# On extensions of hook Weyl modules

Mihalis Maliakas<EMAIL_ADDRESS>Department of Mathematics, University of
Athens  Dimitra-Dionysia Stergiopoulou111Partially supported by Onassis
Foundation grant GZM 065-1<EMAIL_ADDRESS>Department of Mathematics,
University of Athens

###### Abstract

We determine the integral extension groups
$\mathrm{Ext}^{1}(\Delta(\mathrm{h}),\Delta(\mathrm{h}(k)))$ and
$\mathrm{Ext}^{k}(\Delta(\mathrm{h}),$ $\Delta(\mathrm{h}(k)))$, where
$\Delta(\mathrm{h}),\Delta(\mathrm{h}(k))$ are the Weyl modules of the general
linear group $GL_{n}$ corresponding to hook partitions $\mathrm{h}=(a,1^{b})$,
$\mathrm{h}(k)=(a+k,1^{b-k})$.

###### keywords:

Extensions , general linear group , Weyl module , hook

###### MSC:

[2010] 20G05

##  1 Introduction

This paper concerns polynomial representations of the general linear group
$GL_{n}$ over the integers. For a partition $\lambda$, let $\Delta(\lambda)$
denote the Weyl module of $GL_{n}$ of highest weight $\lambda$. The extension
groups $\mathrm{Ext}^{i}(\Delta(\lambda),\Delta(\mu))$ play an important role
in the theory. For example, the $p$-torsion of
$\mathrm{Ext}^{1}(\Delta(\lambda),\Delta(\mu))$ yields the $\mathrm{Hom}$
space between the corresponding modular Weyl modules of $GL_{n}(K)$, where $K$
is an algebraically closed field of characteristic $p>0$, and the dimensions
of the higher modular extensions may be obtained through torsion and
restriction of integral extensions. Jantzen’s sum formula can be viewed and
proved via integral extension groups [4].

There are not many cases where explicit computations of integral extension
groups between Weyl modules have been carried out. In [2] the $GL_{2}$ case
was treated and in [5] the $GL_{3}$ case when $\lambda$ and $\mu$ differ by a
multiple of a single root, both for $i=1$. In [1] the case
$\lambda=(1^{a}),\mu=(a)$ was studied and in [15] the situation where
$\lambda,\mu$ are hooks differing by a single root was considered. In [14] the
case where $\lambda,\mu$ are any partitions differing by a single root was
settled. As the modular extension groups are intimately related to the
integral ones, we mention the result on neighboring Weyl modules [11] Part II
Section 7, and the $SL_{2}$ result in [16] for all $i$ generalizing [9] and
[7]. More can be found in [8].

Let $\mathrm{h},\mathrm{h}(k)$ be hooks,
$\mathrm{h}=(a,1^{b}),\mathrm{h}(k)=(a+k,1^{b-k})$, where $k$ is an integer
such that $1\leq k\leq b$. It follows that
$\mathrm{Ext}^{i}(\Delta(\mathrm{h})),\Delta(\mathrm{h}(k)))=0$ if $i>k$. In
this paper we determine
$\mathrm{Ext}^{1}(\Delta(\mathrm{h}),\Delta(\mathrm{h}(k)))$ and
$\mathrm{Ext}^{k}(\Delta(\mathrm{h})),\Delta(\mathrm{h}(k)))$ (Theorem 3.5 and
Theorem 4.1). Our approach utilizes presentation matrices for various
$\mathrm{Ext}$ groups that we determine from the description of generators and
relations of Weyl modules of [3] ($i=1$) and from the projective resolutions
of [15] ($i=k$). Using these and the degree reduction theorem of Kulkarni
[13], we identify cyclic generators of extension groups of the form
$\mathrm{Ext}^{i}(\Delta(\mathrm{h}),M)$, $(i=1,k)$, where $M$ is a tensor
product of a divided power and an exterior power. Computing the image of these
generators under canonical maps yields the results.

##  2 Recollections

###  2.1 Notation

Let $F$ be a free abelian group of finite rank $n$. Fixing a basis of $F$
yields an identification of general linear groups $GL(F)=GL_{n}(\mathbb{Z})$.
We will be working with homogeneous polynomial representations of
$GL_{n}(\mathbb{Z})$ of degree $r$, or equivalently, with modules over the
Schur algebra $S_{\mathbb{Z}}(n,r)$ [10], Section 2.4. We will write $S(n,r)$
in place of $S_{\mathbb{Z}}(n,r)$. By $DF=\sum_{i\geq 0}D_{i}F$ and $\wedge
F=\sum_{i\geq 0}\wedge^{i}F$ we denote the divided power algebra of $F$ and
the exterior algebra of $F$ respectively. We will usually omit $F$ and write
$D_{i}$ and $\wedge^{i}$.

From [10] or [2], Proposition 2.1, we recall that for each sequence
$a_{1},...,a_{n}$ of non negative integers $a_{i}$ that sum to $r$, the
$S(n,r)$-module $D_{a_{1}}\otimes...\otimes D_{a_{n}}$ is projective.
Throughout this paper all tensor products are over the integers.

For a partition $\lambda$ of $r$ with at most $n$ parts, we denote by
$\Delta(\lambda)$ the corresponding Weyl module for $S(n,r)$. If $\lambda=(a)$
is a partition with one part, then $\Delta(\lambda)=D_{a}$, and if
$\lambda=(1^{b})$, then $\Delta(\lambda)=\wedge^{b}$. A hook $\mathrm{h}$ is a
partition of the form $\mathrm{h}=(a,1^{b})$. The following complex of
$S(n,r)$-modules (which is the dual of the usual Koszul complex) is exact

$0\rightarrow D_{a+b}\rightarrow...\rightarrow
D_{a+1}\otimes\wedge^{b-1}\xrightarrow{\theta_{a}}D_{a}\otimes\wedge^{b}\rightarrow...\rightarrow\wedge^{a+b}\rightarrow
0,$

where $\theta_{a}$ is the composition $\
D_{a+1}\otimes\wedge^{b-1}\xrightarrow{\triangle\otimes 1}D_{a}\otimes
D_{1}\otimes\wedge^{b-1}\xrightarrow{1\otimes m}D_{a}\otimes\wedge^{b},$ where
$\triangle$ (respectively, $m$) is the indicated component of the
comultiplication (resp., multiplication) map of the Hopf algebra $DF$ (resp.,
$\wedge F$). It is well known that if $\mathrm{h}=(a,1^{b})$ is a hook, $b\geq
1$, then $\Delta(\mathrm{h})\simeq cok(\theta_{a})\simeq ker(\theta_{a-1}),$
so that we have the following short exact sequence

$0\rightarrow\Delta(\mathrm{h}(1))\xrightarrow{i}D_{a}\otimes\wedge^{b}\xrightarrow{\pi_{0}}\Delta(\mathrm{h})\rightarrow
0,$ (2.1)

where $\mathrm{h}(1)=(a+1,1^{b-1})$, the map $i$ is induced by $\theta_{a}$ on
generators and the map $\pi_{0}$ is induced by the identity map on generators

Notation: Throughout this paper we use the notation
$\mathrm{h}=(a,1^{b}),\mathrm{h}(k)=(a+k,1^{b-k}),1\leq k\leq b,r=a+b.$

###  2.2 Straightening law

We recall the straightening law and the semi-standard basis theorem for
$\Delta(\mathrm{h})$ ([3], Theorem II.3.16). Fix an ordered basis
$e_{1},..,e_{n}$ of $F$. For simplicity, we denote the element $e_{i}$ by $i$
and accordingly the element $e_{i_{1}}^{(a_{1})}...e_{i_{t}}^{(a_{t})}\otimes
e_{j_{1}}\wedge...\wedge e_{j_{b}}\in D_{a}\otimes\wedge^{b}$ by
${i_{1}}^{(a_{1})}...{i_{t}}^{(a_{t})}\otimes{j_{1}}...{j_{b}}$. The image of
this element under the identification $\Delta(\mathrm{h})\simeq
cok(\theta_{a})$ will be denoted by
${i_{1}}^{(a_{1})}...{i_{t}}^{(a_{t})}|{j_{1}}...{j_{b}}$. Now suppose
$i_{1}<i_{2}<...<i_{t}$ and $j_{1}\leq i_{1}$. Then in $\Delta(\mathrm{h})$ we
have

${i_{1}}^{(a_{1})}...{i_{t}}^{(a_{t})}|{j_{1}}...{j_{b}}=\begin{cases}-\sum\limits_{s\geq
2}{i_{1}}^{(a_{1}+1)}...{i_{s}}^{(a_{s}-1)}...{i_{t}}^{(a_{t})}|{i_{s}}{j_{2}}...{j_{b}},&\mbox{if}\;j_{1}=i_{1}\\\
-\sum\limits_{s\geq
1}j_{1}{i_{1}}^{(a_{1})}...{i_{s}}^{(a_{s}-1)}...{i_{t}}^{(a_{t})}|{i_{s}}{j_{2}}...{j_{b}},&\mbox{if}\;j_{1}<i_{1}.\end{cases}$

A $\mathbb{Z}$ \- basis of $\Delta(\mathrm{h})$ is the set of all
${i_{1}}^{(a_{1})}...{i_{t}}^{(a_{t})}|{j_{1}}...{j_{b}}$, where
$a_{1}+...+a_{t}=a,\;i_{1}<...<i_{t}$ and $i_{1}<j_{1}<...<j_{b}.$

###  2.3 Resolutions of hooks

We will use the the explicit finite projective resolution $P_{*}(a,b)$ of
$\Delta(\mathrm{h})$,

$0\rightarrow...\rightarrow
P_{2}(a,b)\xrightarrow{\theta_{2}(a,b)}P_{1}(a,b)\xrightarrow{\theta_{1}(a,b)}P_{0}(a,b)$

of [15], Theorem 1, which we now recall. For short we denote the tensor
product $D_{a_{1}}\otimes...\otimes D_{a_{m}}$ of divided powers by
$D(a_{1},...,a_{m})$. Let $P_{i}(a,b)=\sum D(a_{1},...,a_{b+1-i})$ where the
sum ranges over all sequences $(a_{1},...,a_{b+1-i})$ of positive integers of
length $b+1-i$ such that $a_{1}+...+a_{b+1-i}=a+b$ and $a\leq a_{1}\leq a+i$.
The differential $\theta_{i}(a,b)$ is defined be sending
$x_{1}\otimes...\otimes x_{b+1-i}\in D(a_{1},...,a_{b+1-i})$ to

$\sum_{j=1}^{s}(-1)^{j+1}x_{1}\otimes...\otimes\triangle(x_{j})\otimes...\otimes
x_{b+1-i}\in D(a_{1},...,u,v,...,a_{s}),$

where $s=b+1-i$ and $\triangle(x_{j})$ is the image of $x_{j}$ under the two-
fold diagonalization $D(a_{j})\rightarrow\sum D(u,v)$, where the sum ranges of
all positive integers $u,v$ such that $u+v=a_{j}$ and
$D(a_{1},...,u,v,...,a_{b+1-i})$ is a summand of $P_{i-1}(a,b)$ with $u$
located at position $j$. We denote by $\triangle_{u,v}:D(a_{j})\rightarrow
D(u,v)$ the indicated component of the two-fold diagonalization
$D(a_{j})\rightarrow\sum D(u,v)$.

