CORRESPONDENCE FUNCTORS AND LATTICES

SERGE BOUC AND JACQUES THÉVENAZ

Abstract. A correspondence functor is a functor from the category of finite sets and correspondences to the category of k-modules, where k is a commutative ring. A main tool for this study is the construction of a correspondence 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 and complete results.

1. Introduction

The present paper is the second in a series which develops the theory of correspondence functors, namely functors from the category of finite sets and correspondences to the category of k-modules, where k is a commutative ring. In the first paper [BT2], we showed that the category of finitely generated correspondence functors is artinian when k is a field. In representation theory, simple modules, or simple functors, are the most basic and important objects to understand. We showed in [BT2] how to parametrize the simple correspondence functors S_{E,R,V} by means of a finite set E, an order relation R on E, and a simple k Aut(E, R)- module V (up to isomorphism).

The present paper establishes a connection between finite lattices and correspondence functors. Associated to any finite lattice T, we construct a correspondence functor F_{T} (Section 4). This is the second indication of the importance of posets and lattices in our work and we describe 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 4.12).

The second main purpose of this paper is to introduce a fundamental functor 𝕊_{E,R} associated to any finite poset (E, R). This is a precursor of each of the simple correspondence functors S_{E,R,V} and it turns out that understanding 𝕊_{E,R} is the key for understanding those simple functors. In particular, the fundamental functors 𝕊_{E,R} will play a crucial role for the determination of dim(S_{E,R,V} (X)) for any finite set X, which will appear in our next paper [BT3]. Actually, the formula for this dimension involves a new invariant associated to lattices which will be introduced in [BT3] and which will give another important motivation for studying the link between finite lattices and correspondence functors.

The fundamental functors can be analyzed by using lattices. If (E, R) is the subposet of irreducible elements in a finite 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 6.5). The kernel of the morphism F_{T} → 𝕊_{E,R^{op}} can be described by a system of linear equations.

Date: June 17, 2017.

1991 Mathematics Subject Classification. AMS Subject Classification: 06B05, 06B15, 06D05, 06D50, 16B50, 18B05, 18B10, 18B35, 18E05.

Key words and phrases. Keywords: finite set, correspondence, functor category, simple functor, poset, lattice.

We show that there is a duality between F_{T} and F_{T^{op}} over any commutative ring k (Theorem 8.9). Moreover, the fundamental functor 𝕊_{E,R} also appears as a subfunctor of F_{T^{op}} (Theorem 9.5). In Section 10, some endomorphisms and idempotents of a lattice T are defined, associated with all possible quotients of T which are total orders. In Section 11, those idempotents are used to describe completely the functor F_{T} in the special case where T is totally ordered and they are also used to find all direct summands associated to total orders in a functor F_{T} corresponding to an arbitrary lattice T.

2. Correspondence functors

In this introductory section, we recall the basic facts we need about correspondence functors (which also appear in [BT2]). We denote by C the category of finite sets and correspondences. Its objects are the finite sets and the set C(Y, X) of morphisms from X to Y is the set of all correspondences from X to Y, namely all subsets of Y × X (using a reverse notation which is convenient for left actions). Given two correspondences R ⊆ Z × Y and S ⊆ Y × X, their composition RS is defined by

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

A correspondence from X to X is also called a relation on X.

Let Σ_{X} be the symmetric group of all permutations of X. Associated with a permutation σ ∈ Σ_{X}, there is a relation on X which we write

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

In particular, Δ_{X} := Δ_{id} is the identity morphism of the object X. If σ, τ ∈ Σ_{X}, then Δ_{στ} = Δ_{σ}Δ_{τ}. The symmetric group Σ_{X} acts on relations by conjugation and we write ^{σ}R = Δ_{σ}RΔ_{σ^{−1}}.

For any commutative ring k, we let kC be the k-linearization of C. The objects are again the finite sets and kC(Y, X) is the free k-module with basis C(Y, X). A correspondence functor is a k-linear functor from kC to k-Mod. We let F_{k} be the category of all correspondence functors (for some fixed commutative ring k). This category has the following feature :

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

We define a minimal set for a correspondence functor F to be a finite set X of minimal cardinality such that F (X) ≠ 0. For a nonzero functor, such a minimal set always exists and is unique up to bijection.

The first instances of correspondence functors are the representable functors kC(−, E), where E is a finite set, and the functors

L_{E,W} := kC(−, E) ⊗_{kC(E,E)} W

where W is a left kC(E, E)-module. The proof of the following result is easy and 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.

2.2. Lemma. Let F_{k} be the category of all correspondence functors and let E be a finite set. The functor

kC(E, E)− Mod ⟶ F_{k} , W ↦ L_{E,W}

is left adjoint of the evaluation functor

F_{k} ⟶ kC(E, E)− Mod , F ↦ F (E) .

The functor L_{E,W} has a subfunctor J_{E,W} defined on any finite set X by

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

2.3. Lemma. Let E be a finite set and let W be a kC(E, E)-module.

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

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

Proof : The result is a slight extension of the first lemma of [Bo1]. The proof is also sketched in Lemma 2.3 of [BST] in the special case of biset functors for finite groups, but it extends without change to the representation theory of an arbitrary category.

Now we want to consider the functor L_{E,W} /J_{E,W} for some specific choices of kC(E, E)-modules. The algebra kC(E, E) of all relations on E was studied in [BT1] and we need a few facts from this approach. A relation R on E is called essential if it does not factor through a set of cardinality strictly smaller than |E|. The k-submodule generated by the set of inessential relations is a two-sided ideal

I_{E} := ∑_{|Y |<|E|} kC(E, Y)kC(Y, E)

and the quotient

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

is called the essential algebra. A large part of its structure has been elucidated in [BT1]. There is a quotient algebra P_{E} = E_{E}/N, where N is a nilpotent twosided ideal defined in [BT1]. 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. The product of two orders R and S in P_{E} is the transitive closure of R ∪ S if this closure is an order, and zero otherwise. This describes completely the algebra structure of P_{E}.

Among the kC(E, E)-modules, there is the fundamental module P_{E}f_{R}, associated to any poset (E, R), where E is a finite set and R denotes the order relation on E which defines the poset structure. Here f_{R} is a suitable idempotent in P_{E}, depending on R, and P_{E}f_{R} is the left ideal generated by f_{R}. Actually, the fundamental module P_{E}f_{R} only depends on the isomorphism type of the poset (E, R), or in other words, for a fixed set E, on the Σ_{E}-conjugacy class of R. More explicitly, if σ ∈ Σ_{E}, then conjugation by σ induces an isomorphism of posets (E, R) ≅ (E, ^{σ}R) and we also have an isomorphism of P_{E}-modules P_{E}f_{R} ≅ P_{E}f_{^{σ}R}, because f_{^{σ}R} = ^{σ}f_{R} (see Lemma 7.1 in [BT1] for details).

The only thing we really need to know about the fundamental module P_{E}f_{R} is its structure as a kC(E, E)-module. This is described in the next result, which combines Corollary 7.3 and Proposition 8.5 of [BT1] (see also Proposition 4.5 of [BT2] and its use).

2.4. 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 kC(E, E).

(b) 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.

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

(d) The action of the algebra of relations kC(E, 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 ^{σ}R = {(σ(e), σ(f)) | (e, f) ∈ R} (or equivalently ^{σ}R = Δ_{σ}RΔ_{σ^{−1}}). (Moreover, τ is unique in the first case.)

Using the (P_{E}, k Aut(E, R))-bimodule structure on P_{E}f_{R}, we define

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

where V is any k Aut(E, R)-module. Then T_{R,V} is a left P_{E}-module, hence also a kC(E, E)-module since P_{E} is a quotient of kC(E, E). The left action of kC(E, E) on T_{R,V} is induced from the action on P_{E}f_{R} described in Proposition 2.4 above. Again, the module T_{R,V} is invariant under Σ_{E}-conjugacy, that is, for a fixed set E and for σ ∈ Σ_{E}, we have an isomorphism of P_{E}-modules T_{^{σ}R,^{σ}V} ≅ T_{R,V}, where σV denotes the conjugate module, namely a module for the group Aut(E, ^{σ}R) = σ Aut(E, R)σ^{−1} (see Theorem 8.1 in [BT1] for details).

The main thing we need to know about T_{R,V} is the following result, which is part of Theorem 8.1 in [BT1].

2.5. Proposition. Assume that k is a field. Let E be a finite set, R an order on E, and V a simple k Aut(E, R)-module. Then T_{R,V} is a simple P_{E}-module (hence also a simple E_{E}-module).

Actually, Theorem 8.1 in [BT1] asserts that every simple E_{E}-module is isomorphic to some module T_{R,V} and that, consequently, 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.

Associated with the above kC(E, E)-modules, we can now define some specific correspondence functors. Using the fundamental module P_{E}f_{R}, we define

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

and we call it the fundamental functor associated with the poset (E, R). Using the module T_{R,V}, we define

S_{E,R,V} := L_{E,T_{R,V}} /J_{E,T_{R,V}} .

Note that 𝕊_{E,^{σ}R} ≅ 𝕊_{E,R} and S_{E,^{σ}R,^{σ}V} ≅ S_{E,R,V}, for any permutation σ ∈ Σ_{E}.

2.6. Proposition.

(a) The set E is a minimal set for 𝕊_{E,R} and 𝕊_{E,R}(E) ≅ P_{E}f_{R} as left kC(E, E)- modules.

(b) The set E is a minimal set for S_{E,R,V} and S_{E,R,V} (E) ≅ T_{R,V} as left kC(E, E)-modules.

(c) If k is a field and V is a simple k Aut(E, R)-module, then S_{E,R,V} is a simple correspondence functor.

Proof : Let W be any P_{E}-module and set S = L_{E,W} /J_{E,W}. Suppose that Y is a finite set such that S(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 P_{E} is a quotient of the essential algebra E_{E}, it follows that W is a module for E_{E} = kC(E, E)/I_{E}, so that the ideal I_{E} acts by zero on W. Therefore ψϕ ∉ I_{E}. But ψϕ factorizes through Y, so we must have |Y | ≥ |E|. Thus E is a minimal set for S. In particular, this holds for S = 𝕊_{E,R} (taking W = P_{E}f_{R}) and also for S = S_{E,R,V} (taking W = T_{R,V}).

Since J_{E,W} vanishes at E by Lemma 2.3, evaluation at E yields

S(E) = L_{E,W} (E)/J_{E,W} (E) = L_{E,W} (E) = kC(E, E) ⊗_{kC(E,E)} W ≅ W ,

and therefore 𝕊_{E,R}(E) ≅ P_{E}f_{R} and S_{E,R,V} (E) ≅ T_{R,V}.

For the proof of (c), notice that T_{R,V} is a simple kC(E, E)-module by Proposition 2.5 and therefore

S_{E,R,V} = L_{E,T_{R,V}} /J_{E,T_{R,V}}

is a simple correspondence functor by Lemma 2.3.

Although we do not need it here, let us mention that more is known about simple correspondence functors, when k is a field. It is proved in [BT2] that any simple functor has the form S_{E,R,V} for some triple (E, R, V) and that 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.

We note that the fundamental functor 𝕊_{E,R} is a precursor of S_{E,R,V}, in the sense of the following lemma.

2.7. Lemma. Suppose that V is a k Aut(E, R)-module generated by a single element v (e.g. a simple module). Consider the surjective morphism of correspondence functors

Φ : L_{E,P_{E}f_{R}} ⟶ L_{E,T_{R,V}}

induced by the surjective homomorphism of P_{E}-modules

Φ_{E} : P_{E}f_{R} ⟶ P_{E}f_{R} ⊗_{k Aut(E,R)} V = T_{R,V} , a ↦ a ⊗ v .

Then Φ induces a surjective morphism of correspondence functors

𝕊_{E,R} ⟶ S_{E,R,V} .

Proof : Let φ ⊗ f_{R} ∈ kC(X, E) ⊗_{kC(E,E)} P_{E}f_{R} = L_{E,P_{E}f_{R}}(X). If φ ⊗ f_{R} ∈ J_{E,P_{E}f_{R}}(X), then for every ψ ∈ C(E, X), we have ψφ·f_{R} = 0. Then Φ_{X}(φ⊗f_{R}) = φ ⊗ f_{R} ⊗ v and we have

ψφ · (f_{R} ⊗ v) = (ψφ · f_{R}) ⊗ v = 0 .

This shows that Φ_{X}(φ⊗f_{R}) ∈ J_{E,T_{R,V}} (X), so that Φ_{X}(J_{E,P_{E}f_{R}}(X)) ⊆ J_{E,T_{R,V}} (X). Therefore Φ induces a morphism of correspondence functors 𝕊_{E,R} ⟶ S_{E,R,V} which remains surjective.

By means of a very detailed analysis of 𝕊_{E,R} which will be carried out in [BT3], we shall show that it is possible to recover S_{E,R,V} from 𝕊_{E,R} by simply tensoring with V. Consequently, mastering 𝕊_{E,R} will be the key for obtaining information about the simple functors S_{E,R,V} and this explains why the fundamental functors play a crucial role throughout our work.

3. Posets and lattices

In this section, we give some definitions, fix some notation, and prove some basic lemmas, which will be used throughout.

By an order R on a finite set E, we mean a partial order relation on E. In other words, (E, R) is a finite poset.

3.1. Notation and definitions. Let (E, R) be a finite poset.

(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.

(e) The opposite order relation R^{op} is defined by the property that a ≤_{R^{op}} b if and only if b ≤_{R} a.

3.2. 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. An element e ≠ ˆ0 is irreducible if and only if the equality e = s ∨ t, for s, t ∈ T, implies 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, R) be a subposet of T. We say that it is a full subposet of T if for all e, f ∈ E we have :

e ≤_{R} f ⟺ e ≤_{T} f .

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.

3.3. Notation. Let (E, R) be a finite poset.

(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}).

3.4. Remark. Let R be a preorder on a finite set E, that is, a relation which is reflexive and transitive. 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 ∼. It is easy to see that the quotient map E → \overline{E} induces an isomorphism of lattices I_{↓}(E, R) ≅ I_{↓}(\overline{E}, \overline{R}).

