SIMPLE AND PROJECTIVE CORRESPONDENCE FUNCTORS

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. We determine exactly which simple correspondence functors are projective. We also determine which simple modules are projective for the algebra of all relations on a finite set. Moreover, we analyze the occurrence of such simple projective functors inside the correspondence functor F associated with a finite lattice and we deduce a direct sum decomposition of F.

1. Introduction

In the present paper, we continue to develop 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. Assuming that k is a field, 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). Here, we determine which of them are projective (or equivalently injective).

We say that a poset (E, R) is a pole poset if it is obtained by stacking posets having either cardinality one or cardinality two with two incomparable elements (see Section 2 for details).

1.1. Theorem. Let k be a field and let S_{E,R,V} be the simple correspondence functor parametrized by a finite set E, an order relation R on E, and a simple k Aut(E, R)-module V. The following conditions are equivalent :

(a) S_{E,R,V} is projective.

(b) The poset (E, R) is a pole poset and V is a projective k Aut(E, R)-module.

(c) Either (E, R) is a totally ordered poset or (E, R) is a pole poset and the characteristic of k is different from 2.

Since the group Aut(E, R) of automorphisms of a pole poset is a 2-group, (b) and (c) are easily seen to be equivalent. However, it requires much more work to prove that (a) implies (b), and also that (b) implies (a) (see Section 4). In the case when (E, R) is totally ordered, the projectivity of S_{E,R,V} was already proved in Corollary 11.11 of [BT3].

Every simple functor S_{E,R,V} has a precursor 𝕊_{E,R}, called the fundamental functor associated with the poset (E, R) (see Proposition 3.4). This functor 𝕊_{E,R} has the advantage of being defined over any commutative base ring k. In analogy with the theorem above, we prove in Section 5 that 𝕊_{E,R} is projective if and only if (E, R) is a pole poset.

Date: June 28, 2021.

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.

As in [BT4], the theory of correspondence functors has consequences for the representation theory of the monoid algebra kR_{X}, where R_{X} is the monoid of all relations on a finite set X. We show in Section 6 that, whenever |E| ≤ |X|, the evaluation S_{E,R,V} (X) is a projective kR_{X}-module if and only if the simple functor S_{E,R,V} is projective. This provides a criterion, again in terms of pole posets, for the projectivity of a simple module for the monoid algebra kR_{X}.

Associated with a finite lattice T, there is a correspondence functor F_{T} which is defined over an arbitrary commutative ring k and which plays a crucial role in the theory, see [BT3, BT4]. We know in particular that F_{T} is projective if and only if the lattice T is distributive, for instance if T is a pole lattice. Also, the assignment T ↦ F_{T} is known to be a fully faithful functor by [BT3].

If T is arbitrary, we show that F_{T} has direct summands corresponding to pole lattices appearing inside T, by means of suitably constructed idempotents in End(F_{T}). Actually, most of the work is done in End(T) (where morphisms between lattices are defined to be k-linear combinations of join-morphisms), and then corresponding results for End(F_{T}) are obtained using the fully faithful functor T ↦ F_{T}. The construction of idempotents in End(T) is quite technical (see Section 7) but it provides an explicit description of the part of End(T) which corresponds to pole lattices (see Section 8).

In Section 9, we analyze the special case of a pole lattice Q (see Theorem 9.1 for details).

1.2. Theorem. Let Q be a pole lattice. Then End(F_{Q}) is isomorphic to a direct sum of matrix algebras

End(F_{Q}) ≅ End(Q) ≅ ⨁_{P} M_{n(Q,P)}(k Aut(P))

where P varies among pole lattices inside Q and n(Q, P) is some explicit integer.

From this, we obtain a decomposition of F_{Q} as a direct sum of projective functors (Theorem 9.4) and each summand is also simple when k is a field (Corollary 9.11). Finally, if T is an arbitrary finite lattice, we describe a projective direct summand of F_{T} corresponding to all pole lattices which appear inside T (Theorem 9.12).

2. Pole posets, pole lattices, and opposite morphisms

We first recall some standard facts about lattices and fix the terminology and the notation. If T is a finite lattice, we denote by ∨ its join, ∧ its meet, and ≤_{T} its order relation. When the context is clear, we simply write ≤ instead of ≤_{T}. The unique minimal element is written ˆ0 and the unique maximal element ˆ1. We let T^{op} denote the opposite lattice, such that

x ≤_{T} y ⟺ y ≤_{T^{op}} x .

A join-irreducible element in T is simply called irreducible. We write Irr(T) for the full subposet of irreducible elements of T. Recall that ˆ0 is an empty join, hence is not irreducible. Similarly ˆ1 is an empty meet. If e ∈ T is irreducible, then the halfopen interval [ ˆ0, e [_{T} has a unique maximal element, written r(e). In other words, r(e) = sup{x ∈ T | x < e}. Similarly, if a is meet-irreducible (i.e. irreducible in the opposite lattice T^{op}), then we define s(a) = inf{x ∈ T | a < x}. Any finite poset A is isomorphic to the full subposet of irreducible elements of a lattice, e.g. the lattice I_{↓}(A) of all subsets of A closed under taking smaller elements.

Now we introduce one of the main concepts for the present a paper. Let A and B be two finite posets. Define A ∗ B to be the poset whose underlying set is the disjoint union A ⊔ B and whose order relation is the union of the order relation of A, the order relation of B, and the requirement that a ≤ b for all a ∈ A and b ∈ B. If A_{1}, …, A_{r} are finite posets, then A_{1} ∗ A_{2} ∗ … ∗ A_{r} is defined inductively.

A pole poset is a poset of the form A_{1} ∗ A_{2} ∗ … ∗ A_{r} where each A_{i} either has cardinality one, or consists of exactly two incomparable elements. If a ∈ A_{i}, then clearly a has height i − 1 (with the usual convention that the minimal elements, that is, those in A_{1}, have height 0). The definition implies that there are two types of elements in a pole poset P :

(a) If A_{i} = {a} has cardinality one, then a is comparable to every element of P.

(b) If A_{i} = {a, b} has cardinality two, then a is comparable to every element of P − {b}. In that case, b will be called the twin of a and written ă. In particular, ˘ ă = a.

Notice that a totally ordered poset is a pole poset (with no twins). We write P_{1} for the set of elements of the first type (the ‘totally ordered’ part of P) and P_{2} for the set of elements of the second type (the ‘twin’ part of P).

A pole lattice is a lattice whose underlying poset is a pole poset. Whenever a and ă are incomparable elements of height i in a pole lattice P, then they are both join-irreducible and meet-irreducible. In this case, there is a single element of height i − 1, namely r(a) = a ∧ ă, and a single element of height i + 1, namely s(a) = a ∨ ă. Clearly r(a) = r(ă) and s(a) = s(ă). Also, ˆ0 is the unique element of height 0 and ˆ1 is the unique element of maximal height. Finally, we note that this discussion easily implies that any pole lattice is distributive.

2.1. Lemma. Let P be a pole lattice and let X = { ˆ0 } ∪ {s(a) | a ∈ P_{2}}. Then P − X is the set of irreducible elements of P.

Proof : This is easy and is left to the reader.

We aim to show that pole posets can be characterized by an internal condition which will be useful later in Section 4. Recall that a relation R on a set X is a subset R ⊆ X × X and that the opposite relation R^{op} is defined by :

(x, y) ∈ R^{op} ⟺ (y, x) ∈ R .

Moreover, the product of two relations S and T is the relation defined by

ST := { (z, x) ∈ X × X | ∃ y ∈ X such that (z, y) ∈ S and (y, x) ∈ T } .

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

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

In particular, we write Δ_{X} := Δ_{id} for the identity morphism of the object X. The map σ ↦ Δ_{σ} is a monoid homomorphism and Δ_{σ} is invertible for every σ ∈ Σ_{X}. The symmetric group Σ_{X} acts on relations by conjugation : we write R^{σ} = Δ_{σ^{−1}}RΔ_{σ} and ^{σ}R = Δ_{σ}RΔ_{σ^{−1}}.

2.2. Proposition. Let P be a finite poset and let R ⊆ P × P be its order relation (i.e. (x, y) ∈ R ⟺ x ≤ y). Let \overline{R} = (P × P) − R. The following are equivalent :

(a) P is pole poset.

(b) There exists a permutation τ of P such that

∀ x, y ∈ P , if x ≰ y , then y ≤ τ(x) .

(c) There exists a permutation τ of P such that \overline{R}^{op}Δ_{τ^{−1}} ⊆ R.

Moreover, if (b) holds, then τ can be chosen to be an automorphism of the poset P and, in that case, it is unique and it satisfies τ(a) = ă for all twins a ∈ P_{2}.

Proof : First note that the equivalence of (b) and (c) follows immediately from the definitions, because

x ≰ y ⟺ (y, x) ∈ \overline{R}^{op}

while we always have (x, τ(x)) ∈ Δ_{τ^{−1}}.

Suppose that (a) holds and define τ to be the permutation that preserves heights and satisfies τ(a) = ă for all twins a ∈ P_{2}. Let x, y ∈ P such that x ≰ y. If x is the unique element of its height, then x is comparable to all elements of P and τ(x) = x. It follows that y < x = τ(x). If x and x̆ are distinct elements of the same height, i.e. twins, then x is comparable to every element of P − {x̆}. Therefore, if y ≠ x̆, then y < x, hence also y < x̆ = τ(x), while if y = x̆, then y = τ(x). This proves that we get y ≤ τ(x) in all cases, hence (b) holds.

We assume now that (b) holds and we want to prove (a). We proceed by induction on the size of P, starting from the obvious case when |P | = 1. Suppose first that P has at least two distinct maximal elements w and z. Since w ≰ z, we have z ≤ τ(w) by (b), hence z = τ(w) by maximality of z. Similarly, w = τ(z). Now if x ≰ w, then w ≤ τ(x) by (b), hence w = τ(x), so that x = τ^{−1}(w) = z. In other words, if x ≠ w and x ≠ z, then x < w. Similarly, if x ≠ w and x ≠ z, then x < z. Therefore w and z are the unique maximal elements of P and P = Q∗{w, z}, where Q = P − {w, z}.

If x, y ∈ Q and x ≰ y, then y ≤ τ(x) by (b). But the permutation τ exchanges w and z, so it restricts to a permutation of Q. Therefore (b) holds for the poset Q and, by induction, Q is a pole poset. It follows that P is a pole poset, as required.

Suppose now that P has a single maximal element w. If x ≠ w, then w ≰ x by maximality, hence x ≤ τ(w) by (b). If τ(w) = w, then τ restricts to a permutation of Q = P − {w} and again we are done by induction.

So we assume now that our single maximal element w satisfies τ(w) ≠ w. The condition x ≤ τ(w) obtained above means that τ(w) is the unique maximal element of P − {w}. Assume by induction that w > τ(w) > … > τ^{i}(w) and that τ^{j}(w) is the unique maximal element of P − {w, …, τ^{j−1}(w)}, for every j = 1, …, i. Then if x ≠ w, τ(w), …, τ^{i}(w), we have τ^{i}(w) ≰ x, hence x ≤ τ^{i+1}(w) by (b). But τ^{i+1}(w) ≠ τ(w), …, τ^{i}(w), otherwise τ^{i}(w) ∈ {w, τ(w), …, τ^{i−1}(w)} which is impossible by our induction assumption. Therefore, either τ^{i+1}(w) is the unique maximal element of P − {w, …, τ^{i}(w)} and we continue our induction argument, or τ^{i+1}(w) = w.

Our induction argument must stop and we let r ≥ 2 be the smallest integer such that τ^{r}(w) = w. Then w > τ(w) > … > τ^{r−1}(w) and τ^{j}(w) is the unique maximal element of P − {w, …, τ^{j−1}(w)}, for every j = 1, …, r − 1. Moreover, setting Q = P − {w, …, τ^{r−1}(w)}, we obtain

P = Q ∗ {τ^{r−1}(w)} ∗ … ∗ {τ(w)} ∗ {w}

and Q must be invariant under τ. By our main induction procedure, Q is a pole poset. It follows that P is a pole poset. This proves (a) and we are done.

In order to prove our additional statement, we continue the analysis of the permutation τ, as above. In the case when P has two maximal elements w and z, then we have seen that τ(w) = z. Moreover τ restricts to a permutation of Q = P − {w, z}. By induction, τ_{|Q} can be replaced uniquely by an automorphism α of the pole poset Q such that α exchanges all the twins of Q. Extending α to P by requiring that α exchanges w and z (as it must, as we have seen for τ), we obtain an automorphism of P having the additional required properties.

In the case when P has a single maximal element w, then we have seen that τ permutes cyclically the subset S = {w, τ(w), …, τ^{r−1}(w)} for some r ≥ 1, and it restricts to a permutation of Q = P − S. By induction, τ_{|Q} can be replaced uniquely by an automorphism α of the pole poset Q such that α exchanges all the twins. Extending α by the identity on S, we obtain an automorphism of P having the additional required properties.

A join-morphism from a lattice T to a lattice T′ is a map f : T → T′ which commutes with joins, i.e. such that

f( ⋁_{a∈A} a) = ⋁_{a∈A} f(a) ,

for any subset A of T. Similarly, a meet-morphism is a map which commutes with meets. It is easy to see that a join-morphism is order-preserving, by considering the join t_{1} ∨ t_{2} in the case where t_{1} ≤_{T} t_{2} in the lattice T. Moreover, the case A = ∅ shows that a join-morphism maps ˆ0 ∈ T to ˆ0 ∈ T′. The following result is well-known.

2.3. Lemma. Let P and T be finite lattices. Suppose that P is distributive and let E = Irr(P). Then any order-preserving map φ : E → T extends uniquely to a join-morphism φ̃: P → T.

Proof : For any p ∈ P, we can write uniquely

p = ⋁_{\begin{matrix}e∈E \\ e≤p \end{matrix}} e

and then define the extension of φ by

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

To check that φ̃ is a join-morphism, we use the fact that, for any e ∈ E and p, p′ ∈ P, we have

e ≤ p ∨ p′ ⟺ e ≤ p or e ≤ p′ .

This is because, if e ≤ p∨p′, then, e = e∧(p∨p′) = (e∧p)∨(e∧p′) by distributivity, hence by irreducibility, either e = e ∧ p, i.e., e ≤ p, or e = e ∧ p′, i.e., e ≤ p′.

2.4. Notation.

(a) We let L be the category whose objects are the finite lattices and where, for any finite lattices P and T, Hom_{L}(P, T) is the set of all join-morphisms from P to T.

(b) We denote by Inj_{L}(P, T) the set of all injective join-morphisms P → T.

(c) We denote by Sur_{L}(T, P) the set of all surjective join-morphisms T → P.

Recall from Section 8 of [BT3] that, for any join-morphism f : T → P, there is an opposite morphism f^{op} : P^{op} → T^{op} defined by

f^{op}(p) = ⋁_{f(t)≤p} t .

2.5. Lemma. Let P and T be finite lattices and let f : T → P be a join-morphism.

(a) f^{op} : P^{op} → T^{op} is a join-morphism. In other words, for any subset A of P,

f^{op}( ⋀_{a∈A} a) = ⋀_{a∈A} f^{op}(a)

(because the meet ∧ is the join in the opposite lattice).

(b) If g : P → Q is a join-morphism, then (gf)^{op} = f^{op}g^{op}.

(c) (f^{op})^{op} = f.

(d) If f is surjective, then ff^{op} = id_{P}. In particular, f^{op} is injective and, for

any p ∈ P,

f^{op}(p) = ⋁_{f(t)=p} t = sup (f^{−1}(p)) .

(e) If f is injective, then f^{op}f = id_{T}. In particular, f^{op} is surjective.

(f) Passing to the opposite induces bijections Inj_{L}(P, T) → Sur_{L}(T^{op}, P^{op}) and Sur_{L}(T, P) → Inj_{L}(P^{op}, T^{op}).

Proof : (a), (b) and (c) are proved in Section 8 of [BT3].

(d) Let p ∈ P. The equality ⋁_{f(t)≤p} t = ⋁_{f(t)=p} t follows from the fact that f is

surjective and order-preserving. Moreover, it is clear that ⋁_{f(t)=p} t = sup (f^{−1}(p)).

Since f is a join-morphism, we get f( sup (f^{−1}(p))) = p, hence ff^{op} = id_{P}.

(e) This follows from (b), (c), and (d) by passing to opposite morphisms.

(f) This follows from (d) and (e).

For later use, we now prove a specific result in the case when P is a pole lattice.

2.6. Lemma. Let T be a finite lattice and let P be a pole lattice. Then there is a bijection between Inj_{L}(P, T) and Sur_{L}(T, P).

Proof : Associated with the pole lattice P, there is the set

E_{2} = {a_{1}, ă_{1}, a_{2}, ă_{2}, …, a_{n}, ă_{n}}

consisting of all the twins a_{i}, ă_{i}, indexed in such a way that a_{1} < a_{2} < … < a_{n}. Here n is a positive integer (which is zero whenever P is totally ordered). We define

w_{i} = a_{i} ∧ ă_{i} ,  v_{i} = a_{i} ∨ ă_{i} ,  (1 ≤ i ≤ n) ,

and we also set v_{0} = ˆ0 and w_{n+1} = ˆ1. Just above the pair of twins a_{i}, ă_{i}, there is a totally ordered interval [v_{i}, w_{i+1}]. Also, we have a totally ordered interval [v_{0}, w_{1}] below the pair a_{1}, ă_{1}, and a totally ordered interval [v_{n}, w_{n+1}] above the pair a_{n}, ă_{n}. Note that we may have v_{i} = w_{i+1}.