If $A,B$ are $S(n,r)$ \- modules, we write $\mathrm{Hom}(A,B)$ and
$\mathrm{Ext}^{i}(A,B)$ in place of $\mathrm{Hom}_{S(n,r)}(A,B)$ and
$\mathrm{Ext}^{i}_{S(n,r)}(A,B)$ respectively.

We recall the recursions

$\displaystyle P_{0}(a,b)=D(a)\otimes P_{0}(1,b-1),$ $\displaystyle
P_{i}(a,b)=P_{i-1}(a+1,b-1)\oplus D(a)\otimes P_{i}(1,b-1),i>0$

and that under these identifications we have the following.

######  Remark 2.1.

If $M$ is a $S(n,r)$-module, the differential
$\mathrm{Hom}(\theta_{i}(a,b),M)$ of the complex $\mathrm{Hom}(P_{*}(a,b),M)$
looks like

${\mathrm{Hom}(P_{i-2}(a+1,b-1),M)}$${\mathrm{Hom}(P_{i-1}(a+1,b-1),M)}$${\mathrm{Hom}(D(a)\otimes
P_{i-1}(1,b-1),M)}$${\mathrm{Hom}(D(a)\otimes
P_{i}(1,b-1),M)}$${\oplus}$${\oplus}$

where the top horizontal map is $\mathrm{Hom}(\theta_{i-1}(a+1,b-1),M)$, the
bottom one is $-\mathrm{Hom}(1\otimes\theta_{i}(1,b-1),M)$ and the restriction
of the diagonal one on the summand $\mathrm{Hom}(D(a,j,a_{2},...,a_{m}),M)$ is
$\mathrm{Hom}(\triangle_{a,j}\otimes 1\otimes...\otimes 1),M).$

For any $S(n,r)$-module $M$ and any sequence $a_{1},...,a_{m}$ of non negative
integers such that $a_{1}+...+a_{m}=r$ and $m\leq n$, we identify the
$\mathbb{Z}$-module

$\mathrm{Hom}(D(a_{1},...,a_{m}),M)$

with the $(a_{1},...,a_{m})$ weight subspace of $M$ (with respect to the
action of $\mathbb{Z}^{n}$) according to [2], eqn. (11) on p. 178. We will use
such identifications freely throughout this paper.

In particular, suppose $M$ is a skew Weyl module for $S(n,r)$ (denoted be
$K_{\lambda/\mu}(F)$ in [3]). Using the $\mathbb{Z}$-basis of $M$ given by the
semi-standard tableaux [3], Theorem II.3.16, we see that the
$\mathbb{Z}$-module $\mathrm{Hom}(D(a_{1},...,a_{m}),M)$ may be identified
with the $\mathbb{Z}$-submodule of $M$ that has basis the semi-standard
tableaux of $M$ that contain the entry $i$ exactly $a_{i}$ times, $i=1,...,m$.
We call this the semi-standard basis of $\mathrm{Hom}(D(a_{1},...,a_{m}),M)$.
(Perhaps we should remark that what we have called semi-standard tableaux are
called ’co-standard’ in [3], Definition II.3.2: the entries in each row are
weakly increasing from left to right and the entries in each column are
strictly increasing from top to bottom.)

We record here a handy computational remark. For $M$ a skew Weyl module and
$T\in\mathrm{Hom}(D(a_{1},...,a_{m}),M)$ a semi-standard basis element, let
$\phi_{t}(T)$, $1\leq t<m$, be the element of
$\mathrm{Hom}(D(a_{1},...,a_{t}+a_{t+1},...,a_{m}),M)$ obtained from $T$ by
replacing each occurrence of $j>t$ by $j-1$. If $t\geq m$, let
$\phi_{t}(T)=0$. By extending linearly, we obtain for each degree $i$ a map of
$\mathbb{Z}$-modules

$\phi_{t}:\mathrm{Hom}(P_{i}(a,b),M)\rightarrow\mathrm{Hom}(P_{i+1}(a,b),M).$

It is clear that only a finite number of these maps are nonzero. From the
definition of the differential of $P_{*}(a,b)$, we obtain the following
description for the differential of $\mathrm{Hom}(P_{*}(a,b),M)$.

######  Remark 2.2.

With the previous notation, $\mathrm{Hom}(\theta_{i}(a,b),M)=\sum_{i\geq
1}(-1)^{t-1}\phi_{t}.$

Let $n\geq b+1$. Then $P_{*}(a,b)$ is a projective resolution of
$\Delta(\mathrm{h})$. We claim that the $\mathbb{Z}$-module
$\mathrm{Ext}^{i}(\Delta(\mathrm{h}),M)$, where $M$ is any skew Weyl
$S(n,r)$-module, is isomorphic to the torsion submodule of the cokernel
$E^{i}(\Delta(\mathrm{h}),M)$ of the map $\mathrm{Hom}(\theta_{i}(a,b),M)$.
Indeed, by the argument of [6], bottom of p. 634 to the top of p. 635, we have

$E^{i}(\Delta(\mathrm{h}),M)\simeq\mathrm{Ext}^{i}(\Delta(\mathrm{h}),M)\oplus
N,$

where $N$ is the image of the map $\mathrm{Hom}(\theta_{i}(a,b),M)$. (The
argument given in loc. cit. is stated for $i=1$ but is valid for any $i\geq
1$.) As a submodule of a free $\mathbb{Z}$-module, $N$ is a free
$\mathbb{Z}$-module. On the other hand, the $\mathbb{Z}$-module
$\mathrm{Ext}^{i}(\Delta(\mathrm{h}),M)$ is torsion for all $i\geq 1$ by [2],
last paragraph of Section 8. Hence from the above isomorphism we obtain that
the torsion submodule of $E^{i}(\Delta(\mathrm{h}),M)$ is isomorphic to
$\mathrm{Ext}^{i}(\Delta(\mathrm{h}),M)$.

###  2.4 The extensions
$\mathrm{Ext}^{i}(\Delta(\mathrm{h}),D_{a+k}\otimes\wedge^{b-k})$

We will use the following lemma several times. Let

$r_{k}=gcd\left(\tbinom{k+1}{1},\dots,\tbinom{k+1}{k}\right)$

and note that $r_{k}=1$ unless $k+1=p^{e}$, $p$ prime, in which case
$r_{k}=p$.

######  Lemma 2.3.

Suppose $n\geq{b+1}$ and $1\leq k<b$. Then

$\mathrm{Ext}^{i}(\Delta(\mathrm{h}),D_{a+k}\otimes\wedge^{b-k})=\mathrm{Ext}^{i}(\wedge^{k+1},D_{k+1}).$

In particular,
$\mathrm{Ext}^{1}(\Delta(\mathrm{h}),D_{a+k}\otimes\wedge^{b-k})=\mathbb{Z}_{2}$
and
$\mathrm{Ext}^{k}(\Delta(\mathrm{h}),D_{a+k}\otimes\wedge^{b-k})=\mathbb{Z}_{r_{k}}.$

###### Proof.

A special case of the main result, Theorem 2, of [13] yields

$\mathrm{Ext}^{i}(\Delta(a,1^{b}),D_{a+k}\otimes\wedge^{b-k})=\mathrm{Ext}^{i}(D_{a-1}\otimes\wedge^{k+1},D_{a+k}).$

Applying contravariant duality [11], p. 209, and [2], Theorem 7.7, we have

$\mathrm{Ext}^{i}(D_{a-1}\otimes\wedge^{k+1},D_{a+k})=\mathrm{Ext}^{i}(\wedge^{a+k},D_{k+1}\otimes\wedge^{a-1}).$

Again by [13],
$\mathrm{Ext}^{i}(\wedge^{a+k},D_{k+1}\otimes\wedge^{a-1})=\mathrm{Ext}^{i}(\wedge^{k+1},D_{k+1})$
and the first equality of the lemma follows.

We have $\mathrm{Ext}^{1}(\wedge^{k+1},D_{k+1})=\mathbb{Z}_{2},$ by [1],
Section 4, and $\mathrm{Ext}^{k}(\wedge^{k+1},D_{k+1})=\mathbb{Z}_{r_{k}},$
according to [15], eqn. (6) p. 2207. ∎

###  2.5 Summary of notation

For the reader’s convenience we gather here some of the notation introduced in
the previous subsections that will be used often.

  * •

$\mathrm{h}=(a,1^{b})$ and $\mathrm{h}(k)=(a+k,1^{b-k})$: hooks, where $1\leq
k\leq b$ and $r=a+b$, (subsection 2.1).

  * •

${i_{1}}^{(a_{1})}...{i_{t}}^{(a_{t})}|{j_{1}}...{j_{b}}$: the image in
$\Delta(\mathrm{h})$, where $a=a_{1}+...+a_{t}$, of the element
${i_{1}}^{(a_{1})}...{i_{t}}^{(a_{t})}\otimes{j_{1}}...{j_{b}}\in
D_{a}\otimes\wedge^{b}$ under the isomorphism
$cok(\theta_{a})\simeq\Delta(\mathrm{h})$, (subsection 2.2).

  * •

$P_{*}(a,b)$ and $\theta_{*}(a,b)$: projective resolution of
$\Delta(\mathrm{h})$ and the differential of $P_{*}(a,b)$ respectively,
(subsection 2.3).

  * •

$E^{i}(\Delta(\mathrm{h}),M)$: the cokernel of the map
$\mathrm{Hom}(\theta_{i}(a,b),M)$, (subsection 2.3).

  * •

$\phi_{t}$: a summand of the differential $\mathrm{Hom}(\theta_{i}(a,b),M)$
(defined before Remark 2.2).

##  3 $\mathrm{Ext}^{1}(\Delta(\mathrm{h}),\Delta(\mathrm{h}(k)))$

In this section we determine
$\mathrm{Ext}^{1}(\Delta(\mathrm{h}),\Delta(\mathrm{h}(k)))$ for $k>1$. The
case $k=1$ was computed in [15], Theorem 6.

