CARTER-PAYNE HOMOMORPHISMS AND BRANCHING RULES FOR ENDOMORPHISM RINGS OF SPECHT MODULES

HARALD ELLERS AND JOHN MURRAY

Abstract. Let Σ_{n} be the symmetric group of degree n, and let F be a field of characteristic p ≠ 2. Suppose that λ is a partition of n+1, that α and β are partitions of n that can be obtained by removing a node of the same residue from λ, and that α dominates β. Let S^{α} and S^{β} be the Specht modules, defined over F, corresponding to α, respectively β. We give a very simple description of a non-zero homomorphism θ : S^{α} → S^{β} and present a combinatorial proof of the fact that dim Hom_{F Σ_{n}}(S^{α}, S^{β}) = 1. As an application, we describe completely the structure of the ring End_{F Σ_{n}}(S^{λ} ↓_{Σ_{n}}). Our methods furnish a lower bound for the Jantzen submodule of S^{β} that contains the image of θ.

1. Introduction

Let n be a positive integer and let Σ_{n} denote the symmetric group of degree n. For any ring R and any partition α of n, the Specht module S_{R}^{α} is defined to be the submodule of the permutation module R_{Σ_{α}}↑^{Σ_{n}} spanned by certain elements called polytabloids, where Σ_{α} is the Young subgroup associated to α and R_{Σ_{α}} is the trivial RΣ_{α}-module. (See [4] for definitions.) Specht modules play a central role in the representation theory of the symmetric group. Now suppose that F is a field. If F has characteristic 0, the Specht modules defined over F are the simple F Σ_{n}-modules; if F has characteristic p the heads of the Specht modules S_{F}^{α}, with α p-regular, are the simple F Σ_{n}-modules.

It would be useful to understand Hom_{F Σ_{n}}(S^{α}, S^{β}) for all partitions α and β of n when F has positive characteristic, but this problem has not been solved in general. In [2], Carter and Payne exhibited some types of partitions α and β for which Hom_{Σ_{n}}(S^{α}, S^{β}) ≠ 0 even though α ≠ β, generalizing earlier results of Carter and Lusztig [1].

Among the examples discovered by Carter and Payne are one-box shift homomorphisms. Suppose that the Young diagram [β] of β can be obtained by moving a single node in the diagram [α]; then we say that β is one-box-shift of α (see (5)). This is equivalent to saying that there exists a partition λ of n + 1 such that α and β can each be obtained by removing a node from the Young diagram of [λ]. Carter and Payne showed that if the node in [α]\[β] has the same p-residue as the node in [β]\[α], and α dominates β, then Hom_{Σ_{n}}(S^{α}, S^{β}) ≠ 0 (the residue of the node in the i-th row and j-th column of a Young diagram is the integer j − i; the p-residue is just j − i taken modulo p).

In fact, in this case dim Hom_{F Σ_{n}}(S^{α}, S^{β}) = 1, when the characteristic of F is greater than 2. This was proved in the context of Iwahori-Hecke algebras by Lyle [8]. For the group algebras, this result seems to have been known to experts in the area for some time. For example, A. Kleshchev claims, apparently incorrectly, in [5, p498], that this is a special case of the main result in [1]. However, we do not believe that a complete proof of this theorem had been published before Lyle’s paper.

Date: January 30, 2009.

1991 Mathematics Subject Classification. 20C20, 20C30.

Key words and phrases. one-box-shift homomorphism, Jucys-Murphy elements.

Let L_{n+1} be the Jucys-Murphy element consisting of the sum in F Σ_{n+1} of all transpositions in Σ_{n+1} that are not in Σ_{n}. We would like to point out that results from [3] make it possible to give the following simple description of one-box shift homomorphisms:

Theorem 1.1. Let F be a field of characteristic p > 0 and let λ be a partition of n + 1. Suppose that α and β can each be obtained by removing a node from λ, with α⊵β, and that the removed nodes have the same p-residues. Identify S_{F}^{α} and S_{F}^{β} as subquotients of S_{F}^{λ} ↓_{Σ_{n}}. Then multiplication by an appropriate polynomial in L_{n+1} induces a non-zero homomorphisms from S_{F}^{α} to S_{F}^{β}

Proof. Assume that λ has t removable nodes, with p-residues r_{1}, …, r_{t}. Let λ_{i} be the partition of n obtained by removing the i-th removable node from [λ]. Following James [4, 9.3], let S_{i}^{λ} be the subspace of S_{F}^{λ} spanned by those polytabloids e_{t} where t is a standard λ-tableau such that the symbol n+1 is in one of the first i removable nodes of t. Then S_{i}^{λ} is a Σ_{n}-submodule of S^{λ}, and S_{i}^{λ}/S_{i−1}^{λ} ≅ S^{λ_{i}}. The Σ_{n}-series

0 ⊂ S_{0}^{λ} ⊂ S_{2}^{λ} ⊂ … ⊂ S_{t}^{λ} = S^{λ}

is called the Specht series of S_{F}^{λ} ↓_{Σ_{n}}.

Suppose that α = λ_{u} and that β = λ_{v}. Then u < v. In [3], we showed that S_{u}^{λ} ∏_{i=u}^{v−1}(L_{n+1} − r_{i}) ⊆ S_{v}^{λ} and that this map induces a well-defined nonzero homomorphism S_{F}^{α} = S_{u}^{λ}/S_{u−1}^{λ} → S_{v}^{λ}/S_{v−1}^{λ} = S_{F}^{β}.

If p > 2, then Hom_{Σ_{n}}(S_{F}^{α}, S_{F}^{β}) is 1-dimensional, and this map must be a nonzero multiple of the homomorphism defined in [2]. □

In order to effectively use this result, we need some information about the action of L_{n+1} on S^{λ}. The following is a variant of (3.3) in [9]:

Lemma 1.2. Let λ be a partition of n + 1 and let t be a λ-tableau (not necessarily standard). Suppose that n + 1 occupies a column of length r in t. Then λ has one removable node (r, c) for some c ≥ 1. Let R be the set of symbols in t whose columns have length strictly smaller that r. Then in S_{ℤ}^{λ} we have

e_{t}L_{n+1} = (c − r)e_{t} + ∑_{i∈R} e_{t}(n + 1, i).

For example, let λ = [4, 3, 1], n = 7 and t = ^{1 2 3 4 5 6 7 8} . Then n + 1 = 8 occupies a node of residue −2 in t. We use Lemma 1.2 to compute e_{t}(L_{8} + 2). For emphasis we replace the symbol n + 1 by •. The above lemma show that:

(1) e_{1 2 3 4 ; 5 6 7 ; •} (L_{8} + 2) = e_{1 • 3 4 ; 5 6 7 ; 2} + e_{1 2 3 4 ; 5 • 7 ; 6} + e_{1 2 • 4 ; 5 6 7 ; 3} + e_{1 2 3 4 ; 5 6 • ; 7} + e_{1 2 3 • ; 5 6 7 ; 4}

We use this to illustrate Theorem 1.1. Let α = [4, 3] and β = [3, 3, 1]. For p = 5 and n = 7 there is a one-box shift θ : S^{α} → S^{β}. Now u = 1, v = 3 and r_{1} = 3, r_{2} = 1 and r_{3} = −2. The relevant polynomial in L_{n+1} is (L_{8} + 2)(L_{8} − 1).

CARTER-PAYNE HOMOMORPHISMS AND BRANCHING RULES FOR ENDOMORPHISMS 3

We find the image of the polytabloid e_{t} with t = ^{1 2 3 4 5 6 7} under θ. We again replace 8 by •. First note that e_{1 2 3 4 ; 5 6 7 ; •} is an element of S_{3}^{λ} whose image is e_{1 2 3 4 5 6 7} , under

S_{3}^{λ}/S_{2}^{λ} ≅ S^{α}. We consider the summands in e_{t}(L_{8} + 2) given in (1) in turn:

\begin{aligned}e_{1 • 3 4 ; 5 6 7 ; 2} (L_{8} − 1) & = & e_{1 4 3 • ; 5 6 7 ; 2} \\ e_{1 2 3 4 ; 5 • 7 ; 6} (L_{8} − 1) & = & e_{1 2 3 • ; 5 4 7 ; 6} \\ e_{1 2 • 4 ; 5 6 7 ; 3} (L_{8} − 1) & = & e_{1 2 4 • ; 5 6 7 ; 3} \\ e_{1 2 3 4 ; 5 6 • ; 7} (L_{8} − 1) & = & e_{1 2 3 • ; 5 6 4 ; 7} \\ e_{1 2 3 • ; 5 6 7 ; 4} (L_{8} − 1) & = & e_{1 2 3 • ; 5 6 7 ; 4} (3 − 1). \end{aligned}

Deleting • from each tableau, we get a tableau of shape β. So

(2)  e_{1 2 3 4 5 6 7} θ = e_{1 4 3 5 6 7 2} + e_{1 2 3 5 4 7 6} + e_{1 2 4 5 6 7 3} + e_{1 2 3 5 6 4 7} + 2 e_{1 2 3 5 6 7 4} .