Let λ ∈ Inj_{L}(P, T). We want to define an injective meet-morphism λ̃ : P → T associated with λ. First we set

λ̃(a_{i}) = λ(a_{i}) ,  λ̃(ă_{i}) = λ(ă_{i})  (1 ≤ i ≤ n) .

Since λ is a join-morphism, we have

λ(v_{0}) = ˆ0 ,  λ(v_{i}) = λ(a_{i}) ∨ λ(ă_{i})  (1 ≤ i ≤ n) .

Note also that

λ(w_{i}) ≤ λ(a_{i}) ∧ λ(ă_{i})  (1 ≤ i ≤ n) ,  λ(w_{n+1}) ≤ ˆ1 .

We have to define λ̃ on each interval [v_{i−1}, w_{i}], 1 ≤ i ≤ n + 1, and there are two cases for each i.

If 1 ≤ i ≤ n, either λ(w_{i}) = λ(a_{i}) ∧ λ(ă_{i}) or λ(w_{i}) < λ(a_{i}) ∧ λ(ă_{i}). In the first case, we simply set

λ̃(x) = λ(x) , ∀ x ∈ [v_{i−1}, w_{i}] ,

while in the second, we set

λ̃(x) = λ(s(x))  ∀ x ∈ [v_{i−1}, w_{i}[ ,  and  λ̃(w_{i}) = λ(a_{i}) ∧ λ(ă_{i}) ,

where s denotes the shift upwards in the totally ordered interval [v_{i−1}, w_{i}], that is, s(x) = inf{y | x < y}.

Similarly, if i = n + 1, either λ(w_{n+1}) = ˆ1 or λ(w_{n+1}) < ˆ1. In the first case, we simply set

λ̃(x) = λ(x) , ∀ x ∈ [v_{n}, w_{n+1}] ,

while in the second, we set

λ̃(x) = λ(s(x))  ∀ x ∈ [v_{n}, w_{n+1}[ ,  and  λ̃(w_{n+1}) = ˆ1 .

It is easy to see that λ̃ is order-preserving and injective, and moreover

λ̃(a_{i} ∧ ă_{i}) = λ̃(w_{i}) = λ(a_{i}) ∧ λ(ă_{i}) = λ̃(a_{i}) ∧ λ̃(ă_{i}) .

In view of the structure of pole lattices, this means that λ̃ : P → T is a meetmorphism, or in other words a join-morphism λ̃ : P^{op} → T^{op}. Therefore λ̃ ∈ Inj_{L}(P^{op}, T^{op}) and this defines a map

Ω_{P,T} : Inj_{L}(P, T) ⟶ Inj_{L}(P^{op}, T^{op}) , λ ↦ λ̃ .

In the other direction, we proceed as follows. The same construction, applied to P^{op} and T^{op}, defines a map

Ω_{P^{op},T^{op}} : Inj_{L}(P^{op}, T^{op}) ⟶ Inj_{L}(P, T)

and it is elementary to check that Ω_{P^{op},T^{op}} maps λ̃ to λ, because the shift upwards x ↦ s(x) in the opposite [v, w]^{op} of a totally ordered interval corresponds to the shift downwards x ↦ r(x) in the original interval [v, w]. In other words the composite Ω_{P^{op},T^{op}} ∘ Ω_{P,T} is the identity. Similarly, Ω_{P,T} ∘ Ω_{P^{op},T^{op}} is the identity and it follows that Ω_{P^{op},T^{op}} is a bijection.

Now it suffices to compose with the bijection Inj_{L}(P^{op}, T^{op}) → Sur_{L}(T, P) of Lemma 2.5 to obtain a bijection between Inj_{L}(P, T) and Sur_{L}(T, P).

3. Correspondence functors

We recall the basic facts we need about correspondence functors and we refer to Sections 2–4 of [BT2] and Section 2 of [BT3] for more details. 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 (using a reverse notation which is convenient for left actions) is the set of all correspondences from X to Y, namely all subsets of Y × X. A correspondence from X to X is also called a relation on X. 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 } ,

and this generalizes the product of relations, defined in Section 2.

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). In particular, kR_{X} := kC(X, X) is the monoid algebra of the monoid R_{X} of all relations on 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). 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 more generally the functors

L_{E,W} := kC(−, E) ⊗_{kR_{E}} W

where W is a left kR_{E}-module. Actually, the functor W ↦ L_{E,W} is left adjoint of the evaluation functor

F_{k} ⟶ kR_{E} - Mod , F ↦ F (E) .

The correspondence 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} .

We shall work with the functor L_{E,W} /J_{E,W} for some specific choices of kR_{E}- modules W.

Recall from Section 5 of [BT1] or Section 3 of [BT4] that, for a suitable two-sided ideal I, there is a quotient algebra kP_{E} = kR_{E}/I, called the algebra of permuted orders because it has a k-basis P_{E} consisting of all relations on E of the form Δ_{σ}R, where σ runs through the symmetric group Σ_{E} and R is an order relation on E. The product of two order relations R and S in the basis P_{E} is the transitive closure of R∪S if this closure is an order relation, and zero otherwise. This product, together with the conjugation action of permutations on relations, describes completely the algebra structure of P_{E}.

Among the kR_{E}-modules, there is the fundamental module kP_{E}f_{R}, associated with a 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 kP_{E}, depending on R, and kP_{E}f_{R} is the left ideal generated by f_{R}. The main thing we need to know about the fundamental module kP_{E}f_{R} is its structure as a bimodule. This is described in the next result, which combines Corollary 7.3 and Proposition 8.5 of [BT1].

3.1. Proposition. Let E be a finite set and R an order relation on E.

(a) The fundamental module kP_{E}f_{R} is a (kR_{E}, k Aut(E, R))-bimodule and the right action of k Aut(E, R) is free.

(b) kP_{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) The action of the algebra of relations kR_{E} on the module kP_{E}f_{R} is given as follows. For any relation Q ∈ R_{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}

Moreover, τ is unique in the first case.

Using the bimodule structure on kP_{E}f_{R}, we define

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

where V is any k Aut(E, R)-module. Then T_{R,V} is a left kR_{E}-module for the action induced from the action of kR_{E} on kP_{E}f_{R} described in Proposition 3.1 above. The main thing we need to know about T_{R,V} is the following result, which is part of Theorem 8.1 in [BT1].

3.2. Proposition. Assume that k is a field. If V is a simple k Aut(E, R)-module, then T_{R,V} is a simple kR_{E}-module.

Associated with the above kR_{E}-modules, we can now define some specific correspondence functors, as in [BT2] and [BT3]. Using the fundamental module kP_{E}f_{R}, we define

𝕊_{E,R} := L_{E,kP_{E}f_{R}}/J_{E,kP_{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}. Our next result is Proposition 2.6 in [BT3].

3.3. Proposition.

(a) The set E is a minimal set for 𝕊_{E,R} and 𝕊_{E,R}(E) ≅ kP_{E}f_{R} as left kR_{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 kR_{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.

It is proved in Theorem 4.7 of [BT2] that, when k is a field, 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 relation 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 result.

3.4. Proposition. Suppose that V is a simple k Aut(E, R)-module, hence in particular generated by a single element v.

(a) Consider the surjective morphism of correspondence functors

Φ : L_{E,kP_{E}f_{R}} ⟶ L_{E,T_{R,V}}

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

Φ_{E} : kP_{E}f_{R} ⟶ kP_{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} .

(b) Φ induces an isomorphsim

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

Proof : (a) is Lemma 2.7 in [BT3], while (b), which is far from being obvious, is Theorem 7.9 in [BT4].

In short, it it is possible to recover S_{E,R,V} from 𝕊_{E,R} by simply tensoring with V. Consequently, the fundamental functors play a crucial role throughout our work.

Another important construction of correspondence functors is obtained from finite lattices (see [BT3] for details).

3.5. Definition. If T is a finite lattice and X is a finite set, define F_{T} (X) = kT^{X}, the free k-module on the set T^{X} of all functions X → T. Given φ : X → T and a correspondence S ∈ C(Y, X), then Sφ : Y → T is defined by the formula

(Sφ)(y) = ⋁_{(y,x)∈S} φ(x) .

Then F_{T} becomes in this way a correspondence functor.

We want to recall two main properties of this construction but we first need some notation. Let L be the category of finite lattices and join-morphisms, as in Notation 2.4. The k-linearization kL of L has the same objects and 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 kL is the k-bilinear extension of the composition in L. The following results appear in Theorems 4.8 and 4.12 of [BT3].

3.6. Theorem.

(a) The assignment T ↦ F_{T} extends to a k-linear functor F_{?} : kL → F_{k}. Moreover, F_{?} is fully faithful.

(b) If T is a finite lattice, then F_{T} is projective in F_{k} if and only if T is distributive. In particular, if P is a pole lattice, then F_{P} is projective.

This provides a fruitful method for handling correspondence functors. Any endomorphism in kL induces an endomorphism in F_{k}. In particular, any idempotent endomorphism of an object T in kL produces a direct summand of the correspondence functor F_{T}. This mechanism will be exploited in Sections 7, 8, and 9.

Our next lemma gives another realization of the functor F_{T} in a special case. Let E be a finite set and R an order relation on E (i.e. (E, R) is a finite poset). As in [BT3], let I_{↓}(E, R) be the lattice of all subsets of E closed under taking smaller elements with respect to R. Then (E, R) is isomorphic to the poset of irreducible elements of I_{↓}(E, R) via the map e ↦ E_{≤e} = {x ∈ E | x ≤ e}. Notice that r(E_{≤e}) = E_{<e} in the lattice I_{↓}(E, R).

3.7. Lemma. Let (E, R) be a finite poset and let T = I_{↓}(E, R). For any finite set X, define a map

ρ_{X} : F_{T^{op}}(X) ⟶ kC(X, E)R , ρ_{X}(φ) = {(x, e) | e ∉ φ(x)} ⊆ X × E ,

where φ : X → T^{op} is any basis element in F_{T^{op}}(X). Then this induces an isomorphism of correspondence functors ρ : F_{T^{op}} ⟶ kC(−, E)R.

Proof : The result can be obtained by combining Proposition 4.5 and Remark 8.7 in [BT3], using the isomorphism, via complementation, I_{↓}(E, R^{op}) ≅ I_{↓}(E, R)^{op}. We provide instead a direct proof.

Since ρ_{X}(φ) is a subset of X ×E, it is an element of C(X, E). It is right invariant by R because if (x, e) ∈ ρ_{X}(φ), i.e. e ∉ φ(x), and if (e, f) ∈ R, then (x, f) ∈ ρ_{X}(φ) because f ∉ φ(x) (otherwise we would have e ∈ φ(x) since φ(x) is closed under taking smaller elements). Hence ρ_{X}(φ) = ρ_{X}(φ)R ∈ C(X, E)R. It is elementary to check that ρ is a morphism of functors. Moreover, it is an isomorphism because there is an inverse morphism mapping S ∈ C(X, E)R to the function φ_{S} : X → T^{op} defined by

φ_{S}(x) = {e ∈ E | (x, e) ∉ S} .

The fact that S is right invariant by R implies that φ_{S}(x) is closed under taking smaller elements. Details are left to the reader.

There is a direct connection between the functors associated with lattices and the fundamental functors. This is Theorem 6.5 in [BT3].

3.8. Theorem. Let (E, R) be a finite poset. There is a unique surjective morphism

Θ : F_{I_{↓}(E,R^{op})} ⟶ 𝕊_{E,R}

mapping the inclusion map j : E → I_{↓}(E, R^{op}) to f_{R} ∈ 𝕊_{E,R}(E) ≅ kP_{E}f_{R}.

We now recall another main result from [BT3], which will be used in Section 4 (and also in Theorem 9.7). Let T = I_{↓}(E, R) and, as in Section 9 of [BT3], consider the element

(3.9)  γ_{T} := ∑_{A⊆E}(−1)^{|A|}η_{A}^{0} ∈ F_{T^{op}}(E) .

Here η_{A}^{0} : E → T^{op} is the map defined by

η_{A}^{0}(e) = \{ \begin{matrix}r(E_{≤e}) = E_{<e} & if e ∈ A , \\ E_{≤e} & if e ∉ A , \end{matrix}

with values in the lattice T, but viewed as elements of T^{op}.

3.10. Theorem. Let (E, R) be a finite poset and let T = I_{↓}(E, R). The subfunctor of F_{T^{op}} generated by γ_{T} is isomorphic to the fundamental functor 𝕊_{E,R}. Moreover, the isomorphism

<γ_{T} >(E) ⟶ 𝕊_{E,R}(E) ≅ kP_{E}f_{R}

maps γ_{T} to f_{R}.

Proof : The first statement is Theorem 9.5 in [BT3]. The second statement can be traced in the proof of that theorem. More precisely, if j : E → I_{↓}(E, R^{op}) denotes the inclusion map, it is shown that γ_{T} ∈ F_{T^{op}}(E) is the image of j ∈ F_{I_{↓}(E,R^{op})}(E) under a morphism

ξ : F_{I_{↓}(E,R^{op})} ⟶ F_{T^{op}} .

On the other hand, by Theorem 3.8 above, there is a surjective morphism

Θ : F_{I_{↓}(E,R^{op})} ⟶ 𝕊_{E,R}

mapping the inclusion map j to f_{R} ∈ 𝕊_{E,R}(E) ≅ kP_{E}f_{R}. Both morphisms ξ and Θ are proved to have the same kernel and this induces the required isomorphism <γ_{T} > ≅ 𝕊_{E,R}. It follows that this isomorphism maps γ_{T} to f_{R}.

4. Characterization of simple projective functors

Throughout this section, assume that the base ring k is a field and let (E, R) be a finite poset. Our aim is to characterize the triples (E, R, V) such that the simple correspondence functor S_{E,R,V} is projective.

Since S_{E,R,V} is isomorphic to a quotient of the fundamental functor 𝕊_{E,R} (see Proposition 3.4), we shall actually work with the latter. We have 𝕊_{E,R} ≅ <γ_{T} > by Theorem 3.10, where T = I_{↓}(E, R) and γ_{T} is defined by (3.9). We let

ζ : <γ_{T} > ⟶ F_{T^{op}}

be the inclusion morphism. We also let

ρ : F_{T^{op}} ⟶ kC(−, E)R

be the isomorphism of correspondence functors described in Lemma 3.7 and we define

δ := ρζ(γ_{T}) = ρ(γ_{T}) ∈ kC(E, E)R .

In view of the isomorphism ρ, the subfunctor <δ> of kC(−, E)R generated by δ is isomorphic to <γ_{T} >, hence to 𝕊_{E,R}. We shall work with δ and we first need its precise description as a linear combination of relations.

4.1. Lemma. Let δ := ρ(γ_{T}) ∈ kC(E, E)R, where γ_{T} is defined by (3.9), and let A be a subset of E.

(a) ρ(η_{A}^{0}) = \overline{R}^{op} ∪ Δ_{A}, where \overline{R} = (E × E) − R and Δ_{A} = \{(a, a) | a ∈ A\} ⊆ E × E.

(b) δ = ∑_{A⊆E}(−1)^{|A|}(\overline{R}^{op} ∪ Δ_{A}).

(c) R(\overline{R}^{op} ∪ Δ_{A}) = \overline{R}^{op} ∪ Δ_{A}.

(d) Rγ_{T} = γ_{T} and Rδ = δ.

Proof : Throughout this proof, we write x ≤ y for (x, y) ∈ R.

(a) By Lemma 3.7, we have

ρ(η_{A}^{0}) = \{(f, e) ∈ E × E | e ∉ η_{A}^{0}(f)\} = \{ \begin{matrix}\{(f, e) ∈ E × E | e ≮ f\} & if e ∈ A , \\ \{(f, e) ∈ E × E | e ≰ f\} & if e ∉ A . \end{matrix}

But \{(f, e) ∈ E × E | e ≰ f\} = \overline{R}^{op}. If e ∈ A, we need to add to \overline{R}^{op} the element (e, e), because e ≮ e. Therefore ρ(η_{A}^{0}) = \overline{R}^{op} ∪ Δ_{A}, as required.

(b) This follows from (a) and the fact that δ = ρ(γ_{T}) = ∑_{A⊆E}(−1)^{|A|}ρ(η_{A}^{0}).

(c) Since Δ_{E} ⊆ R, we have an inclusion

\overline{R}^{op} ∪ Δ_{A} = Δ_{E}(\overline{R}^{op} ∪ Δ_{A}) ⊆ R(\overline{R}^{op} ∪ Δ_{A}) .

In order to prove the reverse inclusion, we let (a, c) ∈ R(\overline{R}^{op} ∪ Δ_{A}). Then there exists b ∈ E such that a ≤ b and (b, c) ∈ \overline{R}^{op} ∪ Δ_{A}.

If (b, c) ∈ \overline{R}^{op}, that is, c ≰ b, then c ≰ a, otherwise we would have c ≤ a ≤ b. Therefore (a, c) ∈ \overline{R}^{op}.

If (b, c) ∈ Δ_{A}, then b = c ∈ A and there are two cases. If a = b, then (a, c) = (a, a) ∈ Δ_{A}. If a ≠ b, then a < b = c, hence c ≰ a, that is, (a, c) ∈ \overline{R}^{op}.

This completes the proof that R(\overline{R}^{op} ∪ Δ_{A}) ⊆ \overline{R}^{op} ∪ Δ_{A}, hence equality.

(d) It follows from (b) and (c) that Rδ = δ, hence also Rγ_{T} = γ_{T} because ρ is an isomorphism mapping γ_{T} to δ. The latter equality was also proved in Lemma 9.3 of [BT3].

We also need some technical computations involving δ.

