element a of E, for which we distinguish Â = ]·, a[_{E} from \overline{A} =]·, a]_{E}.

21.12. Lemma. A ↦ Â is a closure operation on I_{↓}(E, R).

Proof : Note first that A ⊆ Â ⊆ \overline{A} for all A ∈ I_{↓}(E, R). This is clear in case (K2),

while in case (K1), A ⊆ ]·, a[_{E}= Â ⊂ ]·, a]_{E} = \overline{A}. It follows that \overline{Â} = \overline{A} in all cases.

Now we have to prove that ˆ Â = Â. Suppose first that A is in case (K1), so A is not principal, and Â = ]·, a[_{E}. Then \overline{]·, a[_{E}} = \overline{Â} = \overline{A} = ]·, a]_{E}. Moreover ]·, a[_{E} is not principal otherwise ]·, a[_{E} = \overline{]·, a[_{E}} = ]·, a]_{E}, which is a contradiction. Therefore ]·, a[_{E} satisfies the conditions of case (K1) and

ˆÂ = { \atop ]·, a[_{E}} = ]·, a[_{E}= Â .

If A is in case (K2), then Â = \overline{A}, which cannot be in case (K1), otherwise \overline{A} would be nonprincipal and \overline{\overline{A}} would be principal, contrary to the fact that \overline{A} = \overline{\overline{A}}. Therefore Â = \overline{A} is in case (K2) and

ˆÂ = \overline{Â} = \overline{\overline{A}} = \overline{A} = Â .

So we obtain A ⊆ Â = ˆ Â in all cases.

In order to show that the operation is order-preserving, we let A, B ∈ I_{↓}(E, R) with A ⊆ B. If B̂ = \overline{B}, then

Â ⊆ \overline{A} ⊆ \overline{B} = B̂ .

Otherwise B̂ = ]·, b[_{E} and \overline{B} = ]·, b]_{E} for some b ∈ E. Then A ⊆ B ⊆ ]·, b[_{E} and \overline{A} ⊆ \overline{B} = ]·, b]_{E}. Either \overline{A} ≠ ]·, b]_{E}, in which case

Â ⊆ \overline{A} ⊆ ]·, b[_{E} = B̂ ,

or \overline{A} = ]·, b]_{E}, in which case A is in case (K1) and Â = ]·, b[_{E} = B̂.

We define K(E, R) to be the lattice corresponding to the closure operation A ↦ Â (see Lemma 21.3).

21.13. Lemma. The set of irreducible elements of the lattice K(E, R) is equal to E (viewed as a subset of K(E, R)). Moreover, E is a full subposet of K(E, R).

Proof : Since the lattice K(E, R) corresponds to a closure operation, E is a full subposet of K(E, R) and K(E, R) is generated by E (see Lemma 21.3). Therefore, every irreducible element of K(E, R) belongs to E.

Conversely, let a ∈ E and suppose that a is a join in K(E, R), that is, ]·, a]_{E} = A ∨ B = { \atop A ∪ B}, for some -closed subsets A, B ∈ I_{↓}(E, R). Note that \overline{A ∪ B} =

{ \atop A ∪ B} because

\overline{A ∪ B} = \overline{{ \atop A ∪ B}} = \overline{]·, a]_{E}} = ]·, a]_{E} = { \atop A ∪ B} .

Thus A ∪ B cannot be in case (K1) since the -closure and the -closure are not equal in case (K1). As the condition \overline{A ∪ B} = ]·, a]_{E} is satisfied, this means that A∪B must be principal, in particular -closed. Therefore A∪B = \overline{A ∪ B} = ]·, a]_{E}. It follows that a ∈ A ∪ B. If a ∈ A, then A = ]·, a]_{E} (because A is a lower ideal). Similarly, if a ∈ B, then B = ]·, a]_{E}.

This completes the proof that a is irreducible.

21.14. Remark. The case (K1) in the construction of K(E, R) shows that if an element a ∈ E is reducible in E, then it is made irreducible in K(E, R) by distinguishing between ]·, a[_{E} and ]·, a]_{E}. Thus ]·, a]_{E} is irreducible and ]·, a[_{E} is the unique maximal element of ]·, a]_{E}.

We need further properties of the map π_{T} of Proposition 21.4.

21.15. Lemma. Let (E, R) be a finite poset. Let T be a lattice containing E as a full subposet, and suppose that T is generated by E. Let π_{T} : I_{↓}(E, R) → T be the unique map of posets of Proposition 21.4.

(a) If A, B ∈ I_{↓}(E, R) satisfy π_{T} (A) ≤_{T} π_{T} (B), then Ub A ⊇ Ub B. In particular, if π_{T} (A) = π_{T} (B), then Ub A = Ub B.

(b) Let e ∈ E. Then e is irreducible in T if and only if π_{T}^{−1}(e) = {e}.

(c) The set of irreducible elements of T is equal to the whole of E if and only if π_{T}^{−1}(e) is reduced to the singleton {e} for all e ∈ E.

Proof : (a) By Proposition 21.4, T corresponds to a closure operation A ↦ Ã and if π_{T} (A) = t, then Ã = {x ∈ E | x ≤_{T} t}. Now

π_{T} (A) = π_{T} ( ⋃_{a∈A} ]·, a]_{E}) = ⋁_{a∈A} π_{T} ( ]·, a]_{E}) = ⋁_{a∈A} a .

Then for x ∈ E, we obtain

x ∈ Ub A ⟺ x ≥_{E} a , ∀a ∈ A ⟺ x ≥_{T} a , ∀a ∈ A ⟺ x ≥_{T} π_{T} (A)

because π_{T} (A) = ⋁_{a∈A} a. Similarly x ∈ Ub B if and only if x ≥_{T} π_{T} (B). Since

π_{T} (B) ≥_{T} π_{T} (A), we deduce that Ub B ⊆ Ub A.