Our second goal is to complete our analysis (see [3]) of the endomorphism ring of the restriction of the Specht module S^{λ} to Σ_{n}.

Theorem 1.3. Let F be a field of characteristic p not equal to 2 and let λ be a partition of n + 1. Then the blocks of End_{F Σ_{n}}(S^{λ} ↓_{Σ_{n}}) are parametrized by the p-residues of the removable nodes in [λ]. Let r denote the p-residue of a removable node of λ. Then the corresponding block of End_{F Σ_{n}}(S^{λ} ↓_{Σ_{n}}) has the structure of a truncated polynomial ring over F. Its dimension is equal to the number of removable nodes in [λ] that have p-residue r.

The corresponding statement for the induced Specht module is also true:

Theorem 1.4. Let F be a field of characteristic p not equal to 2 and let λ be a partition of n − 1. Then the blocks of End_{F Σ_{n}}(S^{λ} ↑^{Σ_{n}}) are parametrized by the p-residues of the addable nodes in [λ]. Let r denote the p-residue of an addable node of λ. Then the corresponding block of End_{F Σ_{n}}(S^{λ} ↑^{Σ_{n}}) has the structure of a truncated polynomial ring over F. Its dimension is equal to the number of addable nodes in [λ] that have p-residue r.

It is interesting to compare these theorems to a result of Kleshchev, which says that if D is a simple F Σ_{n}-module, then End_{F Σ_{n−1}}(D ↓_{Σ_{n−1}}) is also a direct sum of truncated polynomial algebras, with one summand for each block of F Σ_{n−1}. (See Theorems 11.2.7 and 11.2.8 of [6].)

In later sections, we revisit the dimension of the space of one-box-shift homomorphisms.

Theorem 1.5. Let F be a field of characteristic not equal to 2 and let β be a one-box-shift of α. Then dim Hom_{F Σ_{n}}(S^{α}, S^{β}) ≤ 1.

Our aim is to show that a proof is possible that uses only the relatively elementary combinatorial tools developed by G. James in his monograph [4]. In particular, we do not use the Schur functor, nor do we make use of any of the deeper machinery of algebraic groups. We also appear to avoid some technical difficulties that arise in the context of Iwahori-Hecke algebras c.f. [8].

All three theorems 1.3, 1.4 and 1.5 are false when p = 2. The proof of Theorem 1.5 relies on Lemma 2.1, which fails for p = 2. In the case of Theorems 1.3, and 1.4, when p = 2, we do not even know how to parametrize the blocks of End_{F Σ_{n}}(S^{λ} ↓_{Σ_{n}}) nor those of End_{F Σ_{n}}(S^{λ} ↑^{Σ_{n}}). For these reasons we now fix F as a field of characteristic p ≠ 2.

The methods we use to prove Theorem 1.5 make it possible to explore how one-box-shift homomorphisms interact with Jantzen layers (see Section 7 for definitions).

Theorem 1.6. Let λ be a partition of n + 1 and let F be a field of characteristic p > 0. Suppose that α and β can be obtained by removing a node of residue r_{α} and r_{β}, respectively, from λ, with α ⊵ β. Let θ : S_{F}^{α} → S_{F}^{β} be a Σ_{n}-homomorphism. If p^{i} divides r_{α} − r_{β} then θ(S_{F}^{α}) is contained in the i-th Jantzen submodule of S_{F}^{β}.

2. Notation and basic results

For the rest of the paper, unless stated otherwise, F is a field of characteristic p > 2 and α is a fixed partition of n. We also fix a bijection

t : [α] → {1, …, n}.

The Young diagram [α] is a collection of boxes in the plane that is oriented left to right and top to bottom. This means that the first row is the one at the top and the first column is the one at the left. The (i, j)-th node of [α] is the box in the i-th row and the j-th column.

We refer to any map T : [α] → {1, …, n} as an α-tableau. Informally, T is a filling of the nodes of [α], using elements from {1, …, n}, with repeats allowed. The value of T at a node b = (r, c) is denoted by T_{b} or T (r, c). If i = T_{b}, we say that node b of T contains the symbol i.

Identify Σ_{n} with the group of permutations of {1, …, n}. Permutations and homomorphisms will generally act on the right. For instance Σ_{n} acts on the set of all bijective α-tableau by functional composition. We can use the bijection t to give a contragradient action of Σ_{n} on all α-tableau: if π ∈ Σ_{n} then define the α-tableau T π as the map T π = t ∘ π^{−1}∘ t^{−1}∘ T.

Suppose that α has r nonzero parts [α_{1} ≥ α_{2} ≥ … ≥ α_{r}]. An α-tabloid is an ordered partition of n̂ whose parts have cardinalities α_{1}, α_{2}, …, α_{r}. For instance the sets {t_{ij} | j = 1, …, α_{i}} for i = 1, …, r, determine an α-tabloid that we shall denote by {t}.

Let R_{t} be the row stabilizer and let C_{t} be the column stabilizer of t. Denote by M^{α} the F Σ_{n}-module consisting of all formal F -linear combinations of α-tabloids. Then Σ_{n} acts on α-tabloids in the obvious way: {t}π := {tπ}. James shows that the corresponding Σ_{n}-module M^{α} is isomorphic to the permutation module F_{R_{t}}↑^{Σ_{n}}.

The following element of M^{α} is called a polytabloid:

e_{t} := ∑_{π∈C_{t}} sgn π{tπ}.

The subspace of M^{α} spanned by polytabloids is a Σ_{n}-module called the Specht module S^{α}. For compactness we use X^{−} := ∑_{σ∈X} sgn(σ)σ, whenever X ⊆ Σ_{n}. So in this notation e_{t} = {t}C_{t}^{−}.

The following important important combinatorial result is one half of [4, 13.5].

Lemma 2.1. Suppose that T is an α-tableau that contains repeated entries in one of its columns. Then T C_{t}^{−} = 0.

Proof. By hypothesis there exist integers r, s, c, with r ≠ s, such that T_{r,c} = T_{s,c}. Then π := (t_{r,c}, t_{s,c}) is a transposition in C_{t}. So πC_{t}^{−} = sgn(π) C_{t}^{−} = −C_{t}^{−}. But T π = T. It follows that T C_{t}^{−} = T πC_{t}^{−} = −T C_{t}^{−}. As char(F) ≠ 2, this implies that T C_{t}^{−} = 0. □

3. Branching rules for endomorphism rings

As in [3], we use E_{n} denote the sum, in F Σ_{n}, of all transpositions in Σ_{n}. Each F Σ_{n}-module M comes equipped with a ring homomorphism F Σ_{n} → End_{F} (M).

Theorem 3.1. Let λ be a partition of n. Then the image of E_{n+1} generates End_{F Σ_{n+1}}(S^{λ} ↑^{Σ_{n+1}}) and the image of E_{n−1} generates End_{F Σ_{n−1}}(S^{λ} ↓_{Σ_{n−1}}).

Proof. Let M be the a summand of the restricted or induced module belonging to a single block of the group algebra. We change notation from n + 1 to n in the first case, and from n − 1 to n in the second case, so that M is an F Σ_{n}-module, and E_{n} plays the role of E_{n+1} or of E_{n−1}. Let ϵ denote the image of E_{n} in End_{F} (M). Then [4] and the authors’ paper [3] shows that M has the following properties:

(1) M has a Specht series

0 = M_{0} ⊂ M_{1} ⊂ … ⊂ M_{m} = M,

such that M_{u}/M_{u−1} ≅ S^{λ_{u}} where λ_{u} is a partition of n. Moreover, λ_{v} is a one-box-shift of λ_{u}, for each v ≠ u.

(2) The labelling partitions satisfyλ_{1} ◁ … ◁ λ_{n}.

(3) The endomorphism ϵ has minimal polynomial (x − r)^{m}. Here ϵ acts as a fixed scalar r on each quotient module M_{u}/M_{u−1}, for u = 1, …, m.

It is known that the scalar r is the sum, taken modulo p, of the residues of the nodes of the Young diagram of any partition λ_{u} that is associated with the given p-block of Σ_{n}. For convenience, we define M_{u} := 0, when u ≤ 0.

Let θ be any F Σ_{n}-homomorphism of M. We shall prove by induction that given i ≥ 0 there exists a polynomial f_{i} ∈ F [x] such that M_{u}(θ − f_{i}(ϵ)) ⊆ M_{u−i}, for u = 1, …, m. So θ = f_{m}(ϵ) lies in the subalgebra generated over F by ϵ and 1_{M}, thus proving the Theorem.