4.2. Lemma. As above, consider δ = ∑_{A⊆E}(−1)^{|A|}(\overline{R}^{op} ∪ Δ_{A}). Let S ∈ C(E, E)R

(that is, S ⊆ E × E and S = SR).

(a) Sδ ≠ 0 if and only if there exists a permutation σ ∈ Σ_{E} such that S = Δ_{σ}R.

(b) If S = RS and Sδ ≠ 0, then there exists an automorphism σ ∈ Aut(E, R) such that S = Δ_{σ}R.

(c) If (E, R) is a pole poset and if (\overline{R}^{op} ∪ Δ_{A})δ ≠ 0, then

A = E_{1}  and  \overline{R}^{op} ∪ Δ_{A} = Δ_{τ} R ,

where τ is the automorphism of (E, R) satisfying τ(a) = ă for all a ∈ E_{2} (the twin part of E) and τ(a) = a for all a ∈ E_{1} (the totally ordered part of E).

Proof : (a) The condition Sδ ≠ 0 is equivalent to Sγ_{T} ≠ 0 (because ρ maps γ_{T} to δ), hence also to Sf_{R} ≠ 0 by Theorem 3.10. By Proposition 3.1, we obtain

Sδ ≠ 0 ⟺ Sf_{R} ≠ 0 ⟹ ∃ σ ∈ Σ_{E} such that Δ_{E} ⊆ Δ_{σ}^{−1}S ⊆ R
⟹ ∃ σ ∈ Σ_{E} such that R ⊆ Δ_{σ}^{−1}SR ⊆ R^{2}
⟹ ∃ σ ∈ Σ_{E} such that R ⊆ Δ_{σ}^{−1}S ⊆ R
⟹ ∃ σ ∈ Σ_{E} such that S = Δ_{σ}R ,

using the equalities S = SR and R^{2} = R (by transitivity and reflexivity of R). Conversely, if S = Δ_{σ}R, then, by Lemma 4.1, we obtain

Sδ = Δ_{σ}Rδ = Δ_{σ}δ ≠ 0 ,

because δ ≠ 0 since it generates a nonzero subfunctor.

(b) We have S = Δ_{σ}R by (a) and since S = RS, we obtain RΔ_{σ}R = Δ_{σ}R, or in other words R^{σ}R = R, where R^{σ} = Δ_{σ}^{−1}RΔ_{σ}. Since Δ_{E} ⊆ R, we get R^{σ} ⊆ R^{σ}R = R, hence R^{σ} = R because both relations have the same cardinality. This means that Δ_{σ} commutes with R, that is, σ is an automorphism of the poset (E, R).

(c) By (b) applied to S = \overline{R}^{op} ∪ Δ_{A} (which satisfies S = RS by Lemma 4.1), we have

\overline{R}^{op} ∪ Δ_{A} = Δ_{σ}R

for some automorphism σ ∈ Aut(E, R). Since (E, R) is a pole poset, σ is the identity on E_{1} and interchanges some of the twins e, ĕ ∈ E_{2}, so in particular σ = σ^{−1}.

If e ∈ E_{2} and ĕ is its twin, then ĕ ≰ e, hence (e, ĕ) ∈ \overline{R}^{op} ⊆ Δ_{σ}R. Therefore (e, σ(e)) ∈ Δ_{σ} and (σ(e), ĕ) ∈ R, that is σ(e) ≤ ĕ. This shows that σ(e) cannot be equal to e, i.e. σ(e) = ĕ. Thus σ interchanges all the twins, that is, it is equal to the automorphism τ of the statement.

If e ∈ E_{1}, then (e, e) ∈ Δ_{σ} and (e, e) ∈ R, so (e, e) ∈ Δ_{σ}R. If conversely (e, e) ∈ Δ_{σ}R, then (e, σ(e)) ∈ Δ_{σ} and (σ(e), e) ∈ R, that is, σ(e) ≤ e. This cannot hold if e ∈ E_{2}, because σ(e) = ĕ ≰ e, and therefore e ∈ E_{1}. It follows that

e ∈ E_{1} ⟺ (e, e) ∈ Δ_{σ}R ⟺ (e, e) ∈ \overline{R}^{op} ∪ Δ_{A} ⟺ (e, e) ∈ Δ_{A} ⟺ e ∈ A ,

the third equivalence using the fact that (e, e) ∉ \overline{R}^{op} because e ≤ e. This shows that A = E_{1} and completes the proof.

One of the key parts of the proof of the main result is contained in the next lemma, which will also be used again in Section 5.

4.3. Lemma. Suppose that k is a field. Let 𝕊_{E,R} be the fundamental functor associated with a finite poset (E, R) and let M be a nonzero direct summand of 𝕊_{E,R}. If M is projective, then (E, R) is a pole poset.

Proof : Since 𝕊_{E,R} ≅ <γ_{T} > by Theorem 3.10, we can view M as a direct summand of <γ_{T} > and we let ω : M ⟶ <γ_{T} > be the inclusion morphism. As above, we let ζ : <γ_{T} > ⟶ F_{T^{op}} be the inclusion morphism and ρ : F_{T^{op}} ⟶ kC(−, E)R be the isomorphism of correspondence functors described in Lemma 3.7. Finally let

α : M ⟶ kC(−, E)R

be the composite α = ρζω.

Since M is projective and the base ring k is a field, M is also injective, by Theorem 10.6 in [BT2]. Therefore the injective morphism α splits, that is, there exists a surjective morphism

σ : kC(−, E)R ⟶ M

such that σα = id. Thus ασ is an idempotent endomorphism of kC(−, E)R. Since R ∈ kC(E, E) is a generator of kC(−, E)R, its image c := σ(R) ∈ M(E) is a generator of M. Now γ_{T} generates <γ_{T} >, so we can write ω(c) = vγ_{T} for some v ∈ kC(E, E). We know that Rγ_{T} = γ_{T} by Lemma 4.1 and therefore vγ_{T} = vRγ_{T}. Replacing v by vR, we can assume that v = vR and we do so. Thus v ∈ kC(E, E)R. Note that c ≠ 0, hence vγ_{T} ≠ 0.

Now for any u ∈ kC(X, E), we have

ασ(uR) = u·ασ(R) = u·α(c) = u·ρζω(c) = u·ρζ(vγ_{T}) = uv·ρζ(γ_{T}) = uvδ ,

where δ = ρζ(γ_{T}) = ρ(γ_{T}) as in Lemma 4.1. In particular, using the fact that δ = δR (because δ ∈ kC(E, E)R), we obtain

ασ(δ) = ασ(δR) = δvδ .

Since vγ_{T} is nonzero, so is its image vδ = ρζ(vγ_{T}) under the injective morphism ρζ and therefore

0 ≠ vδ = ασ(R) = (ασ)^{2}(R) = ασ(vδ) = v·ασ(δ) = vδvδ ,

from which it follows that δvδ ≠ 0.

Summarizing, we have proved that, under the assumption that M is projective, the element δ = ρ(γ_{T}) ∈ kC(E, E)R satisfies :

(4.4)  ∃ v ∈ kC(E, E)R with δvδ ≠ 0 .

Our aim is to show that (4.4) implies that (E, R) is a pole poset.

The condition δvδ ≠ 0 implies that there exists a relation S (in the expression of v ∈ kC(E, E)R as a linear combination of relations) such that δSδ ≠ 0. In particular Sδ ≠ 0, hence S = Δ_{τ} R for some τ ∈ Σ_{E}, by Lemma 4.2. In view of the expression of δ obtained in Lemma 4.1, there exists a subset A ⊆ E such that

(\overline{R}^{op} ∪ Δ_{A})Δ_{τ} Rδ ≠ 0 .

Again, this implies that the relation (\overline{R}^{op} ∪ Δ_{A})Δ_{τ} R has the form

(\overline{R}^{op} ∪ Δ_{A})Δ_{τ} R = Δ_{σ}R

for some σ ∈ Σ_{E}, by Lemma 4.2. Since the left hand side is invariant under left multiplication by R (by Lemma 4.1), part (b) of Lemma 4.2 implies that Δ_{σ} commutes with R (i.e. σ ∈ Aut(E, R)). It follows that

(\overline{R}^{op} ∪ Δ_{A})Δ_{τσ^{−1}}R = R .

In particular, we deduce that

\overline{R}^{op}Δ_{ψ} ⊆ R , where ψ := τσ^{−1} .

By the characterization of Proposition 2.2, this implies that (E, R) is a pole poset, as was to be shown.

4.5. Theorem. Let k be a field and let S_{E,R,V} be the simple correspondence functor parametrized by a finite set E, an order relation R on E, and a simple k Aut(E, R)-module V. The following conditions are equivalent.

(a) The simple correspondence functor S_{E,R,V} is projective.

(b) (E, R) is a pole poset and V is a projective k Aut(E, R)-module.

(c) Either (E, R) is a totally ordered poset or (E, R) is a pole poset and the characteristic of k is different from 2.

Proof : (b) ⟺ (c). For a pole poset (E, R), the group Aut(E, R) is a 2-group (elementary abelian), generated by all the possible transpositions of twins. In case (E, R) is totally ordered, this group is trivial and the unique simple k-module k is automatically projective. In case (E, R) is a pole poset but is not totally ordered, then Aut(E, R) is nontrivial and the characteristic of k comes into play. If char(k) ≠ 2, then all simple k Aut(E, R)-module V are projective (by Maschke’s theorem). If char(k) = 2, then the unique simple k Aut(E, R)-module is the trivial module, which is not projective (by the converse of Maschke’s theorem).

(a) ⇒ (b). Since S_{E,R,V} is projective by assumption and isomorphic to a quotient of the fundamental functor 𝕊_{E,R} by Proposition 3.4, it is isomorphic to a direct summand of 𝕊_{E,R}. Therefore Lemma 4.3 can be applied and it follows that (E, R) is a pole poset.

We also have to prove that V is a projective k Aut(E, R)-module. Let

T = T_{R,V} = kP_{E}f_{R} ⊗_{k Aut(E,R)} V

be the simple kC(E, E)-module appearing in the definition S_{E,R,V} := L_{E,T} /J_{E,T}. By adjunction, there is an isomorphism

End_{F_{k}}(L_{E,T}) ≅ Hom_{kC(E,E)}(T, L_{E,T} (E)) ≅ End_{kC(E,E)}(T)

and this is a skew field by Schur’s lemma (it is actually the field k). This has no nontrivial idempotent and so L_{E,T} is indecomposable. But the surjective morphism

π : L_{E,T} ⟶ L_{E,T} /J_{E,T} = S_{E,R,V}

splits because S_{E,R,V} is projective by assumption. Therefore π is an isomorphism, by indecomposability of L_{E,T}, hence L_{E,T} is projective.

Evaluating this projective functor at the finite set E, we obtain a kC(E, E)- module

L_{E,T} (E) = T = kP_{E}f_{R} ⊗_{k Aut(E,R)} V

which must be projective, by Lemma 10.1 in [BT2]. Now kP_{E}f_{R} ⊗_{k Aut(E,R)} V is actually a module for the quotient algebra kP_{E} = kR_{E}/I (see Section 3). It follows that kP_{E}f_{R} ⊗_{k Aut(E,R)} V is a projective module for the algebra kP_{E}, because of the splitting of the composition of surjective homomorphisms

kR_{E} ⟶ kP_{E} ⟶ kP_{E}f_{R} ⊗_{k Aut(E,R)} V .

Finally, by Theorem 7.5 in [BT1], there is an isomorphism of algebras

kP_{E} ≅ ∏_{R} M_{n_{R}}(k Aut(E, R))

for some integers n_{R}, where R runs over all order relations on E up to isomorphism (see also Remark 3.4 in [BT4]). Thus there is a Morita equivalence

kP_{E} - Mod ≅ ∏_{R} k Aut(E, R) - Mod

and the bimodule inducing the equivalence is ⨁_{R} kP_{E}f_{R} (see Remark 7.6 in [BT1]). Therefore kP_{E}f_{R} ⊗_{k Aut(E,R)} V corresponds to the k Aut(E, R)-module V under this equivalence. Since projectivity is preserved by a Morita equivalence, V is a projective k Aut(E, R)-module, as required.

(b) ⇒ (a). We assume that (E, R) is a pole poset and, as before, we write x ≤ y for (x, y) ∈ R. Our aim is to compute δ^{2} and show that it is an idempotent. In view of the expression of δ in Lemma 4.1, we have to consider terms of the form (\overline{R}^{op} ∪ Δ_{A})δ. By Lemma 4.2, this can be nonzero only if A = E_{1} and \overline{R}^{op} ∪ Δ_{A} = Δ_{τ} R, where τ ∈ Aut(E, R) is the automorphism exchanging all twins e, ĕ ∈ E_{2} and fixing E_{1} = E − E_{2} pointwise.

Thus \overline{R}^{op} ∪ Δ_{E_{1}} is the only term which can come into play and we now show that it is indeed equal to Δ_{τ} R. For any a ∈ E_{1}, we have (a, τ(a)) ∈ Δ_{τ} and τ(a) = a ≤ a, hence (a, a) ∈ Δ_{τ} R. Therefore Δ_{E_{1}} ⊆ Δ_{τ} R. Since (E, R) is a pole poset, Proposition 2.2 implies that \overline{R}^{op} ⊆ RΔ_{τ} = Δ_{τ} R, using the fact that τ is an automorphism of (E, R). So we obtain

\overline{R}^{op} ∪ Δ_{E_{1}} ⊆ Δ_{τ} R .

In order to prove the reverse inclusion, we let (a, τ(a)) ∈ Δ_{τ} (since τ = τ^{−1}) and (τ(a), b) ∈ R, i.e. τ(a) ≤ b. If a ∈ E_{2}, then τ(a) = ă, hence ă ≤ b. Then b ≰ a, that is, (a, b) ∈ \overline{R}^{op}. If a ∈ E_{1}, then τ(a) = a, hence a ≤ b. If a = b, then (a, b) ∈ Δ_{E_{1}}, while if a ≠ b, then a < b, hence b ≰ a, that is, (a, b) ∈ \overline{R}^{op}. This shows the required reverse inclusion and therefore

\overline{R}^{op} ∪ Δ_{E_{1}} = Δ_{τ} R ,

as claimed. In particular (\overline{R}^{op} ∪ Δ_{E_{1}})δ = Δ_{τ} Rδ = Δ_{τ} δ by Lemma 4.1. Therefore

δ^{2} = ∑_{A⊆E}(−1)^{|A|}(\overline{R}^{op} ∪ Δ_{A})δ = (−1)^{|E_{1}|}(\overline{R}^{op} ∪ Δ_{E_{1}})δ = (−1)^{|E_{1}|}Δ_{τ} δ .

Since τ permutes all the subsets A ⊆ E and preserves their cardinality, we have Δ_{τ} δ = δΔ_{τ}. Consequently

((−1)^{|E_{1}|}Δ_{τ} δ)^{2} = (−1)^{2|E_{1}|}Δ_{τ}^{2}δ^{2} = δ^{2} = (−1)^{|E_{1}|}Δ_{τ} δ ,

so we obtain an idempotent.

Right multiplication by this idempotent defines an idempotent endomorphism of the correspondence functor kC(−, E)R (notice that both Δ_{τ} and δ commute with R). The image of this endomorphism is the subfunctor generated by the element (−1)^{|E_{1}|}Δ_{τ} δ, that is, the subfunctor generated by δ because Δ_{τ} is invertible. But we know that <δ> is isomorphic to the fundamental functor 𝕊_{E,R}. Therefore 𝕊_{E,R} is isomorphic to a direct summand of kC(−, E)R, hence a direct summand of kC(−, E) because R^{2} = R is idempotent. Since kC(−, E) is a projective functor by Yoneda’s lemma, we conclude that 𝕊_{E,R} is projective.

Our assumption (b) also says that the k Aut(E, R)-module V is projective. By Proposition 3.4, S_{E,R,V} is isomorphic to 𝕊_{E,R}⊗_{k Aut(E,R)}V, which is in turn a direct summand of 𝕊_{E,R} ⊗_{k Aut(E,R)} k Aut(E, R) ≅ 𝕊_{E,R}. Therefore S_{E,R,V} is projective, proving (a).

Another proof of the implication (b) ⇒ (a) will be given later in Corollary 9.11.

5. Projectivity of fundamental functors

Given a poset (E, R), we know from Proposition 3.4 that every simple functor S_{E,R,V} has a precursor 𝕊_{E,R}, called the fundamental functor associated with the poset (E, R). This is actually defined over any commutative base ring k. The main result of this section is analogous to Theorem 4.5.

5.1. Theorem. Let (E, R) be a finite poset. Then 𝕊_{E,R} is a projective functor if and only if (E, R) is a pole poset.

Proof : Assume first that 𝕊_{E,R} is a projective functor. We allow the base ring k to vary and we write a superscript (k) to emphasize that a functor belongs to the category F_{k} of correspondence functors defined over the base ring k. Let m be a maximal ideal of k and let C = k/m be the corresponding field. The scalar extension functor

F_{k} ⟶ F_{C} , F ↦ C ⊗_{k} F

is left adjoint of the scalar ‘restriction’ functor, which is obviously exact. Therefore, scalar extension sends projective objects to projective objects. In particular, we see that C ⊗_{k} 𝕊_{E,R}^{(k)} is projective.