Note that it is proved in Lemma 3.9 of [BT2] that the representable functors kC(−, E)R and kC(−, \overline{E})\overline{R} are isomorphic, but actually we will view them in a new way in Proposition 4.5. These remarks show that, for our purposes, it is enough to consider orders rather than preorders, and we shall do so in the rest of this paper, without loss of generality.

3.5. 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).

(d) For any finite lattice T having (E, R) as poset of irreducible elements, there is a join-preserving surjective map f : I_{↓}(E, R) ⟶ T which sends any lower ideal A ∈ I_{↓}(E, R) to the join ⋁_{e∈A} e in T.

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

3.6. Convention. In the situation of Lemma 3.5, 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.

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 3.6 applied to R^{op}.

We now introduce a notation which will play an important role in our work (and which was already used in the proof of Theorem 9.2 in [BT2]).

3.7. 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 3.6, we obtain

Γ_{φ} = {(x, e) ∈ X × E | e ∈ φ(x)} .

3.8. 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) One direction follows from (a). For the other direction, let S ∈ C(X, E) be such that SR^{op} = S, or equivalently S ∈ C(X, E)R^{op} (because R^{op} is idempotent by reflexivity and transitivity). Then the set

ϕ(x) = {e ∈ E | (x, e) ∈ S}

is a lower R-ideal in E, thus ϕ is a function X → I_{↓}(E, R). Clearly Γ_{ϕ} = S.

(c) This follows from (b) applied to R^{op}, because I^{↑}(E, R) = I_{↓}(E, R^{op}) and (E, R^{op}) is its poset of irreducible elements.

4. 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.

4.1. 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) := ⋁_{\begin{matrix}x∈X \\ (y,x)∈R \end{matrix}} 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, for every

element α = ∑_{R∈C(Y,X)} α_{R}R ∈ kC(Y, X), where α_{R} ∈ k, we set

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

4.2. Proposition. The assignment sending a finite set X to the k-module F_{T} (X) and a morphism α ∈ kC(Y, X) to the k-linear map 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, we have

(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).

4.3. Remark. The definition of F_{T} only uses the join operation in the lattice T. It follows that the definition would work for a join semi-lattice, but it is actually well-known that a finite join semi-lattice has automatically a structure of lattice (the meet operation being uniquely determined from the sole join). This explains our choice of working with lattices. Such a choice will also be useful in Section 8 when we shall work with opposite lattices.

We now establish the link between the action of correspondences on functions φ : X → T (as in Definition 4.1 above) and the correspondences Γ_{φ} defined in Notation 3.7.

4.4. Lemma. Let T be a finite lattice and let (E, R) be the full subposet of its irreducible elements. Assume that T is distributive, so that T ≅ I_{↓}(E, R). 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 3.7.

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.

Now we can give another description of the correspondence functor associated to a distributive lattice.

4.5. 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 3.8.

(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 4.4. Moreover kC(−, E) is a projective functor by Yoneda’s lemma and kC(−, E)R is a direct summand of kC(−, E) because R is idempotent. Therefore kC(−, E)R is projective.

(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. Our aim is to show that the assignment T ↦ F_{T} becomes a k-linear functor from kL to F_{k}, which will have the remarkable property of being full and faithful.

4.6. Definition. Let L and kL denote the following categories :

• The objects of L and kL are the finite lattices.

• For any two lattices T and T′, the set Hom_{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.

• For any two lattices T and T′, the set Hom_{kL}(T, T′) is the free k-module k Hom_{L}(T, T′) with basis Hom_{L}(T, T′).

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

• The composition of morphisms in kL is the k-bilinear extension of the composition in L.

It is easy to see that a morphism in L is order-preserving, by considering the join t_{1} ∨ t_{2} in the case where t_{1} ≤_{T} t_{2} in the lattice T.

4.7. Remark. The case A = ∅ in Definition 4.6 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.

Morphisms in L are morphisms of join semi-lattices (see Remark 4.3), but they are generally not morphisms of lattices in the sense that they need not commute with the meet operation. The choice of not using the meet operation turns out to be important for the next main theorem.

For our next theorem, we need some notation. 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′.

4.8. 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 U ∈ C(Y, X) be a correspondence. Then F_{T′}(U)(F_{f,X}(φ)) = F_{T′}(U)(f ∘ φ) = U(f ∘ φ) and F_{f,Y} (F_{T} (U)(φ)) = F_{f,Y} (Uφ) = f ∘ Uφ. We show that they are equal by evaluating at any y ∈ Y :

U(f ∘ φ)(y) = ⋁_{(y,x)∈U}(f ∘ φ)(x)
= ⋁_{(y,x)∈U} f(φ(x))
= f( ⋁_{(y,x)∈U} φ(x))
= (f ∘ Uφ)(y) ,

hence U(f ∘ φ) = f ∘ Uφ, which proves (a).

(b) It follows that the assignment T ↦ F_{T} is a functor L → F_{k}. Since kL is the klinearization 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

(4.9)  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

(4.10)  Φ_{Y} (β) = Ω_{β} φ .

Hence Φ is entirely determined by φ. Now Condition (4.9) 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 (4.9) 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

(4.11)  ∑_{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 (4.11). 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 (4.11) 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 (4.11) 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 (4.11) is zero (empty sum). In other words u_{λ} = 0 if λ is not a morphism in the category L (and this is where we see the relevance of the definition of morphisms in 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 (4.10). 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 4.8.

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

4.12. 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 4.8 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 Lemma 3.5, any finite distributive lattice T is isomorphic to the lattice I_{↓}(E, R) of lower ideals of a finite poset (E, R). By Proposition 4.5, the associated functor F_{T} is projective in F_{k}. This completes the proof of Theorem 4.12.

5. 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.

5.1. 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).

5.2. 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).

(c) The set E is a minimal set for the functor F_{T} /H_{T} .

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

(5.3)  (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 (5.3) 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}.

(c) If |X| < |E|, then |φ(X)| < |E| and therefore there is no map φ : X → T such that E ⊆ φ(X). Consequently, the k-basis of (b) is empty in that case, so that (F_{T} /H_{T})(X) = {0}. Now if X = E, then the k-basis of (b) consists of all bijections E → E (followed by the inclusion map E → T), so that (F_{T} /H_{T})(E) ≅ kΣ_{E}. This shows that E is a minimal set for F_{T} /H_{T}.

The quotient functor F_{T} /H_{T} plays a important role in our work, in particular in Theorem 6.5 and for the description of the fundamental functors and the simple functors in [BT3]. We now give another characterization of H_{T} (X).

5.4. 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 4.5, 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 4.5, 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 4.5, 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 4.4, 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).

5.5. 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}. We first need a lemma.

5.6. 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.

(a) 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 3.7.

(b) Γ_{ι} = R^{op}.

(c) The correspondence functor F_{T} is generated by ι ∈ F_{T} (E).

Proof : (a) 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 3.8.

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

(c) For every function φ : X → T, we have φ = Γ_{φ}ι by (a). Therefore F_{T} is generated by ι ∈ F_{T} (E).

5.7. 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, so that T ≅ I_{↓}(E, R), 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 : (a) By Lemma 5.6, 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).

Moreover, Φ_{E}(ι) = Γ_{ι}Φ_{E}(ι) = R^{op}Φ_{E}(ι), because Γ_{ι} = R^{op} by Lemma 5.6. Therefore Φ_{E}(ι) is contained in R^{op}M(E).

(b) Since T is distributive, we have

Γ_{Qφ} = QΓ_{φ}

by Lemma 4.4. 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 wish to give some information on simple functors S_{F,Q,V} appearing as quotients of F_{T}. We prove a more general result over an arbitrary commutative ring k, involving the not necessarily simple functors S_{F,Q,V} introduced in Section 2.

5.8. 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, so that T ≅ I_{↓}(E, R). 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 5.7. But F is a minimal set for S_{F,Q,V} by Proposition 2.6, 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 5.7, Φ_{E}(ι) = m ≠ 0 ∈ R^{op}S_{E,Q,V} (E). By Proposition 2.6, we know that

S_{E,Q,V} (E) ≅ T_{Q,V} = 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 2.4. 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 Theorem 6.2 of [BT1].

(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,T_{Q,V}} and the fact that any functor L_{E,W} is generated by L_{E,W} (E) = W by definition.

It is easy to see that S_{E,Q,V} ≅ S_{E,^{σ}Q,^{σ}V} for any σ ∈ Σ_{E}, because T_{Q,V} ≅ T_{^{σ}Q,^{σ}V} by construction (see Theorem 8.1 in [BT1] for more details). 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 Theorem 6.2 of [BT1].

Thus m = f_{Q} ⊗ v ∈ S_{E,Q,V} (E) is invariant under left multiplication by R^{op}. By Proposition 5.7 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}.

The similar question of finding fundamental functors appearing as quotients of F_{T} will be considered later in Theorem 6.9.

6. The fundamental functor associated to a poset

The fundamental functor 𝕊_{E,R} associated to a poset (E, R) was introduced in Section 2. One of our important goals is to give a precise description of its evaluations and use it to deduce a precise description of the evaluations of simple functors, but this will be fully achieved only in [BT3]. We prepare the ground by proving several main results about 𝕊_{E,R}. Recall from Proposition 2.6 that E is a minimal set for 𝕊_{E,R} and that 𝕊_{E,R}(E) is isomorphic to the fundamental module P_{E}f_{R}, which is described in Proposition 2.4.

Since 𝕊_{E,R} = L_{E,P_{E}f_{R}}/J_{E,P_{E}f_{R}} by definition, it is important to know when an element of L_{E,P_{E}f_{R}}(X) belongs to J_{E,P_{E}f_{R}}(X), where X is a finite set. For this analysis, we note that an element of L_{E,P_{E}f_{R}}(X) is written

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

where λ_{S} ∈ k for every S, because the tensor product is over kC(E, E) and P_{E} is a quotient algebra of kC(E, E). But since Rf_{R} = f_{R} (by Proposition 2.4) and the tensor product is over kC(E, E), we can replace S by SR and obtain a sum running only over S ∈ C(X, E)R.

6.1. Lemma. Consider an element

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

This element belongs to J_{E,P_{E}f_{R}}(X) if and only if

∀ U ∈ RC(E, X) , ∑_{\begin{matrix}S∈C(X,E)R \\ US=R \end{matrix}} λ_{S} = 0 .

Proof : By the definition of J_{E,P_{E}f_{R}}(X), we have

\begin{aligned}∑_{S∈C(X,E)R} λ_{S}S ⊗ f_{R} & ∈ J_{E,P_{E}f_{R}}(X) ⟺ \\ & ∀ U ∈ C(E, X) , ∑_{S∈C(X,E)R} λ_{S}US · f_{R} = 0 . \end{aligned}

By Proposition 2.4, the action of the relation US ∈ C(E, E) on f_{R} is given by

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

We claim that

Δ_{E} ⊆ Δ_{τ^{−1}}US ⊆ R ⟺ US = Δ_{τ} R .

If the left hand side holds, then multiply on the right by R and use the fact that SR = S and R^{2} = R (by transitivity and reflexivity of R) to obtain Δ_{τ^{−1}}US = R, hence US = Δ_{τ} R. Conversely, if the right hand side holds, then Δ_{τ^{−1}}US = R, hence

RΔ_{τ^{−1}}US = R^{2} = R

by transitivity and reflexivity of R. In particular, by reflexivity again,

Δ_{E} ⊆ RΔ_{τ^{−1}}US

so that, for any (a, a) ∈ Δ_{E}, there exists b ∈ E with (a, b) ∈ R and (b, a) ∈ Δ_{τ^{−1}}US = R. By antisymmetry of R, it follows that b = a and therefore (a, a) ∈ Δ_{τ^{−1}}US, so that Δ_{E} ⊆ Δ_{τ^{−1}}US. This shows that the left hand side holds, proving the claim.

It follows that our given element belongs to J_{E,P_{E}f_{R}}(X) if and only if

(6.2)  ∀ U ∈ C(E, X) , ∑_{\begin{matrix}S∈C(X,E)R \\ τ∈Σ_{E} \\ US=Δ_{τ} R \end{matrix}} λ_{S}Δ_{τ} f_{R} = 0 .

But by Proposition 2.4, P_{E}f_{R} is a free k-module with basis {Δ_{τ} f_{R} | τ ∈ Σ_{E}}. Therefore (6.2) is equivalent to

∀ U ∈ C(E, X) , ∀ τ ∈ Σ_{E} , ∑_{\begin{matrix}S∈C(X,E)R \\ US=Δ_{τ} R \end{matrix}} λ_{S} = 0 .

Replacing U by Δ_{τ^{−1}}U, this is equivalent to

(6.3)  ∀ U ∈ C(E, X) , ∑_{\begin{matrix}S∈C(X,E)R \\ US=R \end{matrix}} λ_{S} = 0 .

Now we claim that

US = R ⟺ RUS = R .

If the left hand side holds, then multiply on the left by R and use the fact that R^{2} = R. Conversely, if the right hand side holds, then US = Δ_{E}US ⊆ RUS = R. Moreover, Δ_{E} ⊆ R = RUS, so that, for any (a, a) ∈ Δ_{E}, there exists b ∈ E with (a, b) ∈ R and (b, a) ∈ US ⊆ R. By antisymmetry of R, it follows that b = a and therefore (a, a) ∈ US, so that Δ_{E} ⊆ US. Multiplying on the right by R and using the fact that SR = S, we obtain R ⊆ US. This shows that R = US, as required, proving the claim.

It now follows that in the family of equations (6.3), the equation for U is the same as the equation for RU. Therefore we can assume that U ∈ RC(E, X) and (6.3) is equivalent to

∀ U ∈ RC(E, X) , ∑_{\begin{matrix}S∈C(X,E)R \\ US=R \end{matrix}} λ_{S} = 0 ,

proving the lemma.

Before stating the main theorem, we need another lemma.

6.4. 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. For any correspondence W ∈ C(E, X) and any map φ : X → T, the following two statements are equivalent :

W Γ_{φ} = R^{op} ⟺ W φ = id_{E} ,

where Γ_{φ} ∈ C(X, E) is defined in Notation 3.7.