(b) Suppose that e is irreducible in T and let A ∈ π_{T}^{−1}(e). Since E is a full subposet, we get

Ã = {x ∈ E | x ≤_{T} e} = {x ∈ E | x ≤_{R} e} = ]·, e]_{E} .

Moreover e = π_{T} (A) = ⋁_{a∈A} a. Since e is irreducible, we get e = a for some a ∈ A,

that is, e ∈ A. As A is a lower ideal, we obtain

]·, e]_{E} ⊆ A ⊆ Ã = ]·, e]_{E} ,

hence A = Ã = ]·, e]_{E}. Thus π_{T}^{−1}(e) is reduced to the singleton { ]·, e]_{E} }, namely {e} (in view of the usual identification), as required.

Conversely, suppose that π_{T}^{−1}(e) = {e}. Write e = ⋁_{i=1}^{r} f_{i} with f_{i} irreducible for

all i, hence f_{i} ∈ E since T is generated by E. Let A = ⋃_{i=1}^{r} ]·, f_{i}]_{E}. Then

π_{T} (A) = π_{T} ( ⋃_{i=1}^{r} ]·, f_{i}]_{E}) = ⋁_{i=1}^{r} π_{T} ( ]·, f_{i}]_{E}) = ⋁_{i=1}^{r} f_{i} = e ,

that is, A ∈ π_{T}^{−1}(e). Since this fibre is a singleton, we have

A = Ã = {x ∈ E | x ≤_{T} e} .

It follows that e ∈ A, hence e ≤_{R} f_{i} for some i. This implies e = f_{i}, showing that e is irreducible in T.

(c) follows immediately from (b).

We now come to a result analogous to Theorem 21.8, but for lattices whose set of irreducible elements is the whole of E.

21.16. Theorem. Let (E, R) be a finite poset, let T be a lattice containing E as a full subposet, and suppose that the set of irreducible elements of T is equal to E.

(a) There is a unique surjective map of posets

π_{T} : I_{↓}(E, R) ⟶ T

which preserves joins and which is the identity on E (viewed as a subposet of I_{↓}(E, R)).

(b) There is a unique surjective map of posets

ψ_{T} : T ⟶ K(E, R)

which preserves joins and which is the identity on E (viewed as a subposet of K(E, R)).

(c) The composite ψ_{T} ∘ π_{T} is equal to the map π_{K(E,R)}.

Proof : (a) follows from Proposition 21.4.

(b) We want to define ψ_{T} : T → K(E, R) by taking a pre-image in I_{↓}(E, R) and then applying π_{K(E,R)} : I_{↓}(E, R) → K(E, R). Of course we need to see that this is well-defined. So let A, B ∈ I_{↓}(E, R) be such that π_{T} (A) = π_{T} (B). We have to show that π_{K(E,R)}(A) = π_{K(E,R)}(B), that is, Â = B̂.

By Lemma 21.15, we have Ub A = Ub B, hence \overline{A} = \overline{B}. If both A and B are in case (K2), we get Â = \overline{A} = \overline{B} = B̂. If one of them, say A, is in case (K1), then there exists a ∈ E such that

A ⊆ Â = ]·, a[_{E} ⊂ ]·, a]_{E} = \overline{A} .

Suppose that a ∈ B, so that B = ]·, a]_{E} because \overline{B} = \overline{A} = ]·, a]_{E}. Then we have

a = π_{T} ( ]·, a]_{E}) = π_{T} (\overline{B}) = π_{T} (B) = π_{T} (A) ,

hence A ∈ π_{T}^{−1}(a). Now part (c) of Lemma 21.15 applies, because the set of irreducible elements of T is equal to E by assumption. Therefore π_{T}^{−1}(a) is reduced to the singleton \{a\}, that is, \{\overline{A}\} in view of the usual identification. This is impossible because A ∈ π_{T}^{−1}(a) and A ≠ \overline{A}.

This contradiction shows that a ∉ B and therefore

B ⊆ ]·, a[_{E} ⊂ ]·, a]_{E} = \overline{B} .

Moreover, B is not principal, otherwise B = \overline{B}, hence B = ]·, a]_{E}, contrary to the fact that B ⊆ ]·, a[_{E}. Thus we are in case (K1) and B̂ = ]·, a[_{E}= Â.

We have now proved that there is a well-defined map ψ_{T} : T ⟶ K(E, R), as follows. If t ∈ T, write t = π_{T} (A) for some A ∈ I_{↓}(E, R) and set

ψ_{T} (t) = ψ_{T} (π_{T} (A)) = π_{K(E,R)}(A) .

In particular, this proves (c). The fact that ψ_{T} preserves joins follows in a straightforward fashion :

ψ_{T} (π_{T} (A) ∨ π_{T} (B)) = ψ_{T} (π_{T} (A ∪ B)) = π_{K(E,R)}(A ∪ B)
= π_{K(E,R)}(A) ∨ π_{K(E,R)}(B) = ψ_{T} (π_{T} (A)) ∨ ψ_{T} (π_{T} (B)) .

The map ψ_{T} is the identity on E because both π_{T} and π_{K(E,R)} are the identity on E. The surjectivity of ψ_{T} follows from the surjectivity of π_{K(E,R)}.

(c) This has been noticed above.

Theorem 21.16 shows that K(E, R) is the unique minimal lattice having E as its set of irreducible elements. This has some importance for computational purposes, since working with K(E, R) may be considerably less heavy than working with I_{↓}(E, R). This is made clear in the next example.