As p ≠ 2 and λ_{u} ◁ λ_{v} for u < v, it follows from Corollary 13.17 of [4] that there are no non-zero homomorphisms M_{u}/M_{u−1} → M_{v}/M_{v−1}, when u < v. This implies in particular that M_{u}θ ⊆ M_{u}, for each u = 1, …, m (see Lemma 3.2 of [3] for details). This proves the base case for our induction.

Suppose then that i > 0 is an integer such that there exists a polynomial f_{i} ∈ F [x] such that M_{u}(θ − f_{i}(ϵ)) ⊆ M_{u−i}, for u = 1, …, m. Then there exist well-defined Σ_{n}-homomorphisms ψ_{u,i} : M_{u}/M_{u−1} → M_{u−i}/M_{u−i−1} given by (v + M_{u−1})ψ_{u,i} = v(θ − f_{i}(ϵ)) + M_{u−1}, for all v ∈ M_{u}.

Now M_{u}(ϵ − r)^{i} ⊆ M_{u−i}, for all u. As ϵ has minimal polynomial of degree m, the induced maps (ϵ − r)_{u}^{i} : M_{u}/M_{u−1} → M_{u−i}/M_{u−i} are non-zero, for u > i. But λ_{u−i} is a one-box-shift of λ_{u}. So Hom_{F Σ_{n}}(λ_{u}, λ_{u−i}) is a 1-dimensional F - space, according to Corollary 6.7. We deduce that (ϵ − r)_{u}^{i} is a basis for the space Hom_{F Σ_{n}}(λ_{u}, λ_{u−i}), for all u, i.

The previous two paragraphs imply that there exist Λ_{u} ∈ F such that ψ_{u,i} = Λ_{u}(ϵ−r)_{u}^{i}, for u = 1, …, m. Then M_{u}(θ−f_{i+1}(ϵ)) ⊆ M_{u−i}, for u = 1, …, m, where f_{i+1}(x) = f_{i}(x) + Λ_{m}(x − r)^{i}, if we can show that all Λ_{u} are equal to Λ_{m}. Lemma 3.1 of [3] furnishes us with an element τ ∈ M such that τ_{u} := τ ∏_{i=u+1}^{m}(ϵ − r) belongs to M_{u}\M_{u−1}, for each u ∈ {1, …, m}. Thus

Λ_{u}τ_{u−i}(ϵ − r)^{i} = τ_{u}(θ − f_{i}(ϵ)), mod M_{u−i−1},
= τ(ϵ − r)^{m−u}(θ − f_{i}(ϵ))
= τ(θ − f_{i}(ϵ))(ϵ − r)^{m−u}, as θ ∈ End_{F Σ_{n}}(M),
≡ Λ_{m}τ(ϵ − r)^{i}(ϵ − r)^{m−u}, mod M_{m−i−1},
≡ Λ_{m}τ(ϵ − r)^{i}, mod M_{u−i−1}.

We conclude that Λ_{u} = Λ_{m}. This completes the proof of the inductive step, and hence of the Corollary.

Now consider the map from the polynomial algebra F [x] to the summand associated to a single block of the endomorphism algebra that sends x to multiplication by E_{n}. We have shown that this map is surjective and has kernel generated by a power of x − r, where E_{n} acts as the scalar r on each simple module in the block. If we consider a different block, the scalar r will be different. The theorem follows.

□

Proof of Theorem 1.3. Each p-block of Σ_{n} has an associated positive integer w known as its weight and a p-core κ. Here κ is a partition of n − pw that has no hook-lengths divisible by p. A Specht module S^{μ} belongs to the block if and only if the p-core of μ coincides with κ. It is well known that the p-core of a partition is determined by the multi-set of residues (mod p) of the nodes in its Young diagram. Fix the residue r of a removable node in [λ]. Then all partitions that can be obtained by removing a single node of p-residue r from λ belong to the same p-block, call it b, of Σ_{n}.

The main result of [3], Theorem 3.4, shows that S^{λ} ↓_{Σ_{n}}b is indecomposable. As this is an ideal and a direct summand of End_{F Σ_{n}}(S^{λ} ↓_{Σ_{n}}), it is a block of this algebra. Theorem 3.1 implies that End_{F Σ_{n}}(S^{λ} ↓_{Σ_{n}})b is generated by over F by the image of E_{n}. Let m be the number of removable nodes in [λ] that have p-residue r. Then E_{n} has minimal polynomial (x − r)^{m}. It follows that End_{F Σ_{n}}(S^{λ} ↓_{Σ_{n}})b has the structure of the truncated polynomial ring F [x]/(x^{m}). □

The proof of Theorem 1.4 is almost identical, and is omitted.

Finally we note the following fact about Carter-Payne homomorphisms between Specht modules. The proof is omitted, although it can easily be demonstrated by applying the methods used in the proof of Theorem 3.1.

Proposition 3.2. Suppose that λ is a partition of n + 1 and that r is the p-residue of a removable node in [λ]. Let λ_{1} ◁ … ◁ λ_{m} be all the partitions of n whose Young diagram can be obtained by removing a single node of p-residue r from [λ]. Let f_{i} be a non-zero F Σ_{n}-homomorphism S^{λ_{i}} → S^{λ_{i−1}}, for each i = 2, …, m. Then for 1 ≤ j < i ≤ m, the composite homomorphism f_{i} ∘ f_{i−1} ∘ … ∘ f_{j+2} ∘ f_{j+1} spans Hom_{F Σ_{n}}(S^{λ_{i}}, S^{λ_{j}}). In particular, all these homomorphisms are non-zero.

4. Semi-standard homomorphisms

We now discuss homomorphisms between Specht and permutation modules. If β is a partition of n, it is useful to describe the permutation module M^{β} using certain α-tableaux. Following James, an α-tableau T is of type β if it has β_{i} entries equal to i, for each i ≥ 1. The action of Σ_{n} on α-tableau restricts to an action on the α-tableau of type β. Under this action, the F -span of the α-tableau of type β form a permutation module that is isomorphic to M^{β}.

Let S and T be α-tableau of type β. We write

S ≈ T, if S = T π, for some π ∈ R_{t}.

Then ≈ is an equivalence relation of α-tableau. We also say that S is row equivalent to T. James defines the Σ_{n} homomorphism θ_{T} : M^{α} → M^{β} by {t} θ_{T} := ∑_{S≈T} S.

A tableau T is said to be semi-standard if the numbers are nondecreasing along the rows of T and strictly increasing down the columns of T. Let θ̂_{T} denote the restriction of θ_{T} to S^{α} ⊆ M^{α}. James shows in [4, 13.13] that the set

(3)  {θ̂_{T} | T a semi-standard α-tableau of type β}

forms a basis for Hom_{F Σ_{n}}(S^{α}, M^{β}).

A useful result of James identifies S^{β} as an intersection of the kernels of certain Σ_{n}-homomorphisms on M^{β}. Let β be a partition of n that has k parts. Then [4, 17.18] shows that

(4)  S^{β} := ⋂_{i=1}^{k−1} ⋂_{r=0}^{β_{i+1}−1} ker ψ_{i,r}.

Here for each pair i, r, let ν := (β_{1}, β_{2}, …, β_{i−1}, β_{i} + β_{i+1} − r, r, β_{i+2}, …). Then ψ_{i,r} : M^{β} → M^{ν} is defined by setting {t} ψ_{i,r} as the sum of all ν-tabloids {s} such that row (i + 1) of {s} is a subset of size r of row (i + 1) of {t}.

The action of ψ_{i,r} on α-tableaux of type β is particularly easy to describe. If T is one such tableau then T ψ_{i,r} is the sum of all α-tableaux that can be obtained from T by changing all but r symbols i + 1 to i. For example consider the effect of ψ_{1,2} on a (4, 3)-tableau of type (3^{2}, 1):

1 1 1 2 2 2 3  \underrightarrow{ψ_{1,2}}  ^{1 1 1 1 2 2 3} + ^{1 1 1 2 1 2 3} + ^{1 1 1 2 2 1 3} .

In terms of semistandard homomorphisms, this gives

θ_{1 1 1 2 2 2 3} ψ_{1,2} = 4 θ_{1 1 1 1 2 2 3} + θ_{1 1 1 2 1 2 3} .

It is convenient to use the following concise notation

ψ_{i} := ψ_{i,β_{i+1}−1}.

In particular T ψ_{i} is the sum of all α-tableaux of that can be obtained from T by changing one symbol i + 1 to an i.