###  3.1 Matrices $e^{(1)}(a,b,M)$ and a generator

Let $M=D_{a+k}\otimes\wedge^{b-k}$. For the semi-standard basis $B$ of the
domain $\mathrm{Hom}(D(\mathrm{h}),M)$ $=\mathrm{Hom}(D(a,1,...,1),M)$ of the
map $\mathrm{Hom}(\theta_{1}{(a,b)},M)$ we have $B=B_{0}\cup B_{1}$, where
$B_{0}$ (resp., $B_{1}$) consists of those elements of $B$ that have no
(resp., exactly one) appearance of 1 in the $\wedge^{b-k}$ part. We consider
the usual lexicographic ordering on $B$ and note that every element of $B_{0}$
is less than every element of $B_{1}$. We have

$|B|=\tbinom{b+1}{k+1},|B_{0}|=\tbinom{b}{k},|B_{1}|=\tbinom{b}{k+1}.$

Likewise, for the semi-standard bases $B^{1}$, $B^{2},...,B^{b}$ of
$\mathrm{Hom}(D(a+1,1,...,1),M)$, $\mathrm{Hom}(D(a,2,...,1),M)$, … ,
$\mathrm{Hom}(D(a,1,...,2),M)$ respectively, we have $B^{t}=B^{t}_{0}\cup
B^{t}_{1}$, where $B_{0}^{i}$ (resp. $B_{1}^{i}$) consists of those elements
of $B^{i}$ that have no (resp. exactly one) appearance of 1 in the
$\wedge^{b-k}$ part, and thus

$B^{\prime}=B^{1}_{0}\cup B^{1}_{1}\cup...\cup B^{b}_{0}\cup B^{b}_{1}$

is a basis of the codomain of the map $\mathrm{Hom}(\theta_{1}{(a,b)},M)$. We
order each set $B^{i}$ lexicographically and declare that every element of
$B^{i}$ is less than every element of $B^{i+1}$. For each $i$ we have

$|B^{i}|=\tbinom{b}{k},|B_{0}^{i}|=\tbinom{b-1}{k-1},|B_{1}^{i}|=\tbinom{b-1}{k}.$

Consider the matrix $e^{(1)}(a,b,M)\in
M_{b\binom{b}{k}\times\binom{b+1}{k+1}}(\mathbb{Z})$ of the map
$\mathrm{Hom}(\theta_{1}{(a,b)},M)$, with respect to the previous orderings,
partitioned into $b$ row blocks according to $B^{\prime}=B^{1}\cup\cdots
B^{b}$.

In the next Lemma, the missing entries of any matrix are assumed to be equal
to 0.

######  Lemma 3.1.

Let $p=\tbinom{b-1}{k-1},q=\tbinom{b-1}{k}.$ The matrix $e^{(1)}(a,b,M)$ has
the following properties.

  1. 1.

The first block is
$\left(\begin{array}[]{c|c}A(a,b;k)&B(a,b;k)\end{array}\right)$, where

$\displaystyle
A(a,b;k)=diag(\underbrace{a+1,...,a+1}_{p},\underbrace{1,...,1}_{q})\in
M_{\binom{b}{k}}(\mathbb{Z}),$ $\displaystyle
B(a,b;k)=\left(\begin{array}[]{c|cc }&&\\\ \hline\cr
aI_{q}&&\end{array}\right)\in
M_{\binom{b}{k}\times\binom{b}{k+1}}(\mathbb{Z}).$

  2. 2.

The t-th block, $t>1$, is of the form $\left(\begin{array}[]{c|cc}C_{t}&\\\
\hline\cr&D_{t}\end{array}\right)$, where $C_{t}\in
M_{p\times\binom{b}{k}}(\mathbb{Z})$, $D_{t}\in
M_{q\times\binom{b}{k+1}}(\mathbb{Z})$.

  3. 3.

The sum of the elements in any row of the t-th block, where $t\geq 2$, is
$(-1)^{t-1}2.$

  4. 4.

The last row is of the form $(0\;...\;0\;\pm 2)$.

###### Proof.

(1) The set $B_{0}$ consists of all $T=1^{(a)}i_{1}...i_{k}\otimes
j_{1}...j_{b-k}\in B$ such that

$\displaystyle i_{1}<...<i_{k},\;\;\;j_{1}<...<j_{b-k}$
$\displaystyle\\{i_{1},...,i_{k}\\}\cup\\{j_{1},...,j_{b-k}\\}=\\{2,...,b+1\\}$
$\displaystyle\\{i_{1},...,i_{k}\\}\cap\\{j_{1},...,j_{b-k}\\}=\emptyset.$

The definition of $\phi_{1}$ yields

$\phi_{1}(T)=\begin{cases}(a+1)1^{(a+1)}i^{\prime}_{2}...i^{\prime}_{k}\otimes
j^{\prime}_{1}...j^{\prime}_{b-k},&\mbox{if}\;i_{1}=2\\\
\;\;\;\;\;\;\;\;\;\;\;1^{(a)}i^{\prime}_{2}...i^{\prime}_{k}\otimes
1j^{\prime}_{2}...j^{\prime}_{b-k},&\mbox{if}\;i_{1}\neq 2\end{cases},$

where $i^{\prime}=i-1$. From this it easily follows that the matrix of the
restriction of $\phi_{1}$ on the subgroup of $\mathrm{Hom}(D(\mathrm{h}),M)$
spanned by $B_{0}$ is $A(a,b;k)$.

The set $B_{1}$ consists of all $S=1^{(a-1)}i_{1}...i_{k+1}\otimes
j_{1}...j_{b-k}\in B$ such that

$\displaystyle i_{1}<...<i_{k+1},\;\;\;j_{1}<...<j_{b-k},\;\;\;j_{1}=1$
$\displaystyle\\{i_{1},...,i_{k+1}\\}\cup\\{j_{1},...,j_{b-k}\\}=\\{1,...,b+1\\}$
$\displaystyle\\{i_{1},...,i_{k+1}\\}\cap\\{j_{1},...,j_{b-k}\\}=\emptyset.$

Then

$\phi_{1}(S)=\begin{cases}a1^{(a-1)}i^{\prime}_{2}...i^{\prime}_{k}\otimes
1j^{\prime}_{2}...j^{\prime}_{b-k},&\mbox{if}\;j_{2}\neq 2\\\
0,&\mbox{if}\;j_{2}=2\end{cases},$

From this it easily follows that the matrix of the restriction of $\phi_{1}$
on the subgroup of $\mathrm{Hom}(D(\mathrm{h}),M)$ spanned by $B_{1}$ is
$B(a,b;k)$.

(2) If $T=1^{(a)}i_{1}...i_{k}\otimes j_{1}...j_{b-k}\in B_{0}$, then
$j_{1}\geq 2$, and thus for $t\geq 2$, $\phi_{t}(T)$ is a multiple of an
element of $B^{t}$ that does not contain a 1 in the $\wedge^{b-k}$ part. Hence
$\phi_{t}(T)\in spanB^{t}_{0}$. If $S=1^{(a-1)}i_{1}...i_{k+1}\otimes
j_{1}...j_{b-k}\in B_{1}$, then $j_{1}=1$ and thus $\phi_{t}(T)\in
spanB^{t}_{1}$ for all $t$.

(3) Suppose $T^{\prime}\in B^{t}$. Since $t>1$, $t$ appears exactly twice in
$T^{\prime}$.
Case 1. Suppose $T^{\prime}$ is of the form $T^{\prime}=xt^{(2)}y\otimes z.$
Since each element of $B$ has weight $(a,1,...,1)$, from the definition of
$\phi_{t}$, it follows that there is a unique $T\in B$ such that
$\phi_{t}(T)=cT^{\prime},c\neq 0,$ namely $T=xt(t+1)y_{1}\otimes z_{1}$, where
$y_{1}$ and $z_{1}$ are obtained from $y$ and $z$ respectively by replacing
each $i>t$ by $i+1$. We have $\phi_{t}(T)=2T^{\prime}$.
Case 2. Suppose $T^{\prime}$ is of the form $T^{\prime}=xty\otimes ztw.$ Since
each element of $B$ has weight $(a,1,...,1)$, from the definition of
$\phi_{t}$, it follows that there are exactly two $T_{1},T_{2}\in B$ such that
$\phi_{t}(T_{i})=c_{i}T^{\prime},c_{i}\neq 0,$ namely

$T_{1}=x(t+1)y_{1}\otimes ztw_{1},\;T_{2}=xty_{1}\otimes x(t+1)w_{1},$

where $y_{1}$ and $w_{1}$ are obtained from $y$ and $w$ respectively by
replacing each $i>t$ by $i+1$. We have
$\phi_{t}(T_{1})=\phi_{t}(T_{2})=T^{\prime}$.

(4) This follows from case 1 of (3) since the greatest element in $B^{k}$ is
$T^{\prime}=1^{(a-1)}(b-k+1)...(b-1)b^{(2)}\otimes 12...(b-k)$. ∎

Let $1\leq k<b$ and $M=D_{a+k}\otimes\wedge^{b-k}$. From Lemma 2.3 we have
$\mathrm{Ext}^{1}(\Delta(\mathrm{h}),M)$ $=\mathbb{Z}_{2}$. We will determine
a generator of this $\mathrm{Ext}$ group.

We have mentioned that the torsion subgroup of the abelian group
$E^{1}(\Delta(\mathrm{h}),M)$ with presentation matrix $e^{(1)}(a,b,M)$ is
isomorphic to $\mathrm{Ext}^{1}(\Delta(\mathrm{h}),M)$. We denote by $\pi$ the
natural projection $\pi:\mathrm{Hom}(P_{1}(a,b),M)\to
E^{1}(\Delta(\mathrm{h}),M)$.

######  Lemma 3.2.

Let $\mathrm{h}=(a,1^{b})$, $1\leq k<b$ and $M=D_{a+k}\otimes\wedge^{b-k}$. A
cyclic generator of the abelian group $\mathrm{Ext}^{1}(\Delta(\mathrm{h}),M)$
is $\pi(g_{k})$, where

$g_{k}=\tbinom{a+1}{2}\sum\limits_{T\in B_{0}^{1}}T+a\sum\limits_{T\in
B_{1}^{1}}T+\sum\limits_{i=2}^{b}(-1)^{i-1}\Big{(}a\sum\limits_{T\in
B_{0}^{i}}T+\sum\limits_{T\in B_{1}^{i}}T\Big{)}.$

###### Proof.

Let $E_{i}$ be the i-th column of $e^{(1)}(a,b,M)$ and let
$p=\binom{b-1}{k-1},q=\binom{b-1}{k}$. Consider the $e^{(1)}(a,b,M)$
partitioned into $b$ blocks each consisting of $p+q=\binom{b}{k}$ consecutive
rows. From Lemma 3.1 it follows that

$\displaystyle
a\big{(}E_{1}+...+E_{\binom{b}{k}}\big{)}+E_{\binom{b}{k}+1}+...+E_{\binom{b+1}{k+1}}=$
$\displaystyle\Big{(}\underbrace{a(a+1),...,a(a+1)}_{p},\underbrace{2a,...,2a}_{q},\underbrace{-2a,...,-2a}_{p},\underbrace{-2,...,-2}_{q},...,$
$\displaystyle\underbrace{(-1)^{b-1}2a,...,(-1)^{b-1}2a}_{p},\underbrace{(-1)^{b-1}2,...,(-1)^{b-1}2}_{q}\Big{)}^{t}.$
(3.1)

Hence in the cokernel $E^{1}(\Delta(\mathrm{h}),M)$ of the differential
$\mathrm{Hom}(\theta_{1}(a,b),M)$ we have $2\pi(g_{k})=0$. This shows that
$\pi(g_{k})\in\mathrm{Ext}^{1}(\Delta(\mathrm{h}),M)$. We have $\pi(g_{k})\neq
0$, since otherwise the integer matrix-column $\frac{1}{2}X$, where $X$ is the
right hand side of (3.1), would be a $\mathbb{Z}$ \- linear combination of
columns of $e^{(1)}(a,b,M)$. This is not possible because the last entry of
$\frac{1}{2}X$ is $\pm 1$ while all entries of the last row of
$e^{(1)}(a,b,M)$, $k>0$, are even according to Lemma 3.1 (4). Since
$\mathrm{Ext}^{1}(\Delta(\mathrm{h}),M)=\mathbb{Z}_{2}$, it follows that
$\pi(g_{k})$ generates $\mathrm{Ext}^{1}(\Delta(\mathrm{h}),M)$.∎

###  3.2 Proof for
$\mathrm{Ext}^{1}(\Delta(\mathrm{h}),\Delta(\mathrm{h}(k)))$.

We determine $\mathrm{Ext}^{1}(\Delta(\mathrm{h}),\Delta(\mathrm{h}(k))),k\geq
2$ in this subsection. The main computation is done in the next two lemmas the
first of which takes care of the case $k=2$.