21.17. Example. Let E be a nonempty finite set endowed with the equality relation. Then I_{↓}(E, R) is the set of all subsets of E, hence rather big. However, with the usual identification, L(E, R) = {∅} ⊔ E ⊔ {E}, hence much smaller, and K(E, R) = L(E, R) in this case.

21.18. Example. Let E be a nonempty finite set endowed with a total order. Then I_{↓}(E, R) = {∅} ⊔ E (with the usual identification), L(E, R) = E, and K(E, R) = I_{↓}(E, R).

22. Forests associated to lattices

Our aim in this appendix is to prove Theorem 17.9. This will follow from a much more general combinatorial result on idempotents associated to forests. We start with some idempotents associated to sequences of distinct elements in any finite set:

22.1. Proposition. Let T be a finite set. For a sequence \underline{a} = (a_{0}, a_{1}, …, a_{n}) of distinct elements of T, and for 0 ≤ k ≤ n, let [a_{0}, a_{1}, … a_{k}] be the map from T to itself defined by

∀t ∈ T, [a_{0}, a_{1}, … a_{k}](t) = \{ \begin{matrix}a_{i+1} & if t = a_{i} and i ≤ k − 1 \\ t & if t ∉ \{a_{0}, … a_{k−1}\} . \end{matrix}

In particular [a_{0}] is the identity map of T.

Let h\underline{a} be the element of the algebra k(T^{T}) of the monoid T^{T} of maps from T to itself defined by

h\underline{a} = ∑_{i=0}^{n}(−1)^{i}[a_{0}, a_{1}, …, a_{i}] .

(a) If 0 ≤ j ≤ i, then [a_{0}, …, a_{i}] = [a_{0}, …, a_{j}] ∘ [a_{j}, …, a_{i}].

(b) If 1 ≤ k ≤ n and f : T → T is a map such that f(a_{k}) = f(a_{k−1}), then

f ∘ [a_{0}, …, a_{k−1}] = f ∘ [a_{0}, …, a_{k}].
(c) h\underline{a} is an idempotent of k(T^{T}).

Proof : (a) This is straightforward.

(b) The only element t ∈ T such that [a_{0}, …, a_{k−1}](t) ≠ [a_{0}, …, a_{k}](t) is t = a_{k−1}. Moreover

f ∘ [a_{0}, …, a_{k−1}](a_{k−1}) = f(a_{k−1}) = f(a_{k}) = f ∘ [a_{0}, …, a_{k}](a_{k−1}) .

(c) We have

h2 ; \underline{a} = ∑_{0≤i,j≤n} (−1)^{i+j}[a_{0}, …, a_{i}] ∘ [a_{0}, …, a_{j}]
= ∑_{0≤i≤j≤n} (−1)^{i+j}[a_{0}, …, a_{i}] ∘ [a_{0}, …, a_{j}] + ∑_{0≤j<i≤n} (−1)^{i+j}[a_{0}, …, a_{i}] ∘ [a_{0}, …, a_{j}]

Moreover by (a), for j + 1 ≤ i

[a_{0}, …, a_{i}] ∘ [a_{0}, …, a_{j}] = [a_{0}, …, a_{j+1}] ∘ [a_{j+1}, …, a_{i}] ∘ [a_{0}, …, a_{j}] ,

and the maps [a_{j+1}, …, a_{i}] and [a_{0}, …, a_{j}] clearly commute. Hence

[a_{0}, …, a_{i}] ∘ [a_{0}, …, a_{j}] = [a_{0}, …, a_{j+1}] ∘ [a_{0}, …, a_{j}] ∘ [a_{j+1}, …, a_{i}] .

Now the map f = [a_{0}, …, a_{j+1}] is such that f(a_{j}) = f(a_{j+1}). Thus by (c)

[a_{0}, …, a_{j+1}] ∘ [a_{0}, …, a_{j}] = [a_{0}, …, a_{j+1}] ∘ [a_{0}, …, a_{j+1}] ,

and by (a)

[a_{0}, …, a_{i}] ∘ [a_{0}, …, a_{j}] = [a_{0}, …, a_{j+1}] ∘ [a_{0}, …, a_{j+1}] ∘ [a_{j+1}, …, a_{i}]
= [a_{0}, …, a_{j+1}] ∘ [a_{0}, …, a_{i}] .

Hence for all j + 1 ≤ i, i.e. for all j < i

[a_{0}, …, a_{i}] ∘ [a_{0}, …, a_{j}] = [a_{0}, …, a_{j+1}] ∘ [a_{0}, …, a_{i}] .

This gives:

h2 ; \underline{a} = ∑_{0≤i≤j≤n} (−1)^{i+j}[a_{0}, …, a_{i}] ∘ [a_{0}, …, a_{j}] + ∑_{0≤j<i≤n} (−1)^{i+j}[a_{0}, …, a_{i}] ∘ [a_{0}, …, a_{j}]
= ∑_{0≤i≤j≤n} (−1)^{i+j}[a_{0}, …, a_{i}] ∘ [a_{0}, …, a_{j}] + ∑_{0≤j<i≤n} (−1)^{i+j}[a_{0}, …, a_{j+1}] ∘ [a_{0}, …, a_{i}]
= ∑_{0≤i≤j≤n} (−1)^{i+j}[a_{0}, …, a_{i}] ∘ [a_{0}, …, a_{j}] + ∑_{1≤j≤i≤n} (−1)^{i+j−1}[a_{0}, …, a_{j}] ∘ [a_{0}, …, a_{i}]
= ∑_{0=i≤j≤n} (−1)^{i+j}[a_{0}, …, a_{i}] ∘ [a_{0}, …, a_{j}] = ∑_{0≤j≤n}(−1)^{j}[a_{0}] ∘ [a_{0}, …, a_{j}]
= h\underline{a}

as was to be shown.