For the rest of this paper let a < b be integers such that α_{a} > α_{a+1} and α_{b−1} > α_{b}. So (a, α_{a}) is a removable node of α and (b, α_{b} + 1) is an addable node of α. We fix the partition β of n defined by

(5)  β_{i} = \{ \begin{matrix}α_{a} − 1, & if i = a; \\ α_{b} + 1, & if i = b; \\ α_{i}, & if i ≠ a, b. \end{matrix}

We say that β is a one-box-shift of α. Note that α ▷ β in the dominance order.

Our first task is to enumerate all semi-standard α-tableau of type β. So let T be such a tableau. We claim that

(6)  T (i, j) = i, for all (i, j) ∈ [α], unless a ≤ j < b and j = α_{i}.

For, T (i, j) ≥ i, as T is column strict. So all symbols i occur in the top i rows of T. In particular, the symbols 1, …, a − 1 occupy all entries in rows 1, …, a − 1.

Now suppose that a ≤ i < b. We prove by induction on i that T (i, j) = i, for 1 ≤ j < α_{i}. The base case i = a holds because the α_{a} −1 symbols in T must occupy all but the last entry in row a of T (there is no room for them further up). Now let a < i. Then there are −1 + ∑_{u=1}^{i−1} α_{u} symbols in T equal to one of 1, …, i − 1. These occupy all but one of the ∑_{u=1}^{i−1} α_{u} nodes in the first i − 1 rows of T. It follows that at most one symbol i in T does not belong to row i. This proves the inductive step.

Finally, suppose that b ≤ i. Then there are ∑_{u=1}^{i} α_{u} symbols in T equal to one of 1, …, i and ∑_{u=1}^{i−1} α_{u} nodes in the first i rows of T. It follows from this that every entry in the i-th row of T is equal to i. This completes the proof of (6).

Next we define

T̂(u) := T (u, α_{u}), for u = a, …, b − 1.

Suppose that a < i ≤ b. Then there are β_{i}-symbols in T equal to i. At most one of them does not belong to the i-th row of T. Moreover, in view of (6), if a symbol i does not belong to row i, it occupies the end of a row between a and i − 1 inclusive. This shows that there exists a unique u with a ≤ u ≤ i such that T̂(u) = i. It follows that T̂ : {a, …, b − 1} → {a + 1, …, b} is a bijection. Moreover, T̂(i) ≥ i, and T̂(i + 1) > T̂(i), if α_{i+1} = α_{i}. We call any such bijection T̂ the semi-standard α-bijection of type β associated with T.

We can recover the tableau T from the bijection T̂, using property (6).

Let u_{1} := a and inductively define u_{i} := T̂(u_{i−1}), for i = 2, 3 …, if u_{i−1} < b. This gives a finite set of cardinality m ≥ 1:

{T } := {u_{1} = a < u_{2} < … < u_{m}}.

We claim that T̂(u) ≠ u if and only if u ∈ {T }. We prove this by induction on u ∈ {a, …, b − 1}. For, suppose that T̂(u) ≠ u. Then there exists v with a ≤ v < u and T̂(v) = u. The inductive hypothesis implies that v ∈ {T }. So v = u_{i}, for some i ≥ 1, whence u = T̂(u_{i}) = u_{i+1} also belongs to {T }. Note that b = u_{m+1}. If we temporarily extend T̂ to a permutation of the set {a, …, b} via T̂(b) := a, then this shows that T̂ is the cycle permutation (a = i_{1}, i_{2}, …, i_{m}, i_{m+1} = b). So we can recover T̂ from {T }.

In order for a subset X of {a, …, b − 1} to equal {T } for some semi-standard α-tableau of type β, it is necessary and sufficient that a ∈ X, and if u ∈ X and α_{u} = α_{u+1}, then u + 1 ∈ X. We call any such X a semi-standard α-set of type β. We call {T } the semi-standard α-set of type β associated with T (or with T̂). We summarize the above discussion with:

Lemma 4.1. The associations T ⟷ T̂ ⟷ {T } establish mutually inverse bijections between the semi-standard α-tableau of type β, the semi-standard α-bijections of type β, and the semi-standard α-sets of type β.

Given T ↔ T̂ ↔ {T }, we can use θ̂T̂ or θ̂_{{T }} to denote θ̂_{T}. Moreover, we use |T | or |T̂| to denote the cardinality of {T }.

If α has m_{i} parts of length i, then there are ∏_{α_{a}<i<α_{b}}(m_{i} + 1) semistandard α-tableau of type β. As an example let α = (4, 3^{2}, 2, 1) and β = (3^{3}, 2, 1^{2}). So a = 1, b = 6 and m_{3} = 2, m_{2} = 1, m_{1} = 1. Then there are 12 = (3 + 1)(1 + 1)(1 + 1) semistandard α-tableau of type β. We list them along with the semistandard α-sets of type β. We use · to indicate a symbol i in row i:

\begin{matrix}⋯ 6 ; · · 2 ; · · 3 ; · 4 ; 5 & ⋯ 5 ; · · 2 ; · · 3 ; · 4 ; 6 & ⋯ 4 ; · · 2 ; · · 3 ; · 6 ; 5 & ⋯ 4 ; · · 2 ; · · 3 ; · 5 ; 6 & ⋯ 3 ; · · 2 ; · · 6 ; · 4 ; 5 & ⋯ 3 ; · · 2 ; · · 5 ; · 4 ; 6 \\ \{1\} & \{1, 5\} & \{1, 4\} & \{1, 4, 5\} & \{1, 3\} & \{1, 3, 5\} \\ ⋯ 3 ; · · 2 ; · · 4 ; · 6 ; 5 & ⋯ 3 ; · · 2 ; · · 4 ; · 5 ; 6 & ⋯ 2 ; · · 3 ; · · 6 ; · 4 ; 5 & ⋯ 2 ; · · 3 ; · · 5 ; · 4 ; 6 & ⋯ 2 ; · · 3 ; · · 4 ; · 6 ; 5 & ⋯ 2 ; · · 3 ; · · 4 ; · 5 ; 6 \\ \{1, 3, 4\} & \{1, 3, 4, 5\} & \{1, 2, 3\} & \{1, 2, 3, 5\} & \{1, 2, 3, 4\} & \{1, 2, 3, 4, 5\} \end{matrix}

For the rest of the paper we let

{R } := {i | (i, α_{i}) is a removable node of [α], a ≤ i ≤ b − 1}.

The associated semi-standard α-bijection of type β is denoted by R̂, and the tableau by R. A subset of {R} that contains a is called a removable α-set of type β, or simply a removable set. Clearly each removable α-set of type β is also a semistandard α-set of type β. Each semi-standard set {T } contains a largest removable set {N} = {T } ∩ {R}.

We say that a pair of removable nodes are adjacent if there is no removable node of [α] between them. Likewise, if i, j ∈ {R}, we say that i and j are adjacent if the removable nodes (i, α_{i}) and (j, α_{j}) are adjacent; if i < j this means that α_{i+1} = α_{j}.

5. Relations between Semi-standard homomorphisms

Throughout this section T is a fixed semi-standard α-tableau of type β.

The hooks-lengths in the (α_{b} + 1)-th column of [α] are h_{1}, …, h_{b−1}, where

h_{i} := α_{i} − α_{b} + b − i − 1, for 1 ≤ i ≤ b − 1.

For convenience we set h_{i} := 0, for each i ≥ b. Note that

(7)  \begin{aligned}h_{i} − h_{j} & = (α_{i} − α_{j}) + (j − i), & for 1 ≤ i, j ≤ b − 1 and \\ h_{b−1} − h_{b} & = α_{b−1} − α_{b}. \end{aligned}

We begin with a useful technical result.

Lemma 5.1. Let U ≈ T, let i ≥ 0 and let V be an α-tableau that is obtained from U by changing a set of r ≥ 1 symbols i + 1 to i, with at least one of the changes occurring in row (i + 1) of U. Suppose that UC_{t}^{−} ≠ 0 and that V C_{t}^{−} ≠ 0. Then r = 1 and T̂(i) > i + 1 and T̂(i + 1) = i + 1.

Proof. Let c ≥ 1 be such that U(i + 1, c) = i + 1 and V (i + 1, c) = i. As UC_{t}^{−} ≠ 0 and V C_{t}^{−} ≠ 0, separate applications of Lemma 2.1 show that U(i, c) ≠ i + 1 and V (i, c) ≠ i. Thus U(i, c) ≠ i, i+1. As U ≈ T, we deduce that T̂(i) > i+1. It follows from this that T̂(i + 1) = i + 1. In particular row i + 1 contains all α_{i+1}-symbols i + 1 in U. Now (i, c) is the unique node in row i of U that does not contain i. We conclude that r = 1. □

We will repeatedly, and without further comment, use the fact that S^{α} is generated by e_{t} as a Σ_{n}-module. Moreover for all i and r

(8)  e_{t} θ̂_{T} ψ_{i,r} = ∑_{U≈T} Uψ_{i,r} C_{t}^{−},

as both θ̂_{T} and ψ_{i,r} are Σ_{n}-homomorphisms.