We will use the following notation. If $f:M\to N$ is a map of
$S(n,r)$-modules, we have in the usual way various induced maps

$\displaystyle\mathrm{Hom}(P_{i}(a,b),M)$
$\displaystyle\to\mathrm{Hom}(P_{i}(a,b),N),$ $\displaystyle
E^{i}(\Delta(\mathrm{h}),M)$ $\displaystyle\to E^{i}(\Delta(\mathrm{h}),N),$
$\displaystyle\mathrm{Ext}^{i}(\Delta(\mathrm{h}),M)$
$\displaystyle\to\mathrm{Ext}^{i}(\Delta(\mathrm{h}),N)$

of abelian groups which will all will be denoted by $f^{*}$. For an integer
$m$ let $\epsilon_{m}$ be the remainder of the division of $m$ by 2.

In the statement of the next Lemma, we recall that
$\mathrm{\mathrm{Ext}}^{1}(\Delta(\mathrm{h}),D_{a+1}\otimes\wedge^{b-1})=\mathbb{Z}_{2}$
according to Lemma 2.3. Also from [15], Theorem 6, we have that
$\mathrm{Ext}^{1}(\Delta(\mathrm{h}),\Delta(\mathrm{h}(1)))$ is a cyclic
group.

######  Lemma 3.3.

Let $\mathrm{h}=(a,1^{b})$ and $\mathrm{h}(1)=(a+1,1^{b-1}),b\geq 2$. The map

$\mathrm{\mathrm{Ext}}^{1}(\Delta(\mathrm{h}),D_{a+1}\otimes\wedge^{b-1})\xrightarrow{\pi_{0}^{*}}\mathrm{Ext}^{1}(\Delta(\mathrm{h}),\Delta(\mathrm{h}(1)))$

induced by
$D_{a+1}\otimes\wedge^{b-1}\xrightarrow{\pi_{0}}\Delta(\mathrm{h}(1))$ is
multiplication by the integer $\frac{(a+\epsilon_{b}-1)(a+b)}{2}$.

###### Proof.

The matrix of the map
$\mathrm{Hom}(P_{0}(a,b),\Delta(\mathrm{h}(1)))\rightarrow\mathrm{Hom}(P_{1}(a,b),\Delta(\mathrm{h}(1)))$
with respect to the lexicographic order of semi-standard bases is the
following $b\times b$ matrix according to [15], p. 2211,

$\left(\begin{array}[]{cccccc}a+1&-1&1&\dots&&(-1)^{b-1}\\\
-1&-1&0&\dots&&0\\\ 0&1&1&\dots&&0\\\ \vdots&\vdots&\vdots&&&\vdots\\\
0&0&0&\dots&(-1)^{b-1}&(-1)^{b-1}\end{array}\right).$

Let

$S_{1}=1^{(a+1)}|23...b,S_{2}=1^{(a)}2|23...b,...,S_{b}=1^{(a)}b|23...b$

be the semi-standard basis of
$\mathrm{Hom}(P_{1}(a,b),\Delta(\mathrm{h}(1)))$. From the above matrix it
follows that for each $i$ there is an integer $m_{i}$ such that
$\pi(S_{i})=m_{i}\pi(S_{1})$, and hence a cyclic generator of
$E^{1}(\Delta(\mathrm{h}),\Delta(\mathrm{h}(1)))$ is $\pi(S_{1})$. Moreover by
adding the even numbered columns $2,4,...,2[\frac{b}{2}]$ we have the relation

$\sum\limits_{i=2}^{b}(-1)^{i-1}\pi(S_{i})=[\frac{b}{2}]\pi(S_{1})$

where $[\frac{b}{2}]$ is the largest integer less than or equal to
$\frac{b}{2}$.

Using the notation established at the beginning of subsection 3.1, we have
$B_{0}^{1}=\\{T_{1}\\}$ and $B_{1}^{1}=\\{T_{2},..,T_{b}\\}$, where
$T_{1}=1^{(a+1)}\otimes 23...b$ and $T_{j}=1^{(a)}j\otimes
12...\widehat{j}...b,j=2,...,b$, and it is understood that $\widehat{j}$ means
that $j$ is omitted. Now $\pi_{0}^{*}(T_{1})=S_{1}$. Also

$\pi_{0}^{*}(T_{j})=1^{(a)}j|12...\widehat{j}...b=-1^{(a+1)}|j2...\hat{j}...b=(-1)^{j-1}S_{1},$

where in the second equality the straightening law was used. Thus

$\pi_{0}^{*}\Big{(}\tbinom{a+1}{2}\overline{B_{0}^{1}}+a\overline{B_{1}^{1}}\Big{)}=\Big{(}\tbinom{a+1}{2}+a(\epsilon_{b}-1)\Big{)}S_{1},$

where $\overline{X}=\sum_{T\in X}T$, if $X$ is one of the sets
$B_{0}^{i},B_{1}^{i}$. A similar computation for $j=2,..,b$ yields

$\pi_{0}^{*}\Big{(}\overline{B_{0}^{j}}+\overline{B_{1}^{j}}\Big{)}=\big{(}a+\epsilon_{b}-1\big{)}S_{j},$

and therefore in $E^{1}(\Delta(\mathrm{h}),\Delta(\mathrm{h}(1)))$ we have

$\displaystyle\pi_{0}^{*}(\pi(g_{1}))$
$\displaystyle=\Big{(}\tbinom{a+1}{2}+a(\epsilon_{b}-1)\Big{)}\pi(S_{1})+\big{(}a+\epsilon_{b}-1\big{)}\sum\limits_{i=2}^{b}(-1)^{i-1}\pi(S_{i})$
$\displaystyle=\Big{(}\tbinom{a+1}{2}+a(\epsilon_{b}-1)+\big{(}a+\epsilon_{b}-1\big{)}[\frac{b}{2}]\Big{)}\pi(S_{1}).$

It is easy to verify that
$\tbinom{a+1}{2}+a(\epsilon_{b}-1)+\big{(}a+\epsilon_{b}-1\big{)}[\frac{b}{2}]=\frac{(a+\epsilon_{b}-1)(a+b)}{2}$.
We have shown that the map
$\mathrm{Ext}^{1}(\Delta(\mathrm{h}),D_{a+1}\otimes\wedge^{b-1})\xrightarrow{\pi_{0}^{*}}\mathrm{Ext}^{1}(\Delta(\mathrm{h}),\Delta(\mathrm{h}(1)))$
is multiplication by this integer. Since $g_{1}$ is a generator of
$\mathrm{Ext}^{1}(\Delta(\mathrm{h}),D_{a+1}\otimes\wedge^{b-1})$ (Lemma 3.2)
which is the torsion submodule of
$E^{1}(\Delta(\mathrm{h}),D_{a+1}\otimes\wedge^{b-1})$, the result follows.∎

Since multiplication in the divided power algebra is commutative, we will
often denote a semi-standard basis element of the form
$1^{(a)}i_{1}...i_{s}t^{(2)}i_{s+1}...i_{k-2}\otimes j_{1}...j_{b-k}$ by
$1^{(a)}t^{(2)}i_{1}...i_{k-2}\otimes j_{1}...j_{b-k}$ and likewise for
$1^{(a)}i_{1}...i_{s}ti_{s+1}...i_{k-1}\otimes j_{1}...t...j_{b-k-1}$.

The $\mathrm{Ext}$ groups appearing in the next Lemma are both equal to
$\mathbb{Z}_{2}$ by Lemma 2.3. We want to identify a particular map between
these.

######  Lemma 3.4.

Let $\mathrm{h}=(a,1^{b})$ and $1<k<b$. The map

$\mathrm{Ext}^{1}(\Delta(\mathrm{h}),D_{a+k}\otimes\wedge^{b-k})\xrightarrow{\theta^{*}}\mathrm{Ext}^{1}(\Delta(\mathrm{h}),D_{a+k-1}\otimes\wedge^{b-k+1})$

induced by
$D_{a+k}\otimes\wedge^{b-k}\xrightarrow{\theta}D_{a+k-1}\otimes\wedge^{b-k+1}$
is multiplication by $a+\epsilon_{b-k+1}-1$.