By Theorem 6.6 in [BT4], the evaluation 𝕊_{E,R}^{(k)}(X) at a finite set X has an explicit k-basis B_{X}. This basis is defined independently of k, so that it remains a k′-basis for any ring extension k → k′. Therefore, the natural surjection

C ⊗_{k} 𝕊_{E,R}^{(k)}(X) ⟶ 𝕊_{E,R}^{(C)}(X)

is an isomorphism. Since this holds for any X, we have C ⊗_{k} 𝕊_{E,R}^{(k)} ≅ 𝕊_{E,R}^{(C)} and it

follows that 𝕊_{E,R}^{(C)} is projective. Now the functor M := S_{E,R}^{(C)} satisfies the assumptions of Lemma 4.3 and this lemma then asserts that (E, R) is a pole poset, as was to be shown.

For the converse, we use the proof of (b) ⇒ (a) in Theorem 4.5. This proof (except the last paragraph) tells us precisely that, whenever (E, R) is a pole poset, the fundamental functor 𝕊_{E,R} is projective.

Another proof of the projectivity of 𝕊_{E,R} whenever (E, R) is a pole poset will be given later (see Remark 9.9).

6. Simple and projective modules for the algebra of relations

In this section, we assume that k is a field and we use the close link between simple correspondence functors and simple modules for the monoid algebra kR_{X} = kC(X, X), in order to determine all the simple kR_{X}-modules which are projective. Let us first recall this link, which is Theorem 8.1 in [BT4]. 6.1. Theorem. Let X be a finite set and let k be a field.

(a) The set of isomorphism classes of simple R_{X}-modules is parametrized by the set of isomorphism classes of triples (E, R, V), where E is a finite set with |E| ≤ |X|, R is an order relation on E, and V is a simple k Aut(E, R)- module.

(b) The simple module parametrized by the triple (E, R, V) is S_{E,R,V} (X), where S_{E,R,V} is the simple correspondence functor corresponding to the triple (E, R, V).

Pole posets appear again in the main result of this section. 6.2. Theorem. Assume that k is a field. Let X be a finite set and let (E, R, V) be a triple as in Theorem 6.1 above. The following conditions are equivalent.

(a) The simple kR_{X}-module S_{E,R,V} (X) is projective.

(b) The simple correspondence functor S_{E,R,V} is projective.

(c) (E, R) is a pole poset and V is a projective k Aut(E, R)-module.

Proof : By Theorem 4.5, we already know that (b) and (c) are equivalent.

(a) ⟹ (b). Let W = S_{E,R,V} (X). By the adjunction mentioned in Section 3, the identity map W → S_{E,R,V} (X) gives rise to a morphism

π : L_{X,W} ⟶ S_{E,R,V} ,

which is surjective by simplicity of S_{E,R,V}. Since, by assumption, the kR_{X}-module W is projective and indecomposable, the functor L_{X,W} is projective and indecomposable (see Lemma 2.4 in [BT2]) and is therefore the projective cover of S_{E,R,V}. By Theorem 10.7 in [BT2], we have

S_{E,R,V} ≅ L_{X,W} / Rad(L_{X,W}) ≅ Soc(L_{X,W}) .

Suppose that L_{X,W} is not simple. Then the simple functor S_{E,R,V} appears at least twice in a composition series of L_{X,W}. Therefore, the simple kR_{X}-module W = S_{E,R,V} (X) appears at least twice in a composition series of the kR_{X}-module L_{X,W} (X). But L_{X,W} (X) = kR_{X} ⊗_{kR_{X}} W ≅ W is simple and we obtain a contradiction. It follows that L_{X,W} must be simple and so π : L_{X,W} → S_{E,R,V} is an isomorphism. Since L_{X,W} is projective, so is S_{E,R,V}, as was to be shown.

(b) ⟹ (a). The simple functor S_{E,R,V} must be generated by its nonzero evaluation S_{E,R,V} (X) and it is projective by assumption. Therefore, by Lemma 7.3 in [BT2], the kR_{X}-module S_{E,R,V} (X) is projective.

7. Morphisms and idempotents corresponding to pole lattices

In this section, we construct explicit families of morphisms in the category kL of finite lattices, where k is a commutative ring, in particular orthogonal idempotent endomorphisms of a fixed lattice T. They involve a pole lattice P and they have remarkable properties which will be used in Section 8 to obtain structural results about the endomorphism algebra End_{kL}(T). By means of the fully faithful functor T ↦ F_{T}, we will then see in Section 9 how to deduce structural results about the correspondence functor F_{T}, and in particular about projective direct summands of F_{T}.

Our results generalize those obtained in [BT3] in the special case when P is a totally ordered lattice. We follow the same line of development, but with many necessary additions and technical adaptations.

We first fix a pole lattice P and a surjective join-morphism π : T → P. Recall that P_{1} denotes the subset of elements p ∈ P such that p is comparable to every element of P, while P_{2} denotes the subset consisting of all twins. Let E = Irr(P) be the set of irreducible elements of P, described in Lemma 2.1. We write E_{1} = E ∩P_{1} and E_{2} = E ∩ P_{2} (so that in fact E_{2} = P_{2}).

7.1. Notation. We define a notation associated with the surjective join-morphism π : T → P.

(a) For every p ∈ P, let b_{p}^{π} = π^{op}(p) = sup (π^{−1}(p)). Whenever π is fixed, we write simply b_{p} = b_{p}^{π}.

(b) B = Im(π^{op}) = {b_{p} | p ∈ P }. Notice that B is a subposet of T^{op} which is join-closed, hence a subposet of T which is meet-closed and isomorphic to P.

(c) For every e ∈ E_{1}, let b_{e}^{−} = b_{r(e)} and b_{e}^{+} = b_{e}, where r(e) = sup[ˆ0, e[_{P}.

(d) For every e ∈ E_{2} and if ĕ is the twin of e, let b_{e}^{−} = b_{e} and b_{e}^{+} = b_{s(e)}, where s(e) = inf]e, ˆ1]_{P} = e ∨ ĕ.

7.2. Remark. The definition in (c) and (d) is not uniform since we have b_{e} = b_{e}^{+} in one case and b_{e} = b_{e}^{−} in the other. This strange behavior will be explained in Remark 7.10, where a uniform explanation will be given.

For every e ∈ E, choose a_{e} ∈ [b_{e}^{−}, b_{e}^{+}]_{T} (where the subscript T emphasizes that the interval is taken within the lattice T). This defines a family A = (a_{e})_{e∈E} with the following property.

7.3. Lemma. Let A = (a_{e})_{e∈E} be a family of elements of T such that a_{e} ∈ [b_{e}^{−}, b_{e}^{+}]_{T} for every e ∈ E. Then, whenever e, f ∈ E,

e <_{P} f ⟹ a_{e} ≤_{T} a_{f} .

Proof : If e ∈ E_{2} and f ∈ E_{1} with e <_{P} f, then e <_{P} s(e) ≤_{P} r(f) <_{P} f and therefore

a_{e} ≤_{T} b_{e}^{+} = b_{s(e)} ≤_{T} b_{r(f)} = b_{f}^{−} ≤_{T} a_{f} .

The other cases are easier and are left to the reader.

By Lemma 2.3, since P is a distributive lattice, the order-preserving map E → T, e ↦ a_{e}, extends to a join-morphism

j_{A}^{π} : P ⟶ T , p ↦ a_{p} .

Explicitly, we have aˆ0 = ˆ0 and a_{e∨ĕ} = a_{e} ∨ aĕ whenever e ∈ E_{2} with twin ĕ (these are the only non-irreducible elements of P by Lemma 2.1). Note that j_{A}^{π} is not necessarily a section of π (see the beginning of the proof of Proposition 7.9).

Define the family B^{−} = (b_{e}^{−})_{e∈E} and write

μ(B^{−}, A) = ∏_{e∈E} μ(b_{e}^{−}, a_{e})

where μ(−, −) denotes the Möbius function of the lattice T. Allowing the family A to vary (i.e. a_{e} varies in [b_{e}^{−}, b_{e}^{+}]_{T} for each e ∈ E), define

(7.4)  j^{π} = (−1)^{|E_{1}|} ∑_{A} μ(B^{−}, A) j_{A}^{π} .

This is a k-linear combination of join-morphisms, hence an element of kL(P, T). The morphisms j^{π} have remarkable properties, in particular when j^{π} is composed with the surjection π. We are going to explore those properties in a series of propositions. We first start with a lemma.

7.5. Lemma. Let A = (a_{e})_{e∈E} and Ã = (ã_{e})_{e∈E} be two families as above and fix some g ∈ E. Suppose that ã_{e} = a_{e} for all e ∈ E − {g}. Then the following are equivalent :

(a) jπ ; Ã(p) = j_{A}^{π}(p) for all p ∈ P − {g}.

(b) If g ∈ E_{2}, then ã_{g} ∨ ağ = a_{g} ∨ ağ where ğ is the twin of g.

Proof : Suppose that (b) holds. If p = e ∈ E − {g}, then ã_{e} = a_{e} by assumption, that is, jπ ; Ã(e) = j_{A}^{π}(e). If p = ˆ0, then jπ ; Ã(ˆ0) = ˆ0 = j_{A}^{π}(ˆ0). If now p ∈ P − E and p ≠ ˆ0, then p = u∨ŭ for some u ∈ E_{2}, by the definition of a pole lattice. If g ≠ u, ŭ, then

jπ ; Ã(p) = jπ ; Ã(u ∨ ŭ) = ã_{u} ∨ ãŭ = a_{u} ∨ aŭ = j_{A}^{π}(u ∨ ŭ) = j_{A}^{π}(p) .

If g = u, then the assumption (b) implies that

jπ ; Ã(p) = jπ ; Ã(g ∨ ğ) = ã_{g} ∨ ãğ = ã_{g} ∨ ağ = a_{g} ∨ ağ = j_{A}^{π}(g ∨ ğ) = j_{A}^{π}(p) ,

proving (a).

Assume conversely that (a) holds. If g ∈ E_{1}, then condition (b) is empty and there is nothing to prove. So suppose that g ∈ E_{2}. Then

ã_{g} ∨ ağ = ã_{g} ∨ ãğ = jπ ; Ã(g ∨ ğ) = j_{A}^{π}(g ∨ ğ) = a_{g} ∨ ağ ,

proving (b).

7.6. Definition. Associated with the subset E = Irr(P), there is a subfunctor H_{P} of F_{P} defined as follows. For any finite set X, the evaluation H_{P} (X) is the k-submodule of F_{P} (X) generated by all functions φ : X → P such that E ⊈ φ(X).

This subfunctor is important in the theory of correspondence functors (see Section 5 of [BT3] for details).

7.7. Proposition. Let π ∈ Sur_{L}(T, P) and let j^{π} : P → T be the morphism defined in (7.4).

(a) For any finite set X and any function φ : X → P such that E ⊈ φ(X), we have j^{π}φ = 0.

(b) j^{π} induces a morphism F_{j^{π}} : F_{P} → F_{T} vanishing on H_{P}, hence induces in

turn a morphism

\overline{F}_{j^{π}} : F_{P} /H_{P} ⟶ F_{T} .

Proof : Since (b) immediately follows from (a), it suffices to prove (a). We have

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

For a fixed ψ, we have to prove that the inner sum over A is zero. If this inner sum is empty, then the sum is zero and we are done. Otherwise, we can choose A such that j_{A}^{π}φ = ψ. Let g ∈ E be such that g ∉ φ(X). Then we can modify the family A into Ã, by changing only the image j_{A}^{π}(g) = a_{g} ∈ [b_{g}^{−}, b_{g}^{+}]_{T} into jπ ; Ã(g) = ã_{g} ∈ [b_{g}^{−}, b_{g}^{+}]_{T}, with the extra condition that ã_{g} ∨ ağ = a_{g} ∨ ağ in case g ∈ E_{2}. The point of such a modification is that it is precisely the only kind which does not change the equality j_{A}^{π}φ = ψ, by Lemma 7.5. We set A′ = (a_{e})_{e∈E−{g}} and B^{′−} = (b_{e}^{−})_{e∈E−{g}} and we let

j_{A′}^{π} : E − {g} ⟶ T , e ↦ a_{e} ,

which we extend to a join morphism j_{A′}^{π} : P − {g} → T. We obtain

∑_{\begin{matrix}A \\ j_{A}^{π}φ=ψ \end{matrix}} μ(B^{−}, A) = ∑_{\begin{matrix}A′ \\ j_{A′}^{π} φ=ψ \end{matrix}} μ(B^{′−}, A′) ∑ ; ã_{g} μ(b_{g}^{−}, ã_{g}) ,

where the inner sum runs over all ã_{g} ∈ [b_{g}^{−}, b_{g}^{+}]_{T}, with the extra condition that ã_{g} ∨ ağ = a_{g} ∨ ağ in case g ∈ E_{2}.

If g ∉ E_{2}, then the sum runs over all ã_{g} ∈ [b_{g}^{−}, b_{g}^{+}]_{T} and this is zero by the definition of the Möbius function (because b_{g}^{−} = b_{r(g)} <_{T} b_{g} = b_{g}^{+}). If g ∈ E_{2}, then the extra condition is equivalent to ã_{g} ∨ (b_{g} ∨ ağ) = a_{g} ∨ ağ (because b_{g} = b_{g}^{−} ≤ ã_{g}), so ã_{g} runs over the interval [b_{g}, a_{g} ∨ ağ]_{T} with the condition that its join with the fixed element b_{g} ∨ ağ is equal to the top element a_{g} ∨ ağ. By a well-known property of the Möbius function (Corollary 3.9.3 in [St]), the corresponding sum

∑_{\begin{matrix}ã_{g}∈[b_{g},a_{g}∨ağ]_{T} \\ ã_{g}∨(b_{g}∨ağ)=a_{g}∨ağ \end{matrix}} μ(b_{g}^{−}, ã_{g})

is zero, provided the fixed element b_{g} ∨ ağ is not equal to the bottom element b_{g}. But this is indeed the case since b_{g} ∨ ağ ≥_{T} b_{g} ∨ bğ >_{T} b_{g}, the latter inequality coming from the fact that π(b_{g} ∨ bğ) = g ∨ ğ >_{P} g = π(b_{g}). It follows that the coefficient of every ψ is zero, hence j^{π}φ = 0.

Now we want to compute the composite πj^{π}. For any subset Y of E, we define

(7.8)  ρ_{Y} : E ⟶ P ,  ρ_{Y} (e) = \begin{cases}e & if e ∈ Y , \\ r(e) & if e ∈ E_{1}, e ∉ Y , \\ s(e) & if e ∈ E_{2}, e ∉ Y . \end{cases}

It is easy to see that ρ_{Y} is order-preserving (because, if e ∈ E_{2}, f ∈ E_{1}, and e <_{P} f, then e <_{P} s(e) ≤_{P} r(f) <_{P} f, while the other cases are easier). Therefore, by Lemma 2.3, ρ_{Y} extends to a join-morphism ρ_{Y} : P ⟶ P because the pole lattice P is distributive. Note that ρ_{Y} (p) = p for any p ∉ E. This is clear if p = ˆ0. Otherwise p = e ∨ ĕ for some e ∈ E_{2} by Lemma 2.1 and

p = e ∨ ĕ ≤_{P} ρ_{Y} (e) ∨ ρ_{Y} (ĕ) ≤_{P} s(e) ∨ s(ĕ) = p ∨ p = p ,

forcing equality and ρ_{Y} (p) = ρ_{Y} (e) ∨ ρ_{Y} (ĕ) = p.

7.9. Proposition. Let π ∈ Sur_{L}(T, P) and let j^{π} : P → T be the morphism defined in (7.4).

(a) πj^{π} = ∑_{∅⊆Y ⊆E}(−1)^{|E−Y |}ρ_{Y}, where ρ_{Y} is defined by (7.8).

(b) If Y ≠ E, then E ⊈ ρ_{Y} (P).

Proof : For simplicity, we write < instead of <_{P} and ≤ instead of ≤_{P}.

(a) If e ∈ E_{1} and b_{r(e)} < x in T, then r(e) < π(x) because b_{r(e)} = sup{t ∈ T | π(t) = r(e)}. Thus if x ∈]b_{r(e)}, b_{e}]_{T}, we get r(e) < π(x) ≤ e, hence π(x) = e. Similarly, if e ∈ E_{2} and x ∈]b_{e}, b_{s(e)}]_{T}, then π(x) = s(e). It follows that

πj_{A}^{π}(e) = \begin{cases}e & if e ∈ E_{1} and j_{A}^{π}(e) ∈]b_{e}^{−}, b_{e}^{+}]_{T} =]b_{r(e)}, b_{e}]_{T} , \\ e & if e ∈ E_{2} and j_{A}^{π}(e) = b_{e}^{−} = b_{e} , \\ r(e) & if e ∈ E_{1} and j_{A}^{π}(e) = b_{e}^{−} = b_{r(e)} , \\ s(e) & if e ∈ E_{2} and j_{A}^{π}(e) ∈]b_{e}^{−}, b_{e}^{+}]_{T} =]b_{e}, b_{s(e)}]_{T} . \end{cases}

We see that πj_{A}^{π} = ρ_{Y} for a suitable subset Y ⊆ E and therefore

πj^{π} = ∑_{∅⊆Y ⊆E}(−1)^{|E_{1}|}( ∑_{\begin{matrix}A \\ πj_{A}^{π}=ρ_{Y} \end{matrix}} μ(B^{−}, A))ρ_{Y} .

For a fixed subset Y, in order to realize the condition πj_{A}^{π} = ρ_{Y}, we have the following possibilities :

• If e ∈ Y ∩ E_{1}, then j_{A}^{π}(e) can run freely in ]b_{e}^{−}, b_{e}^{+}]_{T}.
• If e ∈ Y ∩ E_{2}, then j_{A}^{π}(e) must be equal to b_{e}^{−} = b_{e}.
• If e ∈ (E − Y) ∩ E_{1}, then j_{A}^{π}(e) must be equal to b_{e}^{−} = b_{r(e)}.
• If e ∈ (E − Y) ∩ E_{2}, then j_{A}^{π}(e) can run freely in ]b_{e}^{−}, b_{e}^{+}]_{T}.

