text:
\begin{document}
	\title{Integral $Ext^2$ between hook Weyl modules}

	\author{Dimitra-Dionysia Stergiopoulou}
	\address{Department of Mathematics, University of Athens}
	\email{dstergiop@math.uoa.gr}

	\subjclass[2010]{Primary 20G05}

	\keywords{Extensions, Weyl modules, general linear group, hook}

	\date{June 5, 2020}

	\maketitle
	\begin{abstract}
		This paper concerns representations of the integral general linear group. The extension groups $Ext^2$ between any pair of hook Weyl modules are determined via a detailed study of cyclic generators and relations associated to certain extensions. As a corollary, the modular extension groups $Ext^1$ between such modules are determined.
	\end{abstract}

	\section {Introduction}	In the study of polynomial representations of the integral general linear group $GL_n$, the Weyl modules $\Delta(\lambda)$, indexed by partitions $\lambda$, play a central role.
	These modules have an explicit construction, enjoy a standard basis theorem and their characters are the classical Schur functions \cite{Gr}. Moreover, the heads of the modules $K\otimes \Delta(\lambda)$ form a complete set of inequivalent polynomial simple modules for $GL_n(K)$, where $K$ is an infinite field.  However, the structure of the modules $\Delta(\lambda)$ is not well understood.

	One of the important problems in the area is to determine the extension groups $Ext^i(\Delta(\lambda),\Delta(\mu))$. Relatively few explicit results are known. Modular extension groups for $GL_n(K)$ were studied in \cite{A}, when $\lambda$ consists of a single column and $\mu$ of a single row. For $SL_2(K)$, all modular extension groups between Weyl modules were described in \cite{Pa}, generalizing \cite{Er} and \cite{CE}. The modular extension groups $Ext^1(K\otimes\Delta(\lambda), K\otimes\Delta(\mu))$ for $GL_n(K)$ were determined in \cite{DG}, when $\mu$ consists of a single row.

	Concerning the integral extension groups, $Ext^1(\Delta(\lambda),\Delta(\mu))$ for $GL_3$ was computed in \cite{BF}, when $\lambda$ and $\mu$ differ by a multiple of a positive root. The groups $Ext^1(\Delta(\lambda),\Delta(\mu))$ for $GL_n$ was determined in \cite{Ku2}, when $\lambda$ and $\mu$ differ by a single positive root. The groups $Ext^i(\Delta(h), \Delta (h(k)))$ were computed in \cite{MS} for $i=1,k$ and any pair of hooks $h =(a,1^b)$ and $h(k)=(a+k, 1^{b-k})$. The main method there was to determine cyclic generators of certain extension groups of the form $Ext^i(\Delta(h),D_{a+k}\otimes\Lambda^{b-k})$, where $D_*$ and $\Lambda^*$ denote the divided power algebra and the exterior algebra, respectively, of the natural $GL_n$-module. Except for some extreme cases, these generators usually have an involved form. The next step in our method was to calculate the images of these under canonical maps. The idea of using the `skew' extensions $Ext^i(\Delta(h),D_{a+k}\otimes\Lambda^{b-k})$, in place of the usual extensions $Ext^i(\Delta(h),\Delta(h(k)))$, comes from the observation that the former seem to have more manageable presentation matrices. The reason for this is that there is no straightening law involved.

	The purpose of the present paper is to study $Ext^2$ between any two hook Weyl modules using the above strategy. The main result is the following.

	\begin{theo} Consider hook partitions $h=(a,1^b)$ and $h(k)=(a-k,1^{b-k})$ for $k \ge 0$. Suppose $n\ge b+1$. Then, $Ext^2(\Delta(h), \Delta(h(k)))=0$, except possibly when $k\in\{2,3,4\},$ in which cases
		$$Ext^2(\Delta(h), \Delta(h(k))) = \begin{cases}  \mathbb{Z}_s, & \mbox {if}\;k=2,
		\\ \mathbb{Z}_{3/t}, & \mbox {if}\;k=3,
		\\ \mathbb{Z}_{t}, & \mbox {if}\;k=4,
		\end{cases} $$
		where $s= {(a+b)}/{gcd(2,a+b)}$, $t={gcd(3,a+b)}$.
	\end{theo}
As a corollary, we determine the modular $Ext^1$ groups between any hook partitions.

In Section 2 we gather necessary preliminaries. In Section 3 we consider the cases $k\neq 4$. In Section 4, which costitutes the main part of the paper, we complete the proof of the theorem. As a corollary, we determine the dimensions of the modular $Ext^1$ groups between any hook Weyl modules.

	\section{Preliminaries}
	We will use the notation of \cite{MS} but, for completeness, we recall the main points here and supplement with additional material.
	\subsection{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 $S_\mathbb{Z}(n,r)$  \cite{Gr}. We will write $S(n,r)$ in place of $S_\mathbb{Z}(n,r)$.
	By $DF=\sum_{i\geq 0}D_iF$ and $\Lambda F=\sum_{i\geq 0}\Lambda^{i}F$
	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 $\Lambda^i$.

 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 $S(n,r)$. A hook $h$ is a partition of the form $h=(a,1^b)$.

	The following complex of $S(n,r)$-modules (which is the dual of the usual Koszul complex) is exact
	$$0 \rightarrow D_{a+b} \rightarrow ... \rightarrow D_{a+1} \otimes \Lambda^{b-1} \xrightarrow{\Theta_{a}}  D_{a} \otimes \Lambda^{b} \rightarrow ... \rightarrow \Lambda^{a+b} \rightarrow 0,$$
	where $\Theta_{a}$ is the composition
	$\ D_{a+1}\otimes \Lambda^{b-1} \xrightarrow{\triangle \otimes 1} D_{a}\otimes D_1 \otimes \Lambda^{b-1}\xrightarrow{1 \otimes m} D_{a} \otimes \Lambda^{b},$
	where $\triangle$ (respectively, $m$) is the indicated component of the comultiplication (resp., multiplication) map of the Hopf algebra $DF$ (resp., $\Lambda F$). It is  well known that if $h=(a,1^b)$ is a hook, $b \geq 1$, then
	$\Delta (h) \simeq cok(\Theta_{a}) \simeq ker(\Theta_{a-1}).$

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

    \subsection{Straightening law and standard basis.} We recall the straightening law and the standard basis theorem for $\Delta(h)$ \cite{ABW}. Fix an ordered basis $e_1,\dots,e_n$ of $F$. For simplicity, we denote the element $e_i$ by $i$ and accordingly the element $e_{i_1}^{(a_1)} \cdots  e_{i_t}^{(a_t)} \otimes e_{j_1} \wedge \dots \wedge e_{j_b} \in D_a \otimes \Lambda^b$ by ${i_1}^{(a_1)}  \cdots  {i_t}^{(a_t)} \otimes {j_1} \cdots {j_b}$. The image of this element under the identification $\Delta (h) \simeq cok(\Theta_{a})$ will be denoted by ${i_1}^{(a_1)}  \cdots  {i_t}^{(a_t)} / {j_1} \cdots {j_b}$. Now suppose $i_1<i_2< \dots <i_t$ and $j_1 \le i_1$. Then in $\Delta(h)$ we have
	$${i_1}^{(a_1)}\cdots {i_t}^{(a_t)} / {j_1} \cdots {j_b} = \begin{cases} -\sum\limits_{s\geq 2}{i_1}^{(a_1+1)}  \cdots {i_s}^{(a_s-1)}\cdots{i_t}^{(a_t)} / {i_s}{j_2} \cdots {j_b}, & \mbox{if}\;j_1=i_1 \\ -\sum\limits_{s\geq 1}j_1{i_1}^{(a_1)}  \cdots {i_s}^{(a_s-1)}\cdots {i_t}^{(a_t)} / {i_s}{j_2} \cdots {j_b}, & \mbox {if}\;j_1<i_1. \end{cases} $$

	\noindent A $\mathbb{Z}$ - basis of $\Delta(h)$ is the set of elements ${i_1}^{(a_1)}  \cdots {i_t}^{(a_t)} / {j_1} \cdots {j_b} $, where $a_1+\cdots+a_t=a$, $i_1<\dots<i_t$ and $i_1 <j_1< \dots <j_b.$
	\subsection{Resolutions of hooks.}	From \cite{Gr} or \cite{AB} we recall that for each sequence $a_1,\dots,a_n$ of non negative integers $a_i$ that sum to $r$, the $S(n,r)$-module $D_{a_1} \otimes \cdots \otimes D_{a_n}$ is projective.

	We will use the explicit finite projective resolution $P_{*}(a,b)$ of $ \Delta (h)$,
	\[0 \longrightarrow \cdots \longrightarrow P_2(a,b) \xrightarrow{\Theta_{2}(a,b)} P_1(a,b) \xrightarrow{\Theta_{1}(a,b)} P_0(a,b)
	\]
	of \cite{Ma} which we now recall. For short, we denote the tensor product $D_{a_1} \otimes \cdots \otimes D_{a_m}$ of divided powers by $D(a_1,\dots,a_m)$. We have
	$P_i(a,b)=\sum D(a_1,\dots,a_{b+1-i})
	$
	where the sum ranges over all sequences $(a_1,\dots,a_{b+1-i})$ of positive integers of length $b+1-i$ such that $a_1+\cdots+a_{b+1-i}=a+b$ and $a \le a_1 \le a+i$. The differential $\Theta_{i}(a,b)$ is defined be sending $x_1 \otimes \cdots \otimes x_{b+1-i} \in D(a_1,\dots,a_{b+1-i})$ to
	\[ \sum_{j=1}^{b+1-i} (-1)^{j+1}x_1 \otimes \cdots \otimes \triangle(x_j) \otimes \cdots  \otimes x_{b+1-i} \in D(a_1,\dots,u,v,\dots,a_{b+1-i}),
	\]
	where  $\triangle$ is the the two-fold diagonalization $D(a_j) \rightarrow \sum D(u,v)$, the sum ranges of all positive integers $u,v$ such that $u+v=a_j$ and $D(a_1,\dots,u,v,\dots,a_{b+1-i})$ is a summand of $P_{i-1}(a,b)$ with $u$ located in position $j$. Let $ \triangle_{u,v} : D(a_j) \rightarrow D(u,v)$ be 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 $Hom(A,B)$ and $Ext^i(A,B)$ in place of $Hom_{S(n,r)}(A,B)$ and $Ext^i_{S(n,r)}(A,B)$ respectively.

