Irreducible representations of the symmetric groups from slash homologies of p-complexes

In the 40s, Mayer introduced a construction of (simplicial) $p$-complex by using the unsigned boundary map and taking coefficients of chains modulo $p$. We look at such a $p$-complex associated to an $(n-1)$-simplex; in which case, this is also a $p$-complex of representations of the symmetric group of rank $n$ - specifically, of permutation modules associated to two-row compositions. In this article, we calculate the so-called slash homology - a homology theory introduced by Khovanov and Qi - of such a $p$-complex. We show that every non-trivial slash homology group appears as an irreducible representation associated two-row partitions, and how this calculation leads to a basis of these irreduicble representations given by the so-called $p$-standard tableaux.


Introduction
The modern notion of p-complexes first appear in the works of Mayer [M1, M2] back in the 40s. A pcomplex is a sequence of groups (C k ) k∈Z along with maps (called p-differentials) (∂ k : C k → C k−1 ) k∈Z such that any composition of p consecutive such maps is zero. A simple example is to take C k = Z for all 0 ≤ k ≤ p, ∂ k = id for all 0 < k ≤ p, and C k = 0 = ∂ k for all other k's.
In [M1], Mayer defined a homology theory on a p-complex C • = (C k , ∂ k ) k given by the groups a H k := Ker(∂ a )/ Im(∂ p−a ), with a ∈ {1, 2, . . . , p − 1}. Moreover, the p-differential ∂ k naturally induces a map a H k → a+1 H k−1 for all a, which gives the total homology k,a a H k a structure of a (p − 1)complex. However, a result of Spanier states that these homology groups can be determined by the classical simplicial homology groups, and so p-complexes never catch much attention to topologists.
Recently, the Khovanov school advocates that p-complexes should provide a foundational tool for categorifying quantum groups at primitive p-th root of unity. As a part of this programme, in [KQ], Khovanov and Qi gave a more detailed account of two alternative homology theories for p-complexes (which are dual to each other). These homology theories are called slash and backslash homology, and have better compatibility with the technology of differential graded categories than Mayer's version. At each degree k, there are p − 1 slash and backslash homology groups, given by H /a k := Ker(∂ a+1 )/(Im(∂ p−a−1 ) + Ker(∂ a )) and H \a k := (Im(∂ a k+a ) ∩ Ker(∂ p−1−a k ))/ Im(∂ a+1 k+a+1 ) respectively, with a ranging in {0, 1, . . . , p − 2}. The p-differentials ∂ also naturally induce a structure of (p − 1)-complex on the total homology; see subsection 2.2 for details.
This article studies the slash homology of a p-complex associated to an (n − 1)-simplex. Let us be more precise about the construction of this p-complex. Fix an integer n and for each k ∈ {0, 1, . . . , n}, let Ω k be the set of k-subsets, i.e. subsets of {1, 2, . . . , n} of size k. Fix a field F of prime characteristic p > 0 and let F Ω k be the F-vector space with basis Ω k . Let ϕ k : F Ω k → F Ω k−1 be the F-linear map that sends a k-subset ω to the formal sum of all (k − 1)-subsets contained in ω, i.e.
It is an easy exercise to check that ϕ k ϕ k−1 · · · ϕ k−p = 0 for any k (or simply ϕ p = 0 by omitting subscripts). Hence, we have a p-complex of the form It should not be difficult to see that in the case of p = 2, F Ω • is the canonical simplicial chain complex associated to an (n − 1)-simplex.
In fact, F Ω • is not just a p-complex of F-vector spaces, it is also a p-complex of S n representations, where S n is the symmetric group of rank n. Indeed, F Ω k is isomorphic to the module induced from the trivial F(S n−k × S k )-module, i.e. the permutation module M (n−k,k) of the two-part composition (n − k, k) of n, and ϕ k 's are also F S n -linear.
Our first main result is the complete description of the (total) slash homology of F Ω • .
Theorem. (Theorem 4.16) For any prime p and any positive integer n, the slash homology (p − 1)complexes H / k at degree k ∈ {0, 1, . . . , n} of F Ω • is non-vanishing if, and only if, p = 2 and n − 2k ∈ 0, p − 2 . Moreover, in the case when H / k is non-vanishing, it takes the form H / k where D (n−k,k) is the simple F S n -module corresponding to the partition (n − k, k).
We remark that knowing the slash homology is enough to determine the backslash homology and Mayer's version of homology of F Ω • as well; see subsection 2.2 and subsection 2.3. Note also that our condition means that H / k vanishes when 2k ≤ n − (p − 1). In particular, all slash homology groups vanish when p = 2. This is just the classical result which says that the chain complex of an (n − 1)-simplex has zero homology, as every student taking a first course in algebraic topology would know.
Let us give some remark on the technique we use to achieve our goal. Inspire by the "α-sequence" combinatorics used by James in studying the permutation modules [J1], we consider a new statistic, called density, on two-row tableaux (which are equivalent to subsets) of n. In fact, if 2k ≤ n (meaning that (n − k, k) is a two-row partition), then density records when a column in the row-standard tableau is standard. Then we define threshold to allow us carving an induction machine to show that "most" elements in Ker(ϕ) lies in Im(ϕ p−1 ).
One particular step in proving the above theorem is to show that H /0 k is a quotient of the Specht module S (n−k,k) when it is non-zero. Recall that every Specht module S λ admits a basis given by the polytabloids e e e t associated to standard tableaux t of shape λ. It turns out the projecting this basis onto H /0 k results in a basis that is indexed by what Kleshchev calls p-standard tablueax in [Kl] (see Definition 5.1, its following remark, and Lemma 5.11).
Theorem. (Theorem 5.2) The simple F S n -module D (n−k,k) , where k satisfies n − (p − 1) < 2k ≤ n admits a basis of the form e e e t + radS (n−k,k) t is a p-standard tableau , We remark that it is shown in [Kl] that dim F D (n−k,k) is the same as the number of p-standard tableaux of shape (n − k, k), but no explicit description of such a basis is given. Moreover, Kleshchev's starting point is to look at restriction of simple F S n -modules to F S n−1 , whereas our deduction comes from understanding the map ϕ k on permutation modules and the classic proof of the standard polytabloid basis of Specht modules.
at the Institute of Mathematical Sciences, National University of Singapore, December 2017; we are thus deeply grateful for the organisers and the IMS. In addition, we are immensely grateful for Mark Wildon's interests, comments, and encouragement.

Preliminaries
Throughout, F is a fixed field of positive characteristic p > 0. We will mostly follow notations in [Wil]. In particular, by module we mean a finitely generated right modules. Maps are also applied to the right of a module.
Unless otherwise stated, we fix a positive integer n throughout the article, and define m := ⌊n/2⌋+1. Define a, b := {a, a + 1, . . . , b} for a < b ∈ Z, and a := 1, a . For a set S, we denote by |S| its cardinality. We use S \ T to denote the complement of the subset T of S.