Proof : Let e, f ∈ E. Since (E, R) is a full subposet of T, the relation (f, e) ∈ R^{op} is equivalent to e ≤ f, where we write ≤ instead of ≤_{T} for simplicity. Suppose that W Γ_{φ} = R^{op}. Then e ≤ f if and only if

∃ x ∈ X with (f, x) ∈ W , (x, e) ∈ Γ_{φ} ⟺ ∃ x ∈ X with (f, x) ∈ W , e ≤ φ(x) .

Now we can write

φ(x) = ⋁_{\begin{matrix}e∈E \\ e≤φ(x) \end{matrix}} e

because any element of T is a join of irreducible elements. It follows that

(W φ)(f) = ⋁_{\begin{matrix}x∈X \\ (f,x)∈W \end{matrix}} φ(x) = ⋁_{\begin{matrix}x∈X \\ (f,x)∈W \end{matrix}} ⋁_{\begin{matrix}e∈E \\ e≤φ(x) \end{matrix}} e = ⋁_{e≤f} e ,

by the equivalence above. Thus (W φ)(f) = f, so that W φ = id_{E}.

Conversely, suppose that W φ = id_{E}. Since φ = Γ_{φ}ι by Lemma 5.6, we have W Γ_{φ}ι = id_{E} and we obtain, for every e ∈ E,

e = (W Γ_{φ}ι)(e) = ⋁_{\begin{matrix}f∈E \\ (e,f)∈W Γ_{φ} \end{matrix}} ι(f) = ⋁_{\begin{matrix}f∈E \\ (e,f)∈W Γ_{φ} \end{matrix}} f .

Since e is irreducible, it follows that e = f for some f ∈ E such that (e, f) ∈ W Γ_{φ}, and so (e, e) ∈ W Γ_{φ}, showing that Δ_{E} ⊆ W Γ_{φ}. On the other hand, we also obtain f ≤ e for every f ∈ E such that (e, f) ∈ W Γ_{φ}, showing that W Γ_{φ} ⊆ R^{op}. Therefore Δ_{E} ⊆ W Γ_{φ} ⊆ R^{op}. Multiplying on the right by R^{op}, we deduce that W Γ_{φ} = R^{op}, thanks to the fact that Γ_{φ}R^{op} = Γ_{φ} by Lemma 3.8.

The following theorem establishes the link between the functor F_{T} associated with a lattice T and the fundamental correspondence functors.

6.5. 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 exists a unique morphism of correspondence functors

Θ_{T} : F_{T} ⟶ 𝕊_{E,R^{op}}

mapping ι ∈ F_{T} (E) to f_{R^{op}} ∈ 𝕊_{E,R^{op}}(E) ≅ P_{E}f_{R^{op}}. Moreover, Θ_{T} is surjective.

(b) The subfunctor H_{T} of F_{T} is contained in the kernel of Θ_{T}. In particular, Θ_{T} induces a surjective morphism \overline{Θ}_{T} : F_{T} /H_{T} ⟶ 𝕊_{E,R^{op}}.

Proof : (a) For any finite set X and any function φ : X → T, we let

Θ̃_{T,X}(φ) = Γ_{φ} ⊗ f_{R^{op}} ∈ kC(X, E) ⊗_{kC(E,E)} f_{R^{op}} = L_{E,P_{E}f_{R^{op}}} (X) ,

where Γ_{φ} ∈ C(X, E) is defined in Notation 3.7. This extends to a k-linear map

Θ̃_{T,X} : F_{T} (X) ⟶ L_{E,P_{E}f_{R^{op}}} (X)

which we compose with the canonical surjection

π_{X} : L_{E,P_{E}f_{R^{op}}} (X) ⟶ L_{E,P_{E}f_{R^{op}}} (X)/J_{E,P_{E}f_{R^{op}}} (X) = 𝕊_{E,R^{op}}(X)

to obtain a k-linear map

Θ_{T,X} : F_{T} (X) ⟶ 𝕊_{E,R^{op}}(X) .

The family of maps Θ̃_{T,X} is not a morphism of functors, but we are going to prove that it becomes so after composition with π : L_{E,P_{E}f_{R^{op}}} → 𝕊_{E,R^{op}}. We have to show that, for any correspondence V ∈ C(Y, X) and any map φ : X → T, we have

Θ_{T,Y} (V φ) = V Θ_{T,X}(φ) , that is, Θ̃_{T,Y} (V φ) − V Θ̃_{T,X}(φ) ∈ J_{E,P_{E}f_{R^{op}}} (Y) .

In other words, we need to prove that

(Γ_{V φ} − V Γ_{φ}) ⊗ f_{R^{op}} ∈ J_{E,P_{E}f_{R^{op}}} (Y) .

Since Γ_{V φ}R^{op} = Γ_{V φ} and Γ_{φ}R^{op} = Γ_{φ} by Lemma 3.8, we can apply Lemma 6.1 to the element (Γ_{V φ} −V Γ_{φ})⊗f_{R^{op}} (with R^{op} instead of R). Since we have a difference of two terms, we only have to show that the additional conditions in the equations occur simultaneously, that is,

(6.6)  ∀ U ∈ R^{op}C(E, Y) , UΓ_{V φ} = R^{op} ⟺ UV Γ_{φ} = R^{op} .

We emphasize that the equation V Γ_{φ} = Γ_{V φ} does not hold in general (it holds if the lattice T is distributive by Lemma 6.6), but it will become correct after left multiplication by U ∈ R^{op}C(E, Y). Using Lemma 6.4, we have to show that, for all U ∈ R^{op}C(E, Y),

U(V φ) = id_{E} ⟺ (UV)φ = id_{E} .

But this is obvious in view of the action of correspondences on φ ∈ F_{T} (X). Therefore Θ_{T} is a morphism of correspondence functors, as was to be shown.

By Lemma 5.6, the image of ι ∈ F_{T} (E) is

Θ_{T,E}(ι) = Γ_{ι} ⊗ f_{R^{op}} = R^{op} ⊗ f_{R^{op}} = Δ_{E} ⊗ R^{op}f_{R^{op}} = Δ_{E} ⊗ f_{R^{op}} ,

because the tensor product is over kC(E, E) and because R^{op}f_{R^{op}} = f_{R^{op}} by the action of relations on f_{R^{op}} (see Proposition 2.4). Now Δ_{E} ⊗ f_{R^{op}} corresponds to f_{R^{op}} under the isomorphism 𝕊_{E,R^{op}}(E) ≅ P_{E}f_{R^{op}}. Therefore Θ_{T} maps ι to f_{R^{op}}.

Since R^{op}f_{R^{op}} = f_{R^{op}}, any generator S ⊗ f_{R^{op}} of L_{E,P_{e}f_{R^{op}}} (X), where S ∈ C(X, E), can be written

S ⊗ f_{R^{op}} = S ⊗ R^{op}f_{R^{op}} = SR^{op} ⊗ f_{R^{op}} = Γ_{φ} ⊗ f_{R^{op}}

for some φ : X → T, in view of Lemma 3.8. This shows that Θ_{T} is surjective.

(b) A generator of H_{T} (X) is a map φ : X → T such that E ⊈ φ(X). We need to prove that φ ∈ Ker Θ_{T}, or in other words Γ_{φ} ⊗ f_{R^{op}} ∈ J_{E,P_{e}f_{R^{op}}} (X). In order to prove this, we apply Lemma 6.1 and we note that we have here a sum with a single term, so we have to check that the additional condition UΓ_{φ} = R^{op} never holds, where U ∈ R^{op}C(E, X). By Lemma 6.4, UΓ_{φ} ≠ R^{op} is equivalent to Uφ ≠ id_{E}. But Uφ ∈ H_{T} (E) because H_{T} is a subfunctor and obviously id_{E} ∉ H_{T} (E), so Uφ ≠ id_{E}, as required.

6.7. Remark. By Lemma 3.5, any lattice T having (E, R) as its full subposet of irreducible elements is isomorphic to a quotient of the distributive lattice I_{↓}(E, R) via the map

π : I_{↓}(E, R) ⟶ T , A ↦ ⋁_{e∈A} e

where the join is taken in the lattice T. Since I_{↓}(E, R) also has (E, R) as its full subposet of irreducible elements, there is a surjective morphism

Θ_{I_{↓}(E,R)} : F_{I_{↓}(E,R)} ⟶ 𝕊_{E,R^{op}}

which is easily seen to correspond, via the isomorphism F_{I_{↓}(E,R)} ≅ kC(−, E)R^{op} of Proposition 4.5, to the morphism

ω : kC(−, E)R^{op} ⟶ 𝕊_{E,R^{op}} , R^{op} ↦ f_{R^{op}} .

Moreover, it is also elementary to check that the composite

\xymatrix{kC(−, E)R^{op} ≅ F_{I_{↓}(E,R)} \ar[r]^{F_{π}} & F_{T} \ar[r]^{Θ_{T}} & 𝕊_{E,R^{op}}}

is equal to ω.

For our next result, we need a well-known result of algebraic K-theory.

6.8. Lemma. Let L and L′ be finitely generated free k-modules of the same rank, where k is a commutative ring. Then any surjective homomorphism of k-modules f : L → L′ is an isomorphism.

Proof : The homomorphism f splits, because L′ is free. Hence there exists a homomorphism g : L′ → L such that f ∘ g = id_{L′}. Let n be the common rank of the free k-modules L and L′. We can view f and g as square matrices of size n with coefficients in k, such that fg = id_{n}. Taking determinants (which makes sense as k is commutative), we get that det(f) det(g) = det(id_{n}) = 1. Hence det(f) is invertible in k, and f is invertible as well.

We now show that the relationship between F_{T} and 𝕊_{E,R^{op}} is very strong, in the sense that 𝕊_{E,R^{op}} is the only fundamental functor appearing as a quotient of F_{T} /H_{T}, where H_{T} is defined in Notation 5.1. Recall that E is a minimal set for the functor F_{T} /H_{T}, by Proposition 5.2.

6.9. Theorem. Let T be a finite lattice and let (E, R) be the full subposet of its irreducible elements. Let (D, Q) be a finite poset such that there exists a surjective morphism Φ : F_{T} /H_{T} → 𝕊_{D,Q}.

(a) |D| = |E| (so we can assume that D = E).

(b) Assuming that D = E, there exists σ ∈ Σ_{E} such that ^{σ}Q = R^{op}. In other words, the posets (E, Q) and (E, R^{op}) are isomorphic.

(c) The kernel of \overline{Θ}_{T} : F_{T} /H_{T} → 𝕊_{E,R^{op}} vanishes at E and

\overline{Θ}_{T,E} : F_{T} (E)/H_{T} (E) ⟶ 𝕊_{E,R^{op}}(E) ≅ P_{E}f_{R^{op}}

is an isomorphism.

Proof : (a) By Proposition 5.7, any morphism F_{T} /H_{T} → 𝕊_{D,Q} is determined by an element of 𝕊_{D,Q}(E). Since our given morphism Φ is surjective, it is nonzero, and therefore 𝕊_{D,Q}(E) ≠ {0}. Since D is a minimal set for 𝕊_{D,Q}, we have |D| ≤ |E|. Now there is a surjective morphism

Φ_{D} : F_{T} (D)/H_{T} (D) → S_{D,Q}(D) ≠ {0} ,

so F_{T} /H_{T} does not vanish at D. Since E is a minimal set for F_{T} /H_{T} by Proposition 5.2, we have |E| ≤ |D|.

(b) We prove that Φ_{E} : F_{T} (E)/H_{T} (E) → S_{E,Q}(E) is an isomorphism. By Proposition 5.1, F_{T} (E)/H_{T} (E) is a free k-module with basis the set of all bijections E → E (followed by the inclusion map E → T). By Proposition 2.4, S_{E,Q}(E) ≅ P_{E}f_{Q} is a free k-module with a basis consisting of the elements Δ_{σ}f_{Q}, where σ runs through the group Σ_{E} of all permutations of E. Thus the two modules under consideration are finitely generated free k-modules with the same rank and we know that the map Φ_{E} is surjective. Therefore Φ_{E} is an isomorphism by Lemma 6.8.

The same argument applies to the surjective morphism \overline{Θ}_{T,E} : F_{T} (E)/H_{T} (E) → 𝕊_{E,R^{op}}(E), which is therefore also an isomorphism. It follows that there are isomorphisms of kC(E, E)-modules

P_{E}f_{Q} ≅ S_{E,Q}(E) ≅ F_{T} (E)/H_{T} (E) ≅ 𝕊_{E,R^{op}}(E) ≅ P_{E}f_{R^{op}} .

This isomorphism maps f_{Q} to af_{R^{op}} for some a ∈ P_{E}. Therefore it maps f_{Q} = Qf_{Q} to af_{R^{op}} = Qaf_{R^{op}} and in particular Qaf_{R^{op}} ≠ 0. By Proposition 2.4, this is possible only if there exists σ, τ ∈ Σ_{E} such that

Δ_{E} ⊆ Δ_{τ^{−1}}Q ⊆ ^{σ}(R^{op}) .

In particular, Δ_{τ} ⊆ Q, forcing τ = id because Q is an order relation. Thus Q ⊆ ^{σ}(R^{op}). Swapping the role of Q and R^{op}, the same argument shows that R^{op} is contained in a conjugate of Q. Therefore Q and R^{op} are conjugate, as was to be shown.

(c) We have just seen that \overline{Θ}_{T,E} : F_{T} (E)/H_{T} (E) ⟶ 𝕊_{E,R^{op}}(E) is an isomorphism. Therefore Ker Θ_{T,E} = {0}. In other words, the subfunctor Ker Θ_{T} vanishes at E.

7. The kernel of Θ_{T}

Theorem 6.5 shows that 𝕊_{E,R^{op}}(X) is isomorphic to a quotient of F_{T} and we want to understand the kernel. We do this in the following result in terms of a system of equations. The solution of this system of equations is quite hard and will only be obtained in [BT3], when we will compute the dimension of the evaluations of fundamental functors and simple functors.