We recall the recursions  \begin{align*}
&P_0(a,b)=D(a) \otimes P_0(1,b-1), \\ &P_i(a,b)=P_{i-1}(a+1,b-1) \oplus D(a) \otimes P_i(1,b-1), i>0 \end{align*}
and that under these identifications we have the following.
	\begin{rem} The differential $Hom(\Theta_i(a,b), M)$ of the complex \\ $Hom(P_*(a,b),M)$ is given by
		\begin{center}
		\begin{tikzcd}
			Hom(P_{i-2}(a+1,b-1),M) \arrow{r}  \arrow[d, phantom, "\oplus"]
			& Hom(P_{i-1}(a+1,b-1),M)\arrow[d, phantom, "\oplus"]\\
			Hom(D(a) \otimes P_{i-1}(1,b-1),M)\arrow{r} \urar[shorten >= 25pt,shorten <= 25pt]{}
			&Hom(D(a) \otimes P_i(1,b-1),M)
		\end{tikzcd}
	\end{center}
		where the top horizontal map is $Hom(\Theta_{i-1}(a+1,b-1),M)$, the bottom one is $-Hom(1 \otimes  \Theta _i(1,b-1),M)$ and the restriction of the diagonal one on the summand $Hom(D(a,j,a_2,\dots,a_m),M)$  is $Hom(\triangle_{a,j} \otimes 1 \otimes \cdots \otimes 1), M) .$
	\end{rem}
We now consider weight spaces. For any $S(n,r)$-module $M$ and any sequence $a_1,\dots,a_m$ of nonnegative integers such that $a_1+\cdots +a_m=r$ and $m \le n$,  we identify the $\mathbb{Z}$-module $Hom(D(a_1,\dots,a_m),M)$ with the $(a_1,\dots,a_m)$ weight subspace of $M$ (with respect to the action of $\mathbb{Z}^n$) \cite{AB}. We will use such identifications freely throughout this paper.
	In particular, suppose  $M$ is a skew Weyl module for $S(n,r)$ (denoted be $K_{\lambda / \mu}$ in \cite{ABW}). Using the $\mathbb{Z}$-basis of $M$  given by standard tableaux of \cite{ABW}, Thm. II.3.16, we see that the $\mathbb{Z}$-module $Hom(D(a_1,\dots,a_m),M)$ may be identified with the $\mathbb{Z}$-submodule of $M$ that has basis the standard tableaux of $M$ that contain the entry $i$ exactly $a_i$ times, $i\in\{1,\dots,m\}$. We call this the standard basis of $Hom(D(a_1,\dots,a_m),M)$. If $a$ is a non negative integer, we denote by $M_a$ the $\mathbb{Z}$-submodule of $M$ given by
	\[M_a=\sum Hom(D(a,a_2,\dots,a_m),M)\]
	where the sum ranges over all nonnegative integers $a_2,\dots,a_m$ such that $a_2+\cdots+a_m$ $=r-a$ and $m \le n$. A  $\mathbb{Z}$- basis of this is the set of standard tableaux of $M$ in which the entry 1 appears exactly $a$ times. $M_a$ is a $GL_{n-1}$ submodule of $M$ with $GL_{n-1}$ acting on the basis elements $2,\dots,n$.

From the definition of $M_a$ it follows that we have the identification $$Hom_A(D(a_1,\dots,a_m),M)=Hom_B(D(a_2,\dots,a_m),M_{a_1})$$ and thus by summing,  $Hom_A(D(a) \otimes P_i(1,b-1),M)=Hom_B(P_i(1,b-1),M_{a}),$ where $a=a_1$, $A=S(n,r), B=S(n-1,b)$. We refer to the standard basis of $Hom_A(D(a_1,\dots,a_m),M)$ as the standard basis of $Hom_B(D(a_2,\dots,a_m),M_a)$.

Under the above identifications it follows from the definition of the complexes, that the maps
	\begin{align*}
	&Hom(1 \otimes  \Theta _i(1,b-1),M): \\
	&Hom_A(D(a) \otimes P_{i-1}(1,b-1),M) \rightarrow Hom_A(D(a) \otimes P_i(1,b-1),M), \\
	&Hom(\Theta _i(1,b-1),M_a): \\
	&Hom_B(P_{i-1}(1,b-1),M_a) \rightarrow Hom_B(P_i(1,b-1),M_a) \end{align*}
	are equal. From now on we will drop from the above $Hom$ modules the subscripts $A,B$.
	\subsection{Additional Preliminaries.}
	We want to consider matrices of the differential of the complex $Hom(P_*(a,b),M)$, where $M$ is a skew Weyl module.

	We consider the usual lexicographic ordering of the elements of the standard basis of $Hom(D(a_1,\dots,a_m),M)$ $=Hom(D(a_2,\dots,a_m),M_{a_1})$. We identify the sequence $(a_1,\dots,a_m)$ of nonnegative integers, where $m \le n$,  with the sequence  $(a_1,\dots,a_m,0,\dots,0) \in \mathbb{Z}^n $. Now if  $(a_1,\dots,a_m)$ is greater than   $(b_1,\dots,b_{m'})$ , where $m,m' \le n$, in the usual lexicographic ordering of sequences, me declare that each element of the standard basis of $Hom(D(a_1,\dots,a_m),M)$ is less than each element of the standard basis of $Hom(D(b_1,\dots,b_{m'}),M)$.

	For $M$ a skew Weyl module, let $E^{i}(\Delta(h),M)$ be the cokernel of the differential $Hom(\Theta_i (a,b),M)$ of the complex $Hom(P_{*}(a,b),M)$. We know that the torsion part of this abelian group is isomorphic to $Ext^{i}(\Delta(h),M)$ \cite{AB}.

Let $e^{(i)}(a,b,M)$ be the matrix of the differential $Hom(\Theta_i(a,b),M)$ with respect to the standard bases for the various spaces $Hom(D(a_1,\dots,a_m),M)$ and the above ordering. Likewise, let $e^{(i)}(1,b-1,M_a)$ be the matrix of the differential $Hom(\Theta_i(1,b-1),M_a)$ with respect to the standard bases for the various $Hom(D(a_2,\dots,a_m),M_a)$. From the previous discussion and Remark 2.1 we get the following statement.
	\begin{lm} \label{canonicalmatrix}
		Suppose $n \ge b+1$ and $M$ is a skew Weyl module for $S(n,r)$. Then for $i=1$
		\[
		e^{(1)}(a,b,M)=
		\left(
		\begin{array}{c}
		B^1(a,b,M)\\
		\hline \;
		-e^{(1)}(1,b-1,M_a)
		\end{array}
		\right)
		\]
		and for $i>1$
		\[
		e^{(i)}(a,b,M)=
		\left(
		\begin{array}{c|c}
		e^{(i-1)}(a+1,b-1,M) &B^i(a,b,M)\\
		\hline \;
		& -e^{(i)}(1,b-1,M_a)
		\end{array}
		\right),
		\]
		where in both cases $B^i(a,b,M)$ is the matrix of  the diagonal map $Hom(D(a) \otimes P_{i-1}(1,b-1),M) \rightarrow Hom(P_{i-1}(a+1,b-1),M)$ of Remark 2.1.
	\end{lm}
	We will also need a lemma for the special case $M=D_{a+k}\otimes\Lambda^{b-k}$ which we now describe. We will need a different order of the basis elements.

	Let $B_i$ be the standard basis of $Hom(P_{i-1}(a+1,b-1,D_{a+k}\otimes\Lambda^{b-k}))$, $B_{i,1}$ the subset of the standard basis of $Hom(D(a) \otimes P_{i}(1,b-1),D_{a+k}\otimes\Lambda^{b-k}))$ consisting of the standard tableaux with one 1 in the $\Lambda^{b-d}$ part and $B_{i,0}$ the subset of the standard basis of $Hom(D(a) \otimes P_{i}(1,b-1),D_{a+k}\otimes\Lambda^{b-k}))$ consisting of the standard tableaux with no 1 in the $\Lambda^{b-k}$ part. We order each of the sets $B_i, B_{i,1}, B_{i,0}$ lexicographically and we declare that the elements of $B_i$ are less than the elements of $B_{i,1}$, and the elements of $B_{i,1}$ are less than the elements of $B_{i,0.}$ Hence we have a total order on the basis $B_{i-1} \cup B_{i-1,1} \cup B_{i-1,0}$ of the domain of $Hom(\Theta_i(a,b),D_{a+d}\otimes\Lambda^{b-k})$ and a total order on the basis $B_{i} \cup B_{i,1} \cup B_{i,0}$ of the codomain of $Hom(\Theta_i(a,b),D_{a+k}\otimes\Lambda^{b-k})$.
	\begin{lm} \label{skewmatrix}
		With respect to the above ordered bases and for $i \ge2$,
		\[e^{(i)}(\Delta(h), D_{a+k}\otimes\Lambda^{b-k})=
		\left(
		\begin{array}{c|c|c}
		A &*&*\\
		\hline \;

		&B& \\
		\hline
		&&C
		\end{array}
		\right) ,
		\]
		where \begin{align*}
		A &=e^{(i-1)}(\Delta(a+1,1^{b-1})
		,D_{a+k}\otimes\Lambda^{b-k}),\\ B&=-e^{(i)}(\Delta(1,1^{b-1}),D_{1+k}\otimes\Lambda^{b-k-1}),\\ C&=-e^{(i)}(\Delta(1,1^{b-1}),D_{k}\otimes\Lambda^{b-k}). \end{align*}
	\end{lm}
	\begin{proof}The claim for the matrix $A$ follows from Lemma 2.2. For $B$ consider the diagram of abelian groups
	\begin{center}
	\begin{tikzcd}
		spanB_{i-1,1} \arrow{r}{a} \arrow[d, "f_{i-1}"]
	&spanB_{i,1}\arrow[d, "f_i"]\\
Hom(P_{i-1}(1,b-1),D_{1+k}\otimes\Lambda^{b-k-1})	\arrow{r}{\beta}
	&Hom(P_{i}(1,b-1),D_{1+k}\otimes\Lambda^{b-k-1})
	\end{tikzcd}
\end{center}
where we regard the modules in the bottom row as $GL_{n-1}$ modules  with $GL_{n-1}$ acting on the basis elements $1,\dots,n-1$. Here,  $f_{i-1}$ is the isomorphism of abelian groups sending each standard basis element $1^{(a-1)}i_2^{(c_2)}\cdots i_q^{(c_q)}\otimes 1j_2\cdots j_{b-k} \in B_{i-1,1}$ to the standard basis element $(i_2-1)^{(c_2)}\cdots (i_q-1)^{(c_q)}\otimes (j_2-1)\cdots (j_{b-k}-1) \in Hom(P_{i-1}(1,b-1),D_{1+k}\otimes\Lambda^{b-k-1})$ and similarly for $f_i$, $\alpha$ is the restriction of $Hom(\Theta_i(a,b),D_{a+k}\otimes\Lambda^{b-k})$ to $spanB_{i-1,1}$ and $\beta = Hom(\Theta_i(1,b-1),D_{1+k}\otimes\Lambda^{b-k-1})$.We claim that the diagram commutes. Using Remark 2.1 and Remark 2.2 it suffices to show that $f_i\circ \theta_s = \theta_{s-1} \circ f_{i-1}$, $t \ge 2$. Indeed, if\[ T=1^{(a-1)}i_2^{(c_2)}\cdots i_q^{(c_q)}\otimes 1j_2\cdots j_{b-k} \in B_{i-1,1},
	\]then \[ \theta_s(T) = 1^{(a-1)}i_2^{(c_2)}\cdots i_u^{(c_u)}{i^{'}_{u+1}}^{(c_{u+1})}\cdots {i^{'}_{q}}^{(c_{q})}\otimes 1j_2\cdots j_vj^{'}_{v+1}\cdots j^{'}_{b-k},
	\]
	where $i_u=max(\{i_2,\dots,i_q\} \cap (\{2,\dots,s\})$, $j_v=max(\{j_2,\dots,i_{b-k}\} \cap (\{2,\dots,s\})$,
	and $i^{'}=i-1$. We note that $\theta_s(T)$ is either 0 (if $j_v=j^{'}_{v+1}$) or a standard basis element (if $j_v \neq j^{'}_{v+1}$ and $i_u \neq i^{'}_{u+1}$ ) or a non-zero multiple of a standard basis element (if $j_v \neq j^{'}_{v+1}$ and $i_u = i^{'}_{u+1}$). Hence, in all cases we have
	\begin{align*}
	 f_i \circ \theta_s(T) =&  (i_2-1)^{(c_2)}\cdots(i_u-1)^{(c_u)}({i^{'}_{u+1}-1})^{(c_{u+1})}\cdots({i^{'}_{q}-1})^{(c_{q})}\otimes (j_2-1)\\
	 &\cdots(j_v-1)(j^{'}_{v+1}-1)\cdots(j^{'}_{b-k}-1).
	\end{align*}
	Similarly, one verifies that
	\begin{align*}
	\theta_{s-1} \circ f_{i-1}  (T) =&  (i_2-1)^{(c_2)}\cdots(i_u-1)^{(c_u)}({i^{'}_{u+1}-1})^{(c_{u+1})}\cdots({i^{'}_{q}-1})^{(c_{q})}\otimes (j_2-1)\\
    &\cdots(j_v-1)(j^{'}_{v+1}-1)\cdots(j^{'}_{b-k}-1)
	\end{align*}
and thus the diagram commutes. The proof for $C$ is similar and thus omitted. Finally, the middle block in the right column block is indeed 0, since for every $T \in B_{i-1,0}$, and every $s \ge 2$ we have $\theta_s(T) $ is of the form $x\otimes j_1\dots j_{b-k}$ with $1 \notin \{ j_1,\dots,j_{b-k} \}$. \end{proof}
\noindent \textit{Remark.} We will apply many times the isomorphism $f_2$ of the previous proof in  section 4 in order to describe specific relations of $E^{2}(\Delta(h),D_{a+k}\otimes \Lambda^{b-k})$.