For a finite set Ψ, we denote by F Ψ the F-vector space with basis Ψ. If Ψ is a finite subset of a module over some algebra, then we denote by F-spanΨ the set of elements spanned by elements of Ψ over F -we do not assume Ψ is a basis of F-spanΨ in this case, unless otherwise stated.
For a finite set X, denote by S X the group of symmetries on X. Clearly, S X is isomorphic to the symmetric group S r of rank r := |X|.
For a set X acted upon by a group G, and a subset Y ⊂ X, the stabiliser (subgroup) of Y in G, or just Y -stabiliser if the context is clear, is given by the set of g ∈ G such that (Y )g = Y (instead of (y)g = y for all y ∈ Y ).
2.1. The permutation modules F Ω k . Denote by Ω k the set of all k-subsets of n . For ω ∈ Ω k , we denote by ω c the complement n \ ω of ω in n . For ω ∈ Ω k , we label its elements in the way so that The symmetric group S n acts naturally on Ω k by permuting elements of ω ∈ Ω k ; hence F Ω k is a F S n -module. In the language of symmetric group representations, this module is the same as the permutation module M (n−k,k) associated to the composition (n − k, k) of n. For x ∈ F Ω k , the support of x, denoted by Supp(x), is the set of k-subsets ω that has non-zero coefficient in x.
Recall from [Wil] that there is a bilinear operation on n k=0 F Ω k given by bilinearly extending Recall from [Wil] the a-step boundary map ϕ (a) k : F Ω k → F Ω k−a with a ∈ {0, 1, . . . , k} and k ∈ {0, 1, . . . , n}. It will be convenient to define ϕ (a) k to be the zero map whenever k / ∈ 0, n . We will omit the lower script so long as the context is clear. By ϕ we simply mean ϕ (1) .
For a ∈ p − 1 , a! is invertible in F, so induction yields Note that ϕ p = 0 but ϕ (p) = 0.
The following fact will be useful throughout the article.
Lemma 2.1 (Splitting rule). [Wil,Lemma 3.5] For v ∈ F Ω k and w ∈ F Ω ℓ such that ν ∩ ω = ∅ for all ν ∈ Supp(v) and ω ∈ Supp(w), we have 2.2. p-complexes and homologies. Within this subsection, F is an arbitrary field, and p a prime number. Suppose A is an F-algebra. Let C • denote the data (C k , ∂ k : C k → C k−1 ) k∈Z , where C k 's are A-modules and ∂ k 's are A-module maps. For a positive integer r ∈ Z + , we use the notation ∂ r k to denote the composition ∂ k • ∂ k−1 • · · · • ∂ k−r+1 ; we also regard ∂ 0 k as the identity map on C • . We say that C • is a p-complex if ∂ p k = 0 for all k ∈ Z; in which case, ∂ k 's are called the p-differentials of C • . Note that we use chain convention in this article instead of the usual cochain setup in [KQ]; hence, the exchange of subscripts and superscripts in the indices.
A very natural generalisation of the homology theory of ordinary complexes in the setting of pcomplexes is the following. Definition 2.2. Let C • := (C k , ∂ k ) k∈Z be a p-complex of A-modules. The r-th p-homology at degree k ∈ Z of C • , where r ∈ p − 1 , is defined as the A-module r H k (C • ) := Ker(∂ r k )/ Im(∂ p−r k+p−r ). It turns out that, at least in the setting of this article, another (equivalent) homology theory of p-complexes is much easier to calculate than p-homologies.
The a-th slash homology group and backslash homology group at degree k ∈ Z of C • , where a ∈ 0, p − 2 , are defined as the A-modules )/ Im(∂ a+1 k+a+1 ), respectively. The (total) slash homology and backslash homology at degree k ∈ Z are the (p − 1)complexes respectively, where ∂ are naturally induced from the restricted maps ∂ : Ker(∂ a ) → Ker(∂ a+1 ) and Note that the (p − 1)-complex H / k (C • ) has terms lying in degree k + p − 1, k + p − 2, . . . , k. If we view the category of p-complexes of vectors spaces as the category of graded modules over F[∂]/(∂ p ), then the slash-a homology group H /a k (C • ) picks up ("slashes through" 1 ) the a-th radical layer of any indecomposable non-projective direct summand of C • occurring at degree k, whereas the blackslash-a homology group picks up the a-th socle layer. In particular, the total slash homology of a complex corresponds to removing projective direct summands. In contrast, a,k a H k (C • ) is almost always much larger than the total slash homology. For example, for Nevertheless, it is clear that In fact, as remarked in [KQ], the three types of homologies are related by the exact sequence for all r ∈ p − 1 , where ι is induced by the natural inclusion ι : Ker(∂ r−1 ) → Ker(∂ r ). The following general lemma will be useful later.
Lemma 2.4. Let C • := (C k , ∂ k ) k∈Z be a p-complex of A-modules. Then the following holds for any k ∈ Z.
(1) (Slash shifting) ∂ a induces an injective map H /a Indeed, the assumption on x means that (x)∂ a ∈ Ker(∂), so the condition of this being in B /0 k−a = Im(∂ p−1 ) means that there is some y so that (x)∂ a = (y)∂ p−1 . Hence, we have x − (y)∂ p−1−a ≡ 0 mod Ker(∂ a ), which implies x ≡ 0 mod B /a k as required.
(2) Note that the condition b < p is needed just so that p − r ≥ 1. We prove the claim by induction on r. The case r = 1 is precisely the assumption The assumption of p−1 H k−r (C • ) = 0 means that we have u ∈ C k−(r−1) such that v = (u)∂ k−(r−1) . In particular, as we have u ∈ Ker(∂ p−(r−1) k−(r−1) ) By induction hypothesis, we have p−1 H k−(r−1) (C • ) = 0, meaning that there is some w ∈ C k such that (w)∂ r−1 k = u. Thus, we have (w)∂ r k = v, i.e. v ∈ Im(∂ r k ), as required.  Let us now consider the effect of applying F-linear duality (−) * := Hom F (−, F) on the p-complex F Ω • . Note first that F Ω k is a self-dual module for all k ∈ 0, n . Moreover, there is a canonical isomorphism F Ω * k ∼ = F Ω n−k which sends the basis vector ω * of F Ω * k dual to ω ∈ Ω k to the complement subset ω c ∈ Ω n−k , and it yields a commutative diagram where the vertical isomorphisms are the canonical ones. In particular, we have an isomorphism F Ω • ∼ = F Ω * • of p-complexes, where F Ω * • is the p-complex whose degree k term is given by F Ω * n−k and pdifferentials given by ϕ * k . Using Lemma 2.5, this means that we have , for all k ∈ 0, n and a ∈ 0, p − 2 . We call the existence of isomorphisms H /a the back/slash duality 2 of F Ω • . This duality says that, the slash homology of F Ω • alone is enough to understand backslash homology, as well as p-homology using (2.2.1).