7.1. 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)

∑_{\begin{matrix}φ:X→T \\ Γ_{ψ}^{op}Γ_{φ}=R^{op} \end{matrix}} λ_{φ} = 0 .

Recall from Notation 3.7 that Γ_{φ} = {(x, e) ∈ X × E | e ∈ φ(x)} ⊆ C(X, E) and Γ_{ψ} = {(x, e) ∈ X × E | e ∈ ψ(x)} ⊆ C(X, E).

Proof : The image of φ : X → T under the map Θ_{T,X} is equal to the class of Γ_{φ} ⊗ f_{R^{op}} ∈ L_{E,P_{E}f_{R^{op}}} (X) in the quotient

𝕊_{E,R^{op}}(X) = L_{E,P_{E}f_{R^{op}}} (X)/J_{E,P_{E}f_{R^{op}}} (X) .

Therefore a 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} ⊗ f_{R^{op}} belongs to J_{E,P_{E}f_{R^{op}}} (X). We apply Lemma 6.1, using the fact

that Γ_{φ}R^{op} = Γ_{φ}. It follows that u ∈ Ker Θ_{T,X} if and only if

∀ U ∈ R^{op}C(E, X) , ∑_{\begin{matrix}φ:X→T \\ UΓ_{φ}=R^{op} \end{matrix}} λ_{φ} = 0 .

Now U ∈ R^{op}C(E, X) if and only if U^{op} ∈ C(X, E)R if and only if U = Γ_{ψ} for some map ψ : X → I^{↑}(E, R) (by Lemma 3.8). Thus the condition becomes

∀ ψ : X → I^{↑}(E, R) ,  ∑_{\begin{matrix}φ:X→T \\ Γ_{ψ}^{op}Γ_{φ}=R^{op} \end{matrix}} λ_{φ} = 0 ,

as was to be shown.

The condition Γ_{ψ}^{op}Γ_{φ} = R^{op} which appears in the system of equations in Theorem 7.1 is the key for the description of the fundamental functor 𝕊_{E,R^{op}}, and consequently for understanding the simple functors S_{E,R,V}. We need to characterize this condition in various useful ways.

We first introduce the following notation.

7.2. 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.

We can now state the various characterizations of the condition which we need.

7.3. 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. 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_{E} ⊆ 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 5.6, we have φ = Γ_{φ}ι.

(b) ⇔ (c). We prove more generally that Qι = ι if and only if Δ_{E} ⊆ Q ⊆ R^{op}, for any relation Q ∈ C(E, E). Suppose first that Δ_{E} ⊆ 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_{E} ⊆ 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) ⇔ (d). If (c) holds, multiply on the right by R^{op} and use the equality Γ_{φ}R^{op} = Γ_{φ} of Lemma 3.8 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 Theorem 7.3.

Condition (d) will play an important role in the proof of Theorem 9.5 below, while conditions (e) and (f) will be main tools used in [BT3].

8. Duality and opposite lattices

In this section, we prove a perfect duality between the functor associated to a lattice T and the functor associated to the opposite lattice T^{op}. We work with an arbitrary commutative base ring k.

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}) .

Recall that L denotes the category of finite lattices and kL its k-linearization (Definition 4.6). 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}}.

8.1. 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.

(8.2)  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,

(8.3)  ∀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 8.3 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 Hom_{L}(T_{1}, T_{2}) to Hom_{L}(T_{2}^{op}, T_{1}^{op}).

Now if f : T_{1} → T_{2} and g : T_{2} → T_{3} are morphisms in L, the adjunction (8.3) 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}.

8.4. 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.

8.5. 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.

8.6. Lemma.

(a) With the notation 8.5, the family of bilinear forms in Definition 8.4 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).

8.7. Remark. Let T = I_{↓}(E, R^{op}) be the distributive 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 4.5

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 (8.4) 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 10.5 of [BT2].

8.8. 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^{⋆} := ∑_{\begin{matrix}ρ:X→T \\ ρ≤φ \end{matrix}} 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, for 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.

Now we can prove that we have a perfect duality.

8.9. Theorem. Let T be a finite lattice.

(a) Let X be a finite set. The bilinear form (8.4) 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,

(8.10)  (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). In particular, Ψ_{T,X} is surjective.

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 8.3, 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 8.9.

8.11. 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 4.12, 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.

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 8.9, 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}.

9. Duality for fundamental functors and simple functors

By Theorem 6.5, any fundamental functor 𝕊_{E,R} is isomorphic to a quotient of some functor associated to a lattice. One of the main purposes of this section is to use duality to realize 𝕊_{E,R} as a subfunctor of some functor associated to another lattice. We also determine what is the dual of a simple functor.

We will study the subfunctor generated by a specific element of F_{T^{op}}(E) which will be defined below. We need some more notation.

9.1. 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[.

9.2. 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 8.5, for the action of a correspondence on evaluations of F_{T^{op}}.

9.3. 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 Notation 8.8).

(b) R ⋆ γ_{T} = γ_{T}.

Proof : (a) By definition, ι^{⋆} = ∑_{ρ≤ι} μ(ρ, ι)ρ^{∘}, where ρ^{∘} denotes the function ρ,

viewed as a map E → T^{op}, and where μ is the Möbius function of the poset of functions from E to T (see Notation 8.8). 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.

9.4. Lemma. Let f : T → T′ be a morphism in L and let (E, R), respectively (E′, R′), be the full subposet of irreducible elements in T, respectively T′. Suppose that the restriction of f to E is an isomorphism of posets f_{|} : (E, R) ⟶^{≅} (E′, R′).

(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 8.2 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 8.9, 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 8.5, for the action of a correspondence on evaluations of F_{T^{op}}. We now come to our main result.

9.5. 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 : We first show that it suffices to prove the result in the case when T is the lattice I_{↓}(E, R). For any other lattice T′ with the same poset (E, R) of irreducible elements, the inclusion E ⊆ T′ extends to a unique surjective map

f : T = I_{↓}(E, R) ⟶ T′

in the category L which induces the identity on E (see Lemma 3.5). Then ⟨γ_{T′}⟩ is isomorphic to ⟨γ_{T} ⟩ by Lemma 9.4, so we now assume that T = I_{↓}(E, R), which is a distributive lattice.

We now apply Proposition 5.7 to the element γ_{T} ∈ F_{T^{op}}(E), using the fact that γ_{T} ∈ R ⋆ F_{T^{op}}(E), because R ⋆ γ_{T} = γ_{T} by Lemma 9.3. We deduce that γ_{T} ∈ F_{T^{op}}(E) determines a unique morphism

ξ : F_{I_{↓}(E,R^{op})} → F_{T^{op}}

such that ξ_{E}(j) = γ_{T}, where j : E → I_{↓}(E, R^{op}) is the inclusion map.

Then for any finite set X and any function ψ : X → I_{↓}(E, R^{op}), we can compute the map ξ_{X} : F_{I_{↓}(E,R^{op})}(X) → F_{T^{op}}(X) as follows :

ξ_{X}(ψ) = ξ_{X}(Γ_{ψ}j) = Γ_{ψ} ⋆ ξ_{E}(j) = Γ_{ψ} ⋆ γ_{T} ,

using the equality ψ = Γ_{ψ}j of Lemma 5.6. In particular, the image of ξ is the subfunctor ⟨γ_{T} ⟩ generated by γ_{T}.

By Theorem 6.5 and the fact that the lattice I_{↓}(E, R^{op}) has (E, R^{op}) as its full subset of irreducible elements, there is a surjective morphism

Θ_{I_{↓}(E,R^{op})} : F_{I_{↓}(E,R^{op})} ⟶ 𝕊_{E,R}

and its kernel is described in Theorem 7.1. We want to prove that, for any finite set X, the kernel of the surjection

ξ_{X} : F_{I_{↓}(E,R^{op})}(X) ⟶ ⟨γ_{T} ⟩(X)

is equal to the kernel of the surjection

Θ_{I_{↓}(E,R^{op}),X} : F_{I_{↓}(E,R^{op})}(X) ⟶ 𝕊_{E,R}(X) ,

from which the isomorphism ⟨γ_{T} ⟩ ≅ 𝕊_{E,R} will follow.

The kernel of the surjection ξ_{X} is the set of all linear combinations

u = ∑_{ψ:X→I_{↓}(E,R^{op})} λ_{ψ}ψ ,

where λ_{ψ} ∈ k, such that

∑_{ψ:X→I_{↓}(E,R^{op})} λ_{ψ}Γ_{ψ} ⋆ γ_{T} = 0 .

Equivalently, using the nondegeneracy of the bilinear form (−, −)_{X} of (8.4), proved in Theorem 8.9,

∀ φ : X → T, (φ, ∑_{ψ} λ_{ψ}Γ_{ψ} ⋆ γ_{T})_{X} = 0 .

By Equation 8.6, this is in turn equivalent to

∀ φ : X → T, ∑_{ψ} λ_{ψ}(Γ_{ψ}^{op}φ, γ_{T})_{E} = 0 .

Now γ_{T} = ι^{⋆} by Lemma 9.3 and we use ι^{⋆} instead. By Equation 8.10, we have

(Γ_{ψ}^{op}φ, γ_{T})_{E} = (Γ_{ψ}^{op}φ, ι^{⋆})_{E} = δ_{Γ_{ψ}^{op}φ,ι}

and therefore we obtain the condition

∀ φ : X → T, ∑_{\begin{matrix}ψ:X→I_{↓}(E,R^{op}) \\ Γ_{ψ}^{op}φ=ι \end{matrix}} λ_{ψ} = 0 .

Since we have assumed that T = I_{↓}(E, R) and since I_{↓}(E, R^{op}) = I^{↑}(E, R), we have maps φ : X → I_{↓}(E, R) and ψ : X → I^{↑}(E, R) and we know from Theorem 7.3 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}ψ is in the kernel of ξ_{X} if and only if

∀ φ : X → I_{↓}(E, R), ∑_{\begin{matrix}ψ:X→I_{↓}(E,R^{op}) \\ Γ_{ψ}^{op}Γ_{φ}=R^{op} \end{matrix}} 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, by passing to the opposite. Moreover I_{↓}(E, R) = I^{↑}(E, R^{op}) and I^{↑}(E, R) = I_{↓}(E, R^{op}). Therefore u = ∑_{ψ:X→I_{↓}(E,R^{op})} λ_{ψ}ψ is in the kernel of ξ_{X} if and only if

∀φ : X → I^{↑}(E, R^{op}), ∑_{\begin{matrix}ψ:X→I_{↓}(E,R^{op}) \\ Γ_{φ}^{op}Γ_{ψ}=R \end{matrix}} λ_{ψ} = 0 .

By Theorem 7.1, this is equivalent to requiring that u ∈ Ker Θ_{I_{↓}(E,R^{op}),X}. It follows that Ker ξ_{X} = Ker Θ_{I_{↓}(E,R^{op}),X}. Consequently, the images of ξ_{X} and Θ_{I_{↓}(E,R^{op}),X} are isomorphic, that is, ⟨γ_{T} ⟩ ≅ 𝕊_{E,R}. This completes the proof of Theorem 9.5.

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.

9.6. 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 (8.4).

(a) ⟨γ_{T} ⟩^{⊥} = Ker Θ_{T}, where Θ_{T} : F_{T} → 𝕊_{E,R^{op}} is the morphism of Theorem 6.5.

(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 8.6)

⟺ ∑_{{φ \atop Q^{op}φ=ι}} λ_{φ} = 0 ∀Q ∈ C(X, E)R (by 8.10 and Lemma 9.3)

⟺ ∑_{{φ \atop Γ_{ψ}^{op}φ=ι}} λ_{φ} = 0 ∀ψ : X → I^{↑}(E, R) (by Proposition 4.5)

⟺ ∑_{{φ \atop Γ_{ψ}^{op}Γ_{φ}=R^{op}}} λ_{φ} = 0 ∀ψ : X → I^{↑}(E, R)  (by Theorem 7.3)

⟺ ∑_{φ:X→T} λ_{φ}φ ∈ Ker Θ_{T,X} (by Theorem 7.1)

Therefore ⟨γ_{T} ⟩(X)^{⊥} = Ker Θ_{T,X}.

(b) This follows immediately from (a) and Theorem 6.5.

(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 9.5 and ⟨γ_{T} ⟩^{⊥⊥} ≅ 𝕊E,R^{op}^{♮} by (c). Thus we obtain a canonical injective

morphism 𝕊_{E,R} ⟶ 𝕊E,R^{op}^{♮}.

9.7. Remark. We will prove in [BT3] that α_{E,R} : 𝕊_{E,R} ⟶ 𝕊E,R^{op}^{♮} is actually an isomorphism. This is easy to prove if k is a field, because the inclusion ⟨γ_{T} ⟩ ⊆ ⟨γ_{T} ⟩^{⊥⊥} must be an equality since the pairing (8.4) is nondegenerate, by Theorem 8.9.

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 the simple correspondence functor (over k) parametrized by (E, R, V). Part (d) of Theorem 9.6 suggests that the index R must become R^{op} after applying duality. We now show that this is indeed the case.

9.8. 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 : For simplicity, write R_{E} = kC(E, E) for the algebra of all relations on E and recall from Section 2 that P_{E} is a quotient algebra of R_{E}. The evaluation S_{E,R,V} (E) is the R_{E}-module

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

(which is actually a simple R_{E}-module by Proposition 2.5).

Clearly the dual SE,R,V^{♮} is again a simple functor and its minimal set is E again. Moreover, by evaluation at E, we find that

SE,R,V^{♮} (E) ≅ S_{E,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^{♮}} .

But a simple functor with minimal set E is completely determined by its evaluation at E, because it is generated by this minimal nonzero evaluation (by simplicity). Since SE,R,V^{♮} and S_{E,R^{op},V^{♮}} have both an evaluation at E isomorphic to T_{R^{op},V^{♮}}, it follows that

SE,R,V^{♮} ≅ S_{E,R^{op},V^{♮}} ,

as required.