The next lemma will be used several times. The first equality follows from the main result of \cite{Ku1}.
	\begin{lm} \label{basiclemma}
		Suppose $n \ge {b+1}$ and $0 \le k < b$. Then $$Ext^i(\Delta (h),D_{a+k} \otimes \Lambda ^{b-k})=Ext^i(\Lambda ^{k+1}, D_{k+1}).$$ In particular,
		$\;$ \begin{enumerate}
		\item[{\rm (a)}] $Ext^1(\Delta (h),D_{a+k} \otimes \Lambda ^{b-k})= \mathbb{Z}_2, \;  k \ge 1,$
		\item[{\rm (b)}] $Ext^2(\Delta (h),D_{a+k} \otimes \Lambda ^{b-k})= \begin{cases}\mathbb{Z}_3, \;  k=2,3   \\ 0, \; \;\; k \ne 2,3.\end{cases}$ \end{enumerate}
	\end{lm}
	\begin{proof} The first statement and (1) are Lemma 2.3 of \cite{MS}. In \cite{To}, Ex. 11.9, $Ext^2(\Lambda ^{k+1}, D_{k+1})$ was determined and hence we have (2). We give below a different proof of Touz\'e's result for the sake of self completeness.

We calculate $Ext^2(\Lambda ^{k+1}, D_{k+1})$ using the projective resolution $P_*(1,k)$ of $\Lambda ^{k+1}$ and the method of \cite{AB}, Section 9. If $k=0,1$, the length $P_*(1,k)$ is less than 2 and hence $Ext^2(\Lambda ^{k+1}, D_{k+1})=0$. Suppose $k \ge 2$. According to Remark 2.1, the map $Hom( \Theta_2(1,k),D_{k+1})$ looks like:
	\begin{center}
	\begin{tikzcd}
	Hom(P_0(2,k-1),D_{k+1}) \arrow{r}  \arrow[d, phantom, "\oplus"]
	& Hom(P_1(2,k-1),D_{k+1})\arrow[d, phantom, "\oplus"]\\
	Hom(D(1) \otimes P_1(1,k-1),D_{k+1})\arrow{r} \urar[shorten >= 25pt,shorten <= 25pt]{}
	&Hom(D(1) \otimes P_2(1,k-1),D_{k+1})
	\end{tikzcd}
\end{center}
	If $a_1,\dots,a_m$ are positive integers such that $a_1+\cdots+a_m=k+1,$ then we have $Hom(D(a_1,\dots,a_m),D_{k+1})= \mathbb{Z}$,
	a generator is simply the multiplication map, which will be denoted by $1^{(a_1)}\cdots m^{(a_m)}$. Thus,  the map $$Hom(D(a_1,\dots,u,v,\dots,a_m),D_{k+1}) \rightarrow Hom(D(a_1,\dots,a_t,\dots,a_m),D_{k+1}),$$ where $u+v=a_t$, induced by $1 \otimes \cdots \otimes \triangle_{u,v} \otimes \cdots \otimes 1: D(a_1,\dots,a_t,\dots,a_m) \rightarrow D(a_1,\dots,u,v,\dots,a_m)$ is multiplication by the binomial coefficient $ \binom{a_t}{u}$.

	We have the following ordered $\mathbb{Z}$ - bases.
	\begin{itemize}\item $Hom(P_0(2,k-1),D_{k+1}) : 1^{(2)}2\cdots k,$
	\item $Hom(D(1) \otimes P_1(1,k-1),D_{k+1}) :12^{(2)}3\cdots k, 123^{(2)}\cdots k, \dots,12\cdots k^{(2)},$
	\item $Hom(P_1(2,k-1),D_{k+1}) : 1^{(3)}2 \cdots (k-1), 1^{(2)}2^{(2)}4 \cdots (k-1),$\\
	$1^{(2)}23^{(2)}4 \cdots (k-1),\dots ,1^{(2)}23\cdots (k-1)^{(2)}.$
	\item $Hom(D(1) \otimes P_2(1,k-1),D_{k+1}) : 12^{(3)}3 \cdots (k-1), 12^{(2)}3^{(2)}4 \cdots (k-1),$\\
	$\dots, 12^{(2)}\cdots (k-1)^{(2)}, \dots ,123^{(3)}\cdots (k-1), 123^{(2)}4^{(2)}\cdots (k-1),$\\
	$\dots,123^{(2)}\cdots (k-1)^{(2)}, \dots , 12\cdots (k-1)^{(3)}.$
	\end{itemize}
	Now with the above ordered bases, a quick computation shows that the matrix of the top horizontal arrow of $Hom( \Theta_2(1,k),D_{k+1})$ is the $(k-1) \times 1$ matrix $$e^{(1)}(2,k-1,D_{k+1}) = (3 \; \; -2 \; \; \; 2 \; \; \; \dots \; \; (-1)^{k-2}2)^t$$ and the matrix of the diagonal arrow is the $(k-1) \times (k-1)$ matrix $$B^2(1,k,D_{k+1})=diag(3,2,\dots,2).$$ Also, it follows that the matrix  $e^{(2)}(1,k,D_{k+1})$ is of size $ \binom{k}{2} \times k$. In the notation of Lemma 2.2, we have $M=D_{k+1}$ and thus $M_1=D_k$. Therefore the recursion for $e^{(2)}(1,k,D_{k+1})$ is
	\[
	e^{(2)}(1,k,D_{k+1})=
	\left(
	\begin{array}{c|c}
	e^{(1)}(2,k-1,D_{k+1})  &B^2(1,k,D_{k+1})\\
	\hline \;
	& -e^{(2)}(1,k-1,D_k)
	\end{array}
	\right)
	\]
	where $k \ge 2$ and $e^{(2)}(1,2,D_3)=(3 \; \; 3)$ (this follows by an immediate computation). Hence, \[e^{(2)}(1,3,D_4)=  \begin{pmatrix}
	3 &3 & \\
	-2 & & 2 \\
	&-3 &-3
	\end{pmatrix}.\] The nonzero invariant factors of these last two matrices are 3 and 1,3 respectively and therefore $Ext^2( \Lambda ^3,D_3)=Ext^2( \Lambda ^4,D_4)= \mathbb{Z}_3$.

	Let $k \ge 4$. In order to show that $Ext^2(\Lambda ^{k+1}, D_{k+1})=0$, it suffices to show that the last nonzero invariant factor of $e^{(2)}(1,k,D_{k+1})$ is equal to 1. This matrix has rank at most $k-1$ since the matrix product $e^{(2)}(1,k,D_{k+1}) \cdot e^{(1)}(1,k,D_{k+1}) $ is zero, due to the fact that $Hom(P_*(1,k),D_{k+1})$ is a complex. (We have $e^{(1)}(1,k,D_{k+1}) \ne 0 $ since $Ext^1(\Lambda^{k+1},D_{k+1}) \ne 0$. In fact here  $e^{(1)}(1,k,D_{k+1})$ is the $k \times 1$ matrix $(2 \; \; -2 \; \; \; 2 \; \; \; \cdots \; \; (-1)^{k-1}2)^t)$.  It is straightforward to verify that the minor of $e^{(2)}(1,k,D_{k+1})$ corresponding to columns $2,3,\dots,k$ and rows $2,3,\dots,k-1, k+1$ is equal to $\pm 2^{k-1}$ and an immediate induction shows that the minor corresponding to columns $1,2,\dots,k-1$ and rows $1, 1+(k-1), 1+(k-1)+(k-2),\dots, \binom{k}{2}$ is equal to $\pm 3^{k-1}$. Since there exist two relatively prime minors of size the rank of the matrix $e^{(2)}(1,k,D_{k+1})$, the last nonzero invariant factor of this matrix is equal to 1.
	\end{proof}

	Next we describe the differential of $Hom(\Theta_i (a,b),M)$ for $M$ a skew Weyl module. For $T \in Hom(D(a_1,\dots,a_m),M)$ a standard basis element, let $\theta_s (T)$, $1 \le s \le m$, be the element of $Hom(D(a_1,\dots,a_s+a_{s+1},\dots,a_m),M)$ obtained from $T$ by replacing each occurrence of $j>s$ by $j-1$. If $s>m$, let $\theta_s (T)=0$. By extending linearly, we obtain for each degree $i$ a map of $\mathbb{Z}$-modules $Hom(P_i(a,b),M) \rightarrow Hom(P_{i+1}(a,b),M)$ which is denoted by $\Theta_i$. It is clear that only a finite number of these maps are nonzero. From the definition of the differential of $P_*(a,b)$,	with the previous notation, we obtain the following fact.
	\begin{rem}
	 $Hom(\Theta_i(a,b),M) = \sum_{s \ge 1}(-1)^{s-1}\theta_s.$
	\end{rem}

    \section{The cases $k=3$ and $k\ge 5$}
	It is well known, for example by \cite{Ja}, B.3 (4), and \cite{Ma} respectively, that  $Ext^2(\Delta(h),\Delta(h))=$ $Ext^2(\Delta(h),\Delta(h(1)))=0$. Theorem 4.1  of \cite{MS} implies that $Ext^2(\Delta(h),h(2)))=\mathbb{Z}_{(a+b)/gcd(2,a+b)}$.