In particular, combining with (2.1.1), the main result (Theorem 4.16) of this article gives an indirect answer to the question raised by Wildon in [Wil] that asks for the structure of Ker(ϕ 2.4. Review on a filtration of F Ω k . In [J1] (see also [J2,Chapter 17]), James shows that a permutation module M µ corresponding to a composition µ admits a filtration {S µ # ,µ } µ # over a certain series of partitions µ # . In this subsection, we will review his construction and some related results in the special case of µ = (n − k, k) with k ∈ 0, n .
Let t be a (Young) tableau. Denote by t i,j the content of the box located at (row, column)= (i, j), with row counting from top to bottom and column counting from left to right. From now on, we will only work with tableaux that have at most two rows. Definition 2.6. Suppose ℓ, k ∈ 0, n are non-negative integers with ℓ ≤ k and ℓ ≤ n − k. For a (n − k, k)-tableau t is row-standard (resp. column-standard ) if the entries in each row (resp. column) are arranged in increasing order. It is called ℓ-standard if it is row-standard and t 1,j < t 2,j for all j ∈ ℓ (in other words, having the first ℓ columns being standard). A k-standard (n − k, k)-tableau is just a standard tableau in the usual sense. Denote by SYT n (k; ℓ) the set of ℓ-standard Young tableaux of shape (n − k, k).
Retaining the notation on t, ℓ, k. Denote by C t,ℓ ≤ S n the stabiliser of the first ℓ columns of t, i.e. the direct product of order 2 subgroups (t 1,j , t 2,j ) over all j ∈ ℓ . Note that C t,k is just the usual column stabliser of t.
Recall the following relation between row-equivalent classes (a.k.a.tabloid ) of (n − k, k)-tableaux and k-subsets: For a tableau t, its associated k-subset, denoted by {t} := {t 2,1 , t 2,2 , . . . , t 2,k } ∈ Ω k , consist of all the elements in its second row. Conversely, given a k-subset ω, then we have a (rowstandard) tableau t ω whose first row consists of elements of ω c and second row consists of elements of ω in increasing order. Clearly, we have ω = {t ω }, and so we call t ω is the tableau associated to a k-subset.
Definition 2.8. For a subgroup G ≤ S n , the signed sum of G is an element in the group algebra given by the sum of sgn(σ)σ over all σ ∈ G. Denote by κ t,ℓ the signed sum of C t,ℓ . Define ℓ-polytabloid associated to t to be the element in F Ω k given by {t}κ t,ℓ . We say that e e e t,ℓ is ℓ-standard if so is t.
Note that a ℓ-polytabloid is dependent on the defining tableau, i.e. row-equivalent but distinct tableaux can define distinct ℓ-polytabloids. Convention. Whenever we use any of the notations or terminologies above, if ℓ is omitted, then it is assumed that ℓ = k. We also replace t by ω ∈ Ω k in the notation whenever we take the canonical tableau t ω associated ω.
Example 2.9. The ℓ-polytabloid associated to ω := k can be explicitly written as e e e k ,ℓ = k κ t,ℓ = k For an even more explicit case, say n = 8, k = 3, ℓ = 2, then we have We define an F S n -submodule of F Ω k by S k,ℓ := F -span{e e e t,ℓ | t a (n − k, k)-tableau}.
Note that, if t ′ = tσ for some σ ∈ S n , then one can easily see that e e e t ′ ,ℓ = e e e tσ,ℓ = e e e t,ℓ σ. In particular, S k,ℓ is generated (over F S n ) by any single ℓ-polytabloid. S k,ℓ is the module S µ # ,µ used in [J1] mentioned before, where µ # = (n − k, ℓ).
Remark 1. If k − ℓ < p, then Theorem 2.10 (1) and (2) (2) is to find the defining spanning set of S k,ℓ to a basis. This is achieved by putting a total order on the set of tabloids (k-subsets in our setting) and generalises the strategy of [Pe], who showed that Specht modules admit a basis given by standard polytabloid.
this defines a partial order on Ω k called the the row-dominance order.
Remark 2. The definition in [J1] is stated differently -they use the backward-reading lexicographical order on k-subsets, i.e. ω < lex ω ′ ∈ Ω k if there is some j ∈ k so that |ω ∩ j | < |ω ′ ∩ j | and |ω ∩ i | = |ω ′ ∩ i for all i ∈ j + 1, k . Our formulation is the opposite of the partial order used in [Sag], which is coarser than the one used in [J1]. With respect to the next result, it does not matter which order one prefer to use.
(1) If t is ℓ-standard, then {t} is minimal in Supp(e e e t,ℓ ) with respect to the row-dominance order.
Part (2) of the theorem in the case when ℓ = k is shown by Peel in [Pe].
We need a refinement of the duality between F Ω k and F Ω n−k mentioned in the previous section.
Proof Recall that the action of g ∈ S n on the dual of a module is given by the action of g −1 on the original module. Since every σ ∈ C k ,ℓ is a commuting product of transpositions, its action on F Ω * k is the same as the original, i.e. e e e * k ,ℓ = ( k κ k ,ℓ ) * = k * κ k ,ℓ . Recall that the duality isomorphism F Ω * k ∼ = F Ω n−k is given by ω * → ω c for any ω * which is the dual basis of ω ∈ Ω k . Hence, e e e * k ,ℓ is mapped to k + 1, n κ k ,ℓ under this map. Observe that C k ,ℓ = C t,ℓ where t is the (k, n − k) tableau given by swapping the two rows of t k , which is the same as t k+1,n . Thus, e e e * k ,ℓ is mapped to e e e k+1,n ,ℓ . Since S k,ℓ is generated by any single ℓ-polytabloid e e e ω,ℓ as an F S n -module (and similarly for S n−k,ℓ ), the argument in the previous paragraph implies the required isomorphism.

Partitioning k-subsets
Recall that m is defined to be ⌊n/2⌋+1. Sometimes for the ease of referencing, we say that k ∈ 0, n is low if k < m (equivalently, 2k ≤ n), respectively high if k ≥ m (equivalently, 2k > n).
In subsection 3.1, we introduce the key tools -called density and threshold -of our investigation, and look at some of their elementary properties. Then in subsection 3.2, we present some ideas how one can modify the proof of exactness of F Ω • in the case when p = 2 to higher prime p.
3.1. Density and threshold. Roughly speaking, we need a book-keeping device to see how far t ω is away from being (ℓ-)standard.
We denote by Ω h k the subset of Ω k consisting of k-subsets with threshold h. It is natural to denote by Ω ≥h k and Ω ≤h k the union of Ω i k over all i ∈ h, m and i ∈ 0, h respectively; similarly for Ω >h k and Ω <h k .
Thus, we have η ∈ Ω 0 k and ω ∈ Ω k k . The following facts about density will be useful.
Proposition 3.3. The following holds for any k-subset ω ∈ Ω k . ( Proof (1): This is straightforward from the definition.