###### Proof.

Consider the semi-standard basis $B^{\prime}=B^{1}_{0}\cup
B^{1}_{1}\cup...\cup B^{b}_{0}\cup B^{b}_{1}$, with the notation as in the
beginning of subsection 3.1, of the codomain of the map
$\mathrm{Hom}(\theta_{1}{(a,b)},D_{a+k}\otimes\wedge^{b-k})$. Likewise we
denote by $C^{\prime}=C^{1}_{0}\cup C^{1}_{1}\cup...\cup C^{b}_{0}\cup
C^{b}_{1}$ the semi-standard basis of the codomain of the map
$\mathrm{Hom}(\theta_{1}{(a,b)},D_{a+k-1}\otimes\wedge^{b-k+1})$. If $X$ is
any one of the sets $B_{0}^{t},B_{1}^{t},C_{0}^{t},C_{1}^{t}$, we let
$\overline{X}=\sum_{T\in X}T.$

We claim that for each $t=1,...,b,$

$\displaystyle\theta^{*}(\overline{B^{t}_{0}})=\epsilon_{b-k+1}\overline{C^{t}_{0}}+\overline{C^{t}_{1}},$
(3.2)
$\displaystyle\theta^{*}(\overline{B^{t}_{1}})=-\epsilon_{b-k}\overline{C^{t}_{1}}.$
(3.3)

Indeed, let $t>1$. We note that $B_{0}^{t}$ consists of all
$1^{(a)}t^{(2)}i_{1}...i_{k-2}\otimes j_{1}...j_{b-k}$ such that

$\displaystyle i_{1}<...<i_{k-2},\;\;j_{1}<...<j_{b-k},$
$\displaystyle\\{i_{1},...,i_{k-2}\\}\cap\\{j_{1},...,j_{b-k}\\}=\emptyset,\;\\{i_{1},...,i_{k-2}\\}\cup\\{j_{1},...,j_{b-k}\\}=\\{2,...,\widehat{t},...,b\\}$

and of all $1^{(a)}ti_{1}...i_{k-1}\otimes j_{1}...t...j_{b-k-1}$ such that

$\displaystyle i_{1}<...<i_{k-1},\;\;j_{1}<...<t<...<j_{b-k-1},$
$\displaystyle\\{i_{1},...,i_{k-1}\\}\cap\\{j_{1},...,j_{b-k-1}\\}=\emptyset,\;\\{i_{1},...,i_{k-1}\\}\cup\\{j_{1},...,j_{b-k-1}\\}=\\{2,...,\widehat{t},...,b\\}$

The definition of $\theta^{*}$ on the above elements yields

$\displaystyle\theta^{*}(1^{(a)}t^{(2)}i_{1}...i_{k-2}\otimes
j_{1}...j_{b-k})=$ $\displaystyle 1^{(a-1)}t^{(2)}i_{1}...i_{k-2}\otimes
1j_{1}...j_{b-k}$ $\displaystyle+1^{(a)}ti_{1}...i_{k-2}\otimes
tj_{1}...j_{b-k}$
$\displaystyle+\sum_{u=1}^{k-2}1^{(a)}t^{(2)}i_{1}...\widehat{i_{u}}...i_{k-2}\otimes
i_{u}j_{1}...j_{b-k}),$ (3.4)
$\displaystyle\theta^{*}(1^{(a)}ti_{1}...i_{k-1}\otimes
j_{1}...t...j_{b-k-1})=$ $\displaystyle 1^{(a-1)}ti_{1}...i_{k-1}\otimes
1j_{1}...t...j_{b-k-1}$
$\displaystyle+\sum_{u=1}^{k-1}1^{(a)}ti_{1}...\widehat{i_{u}}...i_{k-1}\otimes
i_{u}j_{1}...t...j_{b-k-1}),$ (3.5)

where $\widehat{i_{u}}$ means that $i_{u}$ is omitted. We note that each term
in the right hand side of equations (3.4) and (3.5) is of the form $\pm S$,
where $S\in C^{t}$. Moreover, the terms in the right hand side of (3.4) are
distinct and those in the right hand side of (3.5) are distinct. Now let $S\in
C^{t}=C^{t}_{0}\cup C^{t}_{1}.$
(1) Let $S\in C_{0}^{t}$.

  1. (a)

Suppose $S=1^{(a)}tu_{1}...u_{k-2}\otimes v_{1}...t...v_{b-k}$. From (3.4) and
(3.5), it follows that the elements $T_{i}\in B^{t}$ such that $S$ appears
with nonzero coefficient in $\theta^{*}(T_{i})$ are

$\displaystyle T_{0}$ $\displaystyle=1^{(a)}t^{(2)}u_{1}...u_{k-2}\otimes
v_{1}...v_{b-k},$ $\displaystyle T_{i}$
$\displaystyle=1^{(a)}tu_{1}...u_{k-2}v_{i}\otimes
v_{1}...\widehat{v_{i}}...t...v_{b-k},\;i=1,...,b-k.$

Moreover by straightening the $\wedge^{b-k+1}$ part, the coefficient of $S$ in
$\theta^{*}(T_{0})$ is $(-1)^{s}$, $s=\\#\\{i:v_{i}<t\\}$, and the coefficient
of $S$ in $\theta^{*}(T_{i})$ is

$\begin{cases}(-1)^{i-1},&\mbox{if}\;\mbox{$i\leq s$}\\\
(-1)^{i},&\mbox{$i\geq s+1$}.\end{cases}$

Therefore, by summing over $B_{0}^{t}$ be see that the coefficient of $S$ in
$\theta^{*}(\overline{B^{t}_{0}})$ is
$\sum_{i=1}^{b-k+1}(-1)^{i-1}=\epsilon_{b-k+1}.$

  2. (b)

Suppose $S=1^{(a)}t^{(2)}u_{1}...u_{k-3}\otimes v_{1}...v_{b-k+1}$. From (3.4)
and (3.5) it follows that the elements $T_{i}\in B^{t}$ such that $S$ appears
with nonzero coefficient in $\theta^{*}(T_{i})$ are

$\displaystyle T_{i}=1^{(a)}t^{(2)}u_{1}...u_{k-3}v_{i}\otimes
v_{1}...\widehat{v_{i}}...t...v_{b-k},\;i=1,...,b-k.$

Moreover by straightening the $\wedge^{b-k+1}$ part, the coefficient of $S$ in
$\theta^{*}(T_{0})$ is $(-1)^{i-1}$. Thus summing over $B_{0}^{t}$, the
coefficient of $S$ in $\theta^{*}(\overline{B_{0}^{t}})$ is
$\sum_{i=1}^{b-k+1}(-1)^{i-1}=\epsilon_{b-k+1}$.

(2) Let $S\in C_{1}^{t}$.

  1. (a)

Suppose $S=1^{(a-1)}t^{(2)}u_{1}...u_{k-2}\otimes 1v_{1}...v_{b-k}$. From
(3.4) and (3.5), it follows that there is a unique $T\in B_{0}^{t}$ such that
$S$ appears with nonzero coefficient in $\theta^{*}(T)$, namely
$T=1^{(a)}t^{(2)}u_{1}...u_{k-2}\otimes v_{1}...v_{b-k}$, and the coefficient
is 1.

  2. (b)

Suppose $S=1^{(a-1)}tu_{1}...u_{k-1}\otimes 1v_{1}...t...v_{b-k-1}$. From
(3.4) and (3.5), it follows that there is a unique $T\in B_{0}^{t}$ such that
$S$ appears with nonzero coefficient in $\theta^{*}(T)$, namely
$T=1^{(a)}tu_{1}...u_{k-1}\otimes v_{1}...t...v_{b-k-1}$, and the coefficient
is 1.

From the cases (1) and (2), equation (3.2) follows for $t>1.$ The proof of
(3.3), $t>1$, is similar (and a bit shorter) and omitted. Finally, the proof
of (3.2) and (3.3) for $t=1$ is similar (and a bit simpler) and omitted.

We now prove the statement of the Lemma.

$\it{Case\;1.}$ Suppose $b-k+1$ is even. By substituting (3.2) and (3.3) we
obtain

$\theta^{*}(g_{k})=\tbinom{a}{2}\overline{C^{1}_{1}}+(a-1)\sum_{i=2}^{b}(-1)^{i-1}\overline{C_{0}^{i}}$

and thus

$\theta^{*}(g_{k})-(a-1)g_{k-1}=-\tbinom{a}{2}\Big{(}(a+1)\overline{C^{1}_{0}}+\overline{C^{1}_{1}}+2\sum_{i=2}^{b}(-1)^{i-1}\overline{C^{i}_{0}}\Big{)}.$

But
$\pi\big{(}(a+1)\overline{C^{1}_{0}}+\overline{C^{1}_{1}}+2\sum_{i=2}^{b}(-1)^{i-1}\overline{C^{i}_{0}}\big{)}=0$
in $E^{1}(\Delta(\mathrm{h}),D_{a+k-1}\otimes\wedge^{b-k+1})$ because this is
the relation coming from adding the first $\tbinom{b}{k-1}$ columns of the
matrix $e^{(1)}(a,b,D_{a+k-1}\otimes\wedge^{b-k+1})$ according to Lemma 3.1
(1)-(3). Thus in this case the map $\theta^{*}$ is multiplication by $a-1$.

$\it{Case\;2.}$ Suppose $b-k+1$ is odd. By substituting (3.2) and (3.3) we
obtain

$\theta^{*}(g_{k})=\tbinom{a+1}{2}\big{(}\overline{C^{1}_{0}}+\overline{C^{1}_{1}}\big{)}+a\sum_{i=2}^{b}(-1)^{i-1}(\overline{C^{i}_{0}}+\overline{C^{i}_{1}})$

and using this we have

$\theta^{*}(g_{k})-ag_{k-1}=-\tbinom{a}{2}\Big{(}(a+1)\overline{C^{1}_{0}}+\overline{C^{1}_{1}}+2\sum_{i=2}^{b}(-1)^{i-1}\overline{C^{i}_{0}}\Big{)}.$

Thus in this case the map $\theta^{*}$ is multiplication by $a$. ∎

######  Theorem 3.5.

Let $\mathrm{h}=(a,1^{b})$ and $\mathrm{h}(k)=(a+k,1^{b-k})$, where $2\leq
k\leq b$. If $n\geq b+1$, then
$\mathrm{Ext}^{1}(\Delta(\mathrm{h}),\Delta(\mathrm{h}(k))=0$ unless $a+b+k$
is odd in which case
$\mathrm{Ext}^{1}(\Delta(\mathrm{h}),\Delta(\mathrm{h}(k))=\mathbb{Z}_{2}$.

###### Proof.

Applying $\mathrm{Hom}(\Delta(\mathrm{h}),-)$ to the short exact sequence
sequence (2.1) for $\mathrm{h}(k-1)$ in place of $\mathrm{h}$ yields the exact
sequence

$\displaystyle 0$
$\displaystyle\rightarrow\mathrm{Ext}^{1}(\Delta(\mathrm{h}),\Delta(\mathrm{h}(k))\xrightarrow{i*}\mathrm{Ext}^{1}(\Delta(\mathrm{h}),D_{a+k-1}\otimes\wedge^{b-k+1})$
$\displaystyle\xrightarrow{\pi_{0}^{*}}\mathrm{Ext}^{1}(\Delta(\mathrm{h}),\Delta(\mathrm{h}(k-1)))$

because $\mathrm{Hom}(\Delta(\mathrm{h}),\Delta(\mathrm{h}(k-1)))=0$ as
$\mathbb{Q}\otimes\Delta(\mathrm{h})$ and
$\mathbb{Q}\otimes\Delta(\mathrm{h}(k-1))$ are distinct irreducible
representations of $GL_{n}(\mathbb{Q})$.

First let $k=2$. From the above exact sequence, Lemma 2.3 and [15], Theorem 6,
it follows that $\mathrm{Ext}^{1}(\Delta(\mathrm{h}),\Delta(\mathrm{h}(2)))$
is the kernel of the map $\mathbb{Z}_{2}\rightarrow\mathbb{Z}_{a+b}$ which is
multiplication by the integer $\frac{(a+\epsilon_{b}-1)(a+b)}{2}$ according to
Lemma 3.3. Thus the result follows.