It follows that the coefficient (−1)^{|E_{1}|} ∑_{\begin{matrix}A \\ πj_{A}^{π}=ρ_{Y} \end{matrix}} μ(B^{−}, A) is equal to

(−1)^{|E_{1}|} ∏_{e∈Y ∩E_{1}} ( ∑_{a_{e}∈]b_{e}^{−},b_{e}^{+}]_{T}} μ(b_{e}^{−}, a_{e})) · ∏_{e∈(E−Y)∩E_{2}} ( ∑_{a_{e}∈]b_{e}^{−},b_{e}^{+}]_{T}} μ(b_{e}^{−}, a_{e}))

= (−1)^{|E_{1}|} · (−1)^{|Y ∩E_{1}|} · (−1)^{|(E−Y)∩E_{2}|}

= (−1)^{|(E−Y)∩E_{1}|} · (−1)^{|(E−Y)∩E_{2}|}

= (−1)^{|E−Y |} , using the fact that

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

This shows that

πj^{π} = ∑_{∅⊆Y ⊆E}(−1)^{|E−Y |}ρ_{Y} ,

as required.

(b) Suppose that Y is a proper subset of E and let g ∈ E be maximal such that g ∉ Y. We want to prove that g ∉ ρ_{Y} (P). We let p ∈ P and we prove that ρ_{Y} (p) ≠ g.

If p > g, then p ∈ Y and ρ_{Y} (p) = p ≠ g, while if p = g, then ρ_{Y} (g) ≠ g.

Assume now that p ≱ g and g ∈ E_{1}. Then p < g and ρ_{Y} (p) ≤ ρ_{Y} (g) = r(g) < g. Assume now that p ≱ g and g ∈ E_{2}. Then either p < g or p = ğ, the twin of g. If p < g and p ∈ E_{1}, then ρ_{Y} (p) ≤ p < g. If p < g and p ∈ E_{2}, then

ρ_{Y} (p) ≤ s(p) ≤ g. But s(p) is reducible since s(p) = p ∨ p̆, while g is irreducible. Therefore s(p) ≠ g, hence ρ_{Y} (p) < g.

If p = ğ, then p ∈ E_{2} and ρ_{Y} (p) is either p or s(p). But neither p nor s(p) is equal to g.

We have proved that ρ_{Y} (p) ≠ g in all cases, as was to be shown.

7.10. Remark. In the special case when T = P and π = id, we find that j^{id} is a linear combination of the maps ρ_{Y}. It turns out that j^{id} is actually an avatar of the element γ_{P^{op}} ∈ F_{P} (E^{0}) which is defined in (3.9), where E^{0} = Irr(P^{op}). We know that the element γ_{T} plays an important role throughout the theory of correspondence functors (see Section 9 of [BT3] and Section 4 of the present paper). The advantage of γ_{P^{op}} is that it has a uniform definition, contrary to j^{P} (as observed in Remark 7.2).

To make this explicit, let E = Irr(P), viewed as a subposet of P and E^{0} = Irr(P^{op}), viewed also as a subposet of P (so that it is actually (E^{0})^{op} which is the subposet of irreducible elements of P^{op}). Since P is a distributive lattice, it is isomorphic to the lattice I_{↓}(E) of all subsets of E closed under taking smaller elements. The passage to complements induces an isomorphism I_{↓}(E) ≅ I^{↑}(E)^{op}, where I^{↑}(E) is the lattice of all subsets of E closed under taking greater elements. On restriction to E, this induces an order-preserving isomorphism α : E → E^{0}, which turns out to map e ∈ E_{1} to r(e) ∈ E_{1}^{0} (in the totally ordered part) and e ∈ E_{2} to its twin ĕ ∈ E_{2}^{0} (in the twin part).

Now γ_{P^{op}} is a linear combination of maps E^{0} → P and we precompose it with ατ, where τ : E → E exchanges all the twins and fixes all the other elements. We obtain a linear combination of maps E → P and, after an explicit computation, it turns out that

γ_{P^{op}} α τ = ± j^{id} ,

the sign being actually (−1)^{|E_{1}|}. (This computation appears explicitly in the proof of Theorem 9.7, using a bijection ω : E^{0} → E which is actually the inverse of ατ.) The definition of ρ_{Y} in (7.8) was not uniform and, accordingly, j^{id} has a rather strange behavior. However, by means of the isomorphism ατ, the translation of all this in terms of γ_{P^{op}} becomes uniform.

Unfortunately, we need to work with j^{id} rather than γ_{P^{op}}. The reason is that γ_{P^{op}} ∈ F_{P} (E^{0}) is a linear combination of maps E^{0} → P, whereas, after composing with α, we obtain order-preserving maps E → P which are extendible to endomorphisms P → P (because P is a distributive lattice, see Lemma 2.3). The key fact is that endomorphisms are better because they can be composed, in particular it makes sense to consider idempotent endomorphisms.

We can now prove a main result concerning the composite j^{π}π and obtain consequences for the correspondence functor F_{P} associated with the pole lattice P.

7.11. Proposition. Let π ∈ Sur_{L}(T, P) and let j^{π} : P → T be the morphism defined in (7.4). Let q : F_{P} → F_{P} /H_{P} be the canonical surjection, where H_{P} is defined by (7.6).

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

(b) The composite of \overline{F}_{j^{π}} : F_{P} /H_{P} → F_{T} and q F_{π} : F_{T} → F_{P} /H_{P} is the identity morphism of F_{P} /H_{P}.

(c) \overline{F}_{j^{π}} : F_{P} /H_{P} → F_{T} is injective and embeds F_{P} /H_{P} as a direct summand of F_{T}.

(d) F_{j^{π}}F_{π} is an idempotent endomorphism of F_{T} whose image is isomorphic to F_{P} /H_{P}.

Proof : (a) This follows from (d), which is proved below, because the functor F_{?} : kL → F_{k} is fully faithful by Theorem 3.6. Alternatively, it is not difficult to compute directly

j^{π}πj^{π} = j^{π} ∑_{∅⊆Y ⊆E}(−1)^{|E−Y |}ρ_{Y} = j^{π} id_{P} + ∑_{Y ≠E}(−1)^{|E−Y |}j^{π}ρ_{Y} = j^{π} ,

because E ⊈ ρ_{Y} (P) if Y ≠ E by Proposition 7.9, hence j^{π}ρ_{Y} = 0 by Proposition 7.7. Then the equality j^{π}πj^{π} = j^{π} implies that j^{π}π is an idempotent.

(b) By Proposition 7.9, for any finite set X and any function φ : X → P,

F_{π}F_{j^{π}}(φ) = πj^{π}φ = ∑_{∅⊆Y ⊆E}(−1)^{|E−Y |}ρ_{Y} φ = φ + ∑_{Y ≠E}(−1)^{|E−Y |}ρ_{Y} φ .

But E ⊈ ρ_{Y} (P) if Y ≠ E by Proposition 7.9, hence E ⊈ ρ_{Y} φ(X). In other words, ρ_{Y} φ ∈ H_{P} (X), so that

F_{π}F_{j^{π}}(φ) = φ (mod H_{P} (X)) .

Composing with the canonical map q : F_{P} (X) → F_{P} (X)/H_{P} (X) and writing q(φ) = \overline{φ}, we obtain

qF_{π}\overline{F}_{j^{π}}(\overline{φ}) = qF_{π}F_{j^{π}}(φ) = q(φ) = \overline{φ} ,

as was to be shown.

(c) This follows immediately from (b).

(d) This follows immediately from (b) and the obvious equality \overline{F}_{j^{π}}q F_{π} = F_{j^{π}}F_{π}.

One of our aims is to show that the idempotents j^{π}π are orthogonal. In order to understand the product of two idempotents j^{θ}θ and j^{π}π we need to have more information about θj^{π}. This is the purpose of our next three propositions, but we first need a lemma.

7.12. Lemma. Let Q be a pole lattice, let θ ∈ Sur_{L}(T, Q), and let [t_{1}, t_{2}]_{T} be an interval in T. For every q ∈ Q, define

U^{q} = θ^{−1}(q) ∩ [t_{1}, t_{2}]_{T} = {a ∈ [t_{1}, t_{2}]_{T} | θ(a) = q)} ⊆ [t_{1}, t_{2}]_{T} .

Let q_{1} = θ(t_{1}) and assume that U^{q_{1}} is not reduced to the singleton t_{1}. Then for each q ∈ Q, we have ∑_{a∈U^{q}} μ(t_{1}, a) = 0.

Proof : The result is obvious if U^{q} = ∅, so we assume that U^{q} is nonempty. Since all elements of U^{q} have the same image under θ, so has their join and therefore U^{q} has a supremum

u^{q} = sup(U^{q}) ∈ U^{q} .

Now we have [t_{1}, t_{2}]_{T} = ⊔_{q∈Q}U^{q} and, by assumption, U^{q_{1}} = [t_{1}, u^{q_{1}}]_{T} is a nontrivial interval, so that

∑_{a∈U^{q_{1}}} μ(t_{1}, a) = 0 .

This is the starting point of an induction argument. We fix q ∈ Q and we assume by induction that ∑_{a∈U^{r}} μ(t_{1}, a) = 0 for every r ∈ Q such that q_{1} ≤ r < q. Then we

obtain

0 = ∑_{a∈[t_{1},u^{q}]_{T}} μ(t_{1}, a) = ∑_{q_{1}≤r≤q} ∑_{a∈U^{r}} μ(t_{1}, a)
= ∑_{a∈U^{q}} μ(t_{1}, a) + ∑_{q_{1}≤r<q} ∑_{a∈U^{r}} μ(t_{1}, a)
= ∑_{a∈U^{q}} μ(t_{1}, a) ,

using the induction assumption. This completes the proof.

7.13. Proposition. Let π ∈ Sur_{L}(T, P) and θ ∈ Sur_{L}(T, Q), where P and Q are pole lattices, and let j^{π} : P → T be the morphism defined in (7.4). Suppose that θj^{π} ≠ 0. Then the restriction of θ to the subset B = Im(π^{op}) is injective. In particular, |P | ≤ |Q|.

Proof : Let E = Irr(P). By the definition of j^{π}, we have

θj^{π} = (−1)^{|E_{1}|} ∑_{A} μ(B^{−}, A) θj_{A}^{π} = ∑_{ψ:P →Q} ((−1)^{|E_{1}|} ∑_{\begin{matrix}A \\ θj_{A}^{π}=ψ \end{matrix}} μ(B^{−}, A))ψ .

Now fix some morphism ψ : P → Q and, for every e ∈ E and every q ∈ Q, define

U_{e}^{q} = θ^{−1}(q) ∩ [b_{e}^{−}, b_{e}^{+}]_{T} = {a ∈ [b_{e}^{−}, b_{e}^{+}]_{T} | θ(a) = q)} ⊆ [b_{e}^{−}, b_{e}^{+}]_{T} .

Here, we write B = {b_{p} | p ∈ P }, as before. Then, since a join-morphism from P is entirely determined on E = Irr(P), we have

θj_{A}^{π} = ψ ⟺ j_{A}^{π}(e) ∈ U_{e}^{ψ(e)} ∀e ∈ E ⟺ a_{e} ∈ U_{e}^{ψ(e)} ∀e ∈ E .

In particular, if ψ appears in the expression of θj^{π}, then U_{e}^{ψ(e)} ≠ ∅ for every e ∈ E. It follows now that the coefficient of ψ is, up to sign, equal to

∑_{\begin{matrix}A \\ θj_{A}^{π}=ψ \end{matrix}} μ(B^{−}, A) = ∏_{e∈E} ( ∑_{a_{e}∈U_{e}^{ψ(e)}} μ(b_{e}^{−}, a_{e})) .

Suppose that θ_{|B} : B → Q is not injective. Then we want to prove that the coefficient of ψ is zero. This is the case if U_{e}^{ψ(e)} = ∅ for some e ∈ E, because we get an empty sum, which is zero. So we assume that U_{e}^{ψ(e)} ≠ ∅ for every e ∈ E.

The noninjectivity of θ_{|B} implies that there exist two adjacent elements w < y in P such that θ(b_{w}) = θ(b_{y}). There are three cases.

Case 1. y ∈ E_{1} and w = r(y). Then b_{w} = b_{y}^{−} and b_{y} = b_{y}^{+}. Choosing a ∈ U_{y}^{ψ(y)}, we obtain

θ(b_{w}) = θ(b_{y}^{−}) ≤_{Q} θ(a) ≤_{Q} θ(b_{y}^{+}) = θ(b_{y}) ,

hence θ(b_{y}^{−}) = θ(a) = θ(b_{y}^{+}). Since θ(a) = ψ(y), it follows that the whole interval [b_{y}^{−}, b_{y}^{+}]_{T} is mapped to ψ(y) under θ, that is, [b_{y}^{−}, b_{y}^{+}]_{T} = U_{y}^{ψ(y)}. But then

∑_{a_{y}∈U_{y}^{ψ(y)}} μ(b_{y}^{−}, a_{y}) = 0 ,

by the definition of the Möbius function (because b_{y}^{−} ≠ b_{y}^{+}). Therefore the coefficient of ψ is zero.

Case 2. w ∈ E_{2} and y = s(w). Then b_{w} = b_{w}^{−} and b_{y} = b_{w}^{+}. Choosing a ∈ U_{w}^{ψ(w)}, we obtain

θ(b_{w}) = θ(b_{w}^{−}) ≤_{Q} θ(a) ≤_{Q} θ(b_{w}^{+}) = θ(b_{y}) ,

hence θ(b_{w}^{−}) = θ(a) = θ(b_{w}^{+}). Since θ(a) = ψ(w), it follows that the whole interval [b_{w}^{−}, b_{w}^{+}]_{T} is mapped to ψ(w) under θ, that is, [b_{w}^{−}, b_{w}^{+}]_{T} = U_{w}^{ψ(w)}. But then

∑_{a_{w}∈U_{w}^{ψ(w)}} μ(b_{w}^{−}, a_{w}) = 0 ,

and the coefficient of ψ is zero.

Case 3. y ∈ E_{2} and w = r(y). Let z = y̆ be the twin of y, so that b_{z}^{−} = b_{z}. Since w < z, we have b_{w} < b_{z} and

θ(b_{y} ∨ b_{z}) = θ(b_{y}) ∨ θ(b_{z}) = θ(b_{w}) ∨ θ(b_{z}) = θ(b_{w} ∨ b_{z}) = θ(b_{z}) .

Letting q_{1} = θ(b_{z}), we see that U_{z}^{q_{1}} contains both the minimal element b_{z} = b_{z}^{−} of the interval [b_{z}^{−}, b_{z}^{+}]_{T} and another element b_{y} ∨b_{z}, because b_{z} < b_{y} ∨b_{z} ≤ b_{y∨z} = b_{z}^{+}. Thus the assumption of Lemma 7.12 is satisfied and it follows that

∑_{a_{z}∈U_{z}^{ψ(z)}} μ(b_{z}^{−}, a_{z}) = 0 .

Again the coefficient of ψ is zero and we are done.

This completes the proof of the injectivity of θ_{|B} : B → Q. Since π^{op} is injective (by Lemma 2.5), its image B has cardinality |P | and therefore |P | ≤ |Q|.

Let Q be a pole lattice and χ ∈ Sur_{L}(T, Q). Using a slightly abusive notation, it is convenient to define Ker(j^{χ}) to be the kernel of left composition with j^{χ}. More precisely, for our fixed lattice P, we let Ker(j^{χ}) be the kernel of the k-linear map

kL(P, Q) ⟶ kL(P, T) , ψ ↦ j^{χ}ψ .

We use this notation in the rest of the present section.

7.14. Proposition. Let π ∈ Sur_{L}(T, P) and θ, χ ∈ Sur_{L}(T, Q), where P and Q are pole lattices. Suppose that j^{χ}θj^{π} ≠ 0.

(a) There exists a unique isomorphism τ : P → Q of lattices such that

θj^{π} = τ (mod Ker(j^{χ})) .

(b) Moreover, θ(b_{p}) = τ(p), for all p ∈ P (where b_{p} = b_{p}^{π} = π^{op}(p), as before).

Proof : We assume that j^{χ}θj^{π} ≠ 0, and in particular θj^{π} ≠ 0. Let E = Irr(P) and write first

θj^{π} = (−1)^{|E_{1}|} ∑_{A} μ(B^{−}, A) θj_{A}^{π} = ∑_{ψ:P →Q} ((−1)^{|E_{1}|} ∑_{\begin{matrix}A \\ θj_{A}^{π}=ψ \end{matrix}} μ(B^{−}, A))ψ .

Let ψ : P → Q be a map appearing with a nonzero coefficient in the expression of θj^{π} and let A be such that θj_{A}^{π} = ψ. Since j^{χ}θj^{π} ≠ 0, we can also assume that ψ is such that j^{χ}ψ ≠ 0. Proposition 7.7 implies that the function ψ : P → Q must satisfy Irr(Q) ⊆ ψ(P). Since ψ = θj_{A}^{π} is a join-morphism and Irr(Q) generates Q, the map ψ : P → Q must be surjective. By Proposition 7.13, θj^{π} ≠ 0 implies that |P | ≤ |Q|. Therefore |P | = |Q|. It follows that ψ is a bijective join-morphism, hence an isomorphism of lattices.

Proposition 7.13 also asserts that the map θ_{|B} : B → Q is injective. Since |B| = |P | = |Q|, it is a bijection and therefore there is a unique isomorphism τ : P → Q such that

θ(b_{p}) = τ(p) , ∀ p ∈ P .