(2), (2'): We show only (2); the proof of (2') is analogous and left to the keen reader. It follows from (1) that d ω i+1 ≥ d ω i , which is by our assumption at least i + 1. Assume d ω i+1 > i + 1 (otherwise, we are done). Now we just replace i in the statement of by i + 1. Note that it is impossible to have d ω j > j for all i < j < i ′ ; otherwise, we will get d ω i ′ ≥ d ω i ′ −1 ≥ i ′ -a contradiction to our assumption.
This proposition says that d ω , regarded as a function m → N 0 , is increasing and satisfies a discrete version of "rotated" intermediate value theorem.
The following tells us that entries of d ω beyond the threshold "stays either above or below the diagonal".
Since the threshold is h and h < m, we must have d ω m > m. In both cases, the assumption on i implies that we have i < m. So we can apply Proposition 3.3 (2) (resp. (2')) and get that d ω j = j for some i < j < m. But this contradicts the assumption that the threshold of ω is h and h < i.
Let us look at what values of threshold are never attainable.
(Lo): By Proposition 3.3 (1), we have d ω i ≤ k for any i, so there is no ω ∈ Ω k with threshold larger than k, as required.
(Hi): We need to show that there exists i ∈ n − k + 1 so that The following proposition explain why density and threshold somewhat measures how far t ω is from (ℓ−)standard tableaux.
Proposition 3.6. Consider a k-subset ω ∈ Ω k with 2k ≤ n. Then the following holds.
It remains to show that this implies t ω is standard. Indeed, we can prove by induction on k. For k = 1, it is clear. Otherwise, take k minimal so that ω c (k) > ω (k) . By induction hypothesis, the subtableau where the second row takes only k − 1 entries is standard. This means that ω (k−1) ≥ 2(k − 1), and so we have The assumption on threshold says that ω (h) ≤ 2h − 1 and ω (h+1) > 2h + 1, which then implies that ω c We warn that the sense of "far" here is in general not related the partial orders ⊳ r , < lex reviewed in subsection 2.4. Indeed, it is possible that ω ∈ Ω h k with h > 0 but t ω is row-equivalent to a ℓ-standard tableau with ℓ < k − h, as one can see from the following example.
The reason for this "imperfection" is that threshold is defined to be the last index where a nonincreasing column appear in t ω . Threshold are designed solely because it allows us to carve an induction mechanism for calculating Ker(ϕ). There is a refinement of threshold that can be used to give equivalent criteria for ℓ-standardnes; it is given by the α-sequences used by James in [J1].
where [v] denotes the vector corresponding to v ∈ F Ω k with respect to the basis Ω k .
Recall that Ω k is partially ordered by ⊳ r (Definition 2.11). It follows from Theorem 2.10 (1) and Proposition 3.6 (1) that we can write M = [M >0 |M 0 ] as block matrix with M 0 being the identity |Ω 0 k | × |Ω 0 k |-matrix. In particular, M is a lower unitriangular matrix. Consider now the n k × n k -matrix given by stacking the identity matrix on top of M . This matrix defines a change of basis matrix from 3.2. Homogeneity. In the classical case, i.e. when p = 2, the fact that F Ω • is exact can be explained with a one-line proof, namely, for ω ∩{1} = ∅ and ω → 0 otherwise. The displayed formula here no longer holds for p > 2, but it turns out that the key to understanding Ker(ϕ)/ Im(ϕ p−1 ) lies in finding a generalisation of this formulathis is the goal of this subsection, and is achieved in Lemma 3.15 and Lemma 3.16. Let us remark for the moment that the key ingredients in finding the correct generalised formula are (1) Consider ω whose threshold is positive, and (2) use a "partial version" of δ (see Definition 3.11).
Since we only consider the case of non-zero threshold in this subsection, and the threshold of the empty subset ∅ ∈ Ω 0 is 0, throughout this subsection, we will impose the following Assumption: k > 0, unless otherwise specified. Note that we will no longer assume k is low.
Let us start with the definition of the partial version of ϕ.
Definition 3.9. Consider a set I. For any k ∈ 0, n , we define an F-linear map Note that if I ⊃ n , then ϕ I = ϕ n and ϕ I c = 0. Note also that ϕ I is not necessarily a morphism of F S n -modules.
Let us list some elementary properties about ϕ I .
Proposition 3.10. Let I be a subset of n , and ω be a k-subset of n for some positive k ∈ n . ( If ω and I are disjoint, then (ω)ϕ I = 0.
Definition 3.11. Consider a set I. For any k ∈ 0, n , define an F-linear map Similar to ϕ I , δ I is not necessarily a morphism between F S n -modules; also, if I ⊃ n , then δ I = δ n and δ I c = δ ∅ = 0.
The map δ n is the same as δ in the notation of [Wil,Section 3].
Definition 3.12. Let I be a subset of n and h be a non-negative integer. A set S of subsets of n is h-homogeneous on I if |S ∩ I| = h for all S ∈ S .
(1) Any S ⊂ Ω k is k-homogeneous on n .
(2) Any non-empty subset of Ω h k is h-homogeneous on 2h − 1 when Ω h k = ∅. While our definition of ϕ I , δ I , and homogeneity-on-I are made on arbitrary subset I of n ; these are almost always applied in the setting the of second example above, i.e. I = 2h − 1 .
The following lemma roughly says that homogeneity-on-I is preserved by δ I .
Lemma 3.14. Suppose y ∈ F Ω k has Supp(y) being h-homogeneous on some I ⊂ n with h > 0.
Proof We first note the equation holds if m < h, since y = 0 by condition. Without loss of generality, we can assume y = ω ∈ Ω k such that |ω ∩ I| = h. Let I ′ := I \ (ω ∩ I). Then we have Since |ω ∩ I| = h, precisely h terms of the sum on the right is ω, whereas the element i in the other terms is in I \ (ω ∩ I) = I ′ , so we have Finally, as I ′ and ω are disjoint by construction, we have so (3.2.2) minus (3.2.1) gives y(ϕδ I − δ I ϕ) = (2h − m)y, as required.

Determining the slash homolgies
Our goal of this section is to prove the first main result of the article, namely, to calculate the slash homologies of F Ω • . To achieve this, it turns out that knowing a few vanishing conditions for various slash homologies or p-homologies will be immensely helpful. The key result that allows us to obtain these vanishing conditions is Theorem 4.7, whose proof relies on the notion of threshold for k-subsets.
Subsection 4.1 will be devoted to proving Theorem 4.7. We will determine the dimension of all slash homology groups in subsection 4.2, and then finally prove the first main theorem (Theorem 4.16) of the article in subsection 4.3.
4.1. Reduction mechanism. Let us start by introducing a notation for ease of exposition.
Finally, for the sake of completeness, let us see how ϕ 5 kills y (3) : Lemma 4.4. Suppose ω ∈ Ω h k is a k-subset whose threshold is h < k. If for an i ∈ p − 1 the support Supp (ω)ϕ 2h−1 c δ i 2h−1 ϕ i−1 is not empty, then for any θ in this support, d θ j ≥ d ω j holds for all j ∈ h, m with strict inequality in the case when j = h. In particular, the threshold of θ cannot be h.