We now introduce the terminology on forests that we will use here. Part of it may be nonstandard:

22.2. Definition and Notation. A forest is a finite oriented graph without loops nor cycles, such that each vertex has at most one outgoing edge. A root of a forest is a vertex without any outgoing edge. A leaf of a forest is a vertex with an outgoing edge and no incoming edge. A tree is a connected forest.

When G is a forest, we also denote by G the set of its vertices. When x ∈ G, and x is not a root, we denote by d(x) the unique vertex of G such that [figure: x; d(x)] is the outgoing edge at x. When x is a root, we set x = d(x). For n ∈ ℕ, we define d^{n} : G → G by d^{0}(x) = x, for any x ∈ X, and d^{n}(x) = d(d^{n−1}(x)) for n > 0. For x ∈ G, we denote by δ(x) the smallest non negative integer n such that d^{n}(x) = d^{n+1}(x), and we set d^{∞}(x) = d^{δ(x)}(x). Each connected component C of G contains a unique root, equal to d^{∞}(x) for each x ∈ C.

A complete geodesic of a forest G is a sequence of the form (x, d(x), …, d^{n}(x)), where x is a leaf of G, and n = δ(x). A partial geodesic of G starting at some leaf x is a sequence of distinct vertices of the form (x, d(x), …, d^{l}(x)), for some integer l ≥ 1. We say that and edge (y, d(y)) of G belongs to a partial geodesic (x, d(x), …, d^{l}(x)) if there exists an integer j ∈ {0, …, l − 1} such that y = d^{j}(x).

When A is a set of edges of a forest G, we denote by τ_{A} the map from G to itself defined by

∀x ∈ G, τ_{A}(x) = \{ \begin{matrix}d(x) & if (x, d(x)) ∈ A \\ x & otherwise . \end{matrix}

When B is a set of leaves of a forest G, we say that a set A of edges of G is a union of partial geodesics starting in B if there is a subset C of B, and for each x ∈ C, a partial geodesic γ_{x} starting at x, such that A is the union of the sets of edges belonging to the partial geodesics γ_{x}, for x ∈ C. We denote by S_{B} the set of such sets of edges of G.

Finally, we denote by v_{B} the element of k(G^{G}) defined by

v_{B} = ∑_{A∈S_{B}}(−1)^{|A|}τ_{A} .

We will abuse notation and set v_{x} = v_{{x}}, for any leaf x of G.

22.3. Theorem. Let G be a forest.

(a) If x is a leaf of G, then v_{x}^{2} = v_{x}.

(b) If x and y are leaves of G, then v_{x}v_{y} = v_{y}v_{x}.

(c) Let B be a subset of the set of leaves of G. Then v_{B} is an idempotent of k(G^{G}), equal to the product of the (commuting) idempotents v_{x}, for x ∈ B.

(d) Let B and C be two subsets of the set of leaves of G. Then v_{B} ∘v_{C} = v_{B∪C}.

Proof : (a) If there is a single leaf x, and if \underline{a} = (x, d(x), …, d^{n}(x)) is the complete geodesic of G starting at x, then

v_{B} = v_{x} = ∑_{l=0}^{n}(−1)^{l}[x, d(x), …, d^{l}(x)] ,

so v_{x} is equal to the idempotent h\underline{a} introduced in Proposition 22.1.

(b) Let x and y be distinct leaves of G. If d^{∞}(x) ≠ d^{∞}(y), then the complete geodesics starting at x and y lie in different connected components of G. It is clear in this case that v_{x} and v_{y} commute.

Otherwise set r = d^{∞}(x) = d^{∞}(y). The intersection of the complete geodesics starting at x and y is of the form \underline{c} = (c_{0}, …, c_{m} = r), for some integer m ≥ 0. The complete geodesic starting at x is of the form

(x = a_{0}, …, a_{s}, c_{0}, …, c_{m} = r) ,

where s > 0, and the complete geodesic starting at y is of the form

(y = b_{0}, …, b_{t}, c_{0}, …, c_{m} = r) ,

where t > 0.

Now

v_{x} = ∑_{i=0}^{s}(−1)^{i}[a_{0}, …, a_{i}] + (−1)^{s+1} ∑_{i=0}^{m}(−1)^{i}[a_{0}, …, a_{s}, c_{0}, …, c_{i}]
= h\underline{a} + (−1)^{s+1}[a_{0}, …, a_{s}, c_{0}]h\underline{c} ,

where \underline{a} = [a_{0}, …, a_{s}]. Similarly v_{y} = h\underline{b} + (−1)^{t+1}[b_{0}, …, b_{t}, c_{0}]h\underline{c}, where \underline{b} = (b_{0}, …, b_{t}). Clearly h\underline{a} commutes with h\underline{b}, [b_{0}, …, b_{t}, c_{0}], and h\underline{c}, hence it commutes with v_{y}. Similarly h\underline{b} commutes with v_{x}. So proving that that v_{x} and v_{y} commute is equivalent to proving that [a_{0}, …, a_{s}, c_{0}]h\underline{c} and [b_{0}, …, b_{t}, c_{0}]h\underline{c} commute.