For any e ∈ E, we have ψ(e) = θj_{A}^{π}(e) = θ(a_{e}) for some a_{e} ∈ [b_{e}^{−}, b_{e}^{+}]_{T}. If e ∈ E_{1}, then b_{e}^{+} = b_{e}, hence

ψ(e) = θ(a_{e}) ≤_{Q} θ(b_{e}) = τ(e) .

Therefore e ≤_{P} ψ^{−1}τ(e), so that ψ^{−1}τ(e) = e because ψ^{−1}τ is an automorphism of P, hence height-preserving. Similarly, if e ∈ E_{2}, then b_{e}^{−} = b_{e}, hence

τ(e) = θ(b_{e}) ≤_{Q} θ(a_{e}) = ψ(e) ,

so that ψ^{−1}τ(e) ≤_{P} e and ψ^{−1}τ(e) = e. This shows that ψ_{|E} = τ_{|E}, hence ψ = τ. Therefore, whenever A is such that j^{χ}θj_{A}^{π} ≠ 0, then θj_{A}^{π} = τ. It follows that the functions ψ which appear with a nonzero coefficient in the expression of θj^{π} are τ and maps P → Q lying in Ker(j^{χ}).

In the situation of Proposition 7.14, we can replace θ by θ′ := τ^{−1}θ and j^{χ} by j^{χ′} := j^{χ} τ. The effect of this is that we are reduced to the case where Q = P and τ = id_{P}, that is,

θ(b_{p}) = p , ∀ p ∈ P .

For simplicity, we use this reduction in our final result, which is the key for understanding the composition of the morphisms we have introduced.

7.15. Proposition. Let π, θ, χ ∈ Sur_{L}(T, P), where P is a pole lattice. Suppose that θ(b_{p}) = p for all p ∈ P (where b_{p} = b_{p}^{π} = π^{op}(p), as before).

(a) If j^{χ}θj^{π} ≠ 0, then θ = π.

(b) We have

j^{χ}θj^{π} = \begin{cases}j^{χ} & if θ = π , \\ 0 & if θ ≠ π . \end{cases}

Proof : (a) By Proposition 7.14, we have θj^{π} = id_{P} (mod Ker(j^{χ})), because the automorphism τ is the identity by assumption. Moreover, as in the proof of the previous propositions, the coefficient of id_{P} in the expression of θj^{π} is equal to

(−1)^{|E_{1}|} ∑_{\begin{matrix}A \\ θj_{A}^{π}=id_{P} \end{matrix}} μ(B^{−}, A) = (−1)^{|E_{1}|} ∏_{e∈E} ( ∑_{a_{e}∈U_{e}} μ(b_{e}^{−}, a_{e})) ,

where we write simply

U_{e} := U_{e}^{e} = {a ∈ [b_{e}^{−}, b_{e}^{+}]_{T} | θ(a) = e} ⊆ [b_{e}^{−}, b_{e}^{+}]_{T} .

Since the coefficient of id_{P} in the expression of θj^{π} is nonzero (it is 1), every sum ∑_{a_{e}∈U_{e}} μ(b_{e}^{−}, a_{e}) is nonzero, and in particular U_{e} ≠ ∅.

As in the proof of Lemma 7.12, U_{e} has a supremum u_{e} ∈ U_{e}. We also define

V_{e} := {a ∈ [b_{e}^{−}, b_{e}^{+}]_{T} | θ(a) ≤_{P} r(e)}

so that [b_{e}^{−}, u_{e}]_{T} = V_{e} ⊔ U_{e}, because any a ≤_{T} u_{e} satisfies θ(a) ≤_{P} e, hence either θ(a) ≤ r(e) or θ(a) = e. There are two cases.

Case A. V_{e} = ∅. Then U_{e} = [b_{e}^{−}, u_{e}]_{T}. The nonzero sum ∑_{a_{e}∈U_{e}} μ(b_{e}^{−}, a_{e}) forces b_{e}^{−} = u_{e}, hence U_{e} = {b_{e}^{−}}.

Case B. V_{e} ≠ ∅. Then again V_{e} must have a supremum v_{e} ∈ V_{e}, so that V_{e} = [b_{e}^{−}, v_{e}]_{T} and v_{e} <_{T} u_{e}. In that case, we obtain

0 ≠ ∑_{a_{e}∈U_{e}} μ(b_{e}^{−}, a_{e}) = ∑_{a_{e}∈[b_{e}^{−},u_{e}]_{T}} μ(b_{e}^{−}, a_{e}) − ∑_{a_{e}∈V_{e}} μ(b_{e}^{−}, a_{e}) = − ∑_{a_{e}∈V_{e}} μ(b_{e}^{−}, a_{e}) ,

because the sum over [b_{e}^{−}, u_{e}]_{T} is zero since b_{e}^{−} ≤_{T} v_{e} <_{T} u_{e}. Therefore

∑_{a_{e}∈[b_{e}^{−},v_{e}]_{T}} μ(b_{e}^{−}, a_{e}) = ∑_{a_{e}∈V_{e}} μ(b_{e}^{−}, a_{e}) ≠ 0

and this forces b_{e}^{−} = v_{e}, hence U_{e} =]b_{e}^{−}, u_{e}].

By assumption, we know that θ(b_{e}) = e for all e ∈ E, hence b_{e} ∈ U_{e}. If e ∈ E_{2}, then b_{e} = b_{e}^{−}, hence b_{e}^{−} ∈ U_{e}. This forces to be in case A and therefore we obtain :

Case A. U_{e} = {b_{e}} if e ∈ E_{2}.

If e ∈ E_{1}, then b_{e} = b_{e}^{+}, hence b_{e}^{+} ∈ U_{e}. This forces to be in case B with moreover u_{e} = b_{e}^{+}. Since b_{e}^{−} = b_{r(e)}, we get :

Case B. U_{e} =]b_{r(e)}, b_{e}] if e ∈ E_{1}.

Let c_{p} = θ^{op}(p) = sup (θ^{−1}(p)) (that is, c_{p} = b_{p}^{θ} using Notation 7.1). Since we assume that θ(b_{p}) = p, we have b_{p} ≤_{T} c_{p} for all p ∈ P. We now prove that b_{p} = c_{p} by descending induction in the lattice T, starting from the obvious equality bˆ1_{P} = ˆ1_{T} = cˆ1_{P}. For simplicity, we write < and ≤ for the order relation in T. Suppose now that p ∈ P and b_{q} = c_{q} for every q > p. We have to discuss three cases.

Assume that p = r(e) with e ∈ E_{1}. Then b_{p} ≤ c_{p} < c_{e} = b_{e}, hence c_{p} ∈ [b_{r(e)}, b_{e}]_{T} = {b_{p}} ⊔ U_{e} (Case B). But θ(c_{p}) = p ≠ e, so c_{p} ∉ U_{e}. Therefore c_{p} = b_{p}.

Assume that p ∈ E_{2}. Then

b_{p}^{−} = b_{p} ≤ c_{p} < c_{s(p)} = b_{s(p)} = b_{p}^{+} .

Therefore c_{p} ∈ [b_{p}^{−}, b_{p}^{+}]_{T} = U_{p}⊔]b_{p}^{−}, b_{p}^{+}]_{T} (Case A). Since θ(c_{p}) = p, we have c_{p} ∈ U_{p} = {b_{p}}, hence c_{p} = b_{p}.

Assume now that p = e ∧ ĕ where e ∈ E_{2} with twin ĕ. Then b_{e} = c_{e} and bĕ = cĕ. Thus we obtain

b_{p} = b_{e∧ĕ} = b_{e} ∧ bĕ = c_{e} ∧ cĕ = c_{e∧ĕ} = c_{p} ,

as was to be shown. We have now covered all cases, completing the proof that b_{p} = c_{p} for all p ∈ P.

Now we obtain θ^{op}(p) = c_{p} = b_{p} = π^{op}(p) for all p ∈ P, hence θ^{op} = π^{op}. Passing to the opposite, it follows that θ = π, as was to be shown.

(b) We now know by (a) that θ = π whenever j^{χ}θj^{π} ≠ 0. Moreover, in that case, Proposition 7.14 implies that πj^{π} is the sum of id_{P} and an element of Ker(j^{χ}), using our assumption that θ(b_{p}) = p, for all p ∈ P. Applying j^{χ}, it follows that j^{χ}πj^{π} = j^{χ}.

Keeping our fixed finite lattice T, we now allow the pole lattice P to vary.

7.16. Notation.

(a) Pol_{T} is a set of representatives of isomorphism classes of pole lattices P such that Sur_{L}(T, P) is nonempty (hence in particular |P | ≤ |T |, so that Pol_{T} is finite).

(b) For any P ∈ Pol_{T}, the group Aut(P) acts on Sur_{L}(T, P) (by composition) and we let \overline{Sur_{L}(T, P)} be a fixed chosen set of representatives of the orbits.

(c) If χ, θ ∈ \overline{Sur_{L}(T, P)} and τ ∈ Aut(P), we define

f_{χ,τ,θ} = j^{χ}τθ : T ⟶ T .

In particular, f_{π,id_{P} ,π} = j^{π}π is the idempotent of Proposition 7.11.

7.17. Remark.

(a) Let χ′ = σχ ∈ Sur_{L}(T, P) be the image of χ under the action of σ, for some σ ∈ Aut(P). Then j^{χ′} = j^{χ}σ^{−1}. This is proved by going back to Notation 7.1 and using the associated elements b_{p}^{χ} = χ^{op}(p), respectively

b_{p}^{χ′} = χ′^{op}(p) = χ^{op}(σ^{op}(p)) = χ^{op}σ^{−1}(p) = b_{σ^{−1}(p)}^{χ} ,

from which the associated morphism j^{χ}, respectively j^{χ′}, is constructed, as in (7.4). It is then elementary to check that j^{χ′} = j^{χ}σ^{−1}.

(b) Changing the choice of orbit representatives has the following effect. Let χ′ = σχ ∈ Sur_{L}(T, P) and θ′ = ρθ ∈ Sur_{L}(T, P), where σ, ρ ∈ Aut(P). It follows from (a) that we obtain j^{χ′}στρ^{−1}θ′ = j^{χ}τθ.

(c) In particular, f_{π,id_{P} ,π} = j^{π}π is independent of the choice of π in its Aut(P)- orbit.

Now we come to the crucial relations among the endomorphisms f_{χ,τ,θ}.

7.18. Theorem. Let T be a finite lattice and let P, Q ∈ Pol_{T}.

(a) Let χ, θ ∈ \overline{Sur_{L}(T, P)} and τ ∈ Aut(P). Let also π, κ ∈ \overline{Sur_{L}(T, Q)} and σ ∈ Aut(Q). Then

f_{χ,τ,θ} f_{π,σ,κ} = \begin{cases}f_{χ,τσ,κ} & if P = Q and θ = π , \\ 0 & otherwise . \end{cases}

(b) When P varies in Pol_{T} and π varies in \overline{Sur_{L}(T, P)}, the idempotents f_{π,id_{P} ,π} are pairwise orthogonal.

Proof : Let χ′ = τ^{−1}χ ∈ Sur_{L}(T, P), so that j^{χ′} = j^{χ}τ, by Remark 7.17. If P ≠ Q, there is no isomorphism between P and Q, by our choice of Pol_{T}. Therefore we obtain j^{χ′}θj^{π} = 0, by Proposition 7.14. It follows that

f_{χ,τ,θ} f_{π,σ,κ} = j^{χ}τθj^{π}σκ = j^{χ′}θj^{π}σκ = 0 .

So we now assume that P = Q. Suppose that f_{χ,τ,θ} f_{π,σ,κ} ≠ 0. In particular, j^{χ}τθj^{π} ≠ 0, that is, j^{χ′}θj^{π} ≠ 0. By Proposition 7.14, there is a unique isomorphism ρ : P → P such that θj^{π} = ρ (mod Ker(j^{χ′})). Let θ′ = ρ^{−1}θ and χ′′ = ρ^{−1}χ′, hence j^{χ′′} = j^{χ′}ρ = j^{χ}τρ. Then we obtain

0 ≠ j^{χ′}θj^{π} = j^{χ}τθj^{π} = j^{χ}τρρ^{−1}θj^{π} = j^{χ′′}θ′j^{π} .

Moreover, since θj^{π} = ρ + h with j^{χ′}h = 0, we have j^{χ′′}ρ^{−1}h = 0. Therefore

θ′j^{π} = ρ^{−1}θj^{π} = id_{P} +ρ^{−1}h = id_{P} (mod Ker(j^{χ′′})) .

The uniqueness of the automorphism in Proposition 7.14 also implies that we have θ′(b_{p}^{π}) = p for all p ∈ P (where b_{p}^{π} = π^{op}(p), as before).

We are now in the assumptions of Proposition 7.15 for π, θ′, and χ′′. We deduce that θ′ = π, so that θ and π belong to the same orbit under the action of Aut(P). But θ and π belong to a chosen system of representatives \overline{Sur_{L}(T, P)}. Thus we must have ρ = id_{P} and θ = π.

It now follows that we can write θj^{π} = id_{P} +h, where j^{χ′}h = 0, that is, j^{χ}τh = 0. Therefore

f_{χ,τ,θ} f_{π,σ,κ} = j^{χ}τθj^{π}σκ = j^{χ}τ(id_{P} +h)σκ = j^{χ}τσκ = f_{χ,τσ,κ} ,

as was to be shown.

(b) This follows from (a).

8. Subalgebras corresponding to pole lattices

In this section, we show how the results of Section 7 imply some precise information about the structure of the endomorphism algebra End_{kL}(T) of a finite lattice T, where k is a commutative ring.

We continue to use Notation 7.16, so P is a pole lattice running through the set Pol_{T} and \overline{Sur_{L}(T, P)} denotes a set of representatives of Aut(P)-orbits in Sur_{L}(T, P). Let M_{n(T,P)}(k Aut(P)) denote the matrix algebra of size n(T, P) = |\overline{Sur_{L}(T, P)}|, with rows and columns indexed by the set \overline{Sur_{L}(T, P)}, and coefficients in the group algebra k Aut(P). If χ, θ ∈ \overline{Sur_{L}(T, P)} and τ ∈ Aut(P), we let m_{χ,τ,θ} denote the elementary matrix having coefficient τ in position (χ, θ) and zero elsewhere.

With this notation, we can now state a main result, which was already obtained in [BT3] in the special case when P runs over totally ordered lattices.

8.1. Theorem. Let T be a finite lattice. For each P ∈ Pol_{T}, let \overline{Sur_{L}(T, P)} be a set of representatives of the orbits for the action of the group Aut(P) on Sur_{L}(T, P) and let n(T, P) = |\overline{Sur_{L}(T, P)}|.

(a) The map

I_{T} : ⨁_{P ∈Pol_{T}} M_{n(T,P)}(k Aut(P)) ⟶ End_{kL}(T) , m_{χ,τ,θ} ↦ f_{χ,τ,θ}

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 join-morphisms T → T whose image is a pole lattice.

Proof : (a) Let P, Q ∈ Pol_{T}. If P ≠ Q, then m_{χ,τ,θ} and m_{π,σ,κ} are not in the same block, so their product is 0, while the product f_{χ,τ,θ}f_{π,σ,κ} is also zero. If P = Q, then the relations of Theorem 7.18 are the standard relations within a matrix algebra of size n(T, P) with coefficients in the group algebra k Aut(P).

(b) Since the elements m_{χ,τ,θ} form a k-basis of ⨁_{P ∈Pol_{T}} M_{n(T,P)}(k Aut(P)), it suffices to prove that their images f_{χ,τ,θ} are k-linearly independent. Suppose that

∑_{χ,τ,θ} λ_{χ,τ,θ} f_{χ,τ,θ} = 0 ,

where λ_{χ,τ,θ} ∈ k. Multiplying on the left by f_{π,id_{P} ,κ} and on the right by f_{π,σ,π}, we are left with the terms for which χ = κ and θ = π. Therefore we obtain

∑_{τ} λ_{κ,τ,π} f_{π,id_{P} ,κ} f_{κ,τ,π} f_{π,σ,π} = ∑_{τ} λ_{κ,τ,π} f_{π,τσ,π} = 0 .

Now, by Definition 7.4, f_{π,τσ,π} = j^{π}τσπ is a linear combination of distinct maps T → T, one of them being j_{B^{−}}^{π}τσπ, appearing with coefficient ±1, where we use Notation 7.1 and set B^{−} = (b_{e}^{−})_{e∈E}. We claim that the functions j_{B^{−}}^{π}τσπ are pairwise distinct when τ varies. This implies that each coefficient λ_{κ,τ,π} must be zero, proving the required linear independence.

To prove the claim, we write for simplicity ρ = τσ and we allow ρ to vary. The group Aut(P) is isomorphic to C_{2} ×C_{2} ×…×C_{2}, where each C_{2} acts by exchanging two twin elements of E_{2} and fixing the others (where E = Irr(P), as before). So we consider some e ∈ E_{2} and we let ĕ be its twin. Then we get

j_{B^{−}}^{π}ρπ(b_{e}) = j_{B^{−}}^{π}ρ(e) = b_{ρ(e)}^{−} = \begin{cases}b_{e}^{−} & if ρ(e) = e , \\ b− ; ĕ & if ρ(e) = ĕ. \end{cases}

We see that the functions j_{B^{−}}^{π}ρπ are pairwise distinct when ρ varies, proving the claim.

(c) It is clear that E_{T} is a subalgebra. Moreover, every map j_{A}^{χ} is a join-morphism, where A is a family as in Lemma 7.3. Therefore j_{A}^{χ}τθ is a join-morphism whose image is a pole lattice, by construction. It follows that f_{χ,τ,θ} = j^{χ}τθ belongs to E_{T} and hence Im(I_{T}) ⊆ E_{T}.