Next, consider ω 2 ∈ Supp((ω 1 )δ i 2h−1 ). Since 2h−1 < n, ω 2 is obtained by adjoining i new elements from 2h − 1 \ ( 2h − 1 ∩ ω 1 ), so we have d ω2 j = d ω1 j + i for all j ∈ h, m . Finally, by the definition of ϕ, θ is obtained by removing any i − 1 elements from ω 2 , so we have for all j ∈ h, m . Moreover, in the case when j = h, it follows from above and the assumption of h < k that the last inequality in (4.1.3) is strict.
For the case when 2k > n, we have by definition ϕ 2k−1 c = 0, so it follows that y ′ = 0.
From now on, we can assume h < k since the claims (Lo) and (Hi) in the case when h = k follows immediately from (1).
Denote by f i the F-linear map ϕ 2h−1 c δ i 2h−1 ϕ i−1 for each i ∈ p − 1 . To prove the claims (Lo) and (Hi), it suffices to look at the cases when (y h )f i = 0.
For a non-zero (y h )f i , we want to understand the density of the k-subsets in its support. To this end, we need to look at how the density of a k-subset ω ∈ Supp(y h ) changes upon applying the map f i , and so we can assume (ω)f i is non-zero. In particular, we have 2h − 1 c ∩ ω = ∅. In such a case, note that for any a ∈ 2h − 1 c ∩ ω and any ω ∈ Supp(y h ), (Lo): Since y h ∈ F Ω h k , it follows from Lemma 4.4 that any θ ∈ Supp((y h )f i ) satisfies d θ h > h. On the other hand, we have d θ m = |θ ∩ 2m − 1 | = |θ ∩ n | = k < m, where the last inequality follows from the assumption of k being low. Taking (i, j) = (h, m) in Proposition 3.3 (2) yields some j ∈ h + 1, m − 1 such that d θ j = j, meaning that θ ∈ Ω >h k as required. (Hi): As explained above, we only need to consider ω ∈ Supp(y h ) ⊂ Ω h k so that Supp((ω)f i ) is non-empty. For such ω, it follows from Lemma 3.4 (Hi) that d ω j > j for all h < j ≤ m. Moreover, Lemma 4.4 implies that, for any θ ∈ Supp((ω)f i ), we have d θ j ≥ d ω j > j for all h < j ≤ m, and d θ h > h. This implies that θ / ∈ Ω ≥h , hence y ′ / ∈ F Ω ≥h , and the first part of the claim now follows. Finally, if we consider the case when h = 1, then by Lemma 3.5 (Hi), we have Ω 0 k = ∅, so Supp(y ′ ) ⊂ Ω <h k implies that y ′ = 0. Now we are ready to give the key inductive mechanism needed. Lemma 4.6. Consider a non-zero y h ∈ Ker(ϕ k ) that satisfies the following condition.
Using this notation, the assumption says that Supp(y h ) ⊂ Ω k,h and Supp(y h − y (h) ) ⊂ Θ k,h .
Since h is necessarily non-zero, it follows from Lemma 4.2 that y (h) ∈ Ker(ϕ 2h−1 ). Hence, Lemma 4.5 tells us that y x must also be in Ker(ϕ k ). Finally, when k is low and h = k, then x = 0 follows from combining y h = y (h) (which holds by construction) and y ′ = 0 (which is stated in Lemma 4.5 (1)). When k is high and h = 1, similarly, we have y ′ = 0 by Lemma 4.5 (Hi) and y h = y (h) by Ω ≤1 k = Ω 1 k (which in turns follow from Lemma 3.5 (Hi)).
Suppose we have instead 2k > n. By Lemma 3.5 (Hi), any y ∈ F Ω k must have Supp(y) ⊂ Ω ≤a k for some 0 < a ≤ n − k + 1 < k + 1. So we have a chain of congruences y = y a ≡ y a−1 ≡ · · · ≡ y 1 ≡ 0 mod Im(ϕ p−1 ) of elements in Ker(ϕ) with Supp(y i ) ⊂ Ω ≤i k ; here the i-th congruence follows by applying Lemma 4.6 with h = a − i + 1 and using Θ k,h = Ω k,h−1 (in the notation of Lemma 4.6) for all i ∈ a .

Dimensions of homologies.
In this subsection, we determine the dimension of all slash homology groups -along the way, we show various vanishing conditions for some classical p-homology groups.
Since we only look at the p-complex F Ω • , we omit writing F Ω • for the homology groups and just use H Proof Since Theorem 4.7 asserts that H /0 k = 1 H k = 0 whenever 2k > n, the claim follows immediately from Lemma 2.4 (1).
Lemma 4.9. If k is low, i.e. 2k ≤ n, then the following holds.
Since we have p−1 H k−r = 0 for all integers r ∈ min{k, p − 1} , the last part follows immediately from Lemma 2.4 (2).
On one hand, by taking (ℓ, k) in Theorem 2.10 (1) to be (k − 1, n − k) in the current setting, we get that S n−k,k ⊂ Ker(ϕ n−2k+1 n−k ). On the other hand, we can apply F-linear dual to Ker(ϕ n−2k+1 , where the first isomorphism follows from Lemma 2.5 (1) and the second isomorphism follows from the commutative diagram (2.3.1).
Recall that ϕ n−2k+1 n−k+1 induces a map on the slash-cohomologies which sends the (slash-)homology class [x] ∈ H /n−2k+1 n−k+1 of x ∈ F Ω n−k+1 to the homology class of (x)ϕ n−2k+1 n−k+1 in H /0 k . Recall also from slash shifting Lemma 2.4 (1) this induced map is injective. Therefore, it suffices to show that this induced map is zero in order to finish the proof. Indeed, by Lemma 4.9 (1) we have that H /0 k is a quotient of S k,k , so for any x ∈ F Ω n−k+1 , if [(x)ϕ n−2k+1 n−k+1 ] is non-zero in H /0 k , then (x)ϕ n−2k+1 n−k+1 must also be non-zero in S k,k -which is impossible as shown in the claim above.
For the remaining slash homology groups, it turns out they are the same as the slash-0 terms.
Lemma 4.12. For an integer k satisfying n − (p − 1) < 2k ≤ n, and a ∈ 0, n − 2k , ϕ a induces an isomorphism H /a k+a Proof We know from slash-shifting Lemma 2.4 (1) that ϕ a induces an injection H /a k+a ∼ = H /0 k for a ∈ 0, p − 2 . It remains to show that these are surjective for a ∈ 0, n − 2k . Indeed, it follows from Lemma 4.9 (1) that H /0 k is a quotient of S k,k , so let us consider x ∈ S k,k . Since n − 2k < p − 1 by the assumption on k, we have Im(ϕ (a) ) = Im(ϕ a ). So it follows from Theorem 2.10 (2), which says that S k,k = Im(ϕ a k+a | S k+a,k ), that ϕ a k+a is surjective onto S k,k , as required. Our next goal is to show that H /0 k is non-vanishing when n − (p − 1) < 2k ≤ n and p = 2. For this purpose, we consider certain (2-)complexes obtained by "contracting F Ω • ".