	In this Section we will determine the extension groups $Ext^2(\Delta(h),\Delta(h(k)))$ for $k=3$ and $k\geq 5$. We will use certain exact sequences that we now describe.

Consider the short exact sequence
	\begin{equation} \label{s1}
	0\rightarrow\Delta(h(k+1))\xrightarrow{i_k} D_{a+k}\otimes \Lambda^{a-k}\xrightarrow{\pi_k} \Delta(h(k))\rightarrow 0
	\end{equation}
	for every $k\in\mathbb{Z}_{>0}$, where $\pi_k$ is induced by the identity map on generators and $i_k$ is induced by the composition $$D_{a+k+1}\otimes \Lambda^{b-1} \to D_{a+k}\otimes D_1 \otimes \Lambda^{b-1} \to D_{a+k}\otimes \Lambda^{b}$$ of comultiplication in $ D_* $ and multiplication in  $ \Lambda^*$.

	We have $Hom_{S_{\mathbb{Q}}(n,r)}(\Delta(h),\Delta(h(k)))=0$, since $\Delta(h)$ and $\Delta(h(k))$ are distinct irreducible representations of $S_{\mathbb{Q}}(n,r)$. Thus, $Hom(\Delta(h),\Delta(h(k)))=0$. Applying then $Hom(\Delta(h),-)$ to (\ref{s1}) we obtain the exact sequence
	\begin{align}\label{long1}
	0&\longrightarrow Ext^1(\Delta(h),\Delta(h(k+1)))\xrightarrow{i_k^{(1)}}
	Ext^1(\Delta(h),D_{a+k}\otimes\Lambda^{b-k})\nonumber \\
	&\xrightarrow{\pi_k^{(1)}}Ext^1(\Delta(h),\Delta(h(k))) \longrightarrow Ext^2(\Delta(h),\Delta(h(k+1)))\nonumber\\
	&\xrightarrow{i_k^{(2)}}Ext^2(\Delta(h),D_{a+k}\otimes\Lambda^{b-k})
	\xrightarrow{\pi_k^{(2)}} Ext^2(\Delta(h),\Delta(h(k))) \nonumber \\
	&\longrightarrow Ext^3(\Delta(h),\Delta(h(k+1)))\xrightarrow{i_k^{(3)}}Ext^3(\Delta(h),D_{a+k}\otimes\Lambda^{b-k})
	\end{align}
	Using Theorem 3.5 of \cite{MS} and Lemma \ref{basiclemma}, we have the following remark.
	\begin{rem} $\;$\begin{enumerate}
	\item[{\rm (a)}] 	$Ext^1(\Delta(h),\Delta(h(k+1)))=	Ext^1(\Delta(h),D_{a+k}\otimes\Lambda^{b-k})=\mathbb{Z}_2$
	and \\
	 $Ext^1(\Delta(h),\Delta(h(k)))=0$, or
	\item[{\rm (b)}] $Ext^1(\Delta(h),D_{a+k}\otimes\Lambda^{b-k})=Ext^1(\Delta(h),\Delta(h(k)))=\mathbb{Z}_2$
	and \\
	$Ext^1(\Delta(h),\Delta(h(k+1)))=0$.
\end{enumerate}
 \end{rem}
    Thus (\ref{long1}) takes the following form
	\begin{align}\label{long2}
	&0\rightarrow Ext^2(\Delta(h),\Delta(h(k+1)))\xrightarrow{i_k^{(2)}}Ext^2(\Delta(h),D_{a+k}\otimes\Lambda^{b-k})\nonumber \\
	&\xrightarrow{\pi_k^{(2)}}Ext^2(\Delta(h),\Delta(h(k)))\longrightarrow Ext^3(\Delta(h),\Delta(h(k+1)))\nonumber \\
	&\xrightarrow{i_k^{(3)}} Ext^3(\Delta(h),D_{a+k}\otimes\Lambda^{b-k})
	\end{align}

	Then, Theorem 4.1 by \cite{MS} and (3) yield the case $k=3$ of the theorem.
	\begin{prop}
	We have	$Ext^2(\Delta(h),\Delta(h(3)))=\mathbb{Z}_{3/gcd(3,a+b)}.$
	\end{prop}
	\begin{proof}We observe that $Ext^3(\Delta(h),D_{a+2}\otimes\Lambda^{b-2})=0$, as $Ext^3(\Delta(h),D_{a+2}\otimes\Lambda^{b-2})$ $=Ext^3(\Lambda^3, D_3)$, by Lemma 2.4, and $Ext^3(\Lambda^3, D_3)=0$, as $\Lambda^3$ has a projective resolution of length 2 by \cite{A}. So, for $k=2$, (\ref{long2}) has the following form
	\begin{align}\label{glyko}
	&0\rightarrow Ext^2(\Delta(h),\Delta(h(3)))\xrightarrow{i_2^{(2)}}Ext^2(\Delta(h),D_{a+2}\otimes\Lambda^{b-2})\nonumber \\
	&\xrightarrow{\pi_2^{(2)}}Ext^2(\Delta(h),\Delta(h(2)))\longrightarrow Ext^3(\Delta(h),\Delta(h(3)))\xrightarrow{i_2^{(3)}} 0.
	\end{align}
	Using again Lemma 2.4 we have $Ext^2(\Delta(h),D_{a+2}\otimes\Lambda^{b-2})=\mathbb{Z}_3$, which implies that $Ext^2(\Delta(h),\Delta(h(3)))=\mathbb{Z}_3$ or $0$, as $i_2^{(2)}$ is a monomorphism.
	We also know that $Ext^2(\Delta(h),\Delta(h(2)))=\mathbb{Z}_{d_2}$ and $Ext^3(\Delta(h),\Delta(h(3)))=\mathbb{Z}_{d_3}$, by Theorem 4.1 of \cite{MS}, where $d_2=gcd(a+b,\tbinom{a+b}{2})$ and $d_3=gcd(a+b,\tbinom{a+b}{2},\tbinom{a+b}{3})$.	Simple calculations yield that $d_2=d_3$ if and only if $3\notdivides a+b$, so the exact sequence (\ref{glyko}) implies that $i_2^{(2)}$ is an isomorphism if and only if $3\notdivides a+b$. It follows that $Ext^2(\Delta(h),\Delta(h(3)))=\mathbb{Z}_3$ if and only if $3\notdivides a+b$ and $Ext^2(\Delta(h),\Delta(h(3)))=0$ otherwise.
	\end{proof}

	Lemma \ref{basiclemma} (b) also yields that $Ext^2(\Delta(h),D_{a+k-1}\otimes \Lambda^{b-k+1})=0$ for every $k\geq5$ and using again the exact sequence (\ref{long2}) we obtain the following result.
	\begin{prop}
		We have	$Ext^2(\Delta(h),\Delta(h(k)))=0$ for every $5\leq k\leq b$.
	\end{prop}

	\section{The case $k=4$ of the Theorem}
	In this Section we show Theorem 1.1 for $k=4$, which constitutes the main part of the paper. This is done using a refinement of the main strategy of \cite{MS}, namely by first determining a cyclic generator of $Ext^2(\Delta(h),D_{a+3}\otimes \Lambda^{b-3})$, Subsection 4.1, and then computing its image in $Ext^2(\Delta(h),D_{a+2}\otimes \Lambda^{b-2})$, Subsection 4.3, with the aid of the relations that we establish in Subsection 4.2.

	In order to compute $Ext^2(\Delta(h),\Delta(h(4)))$, we use the exact sequence (\ref{long2}) for $k=3$, which takes the following form
	\begin{align*}
	&0\longrightarrow
	Ext^2(\Delta(h),\Delta(h(4)))
	\xrightarrow{i_3^{(2)}}Ext^2(\Delta(h),D_{a+3}\otimes \Lambda^{b-3})\\
	&\xrightarrow{\pi_3^{(2)}}
	Ext^2(\Delta(h),\Delta(h(3))).
	\end{align*}
	By Lemma \ref{basiclemma} (b) we have $Ext^2(\Delta(h),D_{a+3}\otimes \Lambda^{b-3})=\mathbb{Z}_3$.
	In order to determine whether $Ext^2(\Delta(h),\Delta(h(4)))=0$ or $\mathbb{Z}_3$, we consider the following composition
	\begin{equation*}
	Ext^2(\Delta(h),D_{a+3}\otimes \Lambda^{b-3})\xrightarrow{\pi_3^{(2)}} Ext^2(\Delta(h),\Delta(h(3)))\xrightarrow{i_2^{(2)}} Ext^2(\Delta(h),D_{a+2}\otimes \Lambda^{b-2}).
	\end{equation*}
	Let $\phi=i_2^{(2)}\circ\pi_3^{(2)}$. Then, $\phi $ is a map from $\mathbb{Z}_3$ to $\mathbb{Z}_3$ by Lemma \ref{basiclemma}.
	Our aim in this Section is to find a cyclic generator of $Ext^2(\Delta(h),D_{a+3}\otimes \Lambda^{b-3})$ and determine its image under $\phi$ as a multiple of the cyclic generator of $Ext^2(\Delta(h),D_{a+2}\otimes \Lambda^{b-2})$ described by \cite{MS} in Section 4.1.
	\subsection{Cyclic generator of $Ext^2(\Delta (h),D_{a+3}\otimes \Lambda^{b-3})$.}
	In this subsection we will determine a cyclic generator of the group $Ext^2(\Delta (h),D_{a+3}\otimes \Lambda^{b-3})$. This is accomplished using the recursion of Lemma 2.3 for $i=2$, according to which the matrix $e^{(2)}(\Delta(h),D_{a+3}\otimes \Lambda^{b-3}V)$ of the differential $Hom(\Theta_2(a,b),D_{a+3}\otimes\Lambda^{b-3})$ has the following form
		\[
	\left(
	\begin{array}{c|c|c}
	A &*&*\\
	\hline \;

	&B& \\
	\hline
	&&C
	\end{array}
	\right) ,
	\]
	where $A=e^{(1)}(\Delta(a+1,1^{b-1})
	,D_{a+3}\otimes\Lambda^{b-3})$, $B=-e^{(2)}(\Delta(1,1^{b-1}),D_{4}\otimes\Lambda^{b-4})$ and $C=-e^{(2)}(\Delta(1,1^{b-1}),D_{3}\otimes\Lambda^{b-3})$.