Lemma 5.2. Let i ≥ 1 and 1 ≤ r ≤ β_{i+1} − 1 be such that θ̂_{T} ψ_{i,r} ≠ 0. Then a ≤ i < b and r = β_{i+1} − 1.

Proof. Let i ≥ 0, let U ≈ T and let V be a tableau occurring in Uψ_{i,r}.

Suppose that 1 ≤ r < β_{i+1} − 1. As T is semi-standard, Lemma 4.1 implies that at most one symbol i + 1 in U does not belong to row (i + 1). But V is the result of changing β_{i+1} − r ≥ 2 symbols in U from i + 1 to i. So one of the changes occurs in row (i + 1) of U. It then follows from Lemma 5.1 that either U C_{t}^{−} = 0 or V C_{t}^{−} = 0.

Suppose that i < a or b < i. Then all symbols i + 1 in U belong to row (i + 1) of U, and Lemma 5.1 implies that U C_{t}^{−} = 0 or V C_{t}^{−} = 0.

In all cases we have U C_{t}^{−} ψ_{i} = U ψ_{i} C_{t}^{−} = ∑ V C_{t}^{−} = 0. We then use (8) to conclude that θ̂_{T} = 0, which is contrary to hypothesis. □

Lemmas 5.3, 5.4 and 5.5 will show that the image of θ̂_{T} under ψ_{i} is either zero or a specified non-zero integer multiple of a semi-standard homomorphism from S^{α} into the permutation module M^{γ}. Here γ is the composition

γ_{j} = \{ \begin{matrix}β_{i} + 1, & if j = i; \\ β_{i+1} − 1, & if j = i + 1; \\ β_{j}, & otherwise. \end{matrix}

Notice that β ◁ γ ⊴ α.

Lemma 5.3. Let a < i ≤ b − 1 be such that i ∉ {T }. Then

θ̂_{T} ψ_{i} = \{ \begin{matrix}θ̂_{W} , & if i + 1 ∈ \{T \}; \\ 0, & otherwise. \end{matrix}

Here W is obtained from T by changing the single i + 1 in row T̂^{−1}(i + 1) to an i.

Proof. Let U ≈ T. The hypothesis on i implies that the α_{i} symbols i occupy the α_{i} columns of row i in U. So each symbol i + 1 which belongs to row i + 1 of U shares its column with a symbol i in row i of U. Let a ≤ j < b be such that T̂(j) = i + 1.

Suppose first that j = i + 1. Then all symbols i + 1 in U occur in row (i + 1). It then follows from Lemma 5.1 and the first paragraph that U ψ_{i} C_{t}^{−} = 0. We conclude from (8) that θ̂_{T} = 0 in this case.

From now on assume that j ≠ i + 1, whence j < i. Then exactly one of the symbols i + 1 in U occurs in row j, and all the other symbols i + 1 belong to row (i + 1) of U. It then follows from Lemma 5.1 and the first paragraph that either U C_{t}^{−} = 0 or U ψ_{i} C_{t}^{−} = V C_{t}^{−}, where V is the tableau obtained from U by changing the symbol i + 1 in row j to an i. It is obvious that V ≈ W and also that U is the only tableau that is row equivalent to T such that U ψ_{i} C_{t}^{−} = V C_{t}^{−}.

Conversely begin with V ≈ W. Then V has a unique i in row j. Changing this i to an i + 1 produces a tableau U that is row equivalent to T. Moreover clearly U ψ_{i} C_{t}^{−} = V C_{t}^{−}.

These arguments and (8) show that e_{t} θ̂_{T} ψ_{i} = ∑_{V ≈W} V C_{t}^{−} = e_{t} θ̂_{W}. We conclude that in this case θ̂_{T} ψ_{i} = θ̂_{W}. This completes the proof. □

Lemma 5.4. Let a < i ≤ b − 1 be such that α_{T̂^{−1}(i)} = α_{i}. Then

θ̂_{T} ψ_{i} = 0.

Proof. The hypothesis implies that T̂(i − 1) = i and α_{i−1} = α_{i}. Let U ≈ T be such that U C_{t}^{−} ≠ 0. Then one symbol i belongs to row (i − 1) of U, and the remaining α_{i} − 1 symbols i belong to row i of U. It then follows from Lemma 2.1 that each of the first α_{i} columns of U contains exactly one symbol i.

Now either T̂(i) = i + 1 or T̂(i + 1) = i + 1. In either case all symbols i + 1 belongs to the first α_{i} columns of U. It then follows from the previous paragraph that each symbol i + 1 in U belongs to a column that also contains a symbol i. We deduce from Lemma 2.1 that U ψ_{i} C_{t}^{−} = 0. This contradiction means that our supposition that U exists is false.

Using (8), we see that e_{t}θ̂_{T}ψ_{i} = 0. We conclude that θ̂_{T} ψ_{i} = 0. □

Suppose that a ≤ i ≤ b − 1 is such that T̂(i) ≠ i. Let k ≥ 1 be such that (i + k, α_{i+k}) is a removable node of α, and α_{i+1} = α_{i+k}. If T̂(i) = i + 1, set T ∨ i := T. Otherwise, set {T ∨ i} := {T } ∪ {i + 1, …, i + k}. Then {T ∨ i} is the smallest semi-standard α-set of type β that contains i + 1 and {T }. There is a corresponding α-tableau T ∨ i and α-bijection { \atop T ∨ i}:

(9)  { \atop T ∨ i}(r) = \{ \begin{matrix}r + 1, & if i ≤ r < i + k < b; \\ T̂(i), & if r = i + k and T̂(i + k) = i + k; \\ T̂(r), & otherwise. \end{matrix}

Suppose that T is such that T̂(i+k) = i+k. In particular (i, α_{i}) is a removable node of α. The following describe the semi-standard α-sets {S} such that S ∨ i = T ∨ i:

{T }
{T } ∪  {i + k}
{T } ∪  {i + k − 1, i + k}
⋮ ⋮  ⋮
{T } ∪  {i + 2, …, i + k − 1, i + k}
{T } ∪ {i + 1, i + 2, …, i + k − 1, i + k} = {T ∨ i}

Lemma 5.5. Let a ≤ i ≤ b − 1 be such that i = a or α_{T̂^{−1}(i)} ≠ α_{i}. Then

θ̂_{T} ψ_{i} = \{ \begin{matrix}(h_{i} − h_{i+1}) & θ̂_{W} , & if T̂(i) = i + 1; \\ (−1)^{|\{T ∨i\}\\{T \}|} & θ̂_{W} , & otherwise. \end{matrix}

Here W is obtained from T ∨ i by changing the unique symbol i + 1 in row i to an i.

Proof. Let U ≈ T be such that U ψ_{i} C_{t}^{−} ≠ 0.

We consider first the case that T̂(i) = i + 1, or equivalently T = T ∨ i. In particular there exists a column number c so that node (i, c) of U contains the symbol i + 1, and all other symbols i + 1 belong to row (i + 1) of U. Lemma 5.1 then implies that U ψ_{i} C_{t}^{−} = V C_{t}^{−}, where V is obtained from U by changing the i + 1 in node (i, c) to an i. In particular V ≈ W.

Lemma 2.1 implies that column c of U does not contain two symbols i + 1. So either (i, c) is to the right of all nodes in row i + 1 i.e. c ∈ {α_{i+1} + 1, …, α_{i}}, or i + 1 ≠ b and (i + 1, c) is the unique node in row i + 1 of U that does not contain i + 1. This argument and (7) shows that there are α_{i} − α_{i+1} + 1 = h_{i} − h_{i+1} choices for a tableau U such that Uψ_{i} C_{t}^{−} = V C_{t}^{−}. It then follows from (8) that

e_{t} θ̂_{T} ψ_{i} = (h_{i} − h_{i+1}) ∑_{V ≈W} V C_{t}^{−} = (h_{i} − h_{i+1})e_{t} θ̂_{W} .

We conclude that in this case θ̂_{T} ψ_{i} = (h_{i} − h_{i+1})θ̂_{W}.

Now consider the case that T̂(i) ≠ i + 1. Then T̂(i + 1) = i + 1. As T̂(i) ≠ i, this ensures that (i, α_{i}) is a removable node of α. Set j := T̂(i). Lemma 4.1 shows that there exists c ≥ 1 such that node (i, c) of U contains the symbol i+j > i+1, and all other nodes in row i contain the symbol i. Furthermore, all symbols i+1 in U occur in row (i + 1) of U. Then Lemma 2.1 implies that c ≤ α_{i+1} and Uψ_{i}C_{t}^{−} = V C_{t}^{−}, where V is obtained from by changing the i + 1 in node (i + 1, c) of U to an i.