For any k ∈ 0, n and a ∈ p − 1 , consider the (2-)complex C(k, a) • as follows: Note that F Ω r , ϕ r are regarded as zero whenever the index lies outside 0, n . Denote by H m (k, a) the homology of C(k, a) • at the degree where F Ω m appears. Note that, by (2.1.1), H m (k, a) is the same as the homology of the complex given by replacing all ϕ b by ϕ (b) (as we work over a field). By construction, we have Hence, counting the dimensions yields Note that the binomial coefficient n r is zero whenever the integer r is not in 0, n . Readers well-versed in combinatorics may have already realised that the summation on the righthand side enumerates certain prefixes of Dyck paths. For completeness, let us gives more detail to this.
Definition 4.13. Let n ≥ 0 be a non-negative integer. An n-step lattice path is a sequence P of n + 1 points p 0 = (0, 0), p 1 , . . . , p n in Z 2 , so that, for all 1 ≤ i ≤ n, the i-th step s i := p i − p i−1 is either (0, 1) or (1, 0). We also consider such a path as a 1-dimensional object in R 2 by taking the union of all intervals [p i−1 , p i ] ⊂ R 2 over 1 ≤ i ≤ n.
Consider k ∈ n and t ∈ 0, n + 1 . A lattice path is (n − k, k)-open if it has n steps and k of which are (1, 0). Note that such a path necessarily ends on (n − k, k).
For integers s, t ∈ Z ≥0 ∪{∞}, we say that a lattice path is s-bounded-above and t-bounded-below if all the points (x, y) in the path satisfies −t < y − x < s.
Denote by L s,t n−k,k to be the set of all (n − k, k)-open lattice path that is s-bounded-above and t-bounded-below.
Lemma 4.14. For any k ∈ n , we have (4.2.4) j∈Z dim a H pj+k − dim p−a H pj+k−a = |L a,p−a n−k,k |.
In particular, if 2k ≤ n − (p − a) or 2k ≥ n + a, then the summation on the left vanishes Proof It is straightforward to see that, by taking (s, t) = (a, p − a), the right-hand side of (4.2.3) coincide with that of (4.2.2); combining the two gives (4.2.4). For the last statement, it is clear that L s,t n−k,k = ∅ when (n − k, k) lies beyond −s < x − y < t. Hence, in the setting of (s, t) = (a, p − a), L s,t n−k,k = ∅ when the condition −a < n − 2k < p − a, equivalently n − (p − a) < 2k < n + a, does not hold. In other words, there is no lattice path whenever 2k ≤ n − (p − a) or 2k ≥ n + a. The claim now follows by combining this condition with (4.2.3).
For the last statement, we know already from Lemma 4.8 and Lemma 4.10 (2) that H /0 k = 0 whenever n − (p − 1) ≤ 2k or 2k > n. For the other cases, it follows from the previous part of the lemma and Lemma 4.14 that dim H k (k, 1) = |L 1,p−1 n−k,k |. Observe that the lattice path that start with a (1, 0)-step and changes direction in every step is always in L 1,p−1 n−k,k with the exception of p = 2. The claim now follows.

Main result and examples.
Now we can combine all of the information above to obtain the main result of this article.
Before we do that, recall that the Specht module S λ of a two-row partition λ = (n − k, k) has a simple top D λ whenever p = 2.
Theorem 4.16. For any prime p and any positive integer n, the slash-homology (p − 1)-complexes H / k at degree k ∈ 0, n of F Ω • is non-vanishing if, and only if, p = 2 and n − 2k ∈ 0, p − 2 . Moreover, in the case when H / k is non-vanishing, it takes the form Proof If n − 2k / ∈ 0, p − 2 , then H / k = 0 follows from combining Lemma 4.8, 4.10 (3), 4.11. All other cases (non-)vanishing condition, including the form of the non-vanshing slash homologies, follows from combining Lemma 4.15 and Lemma 4.12.
Let us now show that H /0 k ∼ = D (n−k,k) when it is non-zero. By Lemma 4.9 (1), we know that H /0 k is a quotient of S k,k , which is a Specht module corresponding to the partition (n − k, k). Hence, by the fact that S k,k ∼ = S (n−k,k) has a simple top, it suffices to show that dim H /0 k = dim D (n−k,k) . Indeed, since n − (p − 1) < 2k ≤ n, the formula for dim D (n−k,k) can be looked up from [Erd,Example 5.3(3)], as we present below. Note that we replace the notations (m, s, t, d, δ) in loc. cit. by (k, n − 2k, 0, n − 2k + 1, p − (n − 2k + 1)) in our setting for the convenience of the readers.
Using n r = n n−r and reindexing j's, the last two binomial coefficients can be written as n pj+k − n pj+k−1 with j varying over all positive integers. Hence, the right-hand side is precisely which is exactly the same formula for dim H /0 k by Lemma 4.15, as required.
We remark that in the special case of p = 3, 5, the description of H /0 k verifies [Wil,Conjecture 7.5,7.6].
where ζ is the primitive p-th root of unity e 2π √ −1/p .
Note that for 2k ≤ n − (p − 1) or 2k > n, |L 1,p−1 n−k,k | = 0 as we have explained in Lemma 4.14, so one can also write k so that n − (p − 1) < 2k ≤ n instead of k ∈ 0, n in the right-hand side. Proof Similar to dimension counting for (2-)complexes, doing so for p-complexes require multiplying the dimension by ζ, since the Grothendieck ring of the stable category of graded modules over F[x]/(x p ) (the underlying category for slash homologies) is the ring of cyclotomic integers Z[ζ]; c.f. [KQ, Qi]. Then the left-hand side of the claim is just dimension counting for F Ω • , and the right-hand side is dimension counting for the slash homology.

Basis for the slash-0 homologies
Throughout this section, we impose the following Assumption : n − (p − 1) < 2k ≤ n, unless otherwise stated. Recall from Theorem 2.12 that the set SYT n (k) of standard Young (n − k, k)tableaux is an indexing set of a basis (given by standard polytabloid) of S k,k , and from Lemma 4.9 that H /0 k is a quotient of S k,k . Moreover, there is also a well-known bijection between L 1,∞ n−k,k with SYT n (k), so given that dim H /0 k coincide with |L 1,p−1 n−k,k | as shown in the previous section, it is natural to expect the bijection induces a basis for H /0 k . The aim of this subsection is to show that this is indeed the case.
Let us start with the correspondence between SYT n (k) with L 1,∞ n−k,k .