Setting f = [b_{0}, …, b_{t}, c_{0}], we have

h\underline{c}fh\underline{c} = ∑_{0≤i,j≤m} (−1)^{i+j}[c_{0}, …, c_{i}]f[c_{0}, …, c_{j}]
= ∑_{0≤i≤j≤m} (−1)^{i+j}[c_{0}, …, c_{i}]f[c_{0}, …, c_{j}] + ∑_{0≤j<i≤m} (−1)^{i+j}[c_{0}, …, c_{i}]f[c_{0}, …, c_{j}]

Now for 0 ≤ j < i, by Proposition 22.1,

[c_{0}, …, c_{i}]f[c_{0}, …, c_{j}] = [c_{0}, …, c_{j+1}][c_{j+1}, …, c_{i}]f[c_{0}, …, c_{j}]
= [c_{0}, …, c_{j+1}]f[c_{0}, …, c_{j}][c_{j+1}, …, c_{i}] ,

since [c_{j+1}, …, c_{i}] clearly commutes with both f and [c_{0}, …, c_{j}]. Now setting g = [c_{0}, …, c_{j+1}]f, we have that g(c_{j}) = c_{j+1} = g(c_{j+1}), hence g[c_{0}, …, c_{j}] = g[c_{0}, …, c_{j+1}], by Proposition 22.1. It follows that

[c_{0}, …, c_{i}]f[c_{0}, …, c_{j}] = g[c_{0}, …, c_{j}][c_{j+1}, …, c_{i}]
= g[c_{0}, …, c_{j+1}][c_{j+1}, …, c_{i}]
= g[c_{0}, …, c_{i}] ,

thus [c_{0}, …, c_{i}]f[c_{0}, …, c_{j}] = [c_{0}, …, c_{j+1}]f[c_{0}, …, c_{i}] for j < i. This gives

h\underline{c}fh\underline{c} = ∑_{0≤i≤j≤m} (−1)^{i+j}[c_{0}, …, c_{i}]f[c_{0}, …, c_{j}] + ∑_{0≤j<i≤m} (−1)^{i+j}[c_{0}, …, c_{i}]f[c_{0}, …, c_{j}]
= ∑_{0≤i≤j≤m} (−1)^{i+j}[c_{0}, …, c_{i}]f[c_{0}, …, c_{j}] + ∑_{0≤j<i≤m} (−1)^{i+j}[c_{0}, …, c_{j+1}]f[c_{0}, …, c_{i}]
= ∑_{0≤i≤j≤m} (−1)^{i+j}[c_{0}, …, c_{i}]f[c_{0}, …, c_{j}] + ∑_{1≤j≤i≤m} (−1)^{i+j−1}[c_{0}, …, c_{j}]f[c_{0}, …, c_{i}]

= ∑_{0=i≤j≤m} (−1)^{i+j}[c_{0}, …, c_{i}]f[c_{0}, …, c_{j}] = ∑_{j=0}^{n}(−1)^{j}f[c_{0}, …, c_{j}]
= fh\underline{c} .

It follows that

[a_{0}, …, a_{s}, c_{0}]h\underline{c}[b_{0}, …, b_{t}, c_{0}]h\underline{c} = [a_{0}, …, a_{s}, c_{0}][b_{0}, …, b_{t}, c_{0}]h\underline{c} .

Similarly

[b_{0}, …, b_{t}, c_{0}]h\underline{c}[a_{0}, …, a_{s}, c_{0}]h\underline{c} = [b_{0}, …, b_{t}, c_{0}][a_{0}, …, a_{s}, c_{0}]h\underline{c} .

Now [a_{0}, …, a_{s}, c_{0}] and [b_{0}, …, b_{t}, c_{0}] clearly commute, so [a_{0}, …, a_{s}, c_{0}]h\underline{c} and [b_{0}, …, b_{t}, c_{0}]h\underline{c} also commute. This completes the proof of (b).

(c) We will prove that v_{B} = ∏_{x∈B} v_{x} by induction on the cardinality of B. If

B = ∅, then v_{B} = τ_{∅} = id_{G}, which is also equal to an empty product of v_{x}. If B = {x}, then by definition v_{B} = v_{x}. Assume now that (c) holds for any set of leaves of cardinality smaller than |B| and |B| ≥ 2.

We can assume that B is the set of all leaves of G: otherwise we replace G by the full subgraph G′ consisting of the elements d^{j}(x), for x ∈ B and j ≥ 0. The operation f ↦ f̃ extending any function from G′ to itself to a function from G to itself by f̃(y) = y if y ∈ G − G′ is an injective morphism of monoids from G^{′G′} to G^{G}, which sends the element v_{B} computed for the forest G′ to v_{B}.

We can now assume that G is a tree: indeed, if G is a disjoint union of nonempty subforests G_{1} and G_{2}, then its set of leaves B splits as B = B_{1} ⊔ B_{2}, where B_{i} is a set of leaves of G_{i}, for i = 1, 2. Similarly, any set of edges A ∈ S_{B} splits as A = A_{1} ⊔ A_{2}, where A_{i} ∈ S_{B_{i}}, for i = 1, 2. Conversely, if A_{i} ∈ S_{B_{i}}, for i = 1, 2, then the set A = A_{1} ⊔ A_{2} belongs to S_{B}. Moreover the maps τ_{A_{1}} and τ_{A_{2}} clearly commute, and τ_{A_{1}}τ_{A_{2}} = τ_{A}. It follows that

v_{B} = ∑_{{A_{1}∈S_{B_{1}} \atop A_{2}∈S_{B_{2}}}} (−1)^{|A_{1}|+|A_{2}|}τ_{A_{1}}τ_{A_{2}} = v_{B_{1}}v_{B_{2}} = v_{B_{2}}v_{B_{1}} .

By induction hypothesis (c) holds for B_{1} and B_{2}, thus v_{B_{1}} = ∏_{x∈B_{1}} v_{x} and v_{B_{2}} =

∏_{x∈B_{2}} v_{x}, then v_{B} = ∏_{x∈B_{1}⊔B_{2}} v_{x}, so (c) holds for B.