	Consider the following standard basis elements:
	\begin{itemize}
	\item $T_{i,j}^1=1^{(a)}i^{(2)}j\otimes 2\cdots \hat{i} \cdots \hat{j} \cdots b$ \\
	for $i\in\{2,\dots,b-1\}$, $j\in\{i+1,\dots,b\}$ \\
	in $Hom(D_a\otimes {D_1}^{\otimes (i-2)}\otimes D_2 \otimes {D_1}^{\otimes (b-i)}, D_{a+3}\otimes \Lambda^{b-3})$,
	\item $T_{i,j}^k=1^{(a-1)}ki^{(2)}j\otimes 1\cdots\hat{k}\cdots \hat{i} \cdots \hat{j} \cdots b$ \\
	for $k\in\{2,\dots,b-2\}$,
	$i\in\{k+1,\dots,b-1\}$, $j\in\{i+1,\dots,b\}$ \\
	in $Hom(D_a\otimes {D_1}^{\otimes (i-2)}\otimes D_2 \otimes {D_1}^{\otimes (b-i)},D_{a+3}\otimes \Lambda^{b-3})$,
	\end{itemize}
    and the standard basis elements:
    \begin{itemize}
    \item $B_{1,j}^{a,b}=1^{(a+2)}j\otimes 2\cdots \hat{j} \cdots (b-1)$ for $j\in\{2,\dots ,b-1\}$, \\
    in $Hom(D_{a+2}\otimes {D_1}^{\otimes (b-2)},D_{a+3}\otimes \Lambda^{b-3})$, ,
    \item $B_{i,1}^{a,b}=1^{(a)}i^{(3)}\otimes 2\cdots\hat{i}\cdots(b-1)$ for $i\in\{2,\dots,b-1\}$, \\
    in $Hom(D_a\otimes {D_1}^{\otimes (i-2)}\otimes D_3\otimes {D_1}^{\otimes (b-i-1)}, D_{a+3} \otimes \Lambda^{b-3})$,
    \item $B_{i,j}^{a,b}=1^{(a-1)}i^{(3)}j\otimes 1\cdots \widehat{\{i,j\}} \cdots (b-1)$  for $i\in\{2,\dots,b-1\}$ \\ and $j\in\{2,\dots,i-1\}\cup\{i+1,\dots,b-1\}$, \\
    in  $Hom(D_a\otimes {D_1}^{\otimes (i-2)}\otimes D_3\otimes {D_1}^{\otimes (b-i-1)}, D_{a+3} \otimes \Lambda^{b-3})$.
    \end{itemize}

Recall that the cokernel of the differential $Hom(\Theta_i (a,b),M)$ of the complex $Hom(P_{*} (a,b),M)$ is denoted by $E^i (\Delta(h),M)$ (cf. 2.4).

Let $\pi: Hom(P_i (a,b),M)\rightarrow E^{i}(\Delta(h),M)$ be the natural projection.

\smallskip
\noindent \textit{Remark.} In the following proposition and its proof, if the lower bound of the summation index is less than the upper bound, then we regard the sum as zero (empty sum).

\begin{prop} Let
\begin{align*}\Gamma_{a,b}=&\tbinom{a+2} {3}\sum_{j=2}^{b-1}{(-1)}^{j}B_{1,j}^{a,b} -\Big(aB_{2,1}^{a,b}+\sum_{j=3}^{b-1}{(-1)}^{j}B_{2,j}^{a,b}\Big)+ \\
&\sum_{i=3}^{b-2}{(-1)}^{i-1}\Big(aB_{i,1}^{a,b} -\sum_{j=2}^{i-1} {(-1)}^j B_{i,j}^{a,b}+\sum_{j=i+1}^{b-1} {(-1)}^{j} B_{i,j}^{a,b}  \Big)+\\
&{(-1)}^{b}\Big( aB_{b-1,1}^{a,b}-\sum_{j=2}^{b-2}{(-1)}^jB_{b-1,j}^{a,b}\Big).
\end{align*}
Then, for $b\geq 3$, $3\pi(\Gamma_{a,b})=0$ in $E^{2}(\Delta(h),D_{a+3}\otimes \Lambda^{b-3})$ and thus $\pi(\Gamma_{a,b})$ is an element of $Ext^2(\Delta(h),D_{a+3}\otimes \Lambda^{b-3})$.
\end{prop}

\begin{proof} Let \[A_{a,b}=a\sum_{i=2}^{b-1}\sum_{j=i+1}^{b} {(-1)}^{j-i+1}\sum_{s=1}^{b-1}{(-1)}^{s-1}\theta_s(T_{i,j}^1),\] \[C_{a,b}=\sum_{k=2}^{b-2} \sum_{i=k+1}^{b-1} \sum_{j=i+1}^{b} {(-1)}^{j-i-k} \sum_{s=1}^{b-1}{(-1)}^{s-1}\theta_s (T_{i,j}^k).\] We will use induction on $b\geq3$ to show that
	\begin{align}\label{eqnionia}
     A_{a,b}+C_{a,b}=3\Gamma_{a,b}
	\end{align}
	For $b=3$ we observe that $C_{a,3}=0$ as it is an empty sum, and for $T_{2,3}^1=1^{(a)}2^{(2)}3$ in $Hom(D_a\otimes D_2\otimes D_1,D_{a+3})$ we have that \begin{align*}&A_{a,3}+C_{a,3}=A_{a,b}=a(\theta_1(T_{2,3})-\theta_1(T_{2,3}))=a\big({\tbinom{a+2}{2}}\cdot1^{(a+2)}2-3\cdot 1^{(a)}2^{(3)}\big)\\
	&=3{\tbinom{a+2}{3}}\cdot B_{1,2}^{a,3}-3a\cdot B_{2,1}^{a,3}=3\Gamma_{a,3}.
	\end{align*}
    Let $b>3$. For $i=2$ and $j\in\{3,\dots,b\}$ we observe that
	$$\theta_1 (T_{2,j}^1)=\tbinom{a+2}{2} 1^{(a+2)}(j-1)\otimes 2\cdots \widehat{(j-1)} \cdots (b-1)=\tbinom{a+2}{2}B_{1,j-1}^{a,b},$$
	and so we have
	\begin{align} \label{a1}
	A_{a,b}&=a\sum_{i=2}^{b-1}\sum_{j=i+1}^{b} {(-1)}^{j-i+1}\theta_1(T_{i,j}^1) \nonumber \\
	&\,\,\,\,\,\,\,+a\sum_{i=2}^{b-1}\sum_{j=i+1}^{b} {(-1)}^{j-i+1}\sum_{s=2}^{b-1}{(-1)}^{s-1}\theta_s(T_{i,j}^1)\nonumber\\
	&=a\tbinom{a+2}{2}\sum_{j=3}^{b}{(-1)}^{j-1}B_{1,j-1}^{a,b}+
	a\sum_{i=3}^{b-1}\sum_{j=i+1}^{b} {(-1)}^{j-i+1}\theta_1(T_{i,j}^1)\nonumber \\
	&\,\,\,\,\,\,\,+a\sum_{i=2}^{b-1}\sum_{j=i+1}^{b} {(-1)}^{j-i+1}\sum_{s=2}^{b-1}{(-1)}^{s-1}\theta_s(T_{i,j}^1).
	\end{align}
	Consider now the standard basis elements:
	\begin{itemize}
	\item $t_{1,j-1}^0 = 1^{(2)}(j-1)\otimes 2\cdots \widehat{(j-1)} \cdots (b-1)$ in \\
	$Hom(D_2\otimes {D_1}^{\otimes (b-1)},D_3 \otimes \Lambda^{b-1})$, for $i=2$ and $j\in\{3,\dots,b\}$,
	\item $t_{i-1,j-1}^0={(i-1)}^{(2)}(j-1)\otimes 1\cdots \widehat{(i-1)} \cdots \widehat{(j-1)} \cdots (b-1)$ in \\
	$Hom({D_1}^{\otimes (i-2)}$ $\otimes D_2\otimes {D_1}^{\otimes (b+2-i)},D_3 \otimes \Lambda^{b-1})$, for $i\in\{3,\dots,b-1\}$ \\
	and $j\in\{i+1,\dots,b\}$.
	\end{itemize}
	We observe that $\theta_{s-1}(t^0_{i-1,j-1})=f_2(\theta_s(T^1_{i,j}))$ for every $i\in\{2,\dots,b-1\}$, $j\in\{i+1,\dots,b\}$ and $s\in\{2,\dots,b-1\}$, where $f_2$ is the isomorphism described in Lemma \ref{skewmatrix} for $i=2$. By \cite{MS}, Lemma 4.4 for $k=2$, we have the following relation in $Ext^2(\Delta(1,1^{(b-1)}),D_3 \otimes \Lambda^{b-3})$
	\begin{equation}\label{eqms}
	Hom(\Theta_2(a,b),D_{a+2}\otimes \Lambda^{b-2})(\sum_{i=2,j>i}^{b-1,b}{(-1)}^{j-i+1}t_{i-1,j-1}^0)=3\gamma_{1,b-1},
	\end{equation}
	where $\gamma_{1,b-1}=\sum_{i=2}^{b-1}{(-1)}^i\delta_{i-1}^{1,b-1}$,
	\begin{itemize}
		\item $\delta_1^{1,b-1}=1^{(3)}\otimes 2\cdots (b-2)$ in $Hom(D_3\otimes {D_1}^{\otimes (b-3)},D_3\otimes\Lambda^{b-3})$, for $i=2$,
		\item $\delta_{i-1}^{1,b-1}={(i-1)}^3\otimes 1\cdots \widehat{(i-1)} \cdots (b-2)$ in $Hom({D_1}^{\otimes (i-2)}\otimes D_3\otimes {D_1}^ {\otimes (b-i-1)},$ $D_3\otimes\Lambda^{b-3})$, for $i\in\{3,\dots,b-1\}$,
	\end{itemize}
and $\pi(\gamma_{1,b-1})$ is the generator of $Ext^2(\Delta(1,1^{b-1}),D_3\otimes \Lambda^{b-3})$ described by \cite{MS} in Subsection 4.1.
Using Remark 2.5, (\ref{eqms}) is equivalent to
	\begin{equation}\label{rel}
\sum_{i=2}^{b-1}\sum_{j=i+1}^{b} {(-1)}^{j-i+1}\sum_{s=2}^{b-1}{(-1)}^{s}\theta_{s-1}(t_{i-1,j-1}^0)              =3\sum_{i=2}^{b-1}{(-1)}^i\delta_{i-1}^{1,b-1}.
\end{equation}
By the proof of Lemma \ref{skewmatrix}, we have that $\delta_{i-1}^{1,b-1}=f_2(B_{i,1}^{a,b})$ where
\begin{itemize}
	\item $B_{2,1}^{a,b}=1^{(a)}2^{(3)}\otimes 3\cdots (b-1)$ for $i=2$,
	\item $B_{i,1}^{a,b}=1^{(a)}i^{(3)}\otimes 2\cdots \hat{i} \cdots (b-1)$ for $i\in\{3,\dots,b-1\}$.
\end{itemize}
and $f_2$ is the isomorphism described in Lemma \ref{skewmatrix} for $i=2$.