Definition 5.1. Consider an n-step lattice path P ∈ L 1,∞ n−k,k . The k-subset of n associated to P is given by ω P := {a ∈ n | the a-th step in P is (0, 1)} ∈ Ω k . In particular, P → t ωP defines a bijection L 1,∞ n−k,k ↔ SYT n (k). A standard (two-row) tableau t ∈ SYT n (k) is called p-standard if t = t ωP for some ω P ∈ L 1,p−1 n−k,k . Remark 3. The notion of p-standard tableau is introduced in [Kl]. In the case of a tableau t of shape λ = (n − k, k), it is defined as a standard tableau satisfying t 2,j < t 1,j+p−2 for all j ∈ {1, 2, . . . , k}, and n − 2k < p − 1. We show in Lemma 5.11 that this is equivalent to the formulation given above.
To simplify notation, we denote by t P and e e e P the standard tableau t ωP and standard polytabloid e e e t ω P . Likewise, by the (standard) tableau and polytabloid associated to P , we mean t P and e e e P respectively.
From the above correspondence between L 1,∞ n−k,k with standard tableaux, it is natural to expect the following.
In particular, the simple top D (n−k,k) of S (n−k,k) (with n − (p − 1) < 2k ≤ n) admits the same basis where [v] represents v + radS (n−k,k) .
Our aim from now on is to prove this theorem. Note that the final statement follows from combining Theorem 4.16 with Lemma 4.9, which asserts that the radical radS (n−k,k) of the Specht module S (n−k,k) = S k,k is the intersection Im(ϕ p−1 ) ∩ S k,k (as a submodule of the permutation module M (n−k,k) = F Ω k ).
The strategy to prove Theorem 5.2 is to modify the straightening rule -the induction process that is used to show that a Specht module admits a basis given by standard polytabloids in [Pe]. The modification are designed so that we can keep track of how "far" away the tableaux appearing in a Garnir relation is from being a standard tableau associated to a path in L 1,p−1 n−k,k . 5.1. Preliminaries. In this subsection, we present the definition of the partially ordered set that allows us to prove Theorem 5.2 by induction, and also a few techniques that we will use frequently.
We also define L ≥p := L 1,∞ n−k,k \ L 1,p−1 n−k,k . The lattice paths in L 1,p−1 n−k,k are said to be good, and those in L ≥p bad ; it is then natural to use the same adjectives for the associated standard tableau t P and standard polytabloid e e e P = e e e t P .
(1) We say that t almost standard if all of the following hold.
• t is column-standard (i.e. entries increase as we go down each column); • t is second-row-standard (i.e. entries in the second row increases as we go right). Denote by aST := aST n (k) the set of almost standard (n − k, k)-tableaux.
(2) Define a partial order on aST by declaring t t ′ if one of the following conditions are satisfied.
(ii) |{t} ∩ i | ≤ |{t ′ } ∩ i | for all i ∈ n with at least one of the inequalities being strict.
(iii) {t} = {t ′ } and for ω := {t} c , there is some j such that ω (i) lies in the same position in both t, t ′ for all i < j, and ω (j) = t 1,a = t ′ 1,b with a > b. This defines a partial order on aST.
The second condition on is just the row-dominance order of Definition 2.11; it should be regarded as a refinement of the notion of density in Section 3 which gives us a measure on how far a standard tableau is from being t P with P ∈ L 1,p−1 n−k,k . The last condition is used to measure how far an almost standard tableau from being standard, it is the partial order used in the proof of standard polytabloid basis of Specht modules in [Pe,J1] "restricted to the first row".
Note that the standard (n − k, k)-tableaux are totally ordered by the row-dominance order, and so combining with condition (iii), we can see that (aST, ) is a totally ordered set. Note that only the largest tableau t {2,4} is good when p = 3; for any prime p > 3, all standard tableaux are good.
Lemma 5.5. Let t be a (n − k, k)-tableau. For a positive integer j ≤ k, the following equation holds.
Lemma 5.6. The following hold for a (n − k, k)-tableau t.
Colloquially, Lemma 5.7 says that if we replace an entry in the second row of an almost standard tableau by a smaller value, then the new almost standard tableau (after an appropriate column reordering if needed) is larger than the original one, in the -order.
Lemma 5.8. If t ∈ aST is a non-standard almost standard tableau, then e e e t ∈ F-span{e e e t ′ | t ′ ≻ t}.
Proof By non-standardness, we have an integer j ∈ n − k so that t 1,j > t 1,j+1 . If j ≥ k, then it follows from Lemma 5.6 (2) that e e e t = e e e t (t 1,j , t 1,j+1 ). But t(t 1,j , t 1,j+1 ) ≻ t by Definition 5.3 (2) (iii), so the claim follows. Now we can assume j ≤ k. By Lemma 5.5, we have e e e t = e e e u − e e e v , where u = tτ t,j and v = tσ t,j . It is easy to see that u is almost standard and u ≻ t.
Let i be the unique integer so that t 2,i < t 1,j < t 2,i+1 . Note that from the conditions we have t 2,i < t 1,j < t 2,j , so i < j as t has standard second-row. By Lemma 5.6, we have e e e v = e e e v ρ v,i,j . Hence, we have e e e t = e e e u − e e e v ρ v,i,j , so it remains to show that w := vρ v,i,j ≻ t.
By the choice of σ t,j , v is column-standard, and so the choice of i, j now ensures that w ∈ aST. Note also that swapping columns (by ρ v,i,j ) does not change the underlying set {v}, i.e. {w} = {v}. Since t 2,j > t 1,j (by column-standardness of t), {w} = {v} is obtained from replacing t 1,j in {t} by the smaller t 2,j . So it follows from Lemma 5.7 that w ≻ t as required.
The motivation for the previous lemma should be natural: we want to do induction on (aST, ). In contrast, it may be unclear what the motivation of the forthcoming lemma is; we hope the reader can bear with us for the while as its use will be clear in the next subsection (specifically, Lemma 5.15).
For a standard tableau t ∈ SYT n (k), we define In particular, t is the only standard tableau in aST t and is also the maximum in (aST t , ).
Lemma 5.9. For any s ∈ aST t , we have e e e s ∈ e e e t + F-span{e e e t ′ | t ≺ t ′ ∈ aST}.
Proof We prove this by induction (from large to small) on (aST t , ). The claim is trivial for the case when s = t, which is the maximum of (aST t , ). Suppose s = t. We can assume that s 1,k+1 < s 1,k+2 < · · · < s 1,n−k by Lemma 5.6 (2), which means that there exists j ∈ k so that s 1,j > s 1,j+1 . Following the same argument and notation as in the proof of Lemma 5.8 yields e e e s = e e e u − e e e w , where u := sτ s,j and w := sσ s,j ρ with ρ the column reordering permutation ρ that ensures w is almost standard.
On one hand, the construction of u means that it is in aST t and also satisfies u ≻ s, so it follows from induction hypothesis that e e e u ∈ e e e t + F-span{e e e t ′ | t ≺ t ′ ∈ aST}.