Let k ≥ 1 be such that α_{i+k} = α_{i+1} and (i + k, α_{i+k}) is a removable node of α. Let π be the permutation in Σ_{n} that cycles the entries in the nodes of U as follows:

if j > k then π : (i + k, c) → … → (i + 1, c) → (i, c) → (i + k, c);
if j ≤ k then π : (i + j − 1, c) → … → (i + 1, c) → (i, c) → (i + j − 1, c).

Then π belongs to C_{t}. So π C_{t}^{−} = sgn(π) C_{t}^{−}. Also V π ≈ W is strictly increasing down columns, and we obtain each α-tableau that is row equivalent to W as an V π exactly once, by an appropriate choice of U ≈ T.

Note that

\{T ∨ i\} = \{ \begin{matrix}\{T \} ˙∪\{i + 1, …, i + k\}, & if j > k, \\ \{T \} ˙∪\{i + 1, …, i + j − 1\}, & if j ≤ k. \end{matrix}

So sgn(π) = (−1)^{|{T ∨i}\{T }|} does not depend on the choice of U ≈ T. Thus

e_{t} θ̂_{T} ψ_{i} = ∑_{U≈T} U ψ_{i} C_{t}^{−} = (−1)^{|{T ∨i}\{T }|} ∑_{V ≈W} V C_{t}^{−} = (−1)^{|{T ∨i}\{T }|}e_{t} θ̂_{W} .

It follows that in this case θ̂_{T} ψ_{i} = (−1)^{|{T ∨i}\{T }|}θ̂_{W}. □

6. Carter-Payne homomorphisms

We now fix a Σ_{n}-homomorphism θ̂: S^{α} → S^{β}. By (3) we may write

(10)  θ̂ = ∑_{T} Λ_{T} θ̂_{T} ,  for certain Λ_{T} ∈ F,

where T ranges over the semi-standard α-tableaux of type β. We aim to show that, up to a scalar multiple, there is at most one such homomorphism θ̂. Since the image of θ̂ is a submodule of S^{β}, it follows that θ̂ψ_{i} = 0 for all i. Fix an i. Lemmas 5.3, 5.4 and 5.5 show that for all T, θ̂_{T} ψ_{i} is equal to a_{T,i}θ̂_{S_{T,i}}, where a_{T,i} is an integer and S_{T,i} is a semistandard tableau depending on T and i. Maps associated to semistandard tableaus are linearly independent. Thus the equation θ̂ψ_{i} = 0 produces several equations that the coefficients Λ_{T} must satisfy; for any semistandard tableau W, we have ∑ Λ_{T} a_{T,i} = 0, where the sum is over all T such that S_{T,i} = W. We shall collect enough of these equations to show that the space of all Σ_{n}-homomorphisms from S^{α} to S^{β} is at most one-dimensional.

Lemma 6.1. Let {T } be a semi-standard α-set of type β. Then

(−1)^{|{T }|}Λ_{{T }} = (−1)^{|{T }∩{R}|}Λ_{{T }∩{R}}.

Proof. The result is trivial if {T } ⊆ {R}. We prove the result by induction on |{T }\{R}|. So we may assume that {T }\{R} is nonempty. In particular, we may choose i ∈ {a + 1, …, b − 1} such that α_{T̂^{−1}(i)} > α_{i} and α_{i} = α_{i+1}. This forces T̂(i) = i + 1. In terms of tableaux, this can be visualized as follows. The set {T } is associated to a series of vertical strips of entries at the right edge of the tableau, vertical strips whose lowest entry is at a removable node. We are focussing on an entry at the top of one of these strips. The inductive step consists of proving that Λ_{T} + Λ_{{T }\{i}} = 0. Geometrically the end result of this lemma is to show that we need only understand the coefficients of semistandard sets associated to unions of strips of length 1 containing a single removable node.

Lemma 5.5 implies that θ̂_{T}ψ_{i} = θ̂_{W}, where W is the α-tableau that is obtained from T by changing the single i + 1 in row i of T to an i.

Suppose that U is a semi-standard α-tableau of type β such that θ̂_{U}ψ_{i} is a non-zero multiple of θ̂_{W}. Lemmas 5.3, 5.4 and 5.5 imply that either i ∉ {U} or (as α_{i} = α_{i+1}) both i and i + 1 belong to {U}. Moreover {T } and {U} do not differ apart from in the set {i, i + 1}. In the former case {U} = {T }\{i} and hence θ̂_{U}ψ_{i} = θ̂_{W}. In the latter case {U} = {T }. This completes the proof of the inductive step, and hence the proof of the lemma also. □

From now on we concentrate on finding relations among the Λ_{N}, with {N} ⊆ {R} a removable α-set of type β. So we consider N ↔ N̂ ↔ {N} such that {N} is an arbitrary removable α-set of type β.

Lemma 6.2. Let i ∈ {N} and j = N̂(i) be such that (i, α_{i}) and (j, α_{j}) are adjacent removable nodes of [α]. Then

h_{i} Λ_{{N}} + Λ_{{N}\{i}} = h_{j} Λ_{{N}} + Λ_{{N}\{j}}.

The term involving {N}\{i} is omitted if i = a, while the term involving {N}\{j} is omitted if i = b − 1 (and so j = b; recall also that h_{b} = 0).

Proof. We assume the generic case that i ≠ a and j ≠ b. The two exceptional cases follow from similar arguments.

Lemma 5.5 implies that θ̂_{N}ψ_{i} is a non-zero multiple of θ̂_{W}. Here W is the tableau that is obtained by changing the single i + 1 in row i of N ∨ i to an i. Note that {N ∨ i} = {N} ˙∪{i + 1, …, j − 1}. We enumerate all standard α-tableau T of type β such that θ̂_{T} ψ_{i} is a non-zero multiple of θ̂_{W}.

Suppose that T̂(i) = i. Then i ≠ a. Lemma 5.3 forces T̂(i + 1) ≠ i + 1 and θ̂_{T}ψ_{i} = θ̂_{W}. Also {T } = {T ∨ i} = {N ∨ i}\{i}. Thus {T } ∩ {R} = {N}\{i} and |{T }\{T } ∩ {R}| = j − i − 1. Lemma 6.1 then implies that in this case

(11)  Λ_{{T }}θ̂_{T}ψ_{i} = (−1)^{j−i−1}Λ_{{N}\{i}}θ̂_{W} .

From now on we assume that T̂(i) ≠ i. Then Lemma 5.5 forces {T ∨i} = {N ∨i}. There are three cases depending on whether i + 1, j ∈ {T } or not.

Suppose first that i+1 ∈ {T }. Then j ∈ {T } and T = T ∨i, and Lemma 5.5 shows that θ̂_{T}ψ_{i} = (h_{i} −h_{i+1})θ̂_{W}. But {T }∩{R} = {N}. So |{T }\{T }∩{R}| = j −i−1.

We conclude from Lemma 6.1 that in this case

(12)  Λ_{{T }}θ̂_{T}ψ_{i} = (−1)^{j−i−1}(h_{i} − h_{i+1})Λ_{{N}}θ̂_{W} .

There are j −i−1 semi-standard sets {T } such that j ∈ {T } but i+1 ∉ {T }, one for each of the rows i+1, i+3, …, j. For each such {T } we have {T }∩{R} = {N}. In particular |{T ∨ i}\{T } ∩ {R}| = j − i − 1. It then follows from Lemmas 5.5 and 6.1 that

(13)  Λ_{{T }}θ̂_{T} ψ_{i} = (−1)^{j−i+1}Λ_{{N}}θ̂_{W} .

The final possibility is that i + 1, j ∉ {T }. Then {T } = {N}\{j} is a removable set and |{T ∨ i}\{T } ∩ {R}| = j − i. So by Lemmas 5.5 and 6.1 we have

(14)  Λ_{{T }}θ̂_{T}ψ_{i} = (−1)^{j−i}Λ_{{N}\{j}}θ̂_{W} .

We now apply ψ_{i} to θ̂, expand as a linear combination of semi-standard homomorphisms, and examine the coefficient of θ̂_{W} in this expansion. Multiplying (11), (12), (13) and (14) by (−1)^{j−i−1}, we get

Λ_{{N}\{i}} + (h_{i} − h_{i+1})Λ_{{N}} − Λ_{{N}\{j}} + (j − i − 1)Λ_{{N}} = 0.

Using (7) and the fact that α_{i+1} = α_{j}, we can rewrite this as

(h_{i} − h_{j})Λ_{{N}} + Λ_{{N}\{i}} = Λ_{{N}\{j}}.
□

We now prove the main result of this section.