Applying the isomorphism $f_2^{-1}$ to Equation (\ref{rel}) we have,
\begin{equation*}
 \sum_{i=2}^{b-1}\sum_{j=i+1}^{b} {(-1)}^{j-i+1}\sum_{s=2}^{b-1}{(-1)}^{s}f_2^{-1}(\theta_{s-1}(t_{i-1,j-1}^0))              =3\sum_{i=2}^{b-1}{(-1)}^i f_2^{-1}(\delta_{i-1}^{1,b-1})
\end{equation*}
or, equivalently,
\begin{equation}\label{e14e}
 \sum_{i=2}^{b-1}\sum_{j=i+1}^{b} {(-1)}^{j-i+1}\sum_{s=2}^{b-1}{(-1)}^{s}\theta_s(T_{i,j}^1)             =3\sum_{i=2}^{b-1}{(-1)}^iB_{i,1}^{a,b}.
\end{equation}
Using (\ref{e14e}), (\ref{a1}) is equivalent to
\begin{align}\label{a4}
A_{a,b}&=a\tbinom{a+2}{2}\sum_{i=2}^{b-1}{(-1)}^{i}B_{1,i}^{a,b}+a\sum_{i=3}^{b-1}\sum_{j=i+1}^{b} {(-1)}^{j-i+1}\theta_1(T_{i,j}^1)\nonumber\\
&\,\,\,\,\,\,-3a\sum_{i=2}^{b-1}{(-1)}^i B_{i,1}^{a,b}.
\end{align}
We also have
\begin{align}\label{a8}
C_{a,b}&=\sum_{i=3}^{b-1} \sum_{j=i+1}^{b} {(-1)}^{j-i}\theta_1 (T_{i,j}^2) + \sum_{k=3}^{b-2} \sum_{i=k+1}^{b-1} \sum_{j=i+1}^{b} {(-1)}^{j-i-k}\theta_1 (T_{i,j}^k)\nonumber \\
&\,\,\,\,\,\,+\sum_{k=2}^{b-2} \sum_{i=k+1}^{b-1} \sum_{j=i+1}^{b} {(-1)}^{j-i-k} \sum_{s=2}^{b-1}{(-1)}^{s-1}\theta_s (T_{i,j}^k)\nonumber \\
&=\sum_{i=3}^{b-1} \sum_{j=i+1}^{b} {(-1)}^{j-i}\theta_1 (T_{i,j}^2) +\sum_{k=2}^{b-2} \sum_{i=k+1}^{b-1} \sum_{j=i+1}^{b} {(-1)}^{j-i-k} \sum_{s=2}^{b-1}{(-1)}^{s-1}\theta_s (T_{i,j}^k),
\end{align}
as $\theta_1 (T_{i,j}^k)=0$ for every $k\in\{3\dots,b-2\}$, because of the $12$ part of $T_{i,j}^k$ in $\Lambda^{b-3}$.
Consider now the basis elements:
\begin{itemize}
	\item $t_{i-1,j-1}^1=1{(i-1)}^{(2)}(j-1)\otimes 2\cdots\widehat{(i-1)}\cdots\widehat{(j-1)}\cdots (b-1)$ in \\
	$Hom({D_1}^{\otimes (i-2)}\otimes D_2 \otimes {D_1}^{\otimes (b-i)},D_4\otimes\Lambda^{b-4})$, for $k=2$ and $i\in\{3,\dots,b-1\}$, $j\in\{i+1,\dots,b\}$,
	\item $t_{i-1,j-1}^{k-1}=(k-1){(i-1)}^{(2)}(j-1)\otimes 1\cdots\widehat{(k-1)}\cdots\widehat{(i-1)}\cdots \widehat{(j-1)} \cdots (b-1)$ in $Hom({D_1}^{\otimes (i-2)}\otimes D_2 \otimes {D_1}^{\otimes (b-i)},D_4\otimes\Lambda^{b-4})$ , for $k\in\{3,\dots,b-2\}$, $i\in\{k+1,\dots,b-1\}$, $j\in\{i+1,\dots,b\}$.
\end{itemize}
 By induction on $b$, for $a=1$, we have  $A_{1,b-1}+C_{1,b-1}=3\Gamma_{1,b-1},$ or, equivalently,
\begin{align}\label{eq10}
&\sum_{k=2}^{b-2} \sum_{i=k+1}^{b-1} \sum_{j=i+1}^{b} {(-1)}^{j-i-k+1} \sum_{s=2}^{b-1}{(-1)}^{s}\theta_{s-1} (t_{i-1,j-1}^{k-1})\nonumber \\
&=3\sum_{j=3}^{b-1}{(-1)}^{j-1}B_{1,j-1}^{1,b-1} -3\Big(B_{2,1}^{1,b-1}+\sum_{j=4}^{b-1}{(-1)}^{j-1}B_{2,j-1}^{1,b-1}\Big)\nonumber \\
&\,\,\,\,\,\,+3\sum_{i=4}^{b-1}{(-1)}^{i}\Big(B_{i-1,1}^{1,b-1} -\sum_{j=3}^{i-1} {(-1)}^{j-1} B_{i-1,j-1}^{1,b-1}+\sum_{j=i+1}^{b-1} {(-1)}^{j-1} B_{i-1,j-1}^{1,b-1}  \Big)\nonumber \\
&\,\,\,\,\,\,+3{(-1)}^{b-1}\Big(B_{b-2,1}^{1,b-1}-\sum_{j=3}^{b-2}{(-1)}^{j-1}B_{b-2,j-1}^{1,b-1}\Big).
\end{align}
Applying $f_2^{-1}$ to (\ref{eq10}), using that $f_2(B_{i,j}^{a,b})=B_{i-1,j-1}^{1,b-1}$ and $f_2(\theta_s(T_{i,j})^k)=\theta_{s-1}(t_{i-1,j-1}^{k-1})$ for $i\geq 2$, $j\in\{3,\dots,i-1\}\cup\{i+1,\dots,b-2\}$, we obtain
\begin{align}\label{eq11}
&\sum_{k=2}^{b-2} \sum_{i=k+1}^{b-1} \sum_{j=i+1}^{b} {(-1)}^{j-i-k+1} \sum_{s=2}^{b-1}{(-1)}^{s}\theta_{s} (T_{i,j}^{k})\nonumber \\
&=3\sum_{j=3}^{b-1}{(-1)}^{j-1}B_{2,j}^{a,b} -3\Big(B_{3,2}^{a,b}+\sum_{j=4}^{b-1}{(-1)}^{j-1}B_{3,j}^{a,b}\Big)\nonumber \\
&\,\,\,\,\,\,+3\sum_{i=4}^{b-2}{(-1)}^{i}\Big(B_{i,2}^{a,b} -\sum_{j=3}^{i-1} {(-1)}^{j-1} B_{i,j}^{a,b}+\sum_{j=i+1}^{b-1} {(-1)}^{j-1} B_{i,j}^{a,b}  \Big)\nonumber \\
&\,\,\,\,\,\,+3{(-1)}^{b-1}\Big(B_{b-1,2}^{a,b}-\sum_{j=3}^{b-2}{(-1)}^{j-1}B_{b-1,j}^{a,b}\Big).
\end{align}
Equation (\ref{eq11}) implies that (\ref{a8}) is equivalent to
\begin{align}\label{eq12}
C_{a,b}&=\sum_{i=3}^{b-1} \sum_{j=i+1}^{b} {(-1)}^{j-i}\theta_1 (T_{i,j}^2) +3\sum_{j=3}^{b-1}{(-1)}^{j-1}B_{2,j}^{a,b} \nonumber \\
&\,\,\,\,\,\,-3\Big(B_{3,2}^{a,b}+\sum_{j=4}^{b-1}{(-1)}^{j-1}B_{3,j}^{a,b}\Big)\nonumber \\
&\,\,\,\,\,\,+3\sum_{i=3}^{b-2}{(-1)}^{i}\Big(B_{i,2}^{a,b} -\sum_{j=3}^{i-1} {(-1)}^{j-1} B_{i,j}^{a,b}+\sum_{j=i+1}^{b-1} {(-1)}^{j-1} B_{i,j}^{a,b}  \Big)\nonumber \\
&\,\,\,\,\,\,+3{(-1)}^{b-1}\Big(B_{b-1,2}^{a,b}-\sum_{j=3}^{b-2}{(-1)}^{j-1}B_{b-1,j}^{a,b}\Big).
\end{align}
We observe that $\theta_1(T_{i,j}^2)=a\theta_1(T_{i,j}^1)$ for $i\in\{3,\dots,b-1\}$, $j\in\{i+1,\dots,b\}$, as $\theta_1(T_{i,j}^1)=1^{(a)}{(i-1)}^2(j-1)\otimes 1\cdots\widehat{(i-1)}\cdots \widehat{(j-1)}\cdots (b-1)$, and
$\theta_1(T_{i,j}^2)=a1^{(a)}{(i-1)}^2(j-1)\otimes 1\cdots\widehat{(i-1)}\cdots \widehat{(j-1)}\cdots (b-1)$. Using this observation and adding equations  (\ref{a4}) and (\ref{eq12}), we obtain $A_{a,b}+C_{a,b}=3\Gamma_{a,b}$
and we have proved inductively equation (\ref{eqnionia}).

As we have the relations $\sum_{s=1}^{b-1}{(-1)}^{s-1}\pi(\theta_s(T_{i,j}^k))=0$ for every $k\in\{1,\dots,b-2\}$, $i\in\{k+1,\dots,b-1\}$, $j\in\{i+1,\dots,b\}$, it follows that $\pi(A_{a,b})=\pi(C_{a,b})=0$ and so $3\pi(\Gamma_{a,b})=0$.\end{proof}
\begin{prop}
The element $\pi(\Gamma_{a,b})$ of Proposition 3.1 is a cyclic generator of  $Ext^2(\Delta(h),D_{a+3}\otimes\Lambda^{b-3})$.
\end{prop}
\begin{proof}We will first show that $\pi(\Gamma_{a,b})$ is nonzero. Indeed, if $\pi(\Gamma_{a,b})=0$, then $\pi(\Gamma_{a,b})$ would be equal to a linear combination of the relations in the extension group $Ext^2(\Delta(h),D_{a+3}\otimes\Lambda^{b-3})$ corresponding to the columns of the matrix $e^{(2)}(\Delta(h),D_{a+3}\otimes\Lambda^{b-3})$. This implies that the coefficient $-1$ of $\pi(B_{b-1,b-2})$ in $\pi(\Gamma_{a,b})$ should be a linear combination of the nonzero entries of the last row of $e^{(2)}(\Delta(h),D_{a+3}\otimes\Lambda^{b-3})$. These are the coefficients of $\pi(B_{b-1,b-2})$ resulting from the application of $\pi\circ\theta_{b-1}$ on the standard basis elements: $$1^{(a-1)} (b-2) {(b-1)}^{(2)} b\otimes 1\cdots(b-3), \, 1^{(a-1)}(b-2)(b-1)b^{(2)}\otimes 1\cdots(b-3).$$ Indeed, let $\lambda=(\lambda_1,\lambda_2,\dots,\lambda_b)$ be a sequence with $\lambda_1\in\{0,1\},\,\lambda_i\in\{1,2\},\, \sum_{i=1}^{b}\lambda_i$ $=b$, and a standard basis element $T_{\lambda}$ in $Hom(D_{a+\lambda_1}\otimes D_{\lambda_2}\otimes \cdots \otimes  D_{\lambda_b}, D_{a+3}\otimes\Lambda^{b-3})$, such that $\theta_{b-1}(T_{\lambda})=c\cdot B_{b-1,b-2}^{a,b}$ for some $c\in\mathbb{Z}$. Then $\theta_{b-1}(T_{\lambda})$ has weight $(1^{a+\lambda_1},2^{\lambda_2},\dots,{(b-1)}^{\lambda_{b-1}+\lambda_b})$ which must be equal to $(1^{a},2,\dots,(b-2),{(b-1)}^3)$, the weight of $B_{b-1,b-2}^{a,b}$. This implies that $\lambda_1=0$, $(\lambda_{b-1},\lambda_{b})\in\{(1,2),(2,1)\}$ and $$T_{\lambda}\in\{1^{(a-1)} (b-2) {(b-1)}^{(2)} b\otimes 1\cdots(b-3), 1^{(a-1)}(b-2)(b-1)b^{(2)}\otimes 1\cdots(b-3)\}.$$ By straightforward calculations, we have,
	\begin{align*}
	&\pi(\theta_{b-1}(1^{(a-1)} (b-2) {(b-1)}^{(2)} b\otimes 1\cdots(b-3)))\\
	&=\pi(\theta_{b-1}(1^{(a-1)}(b-2)(b-1)b^{(2)}\otimes 1\cdots(b-3)))\\
	&=3 \pi(B_{b-1,b-2}^{a,b}),
	\end{align*}
	which means that $-1$ should be a linear combination of $3$ and $3$, i.e. multiple of $3$. This implies that $\pi(\Gamma_{a,b})$ is nonzero. As $Ext^2(\Delta(h),D_{a+3}\otimes\Lambda^{b-3})= \mathbb{Z}_3$, $3\pi(\Gamma_{a,b})=0$ (by Proposition 4.1) and $\pi(\Gamma_{a,b})\neq 0$, we conclude that $\pi(\Gamma_{a,b})$ is a generator of $Ext^2(\Delta(h),D_{a+3}\otimes\Lambda^{b-3})$. \end{proof}
\subsection{Relations in $Ext^2(\Delta(h),D_{a+2}\otimes \Lambda^{b-2})$.}
In this Subsection we will prove certain relations in $E^2(\Delta(h),D_{a+2}\otimes \Lambda^{b-2})$ using the columns of $e^{(2)}(\Delta(h), D_{a+2}\otimes \Lambda^{b-2})$, i.e. using the equalities $\sum_{s=1}^{b-1}{(-1)}^{s-1}\pi(\theta_s(T))=0$, where $T$ is a standard basis element of the domain of the differential $\Theta_2 (a,b)$. These relations will be described in Lemmas 4.3 and 4.4. They will yield Proposition 4.5 which will be used in the proof of Proposition 4.6, in order to facilitate the computation of the image of the cyclic generator of Proposition 4.2. Proposition 4.6 will then yield the case $k=4$ of Theorem 1.1.