On the other hand, the same argument as in the proof of Lemma 5.8 yields w ≻ s ≻ t (while w is not in aST t ). Therefore, combining with e e e s = e e e u − e e e w , the claim follows. 5.2. The case of bad tableaux. The aim of this subsection is to show a similar statement as Lemma 5.8 but with t being a bad standard tableau and with the polytabloids replaced by their cosets in H /0 k . We start with the following key lemma which provides a relation between certain polytabloids in H /0 k . Lemma 5.10. For a (n − k, k)-tableau t and an integer i ∈ k − (p − 2) , consider the (p − 1)-subset B := {t 1,i , t 1,i+1 , . . . , t 1,i+p−2 }, the following equation holds.
e e e t τ ∈SB τ = (e e e t ′ · K)ϕ p−1 , where t ′ is the tableau obtained from t by removing all the columns containing an element of B, and K is the set consisting of all the entries in these removed columns.
We remark that Figure 2 will be helpful to understand the statement. Proof Let us first set up some notations. Take ℓ := |K ∩ {t}| ∈ 0, p − 2 . In particular, t ′ is a (n − k − (p − 1), k − ℓ)-tableau. We take L to be the set of entries in the first row that are not in B. See Figure 2.
We will prove the lemma by showing the following two equalities: (i) e e e t τ ∈SB τ = −e e e t ′ · (K)ϕ (p−1) .
(ii) −e e e t ′ · (K)ϕ (p−1) = (e e e t ′ · K)ϕ p−1 . Proof of (i): For clarity, what we need to show is the equality Consider a k-subset ω that is of the form {t}στ with σ ∈ S B and τ ∈ C t . Let σ ′ ∈ C t ′ be the unique element given by the restriction of σ on t ′ . Then we can write ω = ({t ′ }σ ′ ) · (K ∩ ω). Using Figure 2, one can observe the following. Firstly, the ℓ-subset K ∩ ω, must appear as a unique support of (K)ϕ (p−1) ℓ+p−1 . Hence, ω must arise in the support in the right-hand side of (5.2.1). Secondly, for every k-subset in the said support, one can always obtain it in the form of {t}στ . Now we need to show that the coefficients λ ω and µ ω of ω in the left-hand side and right-hand side of (5.2.1), respectively, coincide.
Proof of (ii): It is clear from the construction that every k-subset in Supp(e e e t ′ ) is disjoint from K, so it follows from the splitting rule (Lemma 2.1) that (e e e t ′ · K)ϕ (p−1) = p−1 i=0 (e e e t ′ )ϕ (i) · (K)ϕ (p−1−i) .
It then follows from Theorem 2.10 that e e e t ′ ∈ S k−ℓ,k−ℓ is annihilated by ϕ i = i!ϕ (i) for any i > 0, so we have (e e e t ′ · K)ϕ (p−1) = e e e t ′ · (K)ϕ (p−1) . This finishes the proof of (ii).
Before proceeding to the next step, we need an elementary observation on the elements in {t P } that correspond to a (0, 1)-step intersecting or lying below the line y = x − (p − 1).
Lemma 5.11. Let t := t P be a standard tableau associated to a path P , and (x, y) be the starting point of the i-th (0,1)-step of P . Then (x, y) satisfies y ≤ x − (p − 1) if, and only if, we have a strictly descending chain t 2,i > t 1,i+p−2 > t 1,i+p−3 > · · · > t 1,i .
In particular, P is bad if, and only if, there is some i ∈ k such that t 2,i ≥ 2i − 1 + p − 1; in which case, we also have i + p − 2 ≤ n − k.
Proof The if statement follows from t being standard and the correspondence between standard (n − k, k)-tableau and L 1,∞ n−k,k . For the only if part, by definition we have y = i − 1, and so we have x ≥ i + p − 2. Since x counts the number of entries in the first row of t that are less than t 2,i , so x ≥ i + p − 2 implies that t 2,i > t 1,i+p−2 ; the rest of the chain follows from the row-standardness of t.
For the final statement, observe from the condition on the chain that the smallest possible value t 2,i can take is (i + p − 2) + (i − 1) + 1 = 2i − 1 + p − 1, where i + p − 2 is the number of terms in the chain above that appear after t 2,i (hence the inequality i + p − 2 ≤ n − k), and (i − 1) is the number of t 2,j satisfying t 2,j < t 2,i .
Lemma 5.11 allows us to introduce the following terminology.
Definition 5.12. For a bad standard tableau t, we say that t 2,i is a bad entry if t 2,i ≥ 2i − 1 + p − 1.
Lemma 5.13. Let t be a bad standard tableau with a bad entry t 2,i . Then the element τ ∈SB i e e e t τ ∈ S k,k belongs to the submodule Im(ϕ p−1 k+p−1 ). Proof Take B := B i (t 1,i , t 2,i ), then we have (t 1,i , t 2,i )S Bi (t 1,i , t 2,i ) = S B . This implies that τ ∈SB i e e e t τ = σ∈SB e e e t (t 1,i , t 2,i )σ (t 1,i , t 2,i ) = − σ∈SB e e e t σ (t 1,i , t 2,i ), where the last equality follows from the observation that e e e t = −e e e t (t 1,i , t 2,i ). On the other hand, it follows from Lemma 5.10 that σ∈SB e e e t σ ∈ Im(ϕ p−1 ). Hence, the element on the right-hand side is also in Im(ϕ p−1 ), and so is τ ∈SB i e e e t τ .
Lemma 5.14. Let t be a bad standard tableau with a bad entry t 2,i . Then tσ is column-standard for any σ ∈ B i . In particular, tσ is almost standard if (t 2,i )σ = t 2,i .
Proof Column-standardness follows from the existence of the chain in Lemma 5.11. More precisely, for j ∈ p − 2 with i + j ≤ k, we have (tσ) 2,i+j = t 2,i+j > t 2,i ≥ t 1,j+l for all l ∈ p − 2 and so t 2,i+j > (tσ) 1,i+j always. For the i-th column, again the chain in Lemma 5.11 tells us that (t 2,i )σ ≥ t 1,i+1 > t 1,i = (tσ) 1,i . The last statement follows from the fact that t is standard and σ fixes all entries in the second row. Before proceeding, recall our notation [v] := v + Im(ϕ p−1 ) ∈ H in H /0 k . So it suffices to show that [e e e t σ] with non-identity σ ∈ S Bi is in F-span{e e e t ′ | t ′ ≻ t}. We partition S Bi \ {1} into two disjoint subsets S ⊔ T so that S consists of all the non-identity permutations σ that fix t 2,i . By Lemma 5.9, for any σ ∈ S, we have e e e t σ ∈ e e e t + F-span{t ′ ∈ aST | t ′ ≻ t}.
Therefore, we have σ∈S e e e t σ = |S|e e e t + v for some v ∈ F-span{e e e t ′ | t ′ ≻ t}. Note that the subgroup {1} ⊔ S of S Bi that stabilises t 2,i is of order (p − 2)!, which is congruent to 1 mod p by Wilson's theorem. This means that |S| is congruent to 0 mod p, which yields