Now we construct the required pairing. By Proposition 2.4, 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^{♮} = ⨁_{ε∈[Σ_{E}/ Aut(E,R)]} Δ_{ε}f_{R^{op}} ⊗ V^{♮} .

We define the pairing

⟨−, −⟩ : T_{R,V} × T_{R^{op},V^{♮}} ⟶ k , ⟨Δ_{σ}f_{R} ⊗ v , Δ_{ε}f_{R^{op}} ⊗ α⟩ := δ_{σ,ε} α(v) ,

where σ, ε ∈ [Σ_{E}/ Aut(E, R)], v ∈ V, α ∈ V^{♮}.

By choosing dual bases of V and V^{♮}, we easily find dual bases of T_{R,V} and T_{R^{op},V^{♮}}, and it follows that this pairing is nondegenerate. We are left with the proof of the required property of this pairing, namely

⟨Q · x , y⟩ = ⟨x , Q^{op} · y⟩ for all x ∈ T_{R,V} , y ∈ T_{R^{op},V^{♮}}, Q ∈ C(E, E) .

By Proposition 2.4, the action of Q on P_{E}f_{R} is given by:

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}

It follows that

⟨Q · Δ_{σ}f_{R} ⊗ v , Δ_{ε}f_{R^{op}} ⊗ α⟩ = \{ \begin{matrix}α(v) & if ε = τσ and Δ_{E} ⊆ Δ_{τ^{−1}}Q ⊆ ^{σ}R , \\ 0 & otherwise . \end{matrix}

On the other hand

⟨Δ_{σ}f_{R}⊗v , Q^{op}·Δ_{ε}f_{R^{op}}⊗α⟩ = \{ \begin{matrix}α(v) & if ρε = σ and Δ_{E} ⊆ Δ_{ρ^{−1}}Q^{op} ⊆ ^{ε}R^{op} , \\ 0 & otherwise . \end{matrix}

We put ρ = τ^{−1} and we assume that ε = τσ (that is, ρε = σ). We then obtain equivalent conditions:

Δ_{E} ⊆ Δ_{τ} Q^{op} ⊆ ^{ε}R^{op} ⟺ Δ_{E} ⊆ QΔ_{τ^{−1}} ⊆ ^{ε}R  (taking the opposite)
⟺ Δ_{E} ⊆ QΔ_{τ^{−1}} ⊆ ^{τσ}R  (because ε = τσ)
⟺ Δ_{E} ⊆ Δ_{τ^{−1}}Q ⊆ ^{σ}R  (conjugating by Δ_{τ^{−1}})

Therefore, we obtain the required equality

⟨Q · Δ_{σ}f_{R} ⊗ v , Δ_{ε}f_{R^{op}} ⊗ α⟩ = ⟨Δ_{σ}f_{R} ⊗ v , Q^{op} · Δ_{ε}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.

10. Embeddings and idempotents corresponding to total orders

In this section, we construct morphisms of lattices in the category kL between a finite lattice T and a totally ordered lattice. This will be used in Section 11 to obtain detailed information about correspondence functors associated to a total order.

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 distributive 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}.

Let π : T → \underline{n} be a surjective morphism of finite lattices. For every h ∈ \underline{n}, let

b_{h} = sup (π^{−1}(h)) = ⋁_{π(t)=h} t = ⋁_{π(t)≤h} t .

In other words, b_{h} = π^{op}(h) with the notation of Lemma 8.1. Then we have ˆ0 ≤ b_{0} < b_{1} < … < b_{n−1} < b_{n} = ˆ1. If h ∈ Irr(\underline{n}) = [n] = \{1, …, n\}, the whole interval ]b_{h−1}, b_{h}]_{T} is mapped to h under π, while b_{h−1} is mapped to h − 1. The morphism π determines uniquely the totally ordered n-tuple B = (b_{0}, b_{1}, …, b_{n−1}).

Conversely, if B = (b_{0}, b_{1}, …, b_{n−1}) is a strictly increasing n-tuple in T − {ˆ1} and if b_{n} = ˆ1 (so that ˆ0 ≤ b_{0} < b_{1} < … < b_{n−1} < b_{n} = ˆ1), then B determines a unique surjective morphism π^{B} : T → \underline{n} by setting π^{B}(t) = h if t ≤ b_{h} and t ≰ b_{h−1} (hence in particular π^{B}(ˆ0) = 0).

For every h ∈ [n], choose a_{h} ∈ [b_{h−1}, b_{h}]_{T}, and define the n-tuple

A = (a_{1}, a_{2}, …, a_{n}) .

Set also a_{0} = ˆ0. Notice that A is totally ordered : ˆ0 = a_{0} ≤ a_{1} ≤ … ≤ a_{n} ≤ ˆ1, with possible equalities. Define

j_{A}^{B} : \underline{n} ⟶ T , h ↦ a_{h} .

It is easy to see that j_{A}^{B} is a morphism of lattices (because if e < f in \underline{n}, then e ≤ f − 1, hence a_{e} ≤ b_{e} ≤ b_{f−1} ≤ a_{f}). Note that j_{A} is a section of π^{B} if a_{h} ∈]b_{h−1}, b_{h}]_{T} for every h ∈ [n], but not if a_{h} = b_{h−1} for some h.

Write

μ(B, A) = ∏_{h∈[n]} μ_{T} (b_{h−1}, a_{h}) ,

where μ_{T} (−, −) denotes the Möbius function of the lattice T. For simplicity, we write μ = μ_{T} throughout this section and the next. Allowing the n-tuple A to vary (i.e. a_{h} varies in [b_{h−1}, b_{h}]_{T} for each h), define

j^{B} = (−1)^{n} ∑_{A} μ(B, A)j_{A}^{B} ∈ kL(\underline{n}, T) ,

or in other words

j^{B} = (−1)^{n} ∏_{h=1}^{n} ( ∑_{a_{h}∈[b_{h−1},b_{h}]_{T}} μ(b_{h−1}, a_{h})) j_{(a_{1},a_{2},...,a_{n})}^{B} .

10.1. Proposition. Let T be a finite lattice and let B be a strictly increasing n-tuple in T − \{ˆ1\}. Let j^{B} : \underline{n} → T be the corresponding morphism, constructed as above.

(a) For any finite set X and any function φ : X → \underline{n} such that [n] ⊈ φ(X), we have j^{B}φ = 0.

(b) j^{B} induces F_{j^{B}} : F\underline{n} → F_{T} vanishing on H\underline{n}, hence induces in turn a morphism

\overline{F}_{j^{B}} : F\underline{n}/H\underline{n} ⟶ F_{T} .

Proof : (a) We have

j^{B}φ = (−1)^{n} ∑_{A} μ(B, A)j_{A}^{B}φ = ∑_{ψ:X→T}(−1)^{n}( ∑_{\begin{matrix}A \\ j_{A}^{B}φ=ψ \end{matrix}} μ(B, A))ψ .

Let g ∈ [n] be such that g ∉ φ(X). Then we can modify each n-tuple A by changing freely the image j_{A}^{B}(g) = a_{g} ∈ [b_{g−1}, b_{g}]_{T} without changing the equality j_{A}^{B}φ = ψ. This is because φ(x) ≠ g for all x ∈ X. We set A′ = (a_{1}, …, â_{g}, …, a_{n}) and B′ = (b_{0}, …, b̂_{g−1}, …, b_{n−1}) (where denotes omission) and we let

j_{A′}^{B′} : \underline{n} − \{g\} ⟶ T , h ↦ a_{h} .

We obtain

∑_{\begin{matrix}A \\ j_{A}^{B}φ=ψ \end{matrix}} μ(B, A) = ∑_{\begin{matrix}A′ \\ j_{A′}^{B′} φ=ψ \end{matrix}} μ(B′, A′) ∑_{a_{g}∈[b_{g−1},b_{g}]_{T}} μ(b_{g−1}, a_{g}) .

But the inner sum on the right is zero by definition of the Möbius function (and because b_{g−1} < b_{g}). Therefore the coefficient of every ψ is zero, hence j^{B}φ = 0.

(b) This follows from (a).

For any subset Y of [n], define

ρ_{Y} : \underline{n} ⟶ \underline{n} , ρ_{Y} (h) = \begin{cases}0 & if h = 0 , \\ h & if h ∈ Y , \\ h − 1 & if h ∉ Y . \end{cases}

10.2. Proposition. Let T be a finite lattice and let B be a strictly increasing n-tuple in T − \{ˆ1\}. Let π^{B} : T → \underline{n} be the corresponding surjective morphism and j^{B} : \underline{n} → T, constructed as above. Let p : F\underline{n} → F\underline{n}/H\underline{n} be the canonical map.

(a) The composite of \overline{F}_{j^{B}} : F\underline{n}/H\underline{n} → F_{T} and p F_{π^{B}} : F_{T} → F\underline{n}/H\underline{n} is the identity morphism of F\underline{n}/H\underline{n}.

(b) \overline{F}_{j^{B}} : F\underline{n}/H\underline{n} → F_{T} is injective and embeds F\underline{n}/H\underline{n} as a direct summand of F_{T}.

(c) F_{j^{B}}F_{π^{B}} is an idempotent endomorphism of F_{T} whose image is isomorphic to F\underline{n}/H\underline{n}.

(d) j^{B}π^{B} is an idempotent endomorphism of T.

(e) π^{B}j^{B} = (−1)^{n} ∑_{∅⊆Y ⊆[n]}(−1)^{|Y |}ρ_{Y}.

Proof : (a) It is clear that

π^{B}j_{A}^{B}(h) = \begin{cases}0 & if h = 0 , \\ h & if j_{A}^{B}(h) ∈]b_{h−1}, b_{h}]_{T} , \\ h − 1 & if j_{A}^{B}(h) = b_{h−1} . \end{cases}

Therefore

π^{B}j^{B} = (−1)^{n} ∑_{∅⊆Y ⊆[n]} ( ∑_{\begin{matrix}A \\ π^{B}j_{A}^{B}=ρ_{Y} \end{matrix}} μ(B, A))ρ_{Y} .

If h ∈ Y, then j_{A}^{B}(h) runs freely over ]b_{h−1}, b_{h}]_{T}. If h ∉ Y, then j_{A}^{B}(h) = b_{h−1} is fixed. It follows that

∑_{\begin{matrix}A \\ π^{B}j_{A}^{B}=ρ_{Y} \end{matrix}} μ(B, A) = ∏_{h∈Y} ( ∑_{a_{h}∈]b_{h−1},b_{h}]_{T}} μ(b_{h−1}, a_{h})) = ∏_{h∈Y} (−1) = (−1)^{|Y |} ,

using the fact that

0 = ∑_{a_{h}∈ [b_{h−1},b_{h}]_{T}} μ(b_{h−1}, a_{h}) = 1 + ∑_{a_{h}∈]b_{h−1},b_{h}]_{T}} μ(b_{h−1}, a_{h}) .

This shows that

π^{B}j^{B} = (−1)^{n} ∑_{∅⊆Y ⊆[n]}(−1)^{|Y |}ρ_{Y} .

Suppose now that Y is a proper subset of [n] and let g ∈ [n] be maximal such that g ∉ Y. If h > g, then ρ_{Y} (h) = h > g, while if h ≤ g, then ρ_{Y} (h) ≤ g − 1. Therefore g ∉ ρ_{Y} (\underline{n}). This shows that [n] ⊈ ρ_{Y} (\underline{n}), for any proper subset Y of [n].

Now for any function φ : X → \underline{n},

F_{π^{B}}F_{j^{B}}(φ) = π^{B}j^{B}φ = (−1)^{n} ∑_{∅⊆Y ⊆[n]}(−1)^{|Y |}ρ_{Y} φ = φ + ∑_{Y ≠[n]}(−1)^{n+|Y |}ρ_{Y} φ .

But [n] ⊈ ρ_{Y} (\underline{n}) if Y ≠ [n], hence [n] ⊈ ρ_{Y} φ(X). In other words, ρ_{Y} φ ∈ H\underline{n}(X), so that F_{π^{B}}F_{j^{B}}(φ) = φ (mod H\underline{n}(X)). Composing with the canonical map p : F\underline{n}(X) → F\underline{n}(X)/H\underline{n}(X) and writing p(φ) = \overline{φ}, we obtain

pF_{π^{B}} \overline{F}_{j^{B}}(\overline{φ}) = \overline{φ} ,

as was to be shown.

(b) This follows immediately from (a).

(c) This follows immediately from (a) and the obvious equality \overline{F}_{j^{B}}p F_{π^{B}} = F_{j^{B}}F_{π^{B}}.

(d) Since the functor F_{?} : kL → F_{k} is fully faithful by Theorem 4.8, j^{B}π^{B} must be an idempotent because its image F_{j^{B}}F_{π^{B}} is an idempotent. Alternatively, it is not difficult to compute directly

j^{B}π^{B}j^{B} = j^{B}(−1)^{n} ∑_{∅⊆Y ⊆[n]}(−1)^{|Y |}ρ_{Y} = j^{B} id\underline{n} +(−1)^{n} ∑_{Y ≠[n]}(−1)^{|Y |}j^{B}ρ_{Y} = j^{B} ,

because [n] ⊈ ρ_{Y} (\underline{n}) if Y ≠ [n], hence j^{B}ρ_{Y} = 0 by Proposition 10.1.

(e) The required equality has already been proved above.

Our aim is to show that the idempotents j^{B}π^{B} are orthogonal. In order to understand the product of two idempotents j^{C}π^{C} and j^{B}π^{B} we need to have more information about π^{C}j^{B}. This is the purpose of the next two propositions.

10.3. Proposition. Let T be a finite lattice and let

B = (b_{0}, b_{1}, …, b_{n−1})  with ˆ0 ≤ b_{0} < b_{1} < … < b_{n−1} < b_{n} = ˆ1 ,
C = (c_{0}, c_{1}, …, c_{m−1})  with ˆ0 ≤ c_{0} < c_{1} < … < c_{m−1} < c_{m} = ˆ1 .

If π^{C}j^{B} ≠ 0, then the restriction of π^{C} to the subset {b_{0}, b_{1}, …, b_{n−1}, b_{n}} is injective. In particular, n ≤ m.