Theorem 6.3. (−1)^{|N|} Λ_{N} = (−1)^{|R|}( ∏_{u∈{R}\{N}} h_{u}) Λ_{R}.

Proof. We shall prove this theorem by repeated applications of Lemma 6.2, with i ≠ a. To do this, we use a partial order on removable α-tableau of type β. Suppose that {S}, {T } ⊆ {R}, with a ∈ {S}, {T }. We write {S} ◁ {T } if {S} = {T }\{i}, for some i, or if there exists (i, α_{i}) and (j, α_{j}) adjacent removable nodes of [α] such that i > j and {S}\{T } = {i} and {T }\{S} = {j}. Let < be the partial order generated by ◁. We shall prove the theorem by induction on <. The base case is {N} = {R}, when the conclusion is trivially true. Assume then that {N} ⊊ {R}.

Suppose first that b − 1 ∉ {N}. We apply Lemma 6.2 to {N} ∪ {b − 1} to get

Λ_{N} = (−h_{b−1}) Λ_{{N}∪{b−1}}.

But {N} ◁ {N} ∪ {b − 1}. So the inductive hypothesis gives

Λ_{{N}∪{b−1}} = ∏_{u∈{R}\{N}∪{b−1}} (−h_{u}) Λ_{R}.

The claimed expression for Λ_{N} follows in this case.

Suppose then that b − 1 ∈ {N}. Then we can find adjacent removable nodes (i, α_{i}) and (j, α_{j}) of [α] such that i > j > a and i ∈ {N} but j ∉ {N}. Lemma 6.2 implies that

Λ_{{N}} = (h_{j} − h_{i})Λ_{{N}∪{j}} + Λ_{{N}∪{j}\{i}}.

But {N} ◁ {N} ∪ {j} and {N} ◁ {N} ∪ {j}\{i}. So by the induction hypothesis we have

\begin{aligned}(h_{j} − h_{i})Λ_{\{N\}∪\{j\}} & = & (−h_{i} + h_{j}) & ∏_{u∈\{R\}\\{N\}∪\{j\}} (−h_{u}) Λ_{R}; \\ Λ_{\{N\}∪\{j\}\\{i\}} & = & (−h_{j}) & ∏_{u∈\{R\}\\{N\}∪\{j\}} (−h_{u}) Λ_{R}. \end{aligned}

Adding, we obtain the inductive step. □

Our corollary is a precise form of Theorem 1.5.

Corollary 6.4. The F -space Hom_{F Σ_{n}}(S^{α}, S^{β}) is at most one dimensional. Every non-zero Σ_{n}-homomorphism S^{α} → S^{β} is a scalar multiple of

∑_{T} ((−1)^{|T |} ∏_{u∈{R}\{T }}h_{u}) θ̂_{T}.

Proof. This follows at once from Lemma 6.1 and Theorem 6.3. □

From now on we assume that

θ̂ = ∑_{T} ((−1)^{|T |} ∏_{u∈{R}\{T }}h_{u}) θ̂_{T}.

Lemma 6.5. Over ℤ we have θ̂ψ_{i,r} = 0, unless i = a and r = β_{a+1} − 1.

Proof. In view of Lemma 5.2, we need only show that θ̂ψ_{i} = 0, for i = a+1, …, b−1. Fix i with a < i < b, and let W be an α-tableau such that there exists a semistandard α-tableau X of type β such that θ̂_{X}ψ_{i} is a nonzero multiple of θ̂_{W}. We enumerate all such tableau X.

Suppose first that (i, α_{i}) is not a removable node of [α]. Then Lemmas 5.2, 5.3, 5.4 and 5.5 show that there is a semi-standard α-tableau T of type β with i ∉ {T } such that X = T or S, where {S} := {T } ∪ {i}. Moreover, Lemmas 5.3 and 5.5 and the definition of θ show that

(Λ_{T} θ̂_{T} + Λ_{S}θ̂_{S})ψ_{i} = (−1)^{|T |}(1 − 1)( ∏_{u∈{R}\{T }} h_{u})θ̂_{W} = 0.

Suppose then that (i, α_{i}) is a removable node of [α]. Let j > i be such that (j, α_{j}) is a removable node of [α] and α_{j} = α_{i+1}. So (i, α_{i}) and (j, α_{j}) are adjacent removable nodes. Then Lemmas 5.2, 5.3, 5.4 and 5.5 show that there is a semistandard α-tableau T of type β with i, i + 1 ∈ {T }, such that X = T ∨ i or {X} = {T }\{i}. Thus {X} is one of the following semi-standard α-sets:

{T },
{S_{i}} := {T }\{i},
{S_{i+1}} := {T }\{i + 1},
{S_{i+2}} := {T }\{i + 1, i + 2},
⋮  ⋮
{S_{j−1}} := {T }\{i + 1, i + 2, …, j − 1},
{S_{j}} := {T }\{i + 1, i + 2, …, j − 1, j}.

Lemmas 5.3 and 5.5 and show that the coefficient of θ̂_{W} in θ ψ_{a} is given by

(Λ_{T} θ̂_{T}+∑_{u=i}^{j} Λ_{S_{u}} θ̂_{S_{u}})ψ_{i} = (−1)^{|T |}((h_{i}−h_{i+1})−h_{i}+(j−i−1)+h_{j})( ∏_{u∈{R}\{T }}h_{u})θ̂_{W} = 0,

as h_{j} − h_{i+1} = i + 1 − j, by (7) using the fact that α_{j} = α_{i+1}. □

Let γ be the partition corresponding to the following composition of n:

α_{1}, …, α_{a}, α_{a+1} − 1, α_{a+2}, …, α_{b−1}, α_{b} + 1, α_{b+1}, ….

We now deal with the exceptional test function ψ_{a}.

Lemma 6.6. Over ℤ we have Im θ̂ψ_{a} ⊆ h_{a}M^{γ}.

Proof. We retain the notation of W, T from the previous lemma, with i = a and j such that (j, α_{j}) be the highest removable node of [α] below (a, α_{a}). Equivalently α_{j+1} < α_{j} and α_{j} = α_{a+1}. The semi-standard sets that contribute to θ̂_{W} can be enumerated as {T }, {S_{i+1}}, …, {S_{j}} i.e. the same as in Lemma 6.5, but with {S_{i}} = {T }\{a} omitted (because it does not contain a, and hence is not a semistandard α-set of type β). So the coefficient of θ̂_{W} in θ ψ_{a} is given by

\begin{aligned}(Λ_{T} θ̂_{T} + ∑_{u=a+1}^{j}Λ_{S_{u}} θ̂_{S_{u}})ψ_{a} & = (−1)^{|T |}((h_{a}−h_{a+1}) + (j−a−1) + h_{j})( ∏_{u∈\{R\}\\{T \}}h_{u})θ̂_{W} \\ & = (−1)^{|T |}h_{a}( ∏_{u∈\{R\}\\{T \}} h_{u})θ̂_{W} . \end{aligned}

□

We also obtain the following criterion for the existence of a non-zero homomorphism. It is a special case of a result of Carter and Payne [2]. The ‘if’ part is also true over fields of characteristic 2.

Corollary 6.7. Let β be a one-box-shift of α, with β_{a} = α_{a} − 1 and β_{b} = α_{b} + 1, for a < b. Then Hom_{F Σ_{n}}(S^{α}, S^{β}) ≠ 0 if and only if α_{a} − a ≡ β_{b} − b mod p.

Proof. Lemmas 6.5 and 6.6 show that, over the field F of characteristic p, the image of θ belongs to S^{β} if and only if h_{a} ≡ 0 mod p (for the only if, for T = R we have θψ_{a} ≠ 0 if p |̸h_{a} as in that case∏_{u∈{R}\{T }} h_{u} = 1). The result follows, as h_{a} = (α_{a} − α_{b}) + (b − a − 1) = (α_{a} − a) − (β_{b} − b). □

7. Jantzen Filtration

We let ⟨ , ⟩ denote the symmetric bilinear form for which the β-tabloids form an orthonormal basis of M^{β}. This form is Σ_{n}-invariant. Over ℚ we have a direct sum decomposition:

M_{ℚ}^{β} = S_{ℚ}^{β} ⊕ S_{ℚ}^{β⊥},

where S_{ℚ}^{β⊥} = {m ∈ M_{ℚ}^{β} | ⟨m, S_{ℚ}^{β}⟩ = 0} is the perpendicular space to S^{β}. Now S_{ℤ}^{β} ⊕ S_{ℤ}^{β⊥} is a full sublattice of M_{ℤ} (in the sense that both have the same ℤ-rank). However the quotient lattice M_{ℤ}^{β}/S_{ℤ}^{β} ⊕ S_{ℤ}^{β⊥} can be quite large. Over a field F of characteristic p, the restriction of the form ⟨ , ⟩ to S^{β} is generally degenerate. Over ℤ, the module S_{ℤ}^{β} has a sequence of Jantzen submodules:

J^{i}(S_{ℤ}^{β}) := {x ∈ S_{ℤ}^{β} | ⟨x, S_{ℤ}^{β}⟩ ⊆ p^{i}ℤ}.