Let r be the root of G, and let \underline{c} = (c_{0}, …, c_{m} = r) be the intersection of the complete geodesics of G (possibly m = 0 and \underline{c} = (r)). If c_{0} is a leaf, then G is an interval (c_{0}, …, c_{m} = r), but this contradicts |B| ≥ 2. Thus the element c_{0} is not a leaf, so the set d^{−1}(c_{0}) = {y ∈ G | d(y) = c_{0}} is nonempty, and actually of cardinality at least 2: if d^{−1}(c_{0}) = {y}, then y belongs to the intersection of the complete geodesics of G, contradicting the definition of \underline{c}.

Fix y ∈ d^{−1}(c_{0}), and let D = {x ∈ B | ∃j ≥ 0, d^{j}(x) = y}. Then ∅ ≠ D ≠ B, and we set C = B − D. We denote by C^{+} (resp. D^{+}) the set of edges of G which belong to a complete geodesic starting in C (resp D). We also set

\overline{C} = C^{+} − \{(c_{0}, c_{1}), (c_{1}, c_{2}), …, (c_{m−1}, c_{m})\}
D̂ = D^{+} − {(y, c_{0}), (c_{0}, c_{1}), (c_{1}, c_{2}), …, (c_{m−1}, c_{m})} .

We illustrate these definitions with the following schematic diagram representing G, in which all edges are oriented downwards:

[figure: C; •; •; …; •; •; D; •; …; •; arrows ; in \overline{C}; arrows ; in D̂; • _{y}; •_{c_{0}}; •_{c_{1}}; • _{c_{m−1}}; •_{c_{m}}]

Let A ∈ S_{B}. There are two cases:

(1) either (y, c_{0}) ∉ A: in this case A splits as the disjoint union of a set U ∈ S_{C} and a set V ∈ S_{D} such that (y, c_{0}) ∉ V, and this decomposition is unique. Observe that the conditions V ∈ S_{D} and (y, c_{0}) ∉ V are equivalent to the conditions V ∈ S_{D} and V ⊆ D̂. Conversely, if U ∈ S_{C} and V ∈ S_{D} with V ⊆ D̂, then A = U ⊔ V ∈ S_{B}, and (y, c_{0}) ∉ A. Moreover τ_{A} = τ_{U}τ_{V}.

(2) or (y, c_{0}) ∈ A: then A splits a the disjoint union of a set U ∈ S_{C} contained in \overline{C} and a set V ∈ S_{D} such that (y, c_{0}) ∈ V, and this decomposition is again unique. Conversely, if U ∈ S_{C} and U ⊆ \overline{C}, and V ∈ S_{D}, then A = U ⊔ V ∈ S_{B} and (y, c_{0}) ∈ A. Moreover τ_{A} = τ_{U}τ_{V} in this case also.

We now compute the product v_{C}v_{D}:

v_{C}v_{D} = ( ∑_{U∈S_{C}}(−1)^{|U|}τ_{U})( ∑_{V ∈S_{D}}(−1)^{|V |}τ_{V} )
= ∑_{U∈S_{C} ; {V ∈S_{D} \atop V ⊆D̂}} (−1)^{|U|+|V |}τ_{U}τ_{V} + ∑_{U∈S_{C} ; {V ∈S_{D} \atop (y,c_{0})∈V}}(−1)^{|U|+|V |}τ_{U}τ_{V}

= ∑_{U∈S_{C} ; {V ∈S_{D} \atop V ⊆D̂}} (−1)^{|U⊔V |}τ_{U⊔V} + ∑_{U∈S_{C} ; {U⊆\overline{C} \atop (y,c_{0})∈V ∈S_{D}}} (−1)^{|U|+|V |}τ_{U}τ_{V} + ∑_{U∈S_{C} ; {U⊈\overline{C} \atop (y,c_{0})∈V ∈S_{D}}} (−1)^{|U|+|V |}τ_{U}τ_{V}

= ∑_{{A∈S_{B} \atop (y,c_{0})∉A}}(−1)^{|A|}τ_{A} + ∑_{U∈S_{C} ; {U⊆\overline{C} \atop (y,c_{0})∈V ∈S_{D}}} (−1)^{|U⊔V |}τ_{U⊔V} + ∑_{U∈S_{C} ; {U⊈\overline{C} \atop (y,c_{0})∈V ∈S_{D}}} (−1)^{|U|+|V |}τ_{U}τ_{V}

= ∑_{{A∈S_{B} \atop (y,c_{0})∉A}}(−1)^{|A|}τ_{A} + ∑_{{A∈S_{B} \atop (y,c_{0})∈A}}(−1)^{|A|}τ_{A} + ∑_{U∈S_{C} ; {U⊈\overline{C} \atop (y,c_{0})∈V ∈S_{D}}} (−1)^{|U|+|V |}τ_{U}τ_{V}

= v_{B} + ∑_{U∈S_{C} ; {U⊈\overline{C} \atop (y,c_{0})∈V ∈S_{D}}} (−1)^{|U|+|V |}τ_{U}τ_{V}

We claim that the sum

(22.4)  Σ = ∑_{U∈S_{C} ; {U⊈\overline{C} \atop (y,c_{0})∈V ∈S_{D}}} (−1)^{|U|+|V |}τ_{U}τ_{V}

is equal to 0. This claim gives v_{C}v_{D} = v_{B}. As C and D are both strictly smaller than B, we may apply the induction hypothesis to both, and this gives

v_{B} = v_{C}v_{D} = ( ∏_{x∈C} v_{x})( ∏_{x∈D} v_{x}) = ∏_{x∈C⊔D} v_{x} = ∏_{x∈B} v_{x} ,

which completes the induction step in the proof of (c).