Proof : We have

π^{C}j^{B} = (−1)^{n} ∑_{A} μ(B, A)π^{C}j_{A}^{B} = (−1)^{n} ∑_{ψ:\underline{n}→\underline{m}} ( ∑_{\begin{matrix}A \\ π^{C}j_{A}^{B}=ψ \end{matrix}} μ(B, A))ψ .

Now fix some morphism ψ : \underline{n} → \underline{m} and, for every h ∈ [n], define

U_{h} = {a ∈ [b_{h−1}, b_{h}]_{T} | π^{C}(a) = ψ(h)} ⊆ [b_{h−1}, b_{h}]_{T} .

Then the condition π^{C}j_{A}^{B} = ψ is equivalent to j_{A}^{B}(h) ∈ U_{h} for every h ∈ [n], that is, a_{h} ∈ U_{h} for every h ∈ [n]. In particular U_{h} ≠ ∅ in that case. Since all elements of U_{h} have the same image under π^{C}, so has their join and therefore U_{h} has a supremum

u_{h} = sup(U_{h}) ∈ U_{h} .

Define

V_{h} = {a ∈ [b_{h−1}, b_{h}]_{T} | π^{C}(a) ≤ ψ(h) − 1} .

Then we clearly have [b_{h−1}, u_{h}]_{T} = V_{h} ⊔ U_{h} because any a ≤ u_{h} satisfies either π^{C}(a) ≤ ψ(h) − 1 or π^{C}(a) = ψ(h). If V_{h} is nonempty (i.e. if b_{h−1} ∉ U_{h}), then V_{h} has a supremum v_{h} = ⋁_{t∈V_{h}} t and V_{h} = [b_{h−1}, v_{h}]_{T}.

Now, in the expression above for π^{C}j^{B}, the coefficient of ψ is (−1)^{n}z_{ψ}, where

z_{ψ} = ∑_{π^{C}j_{A}^{B}=ψ} μ(B, A) = ( ∑_{a_{1}∈U_{1}} μ(b_{0}, a_{1}))( ∑_{a_{2}∈U_{2}} μ(b_{1}, a_{2})) … ( ∑_{a_{n}∈U_{n}} μ(b_{n−1}, a_{n}))

and we examine one inner sum ∑_{a∈U_{h}} μ(b_{h−1}, a). We have already observed that U_{h} ≠ ∅. Now there are two cases :

(a) If V_{h} = ∅, then U_{h} = [b_{h−1}, u_{h}]_{T} and ∑_{a∈U_{h}} μ(b_{h−1}, a) = 0, unless b_{h−1} = u_{h}.

(b) If V_{h} ≠ ∅, then we claim that ∑_{a∈U_{h}} μ(b_{h−1}, a) = 0, unless b_{h−1} = v_{h}.

To prove the claim, suppose that b_{h−1} < v_{h} (hence b_{h−1} < u_{h} because v_{h} < u_{h}). Then we have [b_{h−1}, u_{h}]_{T} = [b_{h−1}, v_{h}]_{T} ⊔ U_{h} and therefore

∑_{a∈U_{h}} μ(b_{h−1}, a) = ∑_{a∈[b_{h−1},u_{h}]_{T}} μ(b_{h−1}, a) − ∑_{a∈[b_{h−1},v_{h}]_{T}} μ(b_{h−1}, a) = 0 ,

because b_{h−1} < u_{h} and b_{h−1} < v_{h}.

It follows that the sum ∑_{a∈U_{h}} μ(b_{h−1}, a) is nonzero in only two cases :

(a) U_{h} = {b_{h−1}}. Then ∑_{a∈U_{h}} μ(b_{h−1}, a) = μ(b_{h−1}, b_{h−1}) = 1.

(b) V_{h} = {b_{h−1}}. Then U_{h} =]b_{h−1}, u_{h}]_{T} and ∑_{a∈U_{h}} μ(b_{h−1}, a) = −1 because

\begin{aligned}0 = ∑_{a∈[b_{h−1},u_{h}]_{T}} μ(b_{h−1}, a) & = & μ(b_{h−1}, b_{h−1}) + ∑_{a∈]b_{h−1},u_{h}]_{T}} μ(b_{h−1}, a) \\ & = & 1 + ∑_{a∈U_{h}} μ(b_{h−1}, a) . \end{aligned}

If the coefficient z_{ψ} is nonzero, then this sum must be nonzero for every h ∈ [n] and we obtain z_{ψ} = (−1)^{p} where p is the number of times the integer h satisfies case (b) (and then the coefficient of ψ in the expression for π^{C}j^{B} is (−1)^{n+p}).

Now suppose that the restriction of π^{C} to {b_{0}, b_{1}, …, b_{n−1}, b_{n}} is not injective. We want to prove that the coefficient z_{ψ} is zero (for our fixed morphism ψ : \underline{n} → \underline{m}). Since π^{C} is order-preserving and not injective on the subset above, there exists h ∈ [n] such that π^{C}(b_{h−1}) = π^{C}(b_{h}). We know that z_{ψ} = 0 if U_{h} is empty. Thus we can assume that U_{h} contains an element a. Since b_{h−1} ≤ a ≤ b_{h}, we have π^{C}(b_{h−1}) ≤ π^{C}(a) ≤ π^{C}(b_{h}), hence equality π^{C}(b_{h−1}) = π^{C}(a) = π^{C}(b_{h}). This means that both b_{h−1} and b_{h} belong to U_{h} (because π^{C}(a) = ψ(h) by definition of U_{h}). Now in both cases (a) and (b) above, the set U_{h} does not contain simultaneously b_{h−1} and b_{h} (it is either {b_{h−1}} or ]b_{h−1}, u_{h}]_{T}). This shows that the sum ∑_{a∈U_{h}} μ(b_{h−1}, a) cannot be nonzero. Therefore z_{ψ} = 0.

This argument holds for every morphism ψ : \underline{n} → \underline{m} and so every coefficient z_{ψ} is zero. It follows that π^{C}j^{B} = 0, as was to be proved.

10.4. Proposition. Let T be a finite lattice and let

B = (b_{0}, b_{1}, …, b_{n−1})  with ˆ0 ≤ b_{0} < b_{1} < … < b_{n−1} < b_{n} = ˆ1 ,
C = (c_{0}, c_{1}, …, c_{m−1})  with ˆ0 ≤ c_{0} < c_{1} < … < c_{m−1} < c_{m} = ˆ1 ,
D = (d_{0}, d_{1}, …, d_{m−1})  with ˆ0 ≤ d_{0} < d_{1} < … < d_{m−1} < d_{m} = ˆ1 .

Then

j^{D}π^{C}j^{B} = \begin{cases}j^{D} & if C = B , \\ 0 & if C ≠ B . \end{cases}

Proof : We assume that j^{D}π^{C}j^{B} ≠ 0, and in particular π^{C}j^{B} ≠ 0. Write first

π^{C}j^{B} = ∑_{A} μ(B, A)π^{C}j_{A}^{B}

where A = (a_{1}, a_{2}, …, a_{n}) and a_{h} ∈ [b_{h−1}, b_{h}] for every h ∈ [n]. Let ψ : \underline{n} → \underline{m} be a map appearing with a nonzero coefficient in the expression of π^{C}j^{B} ≠ 0 as a linear combination of morphisms in the category L. Let A be such that π^{C}j_{A}^{B} = ψ. Since j^{D}π^{C}j^{B} ≠ 0, we can also assume that ψ is such that j^{D}ψ ≠ 0. Proposition 10.1 implies that the function ψ : \underline{n} → \underline{m} must satisfy [m] ⊆ ψ(\underline{n}). Since ψ is a morphism of lattices, we also have ψ(0) = 0. Therefore ψ : \underline{n} → \underline{m} must be surjective. In particular n ≥ m.

By Proposition 10.3, π^{C}j^{B} ≠ 0 implies that n ≤ m. Therefore n = m. Since the map π^{C}j_{A}^{B} : \underline{n} → \underline{n} is order-preserving and surjective, it must be the identity map. This shows that whenever A is such that j^{D}π^{C}j_{A}^{B} ≠ 0, then π^{C}j_{A}^{B} = id\underline{n}. Therefore, the functions ψ which appear with a nonzero coefficient in the expression of π^{C}j^{B} are id\underline{n} and maps in the kernel of j^{D}.

Now Proposition 10.3 also asserts that the restriction of π^{C}

π^{C} : \{b_{0}, b_{1}, …, b_{n−1}, b_{n}\} ⟶ \underline{m}

is injective. Since n = m, this must be a bijection, and since π^{C} is order-preserving, we must have π^{C}(b_{h}) = h for every h ∈ \underline{n}. As in the proof of Proposition 10.3, associated with the map id\underline{n} : \underline{n} → \underline{n}, there is a corresponding subset

U_{h} = \{a ∈ [b_{h−1}, b_{h}]_{T} | π^{C}(a) = id\underline{n}(h)\} = \{a ∈ [b_{h−1}, b_{h}]_{T} | π^{C}(a) = h\} .

Thus we have b_{h} ∈ U_{h} for every h ∈ \underline{n}.

Since id\underline{n} appears with a nonzero coefficient (−1)^{n}z_{id} in the expression of π^{C}j^{B}, the proof of Proposition 10.3 shows that ∑_{a∈U_{h}} μ(b_{h−1}, a) is nonzero. Therefore we must be in one of the two cases (a) or (b) of that proof. But the first case cannot hold because b_{h} ∉ U_{h} = {b_{h−1}}. So we are in case (b) and we have b_{h} ∈ U_{h} =]b_{h−1}, u_{h}]_{T} ⊆]b_{h−1}, b_{h}]_{T}. This forces u_{h} = b_{h} and U_{h} =]b_{h−1}, b_{h}]_{T}.

The equality U_{h} =]b_{h−1}, b_{h}]_{T} means that π^{C} coincides with π^{B} on this interval. This holds for every h ∈ [n]. We now show, by descending induction, that c_{h} = b_{h} for every h ∈ [n]. First c_{n} = ˆ1 = b_{n}. Now since π^{C}(b_{h}) = h, we have b_{h} ≤ c_{h}, but b_{h} ≰ c_{h−1}, for every h ∈ \underline{n}. Assume that c_{h} = b_{h} for every h ≥ i and suppose for contradiction that b_{i−1} < c_{i−1}. Then b_{i−1} < c_{i−1} < c_{i} = b_{i}, hence c_{i−1} ∈]b_{i−1}, b_{i}], which implies that π^{B}(c_{i−1}) = i. But π^{B} coincides with π^{C} on the interval, so π^{C}(c_{i−1}) = i, which contradicts the definition of π^{C}.

We have now proved that n = m and C = B whenever j^{D}π^{C}j^{B} ≠ 0. Moreover, in that case, we have seen that π^{B}j^{B} is the sum of a multiple of id\underline{n} and morphisms in the kernel of j^{D}. The proof of Proposition 10.3 shows that the coefficient of id\underline{n} is equal to (−1)^{n+p}, where p is the number of times the integer h satisfies case (b). But we have noticed that case (a) cannot occur, so p = n and the coefficient is 1. Thus π^{B}j^{B} = id\underline{n} (mod Ker(j^{D})) and it follows that j^{D}π^{B}j^{B} = j^{D}.

Alternatively, the equality π^{B}j^{B} = id\underline{n} (mod Ker(j^{D})) also follows from Proposition 10.1 and Proposition 10.2, because π^{B}j^{B} = (−1)^{n} ∑_{∅⊆Y ⊆[n]}(−1)^{|Y |}ρ_{Y} and j^{D}ρ_{Y} = 0 whenever Y ≠ [n] since [n] ⊈ ρ_{Y} (\underline{n}).

Given a finite lattice T, let P_{T,n} be the set of all strictly increasing n-tuples (d_{0}, d_{1}, …, d_{n−1}) in T − {ˆ1}, that is, such that ˆ0 ≤ d_{0} < d_{1} < … < d_{n−1} < d_{n} = ˆ1. The set P_{T,n} corresponds bijectively to the set of all surjective morphisms T → \underline{n}. If D, C ∈ P_{T,n}, we define

f_{D,C} = j^{D}π^{C} : T ⟶ T .

In particular, f_{B,B} = j^{B}π^{B} is the idempotent of Proposition 10.1.

10.5. Corollary. Let T be a finite lattice.

(a) Let A, B ∈ P_{T,n} and D, C ∈ P_{T,m}, where n, m ≥ 0 are two integers. Then

f_{D,C}f_{B,A} = \begin{cases}f_{D,A} & if C = B (hence n = m) , \\ 0 & if C ≠ B . \end{cases}

(b) When n ≥ 0 varies and B ∈ P_{T,n} varies, the idempotents f_{B,B} are pairwise orthogonal.

Proof : (a) By Proposition 10.4, we have

f_{D,C}f_{B,A} = j^{D}π^{C}j^{B}π^{A} = j^{D}π^{A}

if C = B, and zero otherwise.

(b) This follows from (a).

Let M_{|P_{T,n}|}(k) denote the matrix algebra of size |P_{T,n}|, with rows and columns indexed by the set P_{T,n}. If D, C ∈ P_{T,n}, we let m_{D,C} denote the matrix with coefficient 1 in position (D, C) and 0 elsewhere.

10.6. Theorem. Let T be a finite lattice, let P_{T,n} denote the set of all strictly increasing n-tuples in T −{ˆ1}, and let N be the maximal length of a strictly increasing sequence in T − {ˆ1}.

(a) The map

I_{T} : ⨁_{n=0}^{N} M_{|P_{T,n}|}(k) ⟶ End_{kL}(T) , m_{D,C} ↦ f_{D,C}

is an algebra homomorphism (without unit elements).

(b) I_{T} is injective.

(c) The image of I_{T} is equal to the subalgebra E_{T} (without unit element) of End_{kL}(T) having a k-basis consisting of all maps whose image is totally ordered (and which are morphisms in the category L).

Proof : (a) If |C| ≠ |B|, then m_{D,C} and m_{B,A} are not in the same block, so their product is 0, while the product f_{D,C}f_{B,A} is also zero. If |C| = |B|, then the relations of Corollary 10.5 are the standard relations within a matrix algebra of size |P_{T,n}|.