Now we want to show that E_{T} has a k-basis consisting of all morphisms φ_{λ,τ,π} described as follows. First we fix P ∈ Pol_{T} and we let

φ_{λ,τ,π} = λτπ ,

where π ∈ \overline{Sur_{L}(T, P)}, τ ∈ Aut(P), λ ∈ \overline{Inj_{L}(P, T)}, and where \overline{Inj_{L}(P, T)} denotes a set of representatives of Aut(P)-orbits in Inj_{L}(P, T). If φ : T → T is a join-morphism whose image is isomorphic to P, it is the composite φ = λ̃π̃ of a surjection π̃ ∈ Sur_{L}(T, P) and an injection λ̃ ∈ Inj_{L}(P, T). By our choice of orbit representatives, we have π̃ = τ_{1}π where π ∈ \overline{Sur_{L}(T, P)} and τ_{1} ∈ Aut(P), and similarly λ̃ = λτ_{2} where λ ∈ \overline{Inj_{L}(P, T)} and τ_{2} ∈ Aut(P). Then φ = λτπ = φ_{λ,τ,π}, where τ = τ_{2}τ_{1}. It follows that

\{φ_{λ,τ,π} | π ∈ \overline{Sur_{L}(T, P)} , λ ∈ \overline{Inj_{L}(P, T)} , τ ∈ Aut(P)\}

is a k-basis of the submodule E_{T,P} generated by all endomorphisms whose image is isomorphic to P. Allowing P to vary in Pol_{T}, we deduce that

B = ⋃_{P ∈Pol_{T}} \{φ_{λ,τ,π} | π ∈ \overline{Sur_{L}(T, P)} , λ ∈ \overline{Inj_{L}(P, T)} , τ ∈ Aut(P)\}

is a k-basis of E_{T} = ⨁_{P ∈Pol_{T}} E_{T,P}.

On the other hand, it follows from (a) and (b) that

B′ = ⋃_{P ∈Pol_{T}} \{f_{χ,τ,θ} | χ, θ ∈ \overline{Sur_{L}(T, P)} , τ ∈ Aut(P)\}

is a k-basis of Im(I_{T}). By Lemma 2.6, there is a bijection between Inj_{L}(P, T) and Sur_{L}(T, P). We can also choose representatives to obtain a bijection between \overline{Inj_{L}(P, T)} and \overline{Sur_{L}(T, P)}, because Aut(P) acts freely on each side. 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. We want to prove that the inclusion Im(I_{T}) ⊆ E_{T} is an equality (which is obvious if k is a field since the dimensions are equal).

We now allow the base ring k to vary and we write a superscript (k) to emphasize the dependence on k. Thus we have an injective algebra homomorphism

I_{T}^{(k)} : ⨁_{P ∈Pol_{T}} M_{n(T,P)}(k Aut(P)) ⟶ E_{T}^{(k)} ⊆ End_{kL}(T)

and we let X^{(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}} & X^{(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 X^{(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 k is right exact, so we obtain an exact sequence

\xymatrix{k ⊗ Im(I_{T}^{(ℤ)}) \ar[r]^{1⊗j_{ℤ}} & k ⊗ E_{T}^{(ℤ)} \ar[r]^{1⊗p_{ℤ}} & k ⊗ X^{(ℤ)} \ar[r] & 0 .}

Using the canonical bases B and B′ of Im(I_{T}^{(k)}) and E_{T}^{(k)} respectively, we see that

k ⊗ Im(I_{T}^{(ℤ)}) ≅ Im(I_{T}^{(k)})  and  k ⊗ E_{T}^{(ℤ)} ≅ E_{T}^{(k)} .

Moreover the map 1 ⊗ j_{ℤ} corresponds, under these isomorphisms, to the inclusion map j_{k}. In particular, considering the prime field 𝔽_{p} for any prime number p, we obtain an exact sequence

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

Since 𝔽_{p} is a field and the dimensions are equal, the inclusion map j_{𝔽_{p}} is an equality. Therefore 𝔽_{p} ⊗ X^{(ℤ)} = {0} and this holds for every prime p. Thus we must have X^{(ℤ)} = {0}, because X^{(ℤ)} is finite, so that the inclusion map j_{ℤ} : Im(I_{T}^{(ℤ)}) → E_{T}^{(ℤ)} is an equality. Tensoring with k, it follows that the inclusion map j_{k} : Im(I_{T}^{(k)}) → E_{T}^{(k)} is an equality as well, as required.

8.2. Remark. Let B and B′ be the two bases of E_{T} = Im(I_{T}) described in the proof. The change of basis from B to B′ is not obvious. By construction, every map j_{A}^{χ}τθ belongs to B, but beware of the fact that if θ and χ belong to Sur_{L}(T, P), then j_{A}^{χ}τθ may be a composite T → P′ → T for some pole lattice P′ smaller than P. This is because, in the construction of j_{A}^{χ}, the family A = (a_{e})_{e∈E} does not necessarily consist of distinct elements (where E = Irr(P) as before).

The image under I_{T} of the identity element of ⨁_{P ∈Pol_{T}} M_{n(T,P)}(k Aut(P)) is an idempotent e_{T} of End_{kL}(T) and e_{T} is an identity element of E_{T}. We now prove more.

8.3. Theorem. For every finite lattice T, let E_{T} = Im(I_{T}) be the subalgebra of End_{kL}(T) appearing in Theorem 8.1, and let e_{T} be the identity element of E_{T}.

(a) e_{T} = ∑_{P ∈Pol_{T}} ∑_{π∈\overline{Sur_{L}(T,P)}} f_{π,id_{P} ,π}.

(b) For any finite lattice T′ and any morphism α ∈ Hom_{kL}(T, T′), we have αe_{T} = e_{T′}α. 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 ⨁_{P ∈Pol_{T}} M_{n(T,P)}(k Aut(P)) is equal to

∑_{P ∈Pol_{T}} ∑_{π∈\overline{Sur_{L}(T,P)}} m_{π,id_{P} ,π} .

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

(b) We have seen in the proof of Theorem 8.1 that every element of the canonical basis B of E_{T} has the form φ_{λ,τ,π} = λτπ, where π ∈ \overline{Sur_{L}(T, P)}, λ ∈ \overline{Inj_{L}(P, T)} and τ ∈ Aut(P). Passing to the opposite, we obtain

φ_{λ,τ,π}^{op} = π^{op}τ^{op}λ^{op} = π^{op}τ^{−1}λ^{op}

with π^{op} ∈ Inj_{L}(P^{op}, T^{op}) and λ^{op} ∈ Sur_{L}(T^{op}, P^{op}). It follows that the opposite of the canonical basis element φ_{λ,τ,π} of E_{T} is the canonical basis element φ_{π^{op},τ^{−1},λ^{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. Therefore (e_{T})^{op} = e_{T^{op}}.

Now if α : T → T′ is a join-morphism (i.e. α is in L), then the image of a pole sublattice of T is a pole sublattice of T′. It follows that composition with α maps e_{T} to a linear combination of join-morphisms with a pole lattice as an image, hence invariant under the idempotent element e_{T′}. In other words, we have

α e_{T} = e_{T′} α e_{T} .

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

e_{T′} α e_{T} = e_{T′} α .

The two displayed equations yield αe_{T} = e_{T′}α. This holds as well if α is replaced by a k-linear combination of join-morphisms (i.e. α is in kL), as was to be shown.

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

(d) This follows immediately from (c).

9. Correspondence functors for pole lattices

In this section, we first consider the special case of the endomorphism algebra of a pole lattice Q. We determine completely the structure of this algebra. Applying then the fully faithful functor T ↦ F_{T}, we deduce a direct sum decomposition of the correspondence functor F_{Q}, providing an explicit description of F_{Q} for any pole lattice Q. In particular, when k is a field of characteristic different from 2, F_{Q} is semi-simple. At the end of the section, we return to an arbitrary finite lattice T and describe direct summands of F_{T} corresponding to pole lattices inside T. The results are generalizations of those obtained in [BT3] in the special case of totally ordered lattices.

9.1. Theorem. Let Q be a pole lattice.

(a) The homomorphism of k-algebras of Theorem 8.1

I_{Q} : ⨁_{P ∈Pol_{Q}} M_{n(Q,P)}(k Aut(P)) ⟶ End_{kL}(Q) , m_{χ,τ,θ} ↦ f_{χ,τ,θ} ,

is an isomorphism.

(b) In particular, if k is a field and if either Q is totally ordered or if k is a field of characteristic different from 2, then End_{kL}(Q) is semi-simple.

Proof : (a) Since any join-morphism φ : Q → Q has an image which is a pole lattice, the subalgebra E_{Q} of End_{kL}(Q) appearing in Theorem 8.1 is the whole of End_{kL}(Q). Therefore, the homomorphism I_{Q} is surjective. By Theorem 8.1, I_{Q} is injective, hence an isomorphism.

(b) If Q is totally ordered, then so is each P and Aut(P) is the trivial group. Thus we get matrix algebras M_{n(Q,P)}(k). If Q is not totally ordered, then each Aut(P) is a 2-group (and at least one of them is nontrivial, namely Aut(Q)). The group algebra k Aut(P) is semi-simple when the characteristic of k is different from 2 (Maschke’s theorem). Therefore any matrix algebra M_{q}(k Aut(P)) is semi-simple and it follows that the direct sum is semi-simple as well.

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

9.2. Notation. For any pole lattice P ∈ Pol_{Q}, set

β_{Q,P} := ∑_{π∈\overline{Sur_{L}(Q,P)}} f_{π,id_{P} ,π} .

In particular, when P = Q, then Sur_{L}(Q, Q) = Aut(Q) and \overline{Sur_{L}(Q, Q)} is a singleton which can be chosen to be {id_{Q}}. We then define

ε_{Q} := β_{Q,Q} = f_{id_{Q},id_{Q},id_{Q}} = j^{id^{Q}} = ∑_{∅⊆Y ⊆E}(−1)^{|E−Y |}ρ_{Y} ,

using Proposition 7.9, with E = Irr(Q) and ρ_{Y} ∈ End_{L}(Q) defined by (7.8).

9.3. Proposition. The elements β_{Q,P}, for P ∈ Pol_{Q}, are orthogonal central idempotents of End_{kL}(Q), and their sum is equal to the identity. In particular, the central idempotent ε_{Q} satisfies

ε_{Q} End_{kL}(Q) ≅ k Aut(Q) .

Proof : For every π ∈ \overline{Sur_{L}(Q, P)}, the inverse image of f_{π,id_{P} ,π} under the algebra isomorphism I_{Q} of Theorem 9.1 is the matrix m_{π,id_{P} ,π} of the component M_{n(Q,P)}(k Aut(P)) indexed by P. Summing over all π ∈ \overline{Sur_{L}(Q, P)}, it follows that the inverse image of β_{Q,P} under I_{Q} is the identity element of M_{n(Q,P)}(k Aut(P)). The first statement follows.

In the case P = Q, we know that \overline{Sur_{L}(Q, Q)} is a singleton, so that the corresponding matrix algebra has size 1. The inverse image of ε_{Q} under I_{Q} is the identity element m_{id_{Q},id_{Q},id_{Q}} of the component M_{1}(k Aut(Q)) ≅ k Aut(Q). Clearly ε_{Q} End_{kL}(Q) ≅ M_{1}(k Aut(Q)) ≅ k Aut(Q).

We want to use the fully-faithful functor F_{?} : kL → F_{k} (see Theorem 3.6) to deduce information on the correspondence functor F_{Q}. We already know that F_{Q} is projective, because the pole lattice Q is distributive (see Theorem 3.6). We apply the functor F_{?} : kL → F_{k} to the map j^{π} ∈ Hom_{kL}(P, Q) defined in (7.4), where π ∈ \overline{Sur_{L}(Q, P)}. By Proposition 7.7 we obtain a morphism

F_{j^{π}} : F_{P} ⟶ F_{Q}

which vanishes on H_{P}, where H_{P} is defined by (7.6). By Proposition 7.11, this induces an injective morphism

\overline{F}_{j^{π}} : F_{P} /H_{P} ⟶ F_{Q}

which embeds F_{P} /H_{P} as a direct summand of F_{Q}, corresponding to the idempotent f_{π,id_{P} ,π} = j^{π}π. In particular, for P = Q, we have f_{id_{Q},id_{Q},id_{Q}} = j^{id^{Q}} = ε_{Q} and we obtain an idempotent endomorphism F_{ε_{Q}} of F_{Q} with kernel H_{Q}.

9.4. Theorem. Let Q be a pole lattice and define 𝕊_{Q} := F_{Q}/H_{Q}, where H_{Q} is defined by (7.6).

(a) 𝕊_{Q} is a projective correspondence functor.

(b) There are isomorphisms of correspondence functors

F_{ε_{Q}}F_{Q} ≅ 𝕊_{Q} ,
F_{β_{Q,P}} F_{Q} ≅ 𝕊_{P}^{n(Q,P)} , for each P ∈ Pol_{Q} ,
F_{Q} ≅ ⨁_{P ∈Pol_{Q}} 𝕊_{P}^{n(Q,P)} .

Proof : (a) Since the pole lattice Q is distributive, F_{Q} is projective (Theorem 3.6). Therefore so is its direct summand 𝕊_{Q}.

(b) Since the functor F_{?} : kL → F_{k} is fully faithful (Theorem 3.6), it induces an isomorphism of k-algebras

End_{kL}(Q) ≅ End_{F_{k}}(F_{Q}) .

Now the idempotents f_{π,id_{P} ,π} of End_{kL}(Q), for π ∈ \overline{Sur_{L}(Q, P)} and P ∈ Pol_{Q}, are orthogonal and their sum is equal to the identity, by Theorem 9.1. It follows that the endomorphisms F_{f_{π,id_{P} ,π}} of F_{Q} are orthogonal idempotents, and their sum is the identity. Hence we obtain a decomposition of correspondence functors

F_{Q} = ⨁_{\begin{matrix}P ∈Pol_{Q} \\ π∈\overline{Sur_{L}(Q,P)} \end{matrix}} F_{f_{π,id_{P} ,π}}(F_{Q}) .

By surjectivity of π : Q → P, the image of F_{f_{π,id_{P} ,π}} = F_{j^{π}}F_{π} : F_{Q} → F_{Q} is equal to the image of F_{j^{π}} : F_{P} → F_{Q}. Therefore F_{f_{π,id_{P} ,π}}(F_{Q}) = F_{j^{π}}(F_{P} ). By Proposition 7.11, the image F_{j^{π}}(F_{P} ) is isomorphic to 𝕊_{P} = F_{P} /H_{P} and it follows that

F_{f_{π,id_{P} ,π}}(F_{Q}) ≅ 𝕊_{P} .

Taking P = Q and f_{id_{Q},id_{Q},id_{Q}} = j^{id^{Q}} = ε_{Q}, we obtain the first isomorphism F_{ε_{Q}}F_{Q} ≅ 𝕊_{Q}. Summing over all π ∈ \overline{Sur_{L}(Q, P)} for a fixed P, we obtain the second isomorphism. Finally, summing over all P ∈ Pol_{Q} and all π ∈ \overline{Sur_{L}(Q, P)}, we obtain the third isomorphism.

9.5. Corollary. Let P and P′ be pole lattices. Then

Hom_{F_{k}}(𝕊_{P} , 𝕊_{P′}) ≅ \{ \begin{matrix}\{0\} & if P ≅ P̸′ , \\ k Aut(P) & if P ≅ P′ . \end{matrix}

Proof : Since 𝕊_{P} ≅ F_{ε_{P}} F_{P}, the case P = P′ follows from Proposition 9.3. Now if P ≅ P̸′, it is easy to choose a large enough pole lattice Q such that Sur_{L}(Q, P) ≠ ∅ and Sur_{L}(Q, P′) ≠ ∅. Using the central idempotents β_{Q,P} and β_{Q,P′} of Proposition 9.3, we obtain

Hom_{F_{k}}(F_{β_{Q,P}} F_{Q}, F_{β_{Q,P′}} F_{Q}) ≅ Hom_{F_{k}}(𝕊_{P} , 𝕊_{P′})^{n(Q,P)·n(Q,P′)} .

Since F_{β_{Q,P}} and F_{β_{Q,P′}} are central idempotents of End_{F_{k}}(F_{Q}), and since they are orthogonal if P ≅ P̸′, it follows that Hom_{F_{k}}(F_{β_{Q,P}} F_{Q}, F_{β_{Q,P′}} F_{Q}) = 0 if P ≅ P̸′, hence Hom_{F_{k}}(𝕊_{P} , 𝕊_{P′}) = {0}.

9.6. Remark. Corollary 9.5 actually holds for the fundamental functors associated with any two finite posets. This more general result will be proved in another paper.

Now we prove that the functor 𝕊_{Q} is actually isomorphic to a fundamental functor and we compute the ranks of all its evaluations.

9.7. Theorem. Let Q be a pole lattice and let R be the order relation on the set E = Irr(Q) of irreducible elements of Q. Let 𝕊_{Q} = F_{Q}/H_{Q}.

(a) 𝕊_{Q} is isomorphic to the fundamental functor 𝕊_{E,R^{op}}.

(b) For any finite set X, the k-module 𝕊_{Q}(X) is free of rank

rank(𝕊_{Q}(X)) = ∑_{i=0}^{|E|}(−1)^{i}\binom{|E|}{i}(|Q| − i)^{|X|} .