So it remains to prove the above claim that the sum Σ defined in 22.4 is equal to 0. If m = 0, the claim is trivial, because the sum Σ is empty. So we assume m ≥ 1. We define two sets

U = \{U ∈ S_{C} | U ⊈ \overline{C}\} = \{U ∈ S_{C} | (c_{0}, c_{1}) ∈ U\}
V = {V ∈ S_{D} | (y, c_{0}) ∈ V } .

With this notation,

Σ = ∑_{(U,V)∈U×V}(−1)^{|U|+|V |}τ_{U}τ_{V} .

Let U ∈ U. If (c_{m−1}, c_{m}) ∈ U, we set l = m. Otherwise there is a well defined integer l ≥ 1 such that (c_{l−1}, c_{l}) ∈ U, but (c_{l}, c_{l+1}) ∉ U. Moreover, the set U′ = U ∩ \overline{C} belongs to the set S_{C}^{♯} of elements of S_{C} contained in \overline{C} and containing at least one edge of the form (z, c_{0}), and

U = U′ ⊔ {(c_{0}, c_{1}), …, (c_{l−1}, c_{l})} .

Moreover τ_{U} = τ_{U′}[c_{0}, c_{1}, …, c_{l}]. Conversely, if U′ ∈ S_{C}^{♯} and l ∈ {1, …, m}, then the set U = U′ ⊔{(c_{0}, c_{1}), …, (c_{l−1}, c_{l})} belongs to U. In this way we get a bijection

\xymatrix{α : S_{C}^{♯} × \{1, …, m\} \ar[rr] &  & U \\  & (U′, l) \ar@{|->}[rr] &  & U = U′ ⊔ \{(c_{0}, c_{1}), …, (c_{l−1}, c_{l})\}}

with the property that τ_{α(U′,l)} = τ_{U′}[c_{0}, c_{1}, …, c_{l}].

Similarly, let V ∈ V. If (c_{m−1}, c_{m}) ∈ V, we set q = m. Otherwise, setting c_{−1} = y, there is a well defined integer q ≥ 0 such that (c_{q−1}, c_{q}) ∈ V but (c_{q}, c_{q+1}) ∉ V. Moreover, the set V′ = V ∩ D̂ belongs to the set S_{D}^{♭} of elements of S_{D} contained in D̂ and containing at least one edge of the form (t, y), and

V = V′ ⊔ {(y, c_{0}), …, (c_{q−1}, c_{q})} .

Moreover τ_{V} = τ_{V′}[y, c_{0}, …, c_{l}]. Conversely, if V′ ∈ S_{D}^{♭} and q ∈ {0, …, m}, then V = V′ ⊔ {(y, c_{0}), …, (c_{q−1}, c_{q})} belongs to V. In this way we get a bijection

\xymatrix{β : S_{D}^{♭} × \{0, …, m\} \ar[rr] &  & V \\  & (V′, q) \ar@{|->}[rr] &  & V = V′ ⊔ \{(y, c_{0}), …, (c_{q−1}, c_{q})\}}

with the property that τ_{β(V′,q)} = τ_{V′}[y, c_{0}, …, c_{q}].

We can now compute Σ:

Σ = ∑_{(U,V)∈U×V}(−1)^{|U|+|V |}τ_{U}τ_{V}
= ∑_{{U′∈S_{C}^{♯} \atop V′∈S_{D}^{♭}}} ∑_{{1≤l≤m \atop 0≤q≤m}}(−1)^{|U′|+l+|V′|+q+1} τ_{U′}[c_{0}, c_{1}, …, c_{l}]τ_{V′}[y, c_{0}, …, c_{q}]

= ∑_{{U′∈S_{C}^{♯} \atop V′∈S_{D}^{♭}}} (−1)^{|U′|+|V′|+1} τ_{U′} Σ(V′) ,

where, for a fixed V′ ∈ S_{D}^{♭},

Σ(V′) = ∑_{{1≤l≤m \atop 0≤q≤m}}(−1)^{l+q}[c_{0}, c_{1}, …, c_{l}]τ_{V′}[y, c_{0}, …, c_{q}] .

Now Σ(V′) = Σ(V′)_{≤} + Σ(V′)_{>}, where

Σ(V′)_{≤} = ∑_{1≤l≤q≤m}(−1)^{l+q}[c_{0}, c_{1}, …, c_{l}]τ_{V′}[y, c_{0}, …, c_{q}]
Σ(V′)_{>} = ∑_{0≤q<l≤m}(−1)^{l+q}[c_{0}, c_{1}, …, c_{l}]τ_{V′}[y, c_{0}, …, c_{q}]

For 0 ≤ q < l ≤ m, by Proposition 22.1, we have

[c_{0}, c_{1}, …, c_{l}]τ_{V′}[y, c_{0}, …, c_{q}] = [c_{0}, c_{1}, …, c_{q+1}][c_{q+1}, …, c_{l}]τ_{V′}[y, c_{0}, …, c_{q}] .

Moreover [c_{q+1}, …, c_{l}] clearly commutes with τ_{V′}, since V′ ⊆ D̂, and it also commutes with [y, c_{0}, …, c_{q}]. Thus

[c_{0}, c_{1}, …, c_{l}]τ_{V′}[y, c_{0}, …, c_{q}] = [c_{0}, c_{1}, …, c_{q+1}]τ_{V′}[y, c_{0}, …, c_{q}][c_{q+1}, …, c_{l}] .