Suppose $k\geq 3$. Then
$\mathrm{Ext}^{1}(\Delta(\mathrm{h}),\Delta(\mathrm{h}(k-1)))$ injects in
$\mathrm{Ext}^{1}(\Delta(\mathrm{h}),D_{a+k-2}\otimes\wedge^{b-k+2})$ and thus
$\mathrm{Ext}^{1}(\Delta(\mathrm{h}),\Delta(\mathrm{h}(k))$ is the kernel of
the composite map

$\psi:\mathrm{Ext}^{1}(\Delta(\mathrm{h}),D_{a+k-1}\otimes\wedge^{b-k+1})\rightarrow\mathrm{Ext}^{1}(\Delta(\mathrm{h}),D_{a+k-2}\otimes\wedge^{b-k+2}).$

This map is induced by
$D_{a+k-1}\otimes\wedge^{b-k+1}\xrightarrow{\theta}D_{a+k-2}\otimes\wedge^{b-k+2}$.
According to Lemma 3.4, $\psi:\mathbb{Z}_{2}\rightarrow\mathbb{Z}_{2}$ is by
multiplication by $a+\epsilon_{b-k}-1$. Hence the result follows. ∎

##  4 $\mathrm{Ext}^{k}(\Delta(\mathrm{h}),\Delta(\mathrm{h}(k)))$

Let $\mathrm{h}=(a,1^{b})$ and $\mathrm{h}(k)=(a+k,1^{b-k})$, where we assume
throughout this section that $1\leq k\leq b$. We will prove the following
result.

######  Theorem 4.1.

If $n\geq b+1$, then
$\mathrm{Ext}^{k}(\Delta(\mathrm{h}),\Delta(\mathrm{h}(k)))=\mathbb{Z}_{d_{k}}$,
where $d_{k}=gcd(\tbinom{a+b}{1},\tbinom{a+b}{2},\dots,\tbinom{a+b}{k})$.

This result is known in the special cases $a=1$, $b=k$ [1], Section 4, and any
$a$, $b=k$ [15], eqn. (6) p. 2207.

According to the following Remark, the above $\mathrm{Ext}$ group is the
highest possible nonzero $\mathrm{Ext}$ group between the hooks $\mathrm{h}$
and $\mathrm{h}(k)$. We thank both H. H Andersen and the referee for pointing
out an error in a previous version of this paper concerning the proof of the
Remark and for suggesting the proof that follows.

######  Remark.

Let $n\geq b+1$. If $i>k$, then
$\mathrm{Ext}^{i}(\Delta(\mathrm{h}),\Delta(\mathrm{h}(k)))=0.$

###### Proof.

We use induction on $k$. For $k=0$ the result follows from the general fact
that no Weyl module has non trivial self extension, see [10], B.4. Remark.
Applying $\mathrm{Hom}(\Delta(\mathrm{h}),-)$ to the short exact sequence

$0\rightarrow\Delta(\mathrm{h}(k+1))\rightarrow\Delta(a+k)\otimes\Lambda(b-k)\rightarrow\Delta(\mathrm{h}(k))\rightarrow
0$

we obtain the exact sequence

$\displaystyle\mathrm{Ext}^{i}(\Delta(\mathrm{h}),\Delta(\mathrm{h}(k)))$
$\displaystyle\rightarrow\mathrm{Ext}^{i+1}(\Delta(\mathrm{h}),\Delta(h(k+1)))$
$\displaystyle\rightarrow\mathrm{Ext}^{i+1}(\Delta(\mathrm{h}),\Delta(a+k)\otimes\Lambda^{b-k}).$

The term on the left is zero by induction. By Lemma 2.3, the term on the right
is

$\mathrm{Ext}^{i+1}(\Delta(\mathrm{h}),\Delta(a+k)\otimes\Lambda^{b-k})=\mathrm{Ext}^{i+1}(\Lambda^{k+1},\Delta(k+1))$

which is zero by induction since $i+1>k$. Hence the middle term is zero. ∎

Recall the following notation from Section 2.3. $E^{i}(\Delta(\mathrm{h}),M)$
is the cokernel of the differential $\mathrm{Hom}(\theta_{i}(a,b),M)$ of the
complex $\mathrm{Hom}(P_{*}(a,b),M)$, where $M$ is a skew Weyl module. The
torsion part of this abelian group is isomorphic to
$\mathrm{Ext}^{i}(\Delta(\mathrm{h}),M)$. Let $\pi$ the natural projection
$\pi:\mathrm{Hom}(P_{i}(a,b),M)\to E^{i}(\Delta(\mathrm{h}),M)$ and
$e^{(i)}(a,b,M)$ the matrix of the map $\mathrm{Hom}(\theta_{i}(a,b),M)$ with
respect to orderings to be specified.

We order lexicographically the semi-standard basis of
$\mathrm{Hom}(D(a_{1},...,a_{m}),\Delta(\mathrm{h}(k))).$ Now if
$(a_{1},...,a_{m})$ is greater than $(b_{1},...,b_{m^{\prime}})$ , where
$m,m^{\prime}\leq n$, in the usual lexicographic ordering of sequences, we
declare that each element of the semi-standard basis of
$\mathrm{Hom}(D(a_{1},...,a_{m}),\Delta(\mathrm{h}(k)))$ is less than each
element of the semi-standard basis of
$\mathrm{Hom}(D(b_{1},...,b_{m^{\prime}}),\Delta(\mathrm{h}(k)))$. With
respect to the above orderings, Remark 2.1 yields the following, where the
missing entries in the bottom left part of the matrix are equal to zero.

######  Lemma 4.2.

For $i>1$ and $b>1,$
$e^{(i)}(a,b,\Delta(\mathrm{h}(k)))=\left(\begin{array}[]{c|c}A&*\\\
\hline\cr&*\end{array}\right),$ where
$A=e^{(i-1)}(a+1,b-1,\Delta(\mathrm{h}(k)))$.

###  4.1 A generator of
$\mathrm{Ext}^{k}(\Delta(\mathrm{h}),D_{a+k}\otimes\wedge^{b-k})$

In this subsection we will identify a generator of
$\mathrm{Ext}^{k}(\Delta(\mathrm{h}),D_{a+k}\otimes\wedge^{b-k})$. We assume
throughout that $1\leq k\leq b.$

Let $\Gamma_{k}\in\mathrm{Hom}(P_{k}(a,b),D_{a+k}\otimes\wedge^{b-k})$,

$\Gamma_{k}=\tbinom{a+k}{k+1}\triangle_{1}-\triangle_{2}+\cdots+{(-1)}^{b-k}\triangle_{q},$
(4.1)

where $q=b-k+1$ and

$\displaystyle\triangle_{1}$ $\displaystyle=1^{(a+k)}\otimes 2\dots
q\in\mathrm{Hom}(D(a+k,1,...,1),D_{a+k}\otimes\wedge^{b-k})$
$\displaystyle\triangle_{i}$ $\displaystyle=1^{(a-1)}i^{(k+1)}\otimes
1\dots\widehat{i}\dots q$
$\displaystyle\,\,\,\,\,\,\in\mathrm{Hom}(D(a,1,\dots,k+1,\dots,1),D_{a+k}\otimes\wedge^{b-k}),$

$i=2,\dots,q,$ where $k+1$ is located at the i-th position. Consider the
natural projection $\pi:\mathrm{Hom}(P_{k}(a,b),D_{a+k}\otimes\wedge^{b-k})\to
E^{k}(\Delta(\mathrm{h}),D_{a+k}\otimes\wedge^{b-k}).$

######  Lemma 4.3.

If $k+1=p^{e}$, $p$ prime, then $\pi(\Gamma_{k})\neq 0.$

###### Proof.

Suppose $\Gamma_{k}$ is equal to a linear combination of the columns of $A$,
where $A=e^{(k)}(a,b,D_{a+k}\otimes\wedge^{b-k})$. Then the coefficient
${(-1)}^{q-1}$ of $\triangle_{q}$, $q=b-k+1$, is a linear combination of the
entries of the last row of $A$, as $\triangle_{q}$ is the last basis element
in $\mathrm{Hom}(P_{k}(a,b),D_{a+k}\otimes\wedge^{b-k})$ with respect to the
lexicographic order. We claim that the set of nonzero elements in the last row
of $A$ is $\\{(-1)^{q-1}\tbinom{k+1}{1},\dots,(-1)^{q-1}\tbinom{k+1}{k}\\}.$

Indeed, let $i\in\\{1,\dots,q\\}$,
$\lambda=(\lambda_{1},\dots,\lambda_{q+1})$, where
$\lambda_{1}\in\\{0,\dots,k-1\\}$, $\lambda_{j}\geq 1$ for
$j\in\\{2,\dots,q+1\\}$, $\sum_{j=1}^{q+1}\lambda_{j}=b$, and consider a semi-
standard basis element
$T\in\mathrm{Hom}(D(a+\lambda_{1},\lambda_{2},\dots,\lambda_{q+1}),D_{a+k}\otimes\wedge^{b-k})$,
such that

$\phi_{i}(T)=c\triangle_{q},\;c\in\mathbb{Z}-\\{0\\}.$

By Remark 2.2, the left hand side has weight

$\left(1^{a+\lambda_{1}},2^{\lambda_{2}},\dots,i^{\lambda_{i}+\lambda_{i+1}},\dots,q^{\lambda_{q+1}}\right),$

which must be equal to the weight $\left(1^{a},2,\dots,q-1,q^{k+1}\right)$ of
the right hand side. So $\lambda_{1}=0$ and if $1<i\leq q-1$ then
$\lambda_{i}+\lambda_{i+1}=1$ which contradicts the hypothesis
$\lambda_{i}\geq 1$ for $i\in\\{2,\dots,q+1\\}$. This implies that $i=q$,
$\lambda_{1}=0$, $\lambda_{j}=1$ for $j\in\\{2,\dots,q-1\\}$ and
$\lambda_{q}+\lambda_{q+1}=k+1$. Hence
$T=1^{(a-1)}{q}^{\lambda_{q}}{(q+1)}^{\lambda_{q+1}}\otimes 1\dots(q-1)$,
where $\lambda_{q}+\lambda_{q+1}=k+1$. For such a $T$ we have
$(-1)^{q-1}\phi_{q}(T)=(-1)^{q-1}\tbinom{k+1}{\lambda_{q}}\triangle_{q},$
which proves the claim.

It follows that $gcd\left(\tbinom{k+1}{1},\dots,\tbinom{k+1}{k}\right)=1$
contradicting the assumption $k+1=p^{e}$, $p$ prime. Hence
$\pi(\Gamma_{k})\neq 0$. ∎

Let $q=b-k+1$. Define
$T_{1,j}\in\mathrm{Hom}(D(a+k-1,1,\dots,1),D_{a+k}\otimes\wedge^{b-k})$,
$j=2,\dots,q+1$,

$T_{1,j}=1^{(a+k-1)}j\otimes 2\dots\widehat{j}\dots(q+1),$

and
$T_{i,j}\in\mathrm{Hom}(D(a,1,\dots,k,\dots,1),D_{a+k}\otimes\wedge^{b-k})$,
where $k$ is at the i th-position, $i=2,\dots,q$, $j=i+1,\dots,q+1,$

$T_{i,j}=1^{(a-1)}i^{(k)}j\otimes
1\dots\widehat{i}\dots\widehat{j}\dots(q+1).$

Let

$A=\tbinom{a+k-1}{k}\sum_{j=2}^{q+1}(-1)^{j}T_{1,j}+\sum_{i=2,j>i}^{q,q+1}(-1)^{j-i-1}T_{i,j}$

and consider $\phi(A)$, where $\phi$ is the differential
$\phi=\mathrm{Hom}(\theta_{k}(a,b),D_{a+k}\otimes\wedge^{b-k})$.