Let  \begin{align*}{b_1}^{(1)}&=1^{(a+2)} \otimes 2\cdots (b-1),\\{b_j}^{(1)}&=1^{(a+1)}j \otimes 1\cdots \hat{j} \cdots (b-1)\end{align*}
be the indicated standard basis elements in $Hom(D_{a+2}\otimes {D_1}^{\otimes (b-2)}, D_{a+2} \otimes \Lambda^{b-2})$ and
\begin{align*}{b_1}^{(i)}=&1^{(a)}i^{(2)} \otimes 2 \cdots i \cdots (b-1),\\{b_j}^{(i)}=&1^{(a-1)} i^{(2)}j \otimes 12\cdots \hat{j} \cdots (b-1)\end{align*}
be the indicated standard basis elements in $Hom(D_a \otimes {D_1}^{\otimes (i-2)} \otimes D_3 \otimes {D_1}^{\otimes (b-i-1)},$ $ D_{a+2}\otimes \Lambda^{b-2})$ for $i\in\{2,\dots ,b-1\}$ and $j\in\{2,\dots,b-1\}$.
\begin{lm} \label{bl1} For every $i\in\{2,\dots,b-1\}$,
		\begin{equation*}
		a\pi(b_1^{(i+1)}) - \sum_{j=2}^{i}{(-1)}^j \pi(b_j^{(i+1)})= a\pi(b_1^{(i)})-\sum_{j=2}^{i-1}{(-1)}^{(j)}\pi(b_j^{(i)}) +3{(-1)}^{i+1}\pi(b_i^{(i)}).
		\end{equation*}
\end{lm}
\begin{proof} We consider the basis elements $T_1^{(i+1)}=1^{(a)}{(i+1)}^{(2)}\otimes 2\cdots \widehat{(i+1)} \cdots b$ and
	$T_j^{(i+1)}=1^{(a-1)}j{(i+1)}^{(2)}\otimes 1\cdots \hat{j}\cdots \widehat{(i+1)}\cdots b$, for $j\in\{2,\dots,i\}$. Let \[D_{a,b}=a\sum_{s=1}^{b-1} {(-1)}^{s-1} \theta_s(T_1^{(i+1)})+\sum_{j=2}^i {(-1)}^j\sum_{s=1}^{b-1} {(-1)}^{s-1} \theta_s(T_j^{(i+1)}).\]
	We know that $\sum_{s=1}^{b-1} {(-1)}^{s-1}\pi( \theta_s(T_j^{(i+1)}))=0$, for $j\in\{1,\dots,i\}$, which implies that $\pi(D_{a,b})=0$.
	We observe that $\theta_s (T_j^{(i+1)})=0$ for every $j\in\{1,\dots,i\}$ and $s\in\{i+2,\dots,b-1\}$ because of the $(i+2)(i+3)\cdots b$ part $T_j^{(i+1)}$ in $\Lambda^{b-2}$. So we obtain
	\begin{equation*}
		a\sum_{s=i+2}^{b-1} {(-1)}^{s-1} \theta_s(T_1^{(i+1)})+\sum_{j=2}^i {(-1)}^j\sum_{s=i+2}^{b-1} {(-1)}^{s-1} \theta_s(T_j^{(i+1)})=0
	\end{equation*}
	and
	\begin{equation}\label{2e}
	D_{a,b}=a\sum_{s=1}^{i+1} {(-1)}^{s-1} \theta_s(T_1^{(i+1)})+\sum_{j=2}^i {(-1)}^j\sum_{s=1}^{i+1} {(-1)}^{s-1} \theta_s(T_j^{(i+1)}).
	\end{equation}
	By immediate calculations we have $\theta_s(T_1^{(i+1)})=0$ for every $s\in\{2,\dots,i-1\}$, $\theta_s(T_j^{(i+1)})=0$ for every $s\in\{1,\dots,j-2\}\cup\{j+1,\dots,i-1\}$, $j\in\{2,\dots,i-1\}$ and
	$\theta_s(T_i^{(i+1)})=0$ for every $s\in\{1,\dots,i-2\}$.
    Hence
    \begin{align} D_{a,b}=&a\sum_{s=1}^{i-1} {(-1)}^{s-1} \theta_s(T_1^{(i+1)})+\sum_{j=2}^i {(-1)}^j\sum_{s=1}^{i-1} {(-1)}^{s-1} \theta_s(T_j^{(i+1)})\nonumber \\
    &+a\sum_{s=i}^{i+1} {(-1)}^{s-1} \theta_s(T_1^{(i+1)})+\sum_{j=2}^i {(-1)}^j\sum_{s=i}^{i+1} {(-1)}^{s-1} \theta_s(T_j^{(i+1)})\nonumber \\
    =& (a\theta_1(T_1^{(i+1)})-\theta_1(T_2^{(i+1)})) + \sum_{s=2}^{i-1}(\theta_s(T_s^{(i+1)})-\theta_s(T_{s+1}^{(i+1)}))\nonumber \\
    &+a\sum_{s=i}^{i+1} {(-1)}^{s-1} \theta_s(T_1^{(i+1)})+\sum_{j=2}^i {(-1)}^j\sum_{s=i}^{i+1} {(-1)}^{s-1} \theta_s(T_j^{(i+1)})\nonumber\\
    =&a\sum_{s=i}^{i+1} {(-1)}^{s-1} \theta_s(T_1^{(i+1)})+\sum_{j=2}^i {(-1)}^j\sum_{s=i}^{i+1} {(-1)}^{s-1} \theta_s(T_j^{(i+1)}),
    \end{align}
 as $a\theta_1(T_1^{(i+1)})=\theta_1(T_2^{(i+1)})=a1^{(a)}i^{(2)}\otimes 1\cdots \hat{i} \cdots (b-1)$ and $\theta_s(T_s^{(i+1)})=\theta_s(T_{s+1}^{(i+1)})=1^{(a-1)}si^{(2)}\otimes 1\cdots\hat{i} \cdots (b-1)$.