Proof : (a) We use the element γ_{Q^{op}} ∈ F_{Q}(E^{0}) defined in (3.9), where E^{0} = Irr(Q^{op}). By a well-known result of lattice theory (Theorem 6.2 in [Ro]), the distributive lattice Q^{op} is isomorphic to I_{↓}(E^{0}, R^{0}), where R^{0} is the order relation on E^{0} viewed as a subset of Q^{op}, so that (E^{0}, R^{0}) is the poset of irreducible elements in Q^{op}. Note that the isomorphism Q^{op} ≅ I_{↓}(E^{0}, R^{0}) can also be checked easily and directly because Q^{op} is a pole lattice. Recall that

γ_{Q^{op}} = ∑_{A⊆E^{0}}(−1)^{|A|}η_{A}^{∘} ,

where η_{A}^{∘} : E^{0} → Q denotes the same map as η : E^{0} → Q^{op} and η is defined by

∀e^{0} ∈ E^{0}, η_{A}(e^{0}) = \{ \begin{matrix}s(e^{0}) & if e^{0} ∈ A , \\ e^{0} & if e^{0} ∉ A , \end{matrix}

because r(e^{0}) in the lattice Q^{op} is equal to s(e^{0}) in the lattice Q.

Now we define ω : E^{0} → Q by

ω(e^{0}) = \{ \begin{matrix}s(e^{0}) & if e^{0} ∈ E_{1}^{0} , \\ e^{0} & if e^{0} ∈ E_{2}^{0} , \end{matrix}

and we notice that ω is actually a bijection between E^{0} and E = Irr(Q), because in a pole lattice we have E_{1} = s(E_{1}^{0}) and E_{2} = E_{2}^{0} (by an easy application of Lemma 2.1). Then ω ∈ F_{Q}(E^{0}) and when we apply the idempotent F_{ε_{Q}} we claim that we obtain

(9.8)  F_{ε_{Q}}(ω) = (−1)^{|E_{1}|}γ_{Q^{op}} .

The definition of ε_{Q} (see Notation 9.2) yields

F_{ε_{Q}}(ω) = ε_{Q} ω = ∑_{Y ⊆E}(−1)^{|E−Y |}ρ_{Y} ω .

The definition of ρ_{Y} in (7.8) splits into two cases. If e^{0} ∈ E_{1}^{0}, then

(ρ_{Y} ω)(e^{0}) = ρ_{Y} (s(e^{0})) = \{ \begin{matrix}s(e^{0}) & if s(e^{0}) ∈ Y , \\ r(s(e^{0})) = e^{0} & if s(e^{0}) ∉ Y . \end{matrix}

If now e^{0} ∈ E_{2}^{0}, then

(ρ_{Y} ω)(e^{0}) = ρ_{Y} (e^{0}) = \{ \begin{matrix}e^{0} & if e^{0} ∈ Y , \\ s(e^{0}) & if e^{0} ∉ Y . \end{matrix}

For each Y ⊆ E, we define A ⊆ E^{0} by

Y ∩ E_{1} = s(A ∩ E_{1}^{0})  and  Y ∩ E_{2} = E_{2} − (A ∩ E_{2}) .

Thus we have decompositions

Y = (Y ∩ E_{1}) ⊔ (Y ∩ E_{2}) ⊆ E  and  A = (A ∩ E_{1}^{0}) ⊔ (A ∩ E_{2}^{0}) ⊆ E^{0}

and A runs through all subsets of E^{0} when Y runs through all subsets of E. If e^{0} ∈ E_{1}^{0}, then

s(e^{0}) ∈ Y ∩ E_{1} ⟺ e^{0} ∈ A ∩ E_{1}^{0}

while if e^{0} ∈ E_{2}^{0}, then

e^{0} ∉ Y ∩ E_{2} ⟺ e^{0} ∈ A ∩ E_{2}^{0} .

Therefore the two cases merge into one and we obtain

(ρ_{Y} ω)(e^{0}) = \{ \begin{matrix}e^{0} & if e^{0} ∉ A , \\ s(e^{0}) & if e^{0} ∈ A , \end{matrix}

so that ρ_{Y} ω = η_{A}^{∘}.

As far as the signs are concerned, we have

|E − Y | = |E_{1} − (Y ∩ E_{1})| + |E_{2} − (Y ∩ E_{2})| = |E_{1}^{0} − (A ∩ E_{1}^{0})| + |A ∩ E_{2}^{0}| ,

hence

(−1)^{|E−Y |} = (−1)^{|E_{1}^{0}|} · (−1)^{|A∩E_{1}^{0}|} · (−1)^{|A∩E_{2}^{0}|} = (−1)^{|E_{1}|} · (−1)^{|A|} .

It now follows that

F_{ε_{Q}}(ω) = ∑_{Y ⊆E}(−1)^{|E−Y |}ρ_{Y} ω
= (−1)^{|E_{1}|} ∑_{A⊆E^{0}}(−1)^{|A|}η_{A}^{∘}
= (−1)^{|E_{1}|}γ_{Q^{op}} .

This proves Claim 9.8 above.

Now F_{Q} is generated by ω ∈ F_{Q}(E^{0}), because it is generated by ι ∈ F_{Q}(E) (where ι : E → Q is the inclusion), hence also by any injection from the set E^{0} to Q, by composing ι with a bijection between E^{0} and E. Since F_{ε_{Q}} is an idempotent endomorphism of the correspondence functor F_{Q}, we see that F_{ε_{Q}}F_{Q} is generated by F_{ε_{Q}}(ω). In other words, in view of Claim 9.8 above, F_{ε_{Q}}F_{Q} is generated by γ_{Q^{op}} ∈ F_{Q}(E^{0}). Now Theorem 3.10 asserts that the subfunctor of F_{Q} generated by γ_{Q^{op}} is isomorphic to 𝕊_{E^{0},R^{0}}, where (E^{0}, R^{0}) is the poset of irreducible elements in Q^{op}. But (E^{0}, R^{0}) ≅ (E, R^{op}) via the map ω : E^{0} → E described above. Therefore, using the isomorphism of Theorem 9.4, we obtain

𝕊_{Q} ≅ F_{ε_{Q}}F_{Q} = ⟨γ_{Q^{op}}⟩ ≅ 𝕊_{E^{0},R^{0}} ≅ 𝕊_{E,R^{op}} .

(b) By Definition 7.6, the canonical k-basis of 𝕊_{Q}(X) = F_{Q}(X)/H_{Q}(X) is the set Z(X) of all maps φ : X → Q such that E ⊆ φ(X) ⊆ Q. Therefore 𝕊_{Q}(X) is free of rank |Z(X)|. The number of maps in Z(X) has been computed in Lemma 8.1 of [BT2] and the formula is actually well-known. The formula asserts that this rank is equal to

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

as required.

9.9. Remark. In view of the projectivity of 𝕊_{Q} (Theorem 9.4), the isomorphism 𝕊_{Q} ≅ 𝕊_{E,R^{op}} provides another proof of the projectivity of the fundamental functor 𝕊_{E,R^{op}} whenever (E, R^{op}) is a pole poset. This was first proved in Theorem 5.1.

9.10. Remark. The formula for the rank in Theorem 9.7 is a special case of the general formula proved in [BT4] for the rank of the evaluation of any fundamental functor. We have given here a direct proof in the case of a pole lattice because it is easy, while the proof in the general case is much more elaborate.

When k is a field, we get even more.

9.11. Corollary. Let k be a field. Let Q be a pole lattice and let (E, R) be the poset of irreducible elements in Q.

(a) For any simple k Aut(Q)-module V, the functor 𝕊_{Q} ⊗_{k Aut(Q)} V is simple, isomorphic to S_{E,R^{op},V}.

(b) The correspondence functor 𝕊_{Q} is projective and injective.

(c) If either Aut(Q) is trivial (which occurs if Q is totally ordered) or if the characteristic of k is different from 2, the correspondence functor S_{E,R^{op},V} is simple, projective, and injective.

(d) Under the assumption of (c), 𝕊_{Q} decomposes as a direct sum of simple (and projective) functors

𝕊_{Q} ≅ ⨁_{V} S_{E,R^{op},V} ,

where V runs over simple k Aut(Q)-modules up to isomorphism.

(e) Under the assumption of (c), F_{Q} decomposes as a direct sum of simple (and projective) functors

F_{Q} ≅ ⨁_{P ∈Pol_{Q}} ⨁_{V_{P}} (S_{E_{P} ,R_{P}^{op},V_{P}})^{n(Q,P)} ,

where (E_{P} , R_{P}) denotes the poset of irreducible elements in P and where V_{P} runs over simple k Aut(P)-modules up to isomorphism.

Proof : (a) Using Lemma 2.1, it is easy to check that Aut(Q) = Aut(E, R) = Aut(E, R^{op}), so V is a k Aut(E, R^{op})-module. Recall that the fundamental correspondence functor 𝕊_{E,R^{op}} has a right k Aut(E, R^{op})-module structure (in the sense that each evaluation 𝕊_{E,R^{op}}(X) is a right k Aut(E, R^{op})-module, in a compatible way with all morphisms, which act on the left). Moreover, by Proposition 3.4, we know that the simple functor S_{E,R^{op},V} is obtained from the fundamental functor 𝕊_{E,R^{op}} by simply tensoring with V :

S_{E,R^{op},V} ≅ 𝕊_{E,R^{op}} ⊗_{Aut(E,R^{op})} V ,  that is,  S_{E,R^{op},V} ≅ 𝕊_{Q} ⊗_{k Aut(Q)} V ,

as required.

(b) 𝕊_{Q} is projective by Theorem 9.4. Since k is a field, it is also injective by Theorem 10.6 in [BT2].

(c) When either Aut(Q) is trivial or the characteristic of k is different from 2, k Aut(Q) is semi-simple and every simple k Aut(Q)-module is projective. Moreover, every simple k Aut(Q)-module has dimension 1 because Aut(Q) is an elementary abelian 2-group (the only roots of unity needed are ±1). Therefore we have an isomorphism of k Aut(Q)-modules

k Aut(Q) ≅ ⨁_{V simple} V ,

where V runs over all simple k Aut(Q)-modules up to isomorphism. It follows that

𝕊_{Q} ≅ 𝕊_{Q} ⊗_{k Aut(Q)} k Aut(Q) ≅ ⨁_{V simple} 𝕊_{Q} ⊗_{k Aut(Q)} V ≅ ⨁_{V simple} S_{E,R^{op},V} .

Since 𝕊_{Q} is projective and injective by (b), so is each of its simple direct summands S_{E,R^{op},V}.

(d) The decomposition of 𝕊_{Q} was proved above.

(e) The decomposition of F_{Q} follows immediately from (d) and Theorem 9.4.

In the special case of totally ordered lattices, the results of Corollary 9.11 were already obtained in Corollary 11.11 of [BT3]. Also, notice that (c) provides another proof of the implication (b) ⇒ (a) in Theorem 4.5.

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_{P} corresponding to a pole lattice P. Recall that e_{T} denotes the central idempotent of End_{kL}(T) which is an identity element for the subalgebra E_{T} (see Theorem 8.3).

9.12. Theorem. Let T be a finite lattice. For every finite set X, let F_{T}^{pole}(X) be the k-submodule of F_{T} (X) generated by all the maps φ : X → T such that φ(X) is a pole subposet of T.

(a) F_{T}^{pole} = F_{e_{T}} (F_{T}) and this is a subfunctor of F_{T}.

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

F_{T}^{pole} ≅ ⨁_{\begin{matrix}P ∈Pol_{T} \\ π∈\overline{Sur_{L}(T,P)} \end{matrix}} 𝕊_{P} = ⨁_{P ∈Pol_{T}} 𝕊_{P}^{n(T,P)} .

(c) If Q is a pole lattice, the image of any morphism F_{Q} → F_{T} in F_{k} is contained in F_{T}^{pole}. In particular, any subfunctor of F_{T} isomorphic to the functor 𝕊_{Q} is contained in F_{T}^{pole}.

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

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

Proof : (a) Let φ : X → T be a map such that φ(X) is a pole subposet of T. Let Q be the join-closure of φ(X), so that φ = jψ, where ψ : X → Q and where j : Q → T is the inclusion map. It is easy to see that Q is a join-closed pole lattice. Thus j ∈ Hom_{kL}(Q, T) and so j = je_{Q} because e_{Q} ∈ End_{kL}(Q) is the identity morphism by Theorem 9.1. Now je_{Q} = e_{T} j by Theorem 8.3, hence j = e_{T} j. 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 where ψ is a k-linear combination of maps X → T. Since e_{T} is, by construction, a linear combination of maps whose image is a pole poset, so is e_{T} ψ, proving that φ ∈ F_{T}^{pole}(X).

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

(b) As in the proof of Theorem 9.4, we apply the fully faithful functor kL → F_{k} defined by T ↦ F_{T}. There is a direct sum decomposition of functors

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

The idempotent e_{T} is the sum of the orthogonal idempotents f_{π,id_{P} ,π} of End_{kL}(T), for π ∈ \overline{Sur_{L}(T, P)} and P ∈ Pol_{T}. It follows that the endomorphisms F_{f_{π,id_{P} ,π}} of F_{T} are orthogonal idempotents with sum F_{e_{T}}. Hence we obtain a direct sum decomposition of correspondence functors

F_{T}^{pole} = F_{e_{T}} (F_{T}) = ⨁_{\begin{matrix}P ∈Pol_{T} \\ π∈\overline{Sur_{L}(T,P)} \end{matrix}} F_{f_{π,id_{P} ,π}}(F_{T} ) .

By Proposition 7.11, the image of F_{f_{π,id_{P} ,π}} = F_{j^{π}π} is isomorphic to F_{P} /H_{P} = 𝕊_{P} and is projective by Theorem 9.4, proving the result.

(c) Let α : F_{Q} → F_{T} be a morphism of correspondence functors where Q is a pole lattice. Since the functor T ↦ F_{T} is full, α is the image of a morphism Q → T in kL, which is in turn a linear combination of join-morphisms f : Q → T. Any such f has an image which is a pole subposet of T. Therefore, for any function φ : X → Q, the image of fφ is a pole subposet of T. It follows that the image of the map F_{f}(X) : F_{Q}(X) → F_{T} (X) is contained in F_{T}^{pole}(X). Therefore, the image of the map F_{f} is contained in F_{T}^{pole} and so the image of α is contained in F_{T}^{pole}.

The special case follows from the fact that 𝕊_{Q} is a subfunctor of F_{Q}, by Theorem 9.4.

(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 8.3, 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}}.

9.13. Corollary. Let F be a correspondence functor and let F^{pole} be the sum of all the images of morphisms F_{P} → F, where P varies among pole lattices.

(a) The subfunctor F^{pole} is the image of an idempotent natural transformation ε_{F} : F → F, so that F^{pole} is a direct summand of F.

(b) Hom_{F_{k}} (F^{pole}, (id −ε_{F})(F)) = {0}.

(c) The idempotent ε_{F} is natural in F. In other words, when F varies among correspondence functors, the family of idempotents ε_{F} is a natural transformation of the identity functor F_{k} → F_{k}.

Proof : We only sketch the main arguments of the proof. By Yoneda’s lemma applied to a set of generators of F, there is some index set I and a surjective morphism from a direct sum of representable functors

π : ⨁_{i∈I} kC(−, E_{i}) ⟶ F

and each kC(−, E_{i}) is projective. Moreover, kC(−, E_{i}) is isomorphic to F_{T_{i}} for some distributive lattice T_{i} (by Lemma 3.7). It follows that there is an exact sequence

\xymatrix{⨁_{j∈J} F_{U_{j}} \ar[r] & ⨁_{i∈I} F_{T_{i}} \ar[r]^{π} & F \ar[r] & 0}

where U_{j} is again a distributive lattice for each j in some index set J. Let us write ε for the direct sum of the idempotent endomorphisms of Theorem 9.12, independently of the lattices involved. Thus we have a commutative diagram

\xymatrix{⨁_{j∈J} F_{U_{j}} \ar[d]^{ε} \ar[r] & ⨁_{i∈I} F_{T_{i}} \ar[d]^{ε} \ar[r]^{π} & F \ar[d]^{ε_{F}} \ar[r] & 0 \\ ⨁_{j∈J} F_{U_{j}} \ar[r] & ⨁_{i∈I} F_{T_{i}} \ar[r]^{π} & F \ar[r] & 0}

where ε_{F} : F → F is induced by ε. It is easy to check that ε_{F} is an idempotent morphism and that Im(ε_{F}) ⊆ F^{pole}, because Im(ε_{F}) = Im(πε) and this is the image under π of correspondence functors associated to pole lattices, by Theorem 9.12. Moreover, any pole lattice P is distributive, so F_{P} is projective. Therefore any morphism F_{P} → F lifts to a morphism F_{P} → ⨁_{i∈I} F_{T_{i}} whose image must be contained in Im(ε). Thus F^{pole} is contained in π(Im(ε)) = Im(ε_{F}).

The proofs of (b) and (c) are similar.

References

[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, J. Algebra 495 (2018), 150–198.

[BT3] S. Bouc, J. Thévenaz. Correspondence functors and lattices, J. Algebra 518 (2019), 453–518.

[BT4] S. Bouc, J. Thévenaz. The algebra of Boolean matrices, correspondence functors, and simplicity, J. Comb. Algebra, to appear.

[Ro] S. Roman. Lattices and ordered sets, Springer, New York, 2008.

[St] R. P. Stanley. Enumerative Combinatorics, Vol. I, Second edition, Cambridge studies in advanced mathematics 49, Cambridge University Press, 2012.

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, Institut de mathématiques, EPFL, Station 8, CH-1015 Lausanne, Switzerland.

jacques.thevenaz@epfl.ch