(b) Since the elements m_{D,C} form a k-basis of ⨁_{n=0}^{N} M_{|P_{T,n}|}(k), it suffices to prove that their images f_{D,C} are k-linearly independent. Suppose that

∑_{n=0}^{N} ∑_{D,C∈P_{T,n}} λ_{D,C}f_{D,C} = 0 ,

where λ_{D,C} ∈ k. Multiply on the left by the idempotent f_{A,A} and on the right by the idempotent f_{B,B}. By Corollary 10.5, we obtain λ_{A,B}f_{A,B} = 0. Since f_{A,B} is a linear combination of distinct maps T → T, one of them appearing with coefficient ±1, we must have λ_{A,B} = 0.

(c) It is clear that E_{T} is a subalgebra. Moreover, every map j_{A}^{D}π^{C} has an image which is totally ordered, by construction. Therefore f_{D,C} = j^{D}π^{C} ∈ E_{T} and hence Im(I_{T}) ⊆ E_{T}.

Now E_{T} has a k-basis consisting of all maps φ_{U,V} : T → T described as follows. First U = (u_{0}, u_{1}, …, u_{n−1}) ∈ P_{T,n}, while V ∈ Y_{T,n}, where Y_{T,n} denotes the set of all strictly increasing n-tuples V = (v_{1}, v_{2}, …, v_{n}) in T − {ˆ0} (that is ˆ0 = v_{0} < v_{1} < v_{2} < … < v_{n} ≤ ˆ1). Define λ^{V} : \underline{n} → T by λ^{V} (i) = v_{i} for every i ∈ \underline{n} and then set φ_{U,V} = λ^{V} π^{U}. Then

B_{n} = {φ_{U,V} | U ∈ P_{T,n} , V ∈ Y_{T,n}}

is a k-basis of the submodule E_{T,n} generated by all endomorphisms whose image is isomorphic to \underline{n}, while B = ⋃_{n=0}^{N} B_{n} is a k-basis of E_{T} = ⨁_{n=0}^{N} E_{T,n}. We have seen that B′ = ⋃_{n=0}^{N}{f_{D,C} | D, C ∈ P_{T,n}} is a k-basis of Im(I_{T}). It is an easy exercise to show that there is a bijection between Y_{T,n} and P_{T,n}. Therefore B and B′ have the same cardinality. In other words Im(I_{T}) and E_{T} are free k-modules of the same rank.

Now we allow the base ring k to vary and we write an index k to emphasize the dependence on k. Thus we have an injective algebra homomorphism

I_{T,k} : ⨁_{n=0}^{N} M_{|P_{T,n}|}(k) ⟶ E_{T,k} ⊆ End_{kL}(T)

and we let Q_{k} := E_{T,k}/ Im(I_{T,k}), so that we have a short exact sequence

\xymatrix{0 \ar[r] & Im(I_{T,k}) \ar[r]^{j_{k}} & E_{T,k} \ar[r]^{p_{k}} & Q_{k} \ar[r] & 0 ,}

where j_{k} is the inclusion map and p_{k} the canonical surjection. In the case of the ring of integers ℤ, we see that Q_{ℤ} is a finite abelian group, because Im(I_{T,ℤ}) and E_{T,ℤ} are free ℤ-modules of the same rank. Tensoring with the prime field 𝔽_{p}, where p is a prime, we obtain

\xymatrix{0 \ar[r] & Tor(𝔽_{p}, Q_{ℤ}) \ar[r] & 𝔽_{p} ⊗ Im(I_{T,ℤ}) \ar[r]^{1⊗j_{ℤ}} & 𝔽_{p} ⊗ E_{T,ℤ} \ar[r]^{1⊗p_{ℤ}} & 𝔽_{p} ⊗ Q_{ℤ} \ar[r] & 0 .}

Using the canonical bases of Im(I_{T,k}) and E_{T,k} respectively, we see that k ⊗ Im(I_{T,ℤ}) ≅ Im(I_{T,k}), and similarly k ⊗ E_{T,ℤ} ≅ E_{T,k}. Moreover the inclusion map 1⊗j_{k} corresponds, under these isomorphisms, to the inclusion map j_{k}. In particular, we obtain

\xymatrix{0 \ar[r] & Tor(𝔽_{p}, Q_{ℤ}) \ar[r] & Im(I_{𝔽_{p}}) \ar[r]^{j_{𝔽p}} & E_{𝔽_{p}} \ar[r]^{1⊗p_{ℤ}} & 𝔽_{p} ⊗ Q_{ℤ} \ar[r] & 0 .}

By (b), we know that j_{𝔽_{p}} is injective, hence Tor(𝔽_{p}, Q_{ℤ}) = {0}. But since this holds for every prime number p and Q_{ℤ} is finite, we must have Q_{ℤ} = {0}. It follows that the inclusion map j_{ℤ} : Im(I_{T,ℤ}) → E_{T,ℤ} is an isomorphism. Tensoring with k, it follows that the inclusion map j_{k} : Im(I_{T,k}) → E_{T,k} is an isomorphism as well. In other words, Im(I_{T,k}) = E_{T,k}, as was to be shown.

10.7. Remark. Let B be the canonical basis of E_{T} described in the proof. The change of basis from B to the basis {f_{D,C}} of Im(I_{T}) is not obvious. By construction, every map j_{A}^{D}π^{C} belongs to B, but beware of the fact that if C and D are n-tuples, then j_{A}^{D}π^{C} may be a composite T → \underline{m} → T for some m < n, because the n-tuple A = (a_{1}, a_{2}, …, a_{n}) is increasing but not necessarily strictly increasing, hence may consist of m distinct elements for some m < n.

The image under I_{T} of the identity element of ⨁_{n=0}^{N} M_{|P_{T,n}|}(k) is an idempotent e_{T} of End_{kL}(T) and e_{T} is an identity element of E_{T}. We now prove that the we actually get central idempotents.

10.8. Theorem. For every finite lattice T, let E_{T} = Im(I_{T}) be the subalgebra of End_{kL}(T) appearing in Theorem 10.6, and let e_{T} be the identity element of E_{T}.

(a) e_{T} = ∑_{n=0}^{N} ∑_{B∈P_{T,n}} f_{B,B}.

(b) For any finite lattice T′ and any morphism θ ∈ Hom_{kL}(T, T′), we have θe_{T} = e_{T′}θ. In other words, the family of idempotents e_{T}, for T ∈ L, is a natural transformation of the identity functor id_{kL}.

(c) e_{T} is a central idempotent of End_{kL}(T).

(d) The subalgebra E_{T} is a direct product factor of End_{kL}(T), that is, there exists a subalgebra D such that End_{kL}(T) = E_{T} × D (where E_{T} is identified with E_{T} × {0} and D with {0} × D, as usual).

Proof : (a) The identity element of ⨁_{n=0}^{N} M_{|P_{T,n}|}(k) is equal to ∑_{n=0}^{N} ∑_{B∈P_{T,n}} m_{B,B}.

Taking its image under I_{T} yields the required formula.

(b) We have seen in the proof of Theorem 10.6 that every element of the canonical basis of E_{T} has the form λ^{V} π^{U}, where U = (u_{0}, u_{1}, …, u_{n−1}) ∈ P_{T,n} is a strictly increasing n-tuple in T − {ˆ1}, while V = (v_{1}, v_{2}, …, v_{n}) belongs to the set Y_{T,n} of all strictly increasing n-tuples in T − {ˆ0}. We now compute their opposite, as in Lemma 8.1.

From the surjective morphism π^{U} : T → \underline{n}, we obtain (π^{U})^{op} : \underline{n}^{op} → T^{op} defined as follows : for every h ∈ \underline{n},

(π^{U})^{op}(h) = ⋁_{π^{U} (t)≤h} t = sup ((π^{U})^{−1}(h)) = u_{h} ,

in view of the way U is associated to π^{U} (see the beginning of Section 10). But u_{h} = λ^{U^{op}}(h), hence (π^{U})^{op} = λ^{U^{op}}.

Applying this to the case U = V^{op}, where V ∈ Y_{T,n}, we obtain (π^{V^{op}})^{op} = λ^{(V^{op})^{op}} = λ^{V}. Taking opposites yields (λ^{V})^{op} = (π^{V^{op}}).

It follows that the opposite of the canonical basis element λ^{V} π^{U} of E_{T} is the canonical basis element λ^{U^{op}}π^{V^{op}} of E_{T^{op}}. Therefore, the opposite of the identity element e_{T} of E_{T} must belong to E_{T^{op}}. Moreover, it must be the identity element of E_{T^{op}}, because taking opposites behaves well with respect to composition, by Lemma 8.1. Therefore (e_{T})^{op} = e_{T^{op}}.

Now if θ : T → T′ is a morphism in L, then the image of a totally ordered subset of T is a totally ordered subset of T′. It follows that composition with θ maps E_{T} to linear combination of maps with a totally ordered image, hence invariant under the idempotent element e_{T′}. In other words, we have

θ e_{T} = e_{T′} θ e_{T} .

Applying this equation to T′^{op}, T^{op}, and the morphism θ^{op} : T′^{op} → T^{op}, we obtain θ^{op} e_{T′}^{op} = e_{T^{op}} θ^{op} e_{T′}^{op}. Passing to opposites and using the above equality (e_{T})^{op} = e_{T^{op}}, we get

e_{T′} θ e_{T} = e_{T′} θ .

The two displayed equations then yield θe_{T} = e_{T′}θ. This holds as well if θ is replaced by a linear combination of morphisms, as was to be shown.

(c) This is a special case of (b).

(d) This follows immediately from (c).

11. The case of a total order

In this section, we consider the case of a totally ordered lattice \underline{n}, where n ∈ ℕ. We determine completely the ring of endomorphisms of \underline{n} in the category kL and we deduce a direct sum decomposition of F\underline{n}. For the lattice \underline{n}, the set P_{\underline{n},r} of all strictly increasing r-tuples (d_{0}, d_{1}, …, d_{r−1}) in \underline{n} − \{n\} is just the set of all subsets of {0, 1, …, n − 1} of size r, because any such subset is totally ordered. Throughout this section, we use subsets of size r instead of strictly increasing rtuples. In particular |P_{\underline{n},r}| = \binom{n}{r}.

11.1. Theorem. Let n ∈ ℕ.

(a) The homomorphism of k-algebras of Theorem 10.6

I\underline{n} : ⨁_{m=0}^{n} M_{|P_{\underline{n},r}|}(k) ⟶ End_{kL}(\underline{n}) , m_{D,C} ↦ f_{D,C} ,

is an isomorphism.

(b) In particular, if k is a field, then End_{kL}(\underline{n}) is semi-simple.

Proof : First note that (b) follows from (a), because any matrix algebra M_{q}(k) is simple, so that the direct sum is semi-simple. Since any map φ : \underline{n} → \underline{n} has an image which is totally ordered, the subalgebra E_{T} of End_{kL}(\underline{n}) appearing in Theorem 10.6 is the whole of End_{kL}(\underline{n}). By Theorem 10.6, the morphism I\underline{n} is surjective and injective, hence an isomorphism.

11.2. Remark. Since E_{T} is the whole of End_{kL}(\underline{n}), the canonical basis B of E_{T} is a basis of End_{kL}(\underline{n}), which also has another basis consisting of the morphisms f_{D,C}. As mentioned in Remark 10.7, the change of basis is not straightforward, but it can be made more explicit in the case of a totally ordered lattice considered here.

Every element of the canonical basis of E_{T} has the form λ^{V} π^{U}, where U = (u_{0}, u_{1}, …, u_{n−1}) ∈ P_{T,n} is a strictly increasing n-tuple in T − {ˆ1}, while V = (v_{1}, v_{2}, …, v_{n}) ∈ Y_{T,n} is a strictly increasing n-tuple in T − {ˆ0}. On the other hand, the morphisms f_{D,C} are parametrized by pairs D, C ∈ P_{T,n}. But we have an obvious bijection from P_{T,n} to Y_{T,n}, mapping an n-tuple (u_{0}, u_{1}, …, u_{n−1}) ∈ P_{T,n} to (u_{0} + 1, u_{1} + 1, …, u_{n−1} + 1) ∈ Y_{T,n}. Thus we can parametrize both bases by the same set P_{T,n} × P_{T,n}. Then it is not hard to see that the matrix of the change of basis is unitriangular. Actually, this provides another proof of the fact that the map I\underline{n} is an isomorphism.

11.3. Remark. Theorem 11.1 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, because it turns out that they do not have the same number of idempotents, even when n = 2. Only the corresponding monoid algebras become isomorphic (over ℂ).

Now we consider the central idempotents of End_{kL}(\underline{n}) corresponding to the above decomposition into matrix algebras.

11.4. Notation. For an integer m with 0 ≤ m ≤ n, set

β_{n,m} := ∑_{B∈P_{\underline{n},m}} f_{B,B} .

In particular, for m = n and B = \{0, 1, …, n−1\}, we have π^{B} = id\underline{n} and we define

ε_{n} := β_{n,n} = f_{B,B} = j^{B}π^{B} = j^{B} = (−1)^{n} ∑_{∅⊆Y ⊆[n]}(−1)^{|Y |}ρ_{Y} ,

where, as before, ρ_{Y} ∈ End_{L}(\underline{n}) is defined by ρ_{Y} (h) = h if h ∈ Y and ρ_{Y} (h) = h−1 otherwise.

11.5. Proposition. The elements β_{n,m}, for 0 ≤ m ≤ n, are orthogonal central idempotents of End_{kL}(\underline{n}), and their sum is equal to the identity. In particular, the central idempotent ε_{n} satisfies

ε_{n} End_{kL}(\underline{n}) = kε_{n} .

Proof : For B ∈ P_{\underline{n},m}, the inverse image of f_{B,B} under the algebra isomorphism I of Theorem 11.1 is the matrix m_{B,B} of the component M_{|P_{\underline{n},m}|}(k) indexed by m. Summing over all B ∈ P_{\underline{n},m}, it follows that the inverse image of β_{n,m} under I is the identity element of M_{|P_{\underline{n},m}|}(k). The first statement follows.