Now the map f = [c_{0}, c_{1}, …, c_{q+1}]τ_{V′} has the property that f(c_{q}) = c_{q+1} = f(c_{q+1}), so f[y, c_{0}, …, c_{q}] = f[y, c_{0}, …, c_{q+1}], by Proposition 22.1 again. Thus

[c_{0}, c_{1}, …, c_{l}]τ_{V′}[y, c_{0}, …, c_{q}] = [c_{0}, c_{1}, …, c_{q+1}]τ_{V′}[y, c_{0}, …, c_{q+1}][c_{q+1}, …, c_{l}]
= [c_{0}, c_{1}, …, c_{q+1}]τ_{V′}[y, c_{0}, …, c_{l}] .

This gives

Σ(V′)_{>} = ∑_{0≤q<l≤m}(−1)^{l+q}[c_{0}, c_{1}, …, c_{q+1}]τ_{V′}[y, c_{0}, …, c_{l}]
= ∑_{1≤q≤l≤m}(−1)^{l+q−1}[c_{0}, c_{1}, …, c_{q}]τ_{V′}[y, c_{0}, …, c_{l}]
= ∑_{1≤l≤q≤m}(−1)^{l+q−1}[c_{0}, c_{1}, …, c_{l}]τ_{V′}[y, c_{0}, …, c_{q}]
= −(Σ(V′)_{≤}) ,

and it follows that Σ(V′) = 0, for any V′ ∈ S_{D}^{♭}. Our claim follows, and this completes the proof of Assertion (c).

(d) This follows from (a), (b), (c).

22.5. Corollary. Let G be a forest, and let B be a set of leaves of G. We say that a set A of edges of G is a union of partial geodesics with support B if A is a union of partial geodesics starting in B and if, moreover, for any x ∈ B, (x, d(x)) ∈ A. We denote by S̊_{B} the set of such sets of edges of G.

We define the element u_{B} of k(G^{G}) by

u_{B} = ∑_{A∈S̊_{B}}(−1)^{|B|+|A|}τ_{A} .

If B contains a single element x, we set u_{x} = u_{B}.

(a) Let x be a leaf of G. Then

u_{x} = ∑_{l=1}^{δ(x)}(−1)^{l−1}[x, d(x), …, d^{l}(x)] .

Thus u_{x} = id_{G} −v_{x}. In particular u_{x}^{2} = u_{x}.

(b) If x and y are leaves of G, then u_{x}u_{y} = u_{y}u_{x}.

(c) Let B be a set of leaves of G. Then u_{B} is an idempotent, equal to the product of the (commuting) idempotents u_{x}, for x ∈ B.

Proof : (a) This is clear from the definition.

(b) This follows from (a), since v_{x} and v_{y} commute.

(c) This follows from the observation that S_{B} = ⨆_{C⊆B} S̊_{C}. Thus

v_{B} = ∑_{C⊆B} w_{C} ,

where w_{C} = ∑_{A∈S̊_{C}}(−1)^{|A|}τ_{A}. By Möbius inversion in the poset of subsets of B, this

gives

w_{B} = ∑_{C⊆B}(−1)^{|B−C|}v_{C}
= ∑_{C⊆B}(−1)^{|B−C|} ∏_{x∈C}(id_{G} −u_{x})
= ∑_{C⊆B}(−1)^{|B−C|} ∑_{D⊆C}(−1)^{|D|} ∏_{x∈D} u_{x}
= ∑_{D⊆B}(−1)^{|D|} ∏_{x∈D} u_{x}( ∑_{D⊆C⊆B}(−1)^{|B−C|})
= (−1)^{|B|} ∏_{x∈B} u_{x} ,

becasue the inner sum is zero if D ⊂ B with D ≠ B. Hence u_{B} = (−1)^{|B|}w_{B} = ∏_{x∈B} u_{x}, as was to be shown.

We conclude this section with a proof of Theorem 17.9.

22.6. Theorem. Let T be a finite lattice, let (E, R) the full subposet of its irreducible elements, let Γ = {a ∈ T | a ∉ E, a < r^{∞}s^{∞}(a)}, and let G(T) be the graph structure on T introduced in Definition 17.8. For a ∈ Γ, let u_{a} be the element of k(T^{T}) introduced in Notation 17.4, and let u_{T} denote the composition of all the elements u_{a}, for a ∈ Γ.

(a) The graph G(T) is a forest, and Γ is the set of its leaves.

(b) For a ∈ Γ, the element u_{a} is an idempotent of k(T^{T}).

(c) u_{a} ∘ u_{b} = u_{b} ∘ u_{a} for any a, b ∈ Γ.

(d) The element u_{T} is an idempotent of k(T^{T}). It is equal to

u_{T} = ∑_{A∈S̊_{Γ}}(−1)^{|Γ|+|A|}τ_{A} ,

where S̊_{Γ} is the set of sets of edges of G(T) which are union of partial geodesics with support Γ, introduced in Corollary 22.5.

Proof : (a) If [figure: x; y] is an edge in G(T), then x <_{T} y. So G(T) has no loops and no oriented cycles. Moreover, for each x ∈ T, there exists at most one edge [figure: x; y] in G(T), so G(T) is a forest. By Definition 17.8, the set of leaves of G(T) is exactly Γ.

(b), (c), and (d) follow from the fact that if a ∈ Γ, then the complete geodesic of G(T) is the reduction sequence associated to a. Hence the element u_{a} introduced in Notation 17.4 coincides with the element with the same name introduced in Corollary 22.5.

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Serge Bouc, CNRS-LAMFA, Université de Picardie - Jules Verne, 33, rue St Leu, F-80039 Amiens Cedex 1, France.

serge.bouc@u-picardie.fr

Jacques Thévenaz, Section de mathématiques, EPFL, Station 8, CH-1015 Lausanne, Switzerland.

Jacques.Thevenaz@epfl.ch