######  Lemma 4.4.

We have $\phi(A)=(k+1)\Gamma_{k}$. Moreover, if $k+1=p^{e}$, $p$ prime, then
$\pi(\Gamma_{k})$ is a a generator of
$\mathrm{Ext}^{k}(\Delta(\mathrm{h}),D_{a+k}\otimes\wedge^{b-k})$.

###### Proof.

For the first statement, it suffices to show that

$\phi_{1}(A)=\tbinom{a+k-1}{k}(a+k)\triangle_{1}\;\;\hbox{and}\;\;\phi_{t}(A)=(k+1)\triangle_{t}$

for $t=2,\dots,q$, since $\binom{a+k-1}{k}(a+k)=(k+1)\binom{a+k}{k+1}.$ Using
Remark 2.2 an immediate calculation in each case shows the following.
$i=1$: Then $\phi_{1}(T_{1,2})=(a+k)\triangle_{1}$.
$i=2$: Then $\phi_{1}(T_{2,j})=\binom{a+k-1}{k}\phi_{1}(T_{1,j})$ for all
$j\geq 3.$
$i>2$: Then $\phi_{1}(T_{i,j})=0$ for all $j>i$.

Upon substituting,

$\displaystyle\phi_{1}(A)=$
$\displaystyle\tbinom{a+k-1}{k}(a+k)\triangle_{1}+\tbinom{a+k-1}{k}\sum_{j=3}^{q+1}(-1)^{j}\phi_{1}(T_{1,j})+\sum_{j=3}^{q+1}(-1)^{j-3}\phi_{1}(T_{2,j})$
$\displaystyle=$ $\displaystyle\tbinom{a+k-1}{k}(a+k)\triangle_{1}.$

Similarly for $t>1$, an immediate calculation in each case yields the
following.
$i<t$: Then $\phi_{t}(T_{i,t})=\phi_{t}(T_{i,t+1})$ and $\phi_{t}(T_{i,j})=0$
if $j\neq t,t+1$.
$i=t$: Then $\phi_{t}(T_{t,t+1})=(k+1)\triangle_{t}$ and
$\phi_{t}(T_{t,j})=\phi_{t}(T_{t+1,j})$ if $j\geq t+2$.
$i\geq t+2$: Then $\phi_{t}(T_{i,j})=0$ for all $j>i$.

Upon substituting,

$\displaystyle\phi_{t}(A)=$
$\displaystyle\tbinom{a+k-1}{k}\left((-1)^{t}\phi_{t}(T_{i,t})+(-1)^{t+1}\phi_{t}(T_{i,t})\right)$
$\displaystyle+\sum_{i=2}^{t-1}\left((-1)^{t}\phi_{t}(T_{i,t})+(-1)^{t+1}\phi_{t}(T_{i,t})\right)$
$\displaystyle+(a+k)\triangle_{t}+\sum_{j=t+2}^{q+1}\left((-1)^{j-t-1}\phi_{t}(T_{t,j})+(-1)^{j-t}\phi_{t}(T_{t,j})\right)$
$\displaystyle=$ $\displaystyle(a+k)\triangle_{t}.$

Let $k+1=p^{e}$, $p$ prime. By Lemma 4.3 and the first part of the present
Lemma, $\pi(\Gamma_{k}$) is a nonzero torsion element of the abelian group
$E^{k}(\Delta(\mathrm{h}),D_{a+k}\otimes\wedge^{b-k})$. Thus it is a nonzero
element of $\mathrm{Ext}^{k}(\Delta(\mathrm{h}),D_{a+k}\otimes\wedge^{b-k})$
which according to Lemma 2.3 is $\mathbb{Z}_{p}$. Hence it is a generator of
$\mathrm{Ext}^{k}(\Delta(\mathrm{h}),D_{a+k}\otimes\wedge^{b-k})$.∎

###  4.2 Relations and a generator of
$E^{k}(\Delta(\mathrm{h}),\Delta(\mathrm{h}(k)))$

Let $q=b-k+1$ and define
$\delta_{1}\in\mathrm{Hom}(D(a+k,1,\dots,1),\Delta(\mathrm{h}(k)))$ and
$\delta_{i,j}\in\mathrm{Hom}(D(a+k-j,1,\dots,j+1,\dots,1),\Delta(\mathrm{h}(k)))$,
where $j+1$ is located at the i-th position, by

$\displaystyle\delta_{1}=1^{(a+k)}|2\dots q,$
$\displaystyle\delta_{i,j}=1^{(a+k-j)}i^{(j)}|2\dots
q,\;i=2,\dots,q,\;j=0,\dots k,$

where it is understood that for $j=0$ we have $\delta_{i,0}=\delta_{1}.$

######  Lemma 4.5.

In $E^{k}(\Delta(\mathrm{h}),\Delta(\mathrm{h}(k)))$ the following relations
hold.

$\pi(\delta_{i,j})=\tbinom{a+k+i-2}{j}\pi(\delta_{1}),\;i=2,\dots
q-1,j=0,\dots k.$

###### Proof.

Let $S_{i,j}\in\mathrm{Hom}(P_{k-1}(a,b),\Delta(\mathrm{h}(k)))$,
$i=1,\dots,q$,

$S_{i,j}=1^{(a+k-j)}(i+1)^{(j)}|2\dots\widehat{(i+1)}\dots(q+1)$

and consider the differential in degree $k$

$\sum_{t\geq
1}(-1)^{t-1}\phi_{t}:\mathrm{Hom}(P_{k-1}(a,b),\Delta(\mathrm{h}(k))\rightarrow\mathrm{Hom}(P_{k}(a,b),\Delta(\mathrm{h}(k)))$

of the complex $\mathrm{Hom}(P_{*}(a,b),\Delta(\mathrm{h}(k)))$. An immediate
calculation in each case using Remark 2.2 (and the straightening law for the
first equality in the case $i>1$) yields the following.

$\displaystyle i=1:\;$
$\displaystyle\phi_{1}(S_{1,j})=\tbinom{a+k}{j}\delta_{1},\;\phi_{2}(S_{1,j})=\delta_{2,j},\;\phi_{j}(S_{1,j})=0,\;j>2.$
$\displaystyle i>1:\;$
$\displaystyle\phi_{1}(S_{i})={(-1)}^{i-1}\delta_{i,j-1},\;\phi_{t}(S_{i,j})=0,\;t\in\\{2,\dots,i-1\\},$
$\displaystyle\phi_{i}(S_{i,j})=\delta_{i,j},\;\phi_{i+1}(S_{i,j})=\delta_{i+1,j},\;\phi_{t}(S_{i,j})=0,\;t\geq
i+2.$

In $E^{k}(a,b,\Delta(\mathrm{h}(k)))$, we have the relations

$\sum_{t=1}^{q}(-1)^{t-1}\pi(\phi_{t}(S_{i,j}))=0,\;i=1,\dots q.$

Substituting the above for $i=1$ yields

$\pi(\delta_{2,j})=\tbinom{a+k}{j}\pi(\delta_{1}),j=0,\dots k$ (4.2)

and substituting the above for $i\geq 2$ yields

$\pi(\delta_{i+1,j})=\pi(\delta_{i,j})+\pi(\delta_{i,j-1}),\;i=2,\dots
q-1,j=1,\dots k.$ (4.3)

The equation of the Lemma follows by induction on $i$ using (4.2) and (4.3). ∎

######  Lemma 4.6.

$E^{k}(\Delta(\mathrm{h}),\Delta(\mathrm{h}(k)))$ is a cyclic group generated
by $\pi(\delta_{1})$.

###### Proof.

Induction on $k$, the case $k=1$ owing to the first paragraph of the proof of
Lemma 3.3.

Let $T\in\mathrm{Hom}(P_{k}(a,b),\Delta(\mathrm{h}(k)))$ be a semi-standard
tableau of weight $\lambda=(\lambda_{1},...,\lambda_{q})$, $q=b-k+1$. If
$\lambda_{1}>a$ then by induction and Lemma 4.2 (since
$\mathrm{h}(k)=(a+1+(k-1),1^{b-1-(k-1)}$), $\pi(T)$ is a multiple of
$\pi(\delta_{1})$.

Suppose $\lambda_{1}=a$ in which case $T=1^{(a)}2^{(\lambda_{2}-1)}\dots
q^{(\lambda_{q}-1)}|2\dots q.$ (In this notation it is understood that if
$\lambda_{j}=1$, then the term $j^{(\lambda_{j}-1)}$ is omitted.) We will show
that there are semi-standard tableaux
$T_{i}\in\mathrm{Hom}(P_{k}(a,b),\Delta(\mathrm{h}(k)))$ and
$a_{i}\in\mathbb{Z}$ such that $\pi(T)=\sum_{j}a_{i}\pi(T_{j})$ and $T_{j}<T$
for every $j$ in the lexicographic ordering of semi-standard tableaux of
$\mathrm{Hom}(P_{k}(a,b),\Delta(\mathrm{h}(k)))$. Since this set is finite and
$\delta_{1}$ is the least element, we obtain by induction on the ordering that
$\pi(T)=a\pi(\delta_{1})$, $a\in\mathbb{Z}$.

There exists an $i\geq 2$ such that $\lambda_{i}\geq 2$ because $k>0.$ Let $m$
be the largest such $i$ and let

$S=1^{(a)}2^{(\lambda_{2}-1)}\dots
m^{(\lambda_{m}-1)}|2\dots\widehat{m}\dots(q+1),$

which is a semi-standard tableau in
$\mathrm{Hom}(P_{k-1}(a,b),\Delta(\mathrm{h}(k))).$ Then for
$t\in\\{1,\dots,q\\}$, straightforward calculations yield the following, where
we assume that $m\geq 3$.

  1. 1.

$\phi_{1}(S)=(-1)^{m-2}\tbinom{a+\lambda_{2}-1}{\lambda_{2}-1}1^{(a+\lambda_{2})}2^{(\lambda_{3}-1)}\dots(m-1)^{(\lambda_{m}-2)}|2\dots
q.$

  2. 2.

$\phi_{t}(S)=0$ if $m\geq 4$, $t=2,...,m-2$.

  3. 3.

$\phi_{m-1}(S)=(-1)^{m}\tbinom{\lambda_{m-1}+\lambda_{m}-2}{\lambda_{m}-2}1^{(a+\lambda_{2})}2^{(\lambda_{3}-1)}\dots(m-1)^{(\lambda_{m-1}+\lambda_{m}-2)}|2\dots
q.$

  4. 4.