In the case m = n, the set P_{\underline{n},m} consists of the singleton B = \{0, 1, …, n − 1\} and the corresponding matrix algebra has size 1. We see that the inverse image of ε_{n} under I is the identity element m_{B,B} of the component M_{1}(k) = k. Clearly ε_{n} End_{kL}(\underline{n}) ≅ M_{1}(k) is a free k-module of rank 1, hence ε_{n} End_{kL}(\underline{n}) = kε_{n}.

We want to use the functor F_{?} : kL → F_{k} of Section 4 to deduce information on the correspondence functor F\underline{n}. By Theorem 4.12, we already know that F\underline{n} is projective, because the total order \underline{n} is a distributive lattice. If n = 0 (in which case [n] = ∅) and if n = 1 (in which case [n] = {1}), we recover the cases already considered in Section 5 of [BT2]. Our purpose is to treat now the general case.

We apply the functor F_{?} : kL → F_{k} to the map j^{B} ∈ Hom_{kL}(\underline{m}, \underline{n}) defined in Section 10, where B ∈ P_{\underline{n},m}. By Proposition 10.1 we obtain a morphism

F_{j^{B}} : F\underline{m} ⟶ F\underline{n}

which vanishes on H\underline{m}. By Proposition 10.2, this induces an injective morphism

\overline{F}_{j^{B}} : F\underline{m}/H\underline{m} ⟶ F\underline{n}

which embeds F\underline{m}/H\underline{m} as a direct summand of F\underline{n}, corresponding to the idempotent f_{B,B} = j^{B}π^{B}. In particular, for m = n, we have f_{B,B} = j^{B} = ε_{n} and we obtain an idempotent endomorphism F_{ε_{n}} of F\underline{n} with kernel H\underline{n}.

11.6. Theorem. Let n ∈ ℕ and let 𝕊_{n} = F\underline{n}/H\underline{n}. There are isomorphisms of correspondence functors

F_{ε_{n}}F\underline{n} ≅ 𝕊_{n} ,
F_{β_{n,m}}F\underline{n} ≅ 𝕊_{m}^{⊕\binom{n}{m}} , for each 0 ≤ m ≤ n ,
F\underline{n} ≅ ⨁_{\begin{matrix}0≤m≤n \\ B∈P_{\underline{n},m} \end{matrix}} 𝕊_{|B|} .

Proof : By Theorem 4.8, the functor F_{?} induces an isomorphism of k-algebras

End_{kL}(\underline{n}) ≅ End_{F_{k}}(F\underline{n}) .

Now the idempotents f_{B,B} of End_{kL}(\underline{n}), for B ∈ P_{\underline{n},m} and 0 ≤ m ≤ n, are orthogonal and their sum is equal to the identity, by Theorem 11.1. It follows that the endomorphisms F_{f_{B,B}} of F\underline{n} are orthogonal idempotents, and their sum is the identity. Hence we obtain a decomposition of correspondence functors

F\underline{n} = ⨁_{\begin{matrix}0≤m≤n \\ B∈P_{\underline{n},m} \end{matrix}} F_{f_{B,B}}(F\underline{n}) .

By surjectivity of π^{B} : \underline{n} → \underline{m}, the image of F_{f_{B,B}} = F_{j^{B}}F_{π^{B}} : F\underline{n} → F\underline{n} is equal to the image of F_{j^{B}} : F\underline{m} → F\underline{n}. Therefore F_{f_{B,B}}(F\underline{n}) = F_{j^{B}}(F\underline{m}). By Proposition 10.2, the image F_{j^{B}}(F\underline{m}) is isomorphic to 𝕊_{m} = F\underline{m}/H\underline{m} and it follows that

F_{f_{B,B}}(F\underline{n}) ≅ 𝕊_{m} .

Taking m = n and f_{B,B} = j^{B} = ϵ_{n}, we obtain the first isomorphism F_{ε_{n}}F\underline{n} ≅ 𝕊_{n}. Summing over all B ∈ P_{\underline{n},m} for a fixed m, we obtain the second isomorphism, because |P_{\underline{n},m}| = \binom{n}{m}. Finally, summing over all 0 ≤ m ≤ n and all B ∈ P_{\underline{n},m}, we obtain the third isomorphism.

11.7. 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 Proposition 11.5. Now for integers l, m ∈ {0, …, n}, we use the central idempotents β_{n,l} of Proposition 11.5 and we obtain

Hom_{F_{k}}(F_{β_{n,l}}F\underline{n}, F_{β_{n,m}}F\underline{n}) ≅ Hom_{F_{k}}(𝕊_{l}, 𝕊_{m})^{⊕\binom{n}{l}\binom{n}{m}} .

Since F_{β_{n,l}} and F_{β_{n,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_{n,l}}F\underline{n}, F_{β_{n,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 ranks of all its evaluations.

11.8. Theorem. Let 𝕊_{n} = F\underline{n}/H\underline{n}.

(a) 𝕊_{n} is isomorphic to the fundamental functor 𝕊_{[n],tot}, where tot denotes the total order on [n].

(b) 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) We are going to use the results of Section 8 applied to the lattice T = \underline{n}^{op}. The set of its irreducible elements is

E = {0, 1, …, n−1} ,

with a total order tot^{op} 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 9.2 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

(11.9)  F_{ε_{n}}(ω) = (−1)^{n}γ\underline{n}^{op} .

The definition of ε_{n} yields

F_{ε_{n}}(ω) = (−1)^{n} ∑_{Y ⊆[n]}(−1)^{|C|}ρ_{Y} ω

and the definition of ρ_{Y} gives

(ρ_{Y} ω)(e) = ρ_{Y} (e + 1) = \{ \begin{matrix}e + 1 & if e + 1 ∈ Y , \\ e & if e + 1 ∉ Y . \end{matrix}

Setting Y = A + 1 for each A ⊆ E, we see that ρ_{Y} ω = η_{A}^{∘} and it follows that

F_{ε_{n}}(ω) = (−1)^{n} ∑_{Y ⊆[n]}(−1)^{|Y |}ρ_{Y} ω
= (−1)^{n} ∑_{A⊆E}(−1)^{|A|}η_{A}^{∘}
= (−1)^{n}γ\underline{n}^{op} .

This proves Claim 11.9 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 11.9 above, F_{ε_{n}}F\underline{n} is generated by γ\underline{n}^{op} ∈ F\underline{n}(E). Now Theorem 9.5 asserts that the subfunctor of F\underline{n} generated by γ_{T} = γ\underline{n}^{op} is isomorphic to 𝕊_{E,tot^{op}}. But (E, tot^{op}) ≅ ([n], tot) via the map e ↦ n − e. Therefore, using the isomorphism of Theorem 11.6, we obtain

𝕊_{n} ≅ F_{ε_{n}}F\underline{n} = ⟨γ\underline{n}^{op}⟩ ≅ 𝕊_{E,tot^{op}} ≅ 𝕊_{[n],tot} .

(b) The canonical k-basis of the k-module 𝕊_{n}(X) = F\underline{n}(X)/H\underline{n}(X) is the set Z_{n}(X) 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)|. The number of maps in Z_{n}(X) has been computed in Lemma 9.1 of [BT2] and the formula is actually well-known. The formula 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.

11.10. Remark. We shall see in [BT3] 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 11.7 holds more generally for fundamental functors and the general case will be proved in [BT3].

When k is a field, we get even more. 11.11. 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. Recall from Section 2 that the fundamental functor 𝕊_{E,R} and the simple functor S_{E,R,V} are given by

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

where P_{E}f_{R} is the fundamental module corresponding to (E, R) and T_{R,V} = P_{E}f_{R} ⊗_{k Aut(E,R)} V. Here E = [n], R = tot, k Aut(E, R) = k, and V = k, hence

T_{tot,k} = P_{[n]}f_{tot} ⊗_{k} k ≅ P_{[n]}f_{tot} .

Therefore 𝕊_{[n],tot} ≅ S_{[n],tot,k}.

(b) Since \underline{n} is a distributive lattice, F\underline{n} is projective and injective by Corollary 8.11. Therefore so is its direct summand 𝕊_{n}. It follows that 𝕊_{n} is simple, projective, and injective.

Our last purpose in this section is to find, for any finite lattice T, all the direct summands of F_{T} isomorphic to a functor 𝕊_{n} corresponding to a total order. Recall that e_{T} denotes the central idempotent of End_{kL}(T) which is an identity element for the subalgebra E_{T} (see Theorem 10.8).

11.12. Theorem. Let T be a finite lattice and let N be the maximal length of a strictly increasing sequence in T. For every finite set X, let F_{T}^{tot}(X) be the ksubmodule of F_{T} (X) generated by all the maps φ : X → T such that φ(X) is a totally ordered subset of T.

(a) F_{T}^{tot} is a subfunctor of F_{T}, equal to F_{e_{T}} (F_{T}).

(b) F_{T}^{tot} is a direct summand of F_{T}, isomorphic to

F_{T}^{tot} ≅ ⨁_{\begin{matrix}0≤m≤N \\ B∈P_{T,m} \end{matrix}} 𝕊_{|B|} .

(c) The image of any morphism F\underline{n} → F_{T} is contained in F_{T}^{tot}. In particular, any subfunctor of F_{T} isomorphic to a functor 𝕊_{n} is contained in F_{T}^{tot}.

(d) Hom_{F_{k}} (F_{T}^{tot}, F_{id −e_{T}} (F_{T})) = {0} and Hom_{F_{k}} (F_{id −e_{T}} (F_{T}), F_{T}^{tot}) = {0}.

(e) The splitting of the surjection F_{T} → F_{T}^{tot} is natural in T.

Proof : (a) Let φ ∈ F_{T}^{tot}(X). Writing Im(φ) = {t_{1}, t_{2}, …, t_{n}} in increasing order, we can write φ = jψ, where ψ : X → \underline{n} is the map defined by ψ(x) = i if φ(x) = t_{i}, and where j : \underline{n} → T is the map defined by j(i) = t_{i} for 1 ≤ i ≤ n and j(0) = ˆ0. Clearly j is a morphism in the category L. By Theorem 10.8, we have e_{T} j = je\underline{n} = j, because e\underline{n} is the identity element of End_{kL}(\underline{n}) by Theorem 11.1. Therefore

φ = jψ = e_{T} jψ = e_{T} φ = F_{e_{T}} (φ) ,

proving that φ ∈ F_{e_{T}} (F_{T} (X)).

Conversely, if φ ∈ F_{e_{T}} (F_{T} (X)), then we can write φ = F_{e_{T}} (ψ) = e_{T} ψ for some map ψ : X → T. Since e_{T} is, by construction, a linear combination of maps with a totally ordered image, so is e_{T} ψ, proving that φ ∈ F_{T}^{tot}(X).

This shows that F_{T}^{tot} = F_{e_{T}} (F_{T}) and the latter is a subfunctor of F_{T}.

(b) As in the proof of Theorem 11.6, we apply the fully faithful functor kL → F_{k} defined by T ↦ F_{T}. There is direct sum decomposition of functors

F_{T} = F_{e_{T}} (F_{T}) ⊕ F_{id −e_{T}} (F_{T}) = F_{T}^{tot} ⊕ F_{id −e_{T}} (F_{T}) .

The idempotent e_{T} is the sum of the orthogonal idempotents f_{B,B} of End_{kL}(T), for B ∈ P_{T,m} and 0 ≤ m ≤ N. It follows that the endomorphisms F_{f_{B,B}} of F_{T} are orthogonal idempotents with sum F_{e_{T}}. Hence we obtain a direct sum decomposition of correspondence functors

F_{T}^{tot} = F_{e_{T}} (F_{T}) = ⨁_{\begin{matrix}0≤m≤n \\ B∈P_{T,m} \end{matrix}} F_{f_{B,B}}(F_{T} ) .

By Proposition 10.2, the image of F_{f_{B,B}} is isomorphic to F\underline{m}/H\underline{m} = 𝕊_{m}, where m = |B|, proving the result.

(c) Let α : F\underline{n} → F_{T} be a morphism of correspondence functors. Since the functor T ↦ F_{T} is full, α is the image of a morphism \underline{n} → T in kL, which is in turn a linear combination of order-preserving maps f : \underline{n} → T. For such a map f and for any function φ : X → \underline{n}, the image of fφ is a totally ordered subset of T. It follows that the image of the map F_{f} : F\underline{n}(X) → F_{T} (X) is contained in F_{T}^{tot}(X).

The special case follows from the fact that 𝕊_{n} is a quotient of F\underline{n}, by Theorem 11.6.

(d) The first statement is a consequence of (b) and (c), while the second one follows from a dual argument. Details are left to the reader.

(e) By Theorem 10.8, the family of idempotents e_{T}, for T ∈ L, is a natural transformation of the identity functor id_{kL}. Therefore the family of idempotents F_{e_{T}}, for T ∈ L, is a natural transformation of the identity functor id_{F_{k}}.

References

[Bo1] S. Bouc. Foncteurs d’ensembles munis d’une double action, J. Algebra 183 (1996), 664–736.

[Bo2] S. Bouc. Biset functors for finite groups, Springer Lecture Notes in Mathematics no. 1990 (2010).

[BST] S. Bouc, R. Stancu, J. Thévenaz. Simple biset functors and double Burnside ring, J. Pure Appl. Alg. 217 (2013), 546–566.

[BT1] S. Bouc, J. Thévenaz. The algebra of essential relations on a finite set, J. reine angew. Math. 712 (2016), 225–250.

[BT2] S. Bouc, J. Thévenaz. Correspondence functors and finiteness conditions, preprint, 2017.

[BT3] S. Bouc, J. Thévenaz. The algebra of Boolean matrices, correspondence functors, and simplicity, in preparation.

[FHH] D. Flath, T. Halverson, K. Herbig. The planar rook algebra and Pascal’s triangle, Enseign. Math. 55 (2009), no. 1-2, 77–92.

[St] R. P. Stanley. Enumerative Combinatorics, Vol. I, Wadsworth & Brooks/Cole, Monterey, 1986.

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