The images of these modules under the decomposition map give a corresponding filtration of the Specht module S_{F}^{β} = S_{ℤ}^{β}/pS_{ℤ}^{β}. We note that

(15)  J^{i}(S_{ℤ}^{β}) = S_{ℤ}^{β} ∩ (p^{i}M_{ℤ}^{β} + S_{ℤ}^{β⊥}).

See [7, 1.2] for example. We prove Theorem 1.6 in the following form:

Theorem 7.1. Let α, β be partitions of n, with β_{a} = α_{a} − 1 and β_{b} = α_{b} + 1, for a < b, and β_{j} = α_{j}, for j ≠ a, b. Suppose that p^{i}|(α_{a} − β_{b} + b − a), for some i ≥ 0. Then each θ̂ ∈ Hom_{Σ_{n}}(S^{α}, S^{β}) satisfies Im(θ̂) ⊆ J^{i}(S^{β}).

Proof. In view of Corollary 6.4, we may assume that θ̂ is the map of the previous section, defined over ℤ. Recall that the polytabloid e_{t} := {t}C_{t}^{−} generates S^{α}. Here t is a fixed α-tableau, and C_{t}^{−} is the signed column stabilizer sum of t. Also h_{a} := (α_{a} − β_{b} + b − a). The strategy of this proof is very simple. In (17) below we produce an ‘error term’ E_{t} ∈ M^{β} and show that

e_{t}θ̂ − h_{a}E_{t} ∈ S_{ℤ}^{β}.

Now over ℤ, the image of θ̂ is contained in S_{ℤ}^{β⊥}. So by (15), e_{t}θ̂− h_{a}E_{t} belongs to J^{i}(S_{ℤ}^{β}), which is what we need to prove.

If T is a semistandard α-tableau of type β then there is a corresponding semistandard homomorphism θ̂_{T} : S^{α} → M^{β} such that

e_{t}θ̂_{T} = ∑_{U≈T} UC_{t}^{−}.

By Lemma 2.1, we may assume that U ranges over all tableaux that are row equivalent to T, such that no column of U contains a repeated entry.

Suppose that U, V are α-tableaux. Write U < V if U ≈ V and all nodes (a, c) in U with c > β_{a+1} contain the symbol a. We say that U is right-justified. Now define the following linear combination of α-tableaux (which will belong to M^{γ}, where γ is the type of V):

θ̃_{V} := ∑_{U<V} UC_{t}^{−}.

Now recall that T is a semi-standard α-tableau of type β. The analogues of Lemmas 5.2, 5.3, 5.4 and 5.5 hold for θ̃_{T}, with the following exception:

(16)  θ̃_{T}ψ_{a} = θ̃_{W} , if T̂(a) = a + 1,

and not (h_{a} − h_{a+1})θ̃_{W}. This can be seen by inspecting the original proof. The key point is that there are α_{a} − α_{a+1} + 1 α-tableaux U such that U ≈ T and Uψ_{a} = W. However, only one of these U satisfies U < T.

Now define the following element of M^{β}:

(17)  E_{t} := ∑_{T̂(a)=a+1} ((−1)^{|T |} ∏_{u∈{R}\{T }}h_{u})θ̃_{T}.

So T runs over all semi-standard α-tableaux of type β which contain a (unique) symbol a + 1 in row a. Lemma 6.5 states that θ̂ψ_{i,r} = 0, unless i = a, r = β_{a+1} − 1. In view of the previous paragraph, the analogous result holds for E_{t}:

E_{t} ψ_{i,r} = 0, unless i = a and r = β_{a+1} − 1.

In view of this, and (4), we see that e_{t}θ̂−h_{a}E_{t} ∈ S^{β}, if and only if (e_{t}θ̂−h_{a}E_{t})ψ_{a} = 0. But this equality holds, by Lemma 6.6 and (16). □

As an example, let p = 5 and α = (4, 3) and β = (3^{2}, 1). So a = 1, b = 3 and h_{1} = 5, h_{2} = 3. There are two semi-standard α-tableaux of type β:

1 1 1 2 2 2 3 and ^{1 1 1 3 2 2 2} .

corresponding to the semi-standard α-sets {1, 2} and {1}, respectively. Our linear combination θ : S_{ℤ}^{α} → M_{ℤ}^{β} is then

θ = { \atop θ_{1 1 1 2 2 2 3}} − 3{ \atop θ_{1 1 1 3 2 2 2}} .

The error term is

E_{t} = (^{2 1 1 1 3 2 2} + ^{1 2 1 1 2 3 2} + ^{1 1 2 1 2 2 3} )C_{t}^{−}.

Now consider the following standard α-tableau t and its signed column sum C_{t}^{−}:

t = 1 2 3 4 5 6 7 , C_{t}^{−} = (1 − (1, 5))(1 − (2, 6))(1 − (3, 7)).

We can write e_{t}θ − h_{a}E_{t} as a linear combination of α-tabloids:

e_{t}θ − 5E_{t} = ( \begin{matrix}^{1 1 1 2 2 2 3} + ^{1 1 1 2 2 3 2} + ^{1 1 1 2 3 2 2} + ^{1 1 2 1 2 2 3} + ^{1 2 1 1 2 3 2} + ^{2 1 1 1 3 2 2} \\ −3 (^{1 1 1 3 2 2 2} + ^{1 1 3 1 2 2 2} + ^{1 3 1 1 2 2 2} + ^{3 1 1 1 2 2 2} ) \\ −5 (^{2 1 1 1 3 2 2} + ^{1 2 1 1 2 3 2} + ^{1 1 2 1 2 2 3} ) \end{matrix} ) C_{t}^{−}

= ( \begin{matrix}^{1 1 1 2 2 2 3} + ^{1 1 3 1 2 2 2} − ^{1 1 1 3 2 2 2} \\ ^{1 1 1 2 2 3 2} + ^{1 3 1 1 2 2 2} − ^{1 1 1 3 2 2 2} \\ ^{1 1 1 2 3 2 2} + ^{3 1 1 1 2 2 2} − ^{1 1 1 3 2 2 2} \end{matrix} ) C_{t}^{−}.

= (^{1 1 1 2 2 2 3} + ^{1 1 3 1 2 2 2} + ^{1 1 2 3 2 2 1} ) C_{t}^{−} (1 + (2, 3)(6, 7) + (1, 3)(5, 7)) .

Now

(^{1 1 1 2 2 2 3} + ^{1 1 3 1 2 2 2} + ^{1 1 2 3 2 2 1} ) C_{t}^{−} = (^{3 2 1 4 6 5 7} + ^{4 2 1 7 6 5 4} + ^{7 2 1 3 6 5 4} ) C_{t}^{−} = e_{3 2 1 4 6 5 7} .

Multiplying through by (1 + (2, 3)(6, 7) + (1, 3)(5, 7)), we get

e_{t}θ − 5E_{t} = e_{3 2 1 4 6 5 7} + e_{2 3 1 4 7 5 6} + e_{1 2 3 4 6 7 5} .

We wish to compare this with the expression given by (2). So we use the equalities

\begin{aligned}e_{3 2 1 4 6 5 7} & = & e_{1 2 3 5 6 4 7} + e_{1 2 4 5 6 7 3} − e_{1 2 3 5 6 7 4} \\ e_{2 3 1 4 7 5 6} & = & e_{1 3 2 5 7 4 6} + e_{1 3 4 5 7 6 2} − e_{1 2 3 5 6 7 4} \\ e_{1 2 3 4 6 7 5} & = & −e_{1 2 3 5 6 7 4} . \end{aligned}

Thus

e_{t}θ − 5E_{t} = e_{1 2 3 5 6 4 7} + e_{1 2 4 5 6 7 3} + e_{1 3 2 5 7 4 6} + e_{1 3 4 5 7 6 2} − 3 e_{1 2 3 5 6 7 4} .

8. Acknowledgement

Part of this paper was written while Dr. Ellers was visiting the National University of Ireland Maynooth in July/August 2004. The visit was substantially funded by a New Researcher Award from the National University of Ireland Maynooth. We gratefully acknowledge this assistance.

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Department of Mathematics, Allegheny College Meadville, PA 16335, USA

   E-mail address: hellers@allegheny.edu

Department of Mathematics, National University of Ireland Maynooth, Co. Kildare, Ireland

   E-mail address: John.Murray@maths.nuim.ie