$\phi_{m}(S)={(-1)}^{m-1}T$.

  5. 5.

$\phi_{t}(S)=0$, if $j\geq m+1$.

The tableaux in the right-hand sides of equations (1) and (3) are semi-
standard and less than $T$ in our ordering since $\lambda_{2}>0$, and the
coefficient of $T$ in the right hand side of (4) is $\pm 1$. Hence the desired
result for $m\geq 3$ follows from $\sum_{t=1}^{q}(-1)^{t-1}\pi\phi_{t}(S)=0$.

Let $m=2$. Then similarly,
$0=\sum_{t=1}^{q}(-1)^{t-1}\pi\phi_{t}(S)=\tbinom{a+\lambda_{2}-1}{\lambda_{2}-1}\pi(\delta_{1})-\pi(T)$
and the result follows.∎

###  4.3 Proof of Theorem 4.1

We prove Theorem 4.1 by induction on $k$, the case $k=1$ owing to [15],
Theorem 6. Applying $\mathrm{Hom}(\Delta(\mathrm{h}),-)$ to the short exact
sequence

$0\rightarrow\Delta(\mathrm{h}(k+1))\xrightarrow{i_{k}}D_{a+k}\otimes\wedge^{b-k}\xrightarrow{\pi_{k}}\Delta(\mathrm{h}(k))\rightarrow
0$

yields the exact sequence

$\displaystyle\cdots$
$\displaystyle\rightarrow\mathrm{Ext}^{k}(\Delta(\mathrm{h}),D_{a+k}\otimes\wedge^{b-k})\xrightarrow{\pi^{*}_{k}}\mathrm{Ext}^{k}(\Delta(\mathrm{h}),\Delta(\mathrm{h}(k)))\rightarrow$
$\displaystyle\rightarrow\mathrm{Ext}^{k+1}(\Delta(\mathrm{h}),\Delta(\mathrm{h}(k+1)))\rightarrow
0$ (4.4)

because from Lemma 2.3,
$\mathrm{Ext}^{k+1}(\Delta(\mathrm{h}),D_{a+k}\otimes\wedge^{b-k})=\mathrm{Ext}^{k+1}(\wedge^{k+1},D_{k+1})=0$
as the length of the projective resolution $P_{*}(1,k)$ of $\wedge^{k+1}$ is
less than $k+1$. By Lemma 2.3 we have
$\mathrm{Ext}^{k}(\Delta(\mathrm{h}),D_{a+k}\otimes\wedge^{b-k})=\mathbb{Z}_{r_{k}}.$

If $\mathrm{Ext}^{k}(\Delta(\mathrm{h}),\Delta(\mathrm{h}(k)))=0$, then by
induction $d_{k}=1$. Since $d_{k+1}|d_{k}$, we have $d_{k+1}=1.$ Moreover,
$\mathrm{Ext}^{k+1}(\Delta(\mathrm{h}),\Delta(h(k+1)))=0$ by (4.4) and hence
the result holds in this case.

We may assume that
$\mathrm{Ext}^{k}(\Delta(\mathrm{h}),\Delta(\mathrm{h}(k)))\neq 0$. Since
$E^{k}(\Delta(\mathrm{h}),\Delta(\mathrm{h}(k)))$ is a cyclic
$\mathbb{Z}$-module according to Lemma 4.6, and its torsion subgroup is
nonzero, we have $E^{k}(\Delta(\mathrm{h}),\Delta(\mathrm{h}(k)))$
$=\mathrm{Ext}^{k}(\Delta(\mathrm{h}),\Delta(\mathrm{h}(k)))$.
Case 1: Let $k+1=p^{e}$, $p$ prime. By Lemma 4.4, $\pi(\Gamma_{k})$ is a
generator of
$\mathrm{Ext}^{k}(\Delta(\mathrm{h}),D_{a+k}\otimes\wedge^{b-k})$. We compute
its image under the map $\pi_{k}^{*}$ of (4.4). With the notation established
at the beginning of subsections 4.1 and 4.2 and using the straightening law
and Lemma 4.5 we have

$\displaystyle\pi^{*}_{k}(\pi(\triangle_{i}))$
$\displaystyle=\pi(1^{(a-1)}i^{(k+1)}|1\dots\widehat{i}\dots
q)={(-1)}^{i+1}\pi(\delta_{i,k})$
$\displaystyle={(-1)}^{i+1}\tbinom{a+k+i-2}{k}\pi(\delta_{1}).$

By substituting in (4.1) and using the binomial coefficient identity

$\tbinom{a+k}{k+1}+\sum_{i=2}^{q}\tbinom{a+k+i-2}{k}=\tbinom{a+b}{k+1},$

we obtain $\pi^{*}_{k}(\Gamma_{k})=\tbinom{a+b}{k+1}\pi(\delta_{1}).$ Since,
by Lemma 4.6, $\pi(\delta_{1})$ is a generator of
$\mathrm{Ext}^{k}(\Delta(\mathrm{h}),\Delta(\mathrm{h}(k)))$, we obtain from
(4.4) and the induction hypothesis that
$\mathrm{Ext}^{k+1}(\Delta(\mathrm{h}),\Delta(h(k+1)))=\mathbb{Z}_{d_{k+1}}$.
Case 2: Suppose $k+1$ is divisible by two distinct primes, whence according to
Lemma 2.3,
$\mathrm{Ext}^{k}(\Delta(\mathrm{h}),D_{a+k}\otimes\wedge^{b-k})=0$. From
(4.4) it suffices to show that $d_{k}=d_{k+1}.$

By Theorem 1 of [12], $d_{k}=\frac{a+b}{l_{k}}$, where
$l_{k}=lcm(1^{\eta_{1}},2^{\eta_{2}},\dots,k^{\eta_{k}})$ and $\eta_{i}=1$ if
$i|a+b$, and $\eta_{i}=0$ otherwise. If $k+1\not|a+b$, then $\eta_{k+1}=0$ and
hence $l_{k}=l_{k+1}$. If $k+1|a+b$, then, since $k+1$ is divisible by two
distinct primes, every prime power factor of $k+1$ is less than $k+1$. Hence
$l_{k+1}=l_{k}$. ∎

Let $\mathbb{K}$ be an infinite field of characteristic $p>0$,
$S_{\mathbb{K}}(n,r)$ the Schur algebra for $GL_{n}(\mathbb{K})$ and
$\Delta_{\mathbb{K}}(\lambda)$ the Weyl module for $S_{\mathbb{K}}(n,r)$
corresponding to a partition $\lambda$ of $r$ with at most $n$ parts. Then
$S_{\mathbb{K}}(n,r)$ =$\mathbb{K}\otimes S(n,r)$ and
$\Delta_{\mathbb{K}}(\lambda)$ =$\mathbb{K}\otimes\Delta(\lambda)$. From this
and the universal coefficient theorem [2], Theorem 5.3, our results yield the
following.

######  Corollary 4.7.

Let $\mathbb{K}$ be an infinite field of characteristic $p>0$ and $n\geq b+1$.

  1. 1.

Let $2\leq k\leq b$. Then
$\mathrm{Hom}_{S_{\mathbb{K}}(n,r)}(\Delta_{\mathbb{K}}(\mathrm{h}),\Delta_{\mathbb{K}}(\mathrm{h}(k)))=0$,
unless $p=2$ and $a+b+k$ is odd, in which case
$\mathrm{Hom}_{S_{\mathbb{K}}(n,r)}(\Delta_{\mathbb{K}}(\mathrm{h})),\Delta_{\mathbb{K}}(\mathrm{h}(k)))=\mathbb{K}$.

  2. 2.

Let $1\leq k\leq b$. Then
$\mathrm{Ext}^{k}_{S_{\mathbb{K}}(n,r)}(\Delta_{\mathbb{K}}(\mathrm{h}),\Delta_{\mathbb{K}}(\mathrm{h}(k)))=0$,
unless $p|\tbinom{a+b}{i},i=1,...,k$, in which case
$\mathrm{Ext}^{k}_{S_{\mathbb{K}}(n,r)}(\Delta_{\mathbb{K}}(\mathrm{h}),\Delta_{\mathbb{K}}(\mathrm{h}(k)))=\mathbb{K}$.

##  5 Acknowledgments

We thank H. H. Andersen for various helpful discussions and for pointing out
an error in the Remark after Theorem 4.1 and suggesting a proof. We thank the
reviewer for detailed constructive comments and suggestions that helped
improve the presentation and clarity of the paper and for pointing out an
error in the Remark after Theorem 4.1 and suggesting a proof.

## References

  * [1] Akin K., Extensions of symmetric tensors by alternating tensors, J. Algebra 121 (1989), 358-363.
  * [2] Akin K. and Buchsbaum D., Characteristic-free representation theory of the general linear group II: Homological considerations, Adv. Math. 72 (1988), 172-210.
  * [3] Akin K., Buchsbaum D. and Weyman J., Schur functors and Schur complexes, Adv. Math. 44 (1982), 207-278.
  * [4] Andersen, H., H. and Kulkarni, U., Sum formulas for reductive algebraic groups, Adv. Math. 217 (2008), 419-447.
  * [5] Buchsbaum D. and Flores de Chela D., Intertwining numbers: the three rowed case, J. Algebra 183 (1996), 605-635.
  * [6] Chela D. F., On intertwining numbers, J. Algebra 171 (195), 631-653.
  * [7] Cox A. and Erdmann K., $\mathrm{Ext}^{2}$ between Weyl modules for quantum $GL_{n}$, Math. Proc. Cambridge Philos. Soc. 128 (2000), 441-463.
  * [8] Cox A. and Parker A., Homomorphisms and higher extensions for Schur algebras and symmetric groups, J. Algebra Appl. 04 (2005), 645-670.
  * [9] Erdmann, K., $\mathrm{Ext}^{1}$ for Weyl modules of $SL_{2}(K)$, Math. Z. 218 (1995), 447-459.
  * [10] Green, J. A., Polynomial Representations of GLn, 2nd edition, Springer (2007).
  * [11] Jantzen J. C., Representations of algebraic groups, volume 107 AMS, Providence, RI, second edition, 2003.
  * [12] Joris H., Oestreicher C. and Steinig J., The greatest common divisor of certain sets of binomial coefficients, J. Number Theory 21 (1985), 101-119.
  * [13] Kulkarni U., Skew Weyl modules for $GL_{n}$ and degree reduction for Schur algebras, J. Algebra 224 (2000), 248-262.
  * [14] Kulkarni U., On the $\mathrm{Ext}$ groups between Weyl modules for $GL_{n}$, J. Algebra 304 (2006), 510-542.
  * [15] Maliakas M., Resolutions, homological dimensions, and extension of hook representations, Commun. Algebra 19 (1991), 2195-2216.
  * [16] Parker, A., Higher extensions between modules for $SL_{2}$, Adv. in Math. 209 (2007), 381-405.