 It follows that
\begin{equation*}
\pi(D_{a,b})=a\sum_{s=i}^{i+1} {(-1)}^{s-1} \pi(\theta_s(T_1^{(i+1)}))+\sum_{j=2}^i {(-1)}^j\sum_{s=i}^{i+1} {(-1)}^{s-1}\pi(\theta_s(T_j^{(i+1)}))=0,
\end{equation*}
	which is equivalent to
	\begin{equation} \label{6e}
	a\pi(\theta_i(T_1^{(i+1)}))-\sum_{j=2}^{i}{(-1)}^j\pi(\theta_i(T_j^{(i+1)})) =
	a\pi(\theta_{i+1}(T_1^{(i+1)}))-\sum_{j=2}^{i}{(-1)}^j\pi(\theta_{i+1}(T_j^{(i+1)})).
	\end{equation}
	More straightforward calculations yield:
	\begin{itemize}
		\item $\theta_i(T_1^{(i+1)})=1^{(a)}i^{(2)}\otimes 2\cdots (b-1)=b_1^{(i)}$,
		\item $\theta_i(T_j^{(i+1)})=1^{(a-1)}ji^{(2)}\otimes 1\cdots \hat{j} \cdots (b-1)=b_j^{(i)}$ for every $j\in\{2,\dots ,i-1\}$,
		\item $\theta_i(T_i^{(i+1)})=3\cdot 1^{(a-1)}i^{(3)}\otimes 1\cdots \hat{i} \cdots (b-1)=3 \cdot b_i^{(i)}$
	\item $\theta_{i+1}(T_1^{(i+1)})=1^{(a)}{(i+1)}^{(2)}\otimes 2\cdots (b-1) = b_1^{(i+1)}$,
	\item 	$\theta_{i+1}(T_j^{(i+1)})=1^{(a-1)}j{(i+1)}^2\otimes 1 \cdots \hat{j} \cdots (b-1)=b_j^{(i+1)}$ for every \\
	$j\in\{2,\dots ,i\}$,
\end{itemize}
so (\ref{6e}) is equivalent to
\begin{equation*}
a\pi(b_1^{(i+1)}) - \sum_{j=2}^{i}{(-1)}^j \pi(b_j^{(i+1)})= a\pi(b_1^{(i)})-\sum_{j=2}^{i-1}{(-1)}^{j}\pi(b_j^{(i)}) +3{(-1)}^{i+1}\pi(b_i^{(i)}).
\end{equation*}
\end{proof}
With the notation established before Lemma 4.3, we have the following statement.
\begin{lm} \label{1q1q1} For every $i=2,\dots ,b-2$, \[\sum_{j=i+2}^{b-1}{(-1)}^{j-i}\pi(b_j^{(i+1)})-3\pi(b_{i+1}^{(i+1)})=\sum_{j=i+1}^{b-1}{(-1)}^{j-i}\pi(b_j^{(i)}).\]
\end{lm}
\begin{proof} Consider the standard basis elements $$T_j^{(i+1)}=1^{(a-1)}(i+1)^{(2)}j\otimes 1\cdots \widehat{(i+1)} \cdots \hat{j} \cdots b$$
in $Hom(D_a \otimes {D_1}^{\otimes (i-1)}\otimes D_2 \otimes {D_1}^{\otimes (b-i-1)}, D_{a+2} \otimes \Lambda^{b-2})$
for $j\in\{i+2,\dots,b\}$. We know that $\sum_{s=1}^{b-1}{(-1)}^{s-1}\pi(\theta (T_j^{(i+1)}))=0$ for $j\in\{i+2,\dots,b\}$. This implies that
\begin{equation}\label{t1}
\sum_{j=i+2}^b {(-1)}^{j}\sum_{s=1}^{b-1}{(-1)}^{s-1}\pi(\theta_s(T_j^{(i+1)}))=0.
\end{equation}
But $\theta_s(T_j^{(i+1)})=0$ for every $s\in\{1,\dots , i-1\}$, for every $j\in\{i+2 ,\dots ,b\}$ because of the $1\cdots i$ part of $T_j^{(i+1)}$
in $\Lambda^{b-2}$. So (\ref{t1}) is equivalent to
\begin{equation}\label{t2}
\sum_{j=i+2}^b {(-1)}^{j}\sum_{s=i}^{i+1}{(-1)}^{s-1}\pi(\theta_s(T_j^{(i+1)}))+
\sum_{j=i+2}^b {(-1)}^{j}\sum_{s=i+2}^{b-1}{(-1)}^{s-1}\pi(\theta_s(T_j^{(i+1)}))=0.
\end{equation}
By straightforward calculations we have
$\theta_s(T_{i+2}^{(i+1)})=0$ for every $s\in\{i+3,\dots,b-1\}$,
$\theta_s(T_j^{(i+1)})=0$ for every $s\in\{i+2,\dots ,j-2\}\cup \{j+1, \dots ,b-1\}$ for every $j\in\{i+3, \dots , b-1\}$ and $\theta_s(T_b^{(i+1)})=0$ for every $s\in\{i+2, \dots ,b-2\}$.
So (\ref{t2}) yields
\begin{align}\label{t3}
\sum_{j=i+2}^b {(-1)}^{j}\sum_{s=i}^{i+1}{(-1)}^{s-1}\pi(\theta_s(T_j^{(i+1)}))-
\sum_{j=i+2}^{b-1}(\pi(\theta_j(T_j^{(i+1)}))-\pi(\theta_j(T_{j+1}^{(i+1)})))=0.
\end{align}
We observe now that $\theta_j(T_j^{(i+1)})=\theta_j(T_{j+1}^{(i+1)})$, so (\ref{t3}) is equivalent to
$$\sum_{j=i+2}^b {(-1)}^{j-i}(\pi(\theta_{i+1}(T_j^{(i+1)}))-\pi(\theta_i(T_j^{(i+1)})))=0,$$
or
\begin{equation}\label{t6}
\sum_{j=i+2}^{b-1}{(-1)}^{j-i} \pi(\theta_i(T_j^{(i+1)}))=
\sum_{j=i+3}^{b-1}{(-1)}^{j-i} \pi(\theta_{i+1}(T_j^{(i+1)})) + \pi(\theta_{i+1}(T_{i+2})),
\end{equation}
where
\begin{itemize}
\item $\theta_i(T_j^{(i+1)})=b_{j-1}^{(i)}$ for every $j\in\{i+2,\dots ,b\}$,
\item $\theta_{i+1}(T_{i+2})= 3\cdot b_{i+1}^{(i+1)}$ and $\theta_{i+1}(T_j^{(i+1)})=b_{j-1}^{(i+1)}$ for every $j\in\{i+3, \dots ,b\}$.
\end{itemize}
\noindent Equation (\ref{t6}) is now equivalent to
\begin{equation*}
\sum_{j=i+2}^{b-1}{(-1)}^{j-i}\pi(b_j^{(i+1)})-3\pi(b_{i+1}^{(i+1)})=\sum_{j=i+1}^{b-1}{(-1)}^{j-i}\pi(b_j^{(i)})
\end{equation*}
\noindent for every $i\in\{2,\dots ,b-2\}$. \end{proof}
We are now ready to prove the main result of the present subsection, which provides usefull relations that will be used in Subsection 4.3.
\begin{prop} \label{l1} $\;$ \begin{enumerate}
			\item[{\rm (a)}] For $i=1$, \[\sum_{j=2}^{b-1} (-1)^j \pi(b_j^{(1)})= (a+2)\pi(b_1^{(1)}).\]
	       \item[{\rm (b)}] For every $i\in\{2,\dots, b-1\}$, \[a(-1)^i \pi(b_1^{(i)})+ (-1)^{i}\sum_{j=2}^{i-1} (-1)^{j-1} \pi(b_j^{(i)})+{(-1)}^i \sum_{j=i+1}^{b-1} (-1)^{j-1} \pi(b_j^{(i)}) = 3\pi(b_i^{(i)}).\]
		 \end{enumerate} \end{prop}
\begin{proof} For $i=1$ we consider the standard basis elements $T_j^{(1)}=1^{(a+1)}j\otimes 2\cdots\hat{j} \cdots b$ of $Hom(D_{a+1}\otimes {D_1}^{\otimes (b-1)}, D_{a+2} \otimes \Lambda^{b-2})$ for $j\in\{2,\dots , b\}$.
Then we have the relations $\sum_{s=1}^{b-1} (-1)^{s-1}\pi(\theta_s (T_j^{(1)}))=0$, for every $j\in\{2,\dots ,b\}$, and so
\begin{equation} \label{e1}
\sum_{j=2}^b (-1)^{j-1} \pi(\theta_1 (T_j^{(1)})) + \sum_{s=2}^{b-1} (-1)^{s-1} \sum_{j=2}^b (-1)^{j-1}\pi(\theta_s (T_j^{(1)})) =0.
\end{equation}
But $\theta_2 (T_j^{(1)})=0$ for $j\geq 4$, $\theta_s (T_j^{(1)})=0$ for
$j \in \{2,\dots,s-1\}\cup \{s+1,\dots b\}$, and $\theta_{b-1}(T_j^{(1)})=0$ for $j\leq b-2$. So $(\ref{e1})$ is equivalent to
\begin{equation} \label{e2}
\sum_{j=2}^b (-1)^{j-1} \pi(\theta_1 (T_j^{(1)})) + \sum_{s=2}^{b-1} (\pi(\theta_s (T_s^{(1)})) -\pi(\theta_s (T_{s+1}^{(1)}))) =0.
\end{equation}
We observe now that $\theta_s (T_s^{(1)}) = \theta_s (T_{s+1}^{(1)}) = 1^{(a+1)}s \otimes 2\cdots (b-1)$ and $\theta_1 (T_2^{(1)})= (a+2) b_1 ^{(1)}$ and $\theta_1(T_j^{(1)})=b_{j-1}^{(1)}$ for $j\in\{3,\dots , b\}$, so $(\ref{e2})$ is equivalent to
\begin{equation*}
\sum_{j=2}^{b-1} (-1)^{j} \pi(b_j^{(1)}) = (a+2)\pi(b_1^{(1)}).
\end{equation*}
For $i=2$, we consider the standard basis elements $T_1^{(2)}=1^{(a)}2^{(2)}\otimes 3\dots b$ and $T_j^{(2)}=1^{(a-1)}2j\otimes 1\hat{2}\dots \hat{j}\dots b$ in $Hom(D_a \otimes D_2 \otimes D_1 \otimes \dots \otimes D_1,D_{a+2}\otimes \Lambda^{b-2})$, for $j\in\{3,\dots ,b\}$. Then $\sum_{s=1}^{b-1} {(-1)}^{s-1}\pi(\theta_s (T_1^{(2)}))=0$ and $\sum_{s=1}^{b-1}{(-1)}^{s-1} \pi(\theta_s (T_j^{(2)}))=0$ for $j\in\{3,\dots ,b\}$. So we obtain
\begin{equation*} \label{e3} \sum_{j=3}^{b}{(-1)}^{j-1}\sum_{s=1}^{b-1}{(-1)}^{s-1} \pi(\theta_s (T_j^{(2)}))=0,
\end{equation*}
which is equivalent to
\begin{align}\label{e7}
&\sum_{j=3}^{b} {(-1)}^{j-1} \pi(\theta_1 (T_j^{(2)})) - \sum_{j=3}^{b} {(-1)}^{j-1} \pi(\theta_2 (T_j^{(2)})) \nonumber \\
&+ \sum_{s=3}^{b-1}{(-1)}^{s-1} \sum_{j=3}^{b}{(-1)}^{j-1}\pi(\theta_s (T_j^{(2)}))=0.
\end{align}
But $\theta_s (T_3^{(2)})=0$ for every $s\in\{4,\dots,b-1\}$, $\theta_s(T_b^{(2)})=0$ for every $s\in\{3,\dots b-2\}$, and $\theta_s (T_j^{(2)})=0$ for $s\in \{3,\dots,j-2\}\cup \{j+1,\dots,b-1\}$ for every $j\in\{4, \dots, b-1\}$ and also we have $\theta_s(T_s^{(2)})=\theta_s (T_{s+1}^{(2)})=1^{(a-1)}2^{(2)}s\otimes 1\hat{2}\dots \hat{s}\dots (b-1)$. As a result, $(\ref{e7})$ is equivalent to
\begin{equation} \label{e9}
\sum_{j=3}^{b} {(-1)}^{j-1} \pi(\theta_1 (T_j^{(2)})) - \sum_{j=3}^{b} {(-1)}^{j-1} \pi(\theta_2 (T_j^{(2)})) =0.
\end{equation}
But $\theta_1(T_j^{(2)})= \tbinom{a+1}{2}b_{j-1}^{(1)}$, for $j=3,\dots,b$, and thus
\begin{align}\label{e10}
&\sum_{j=3}^{b} {(-1)}^{j-1} \pi(\theta_1 (T_j^{(2)}))=\sum_{j=3}^{b} {(-1)}^{j-1}\tbinom{a+1}{2}\pi(b_{j-1}^{(1)})
= \tbinom{a+1}{2}\sum_{j=2}^{b-1}{(-1)}^j \pi(b_j^{(1)}) \nonumber \\
&=\tbinom{a+1}{2}(a+2)\pi(b_1^{(1)})=a\tbinom{a+2}{2}\pi(b_1^{(1)}),
\end{align}
where we used part (a) of the Proposition 4.5 in the second equality. We observe that $\theta_2(T_3^{(2)})=3b_2^{(2)}$ and $\theta_2(T_j^{(2)})=b_{j-1}^{(2)}$ for $j\in\{4,\dots,b\}$, and so
\begin{equation}\label{e11}
 \sum_{j=3}^{b} {(-1)}^{j-1} \pi(\theta_2 (T_j^{(2)})) = 3\pi(b_2^{(2)}) + \sum_{j=4}^{b} {(-1)}^{j-1} \pi(b_{j-1}^{(2)}) =3\pi(b_2^{(2)}) + \sum_{j=3}^{b-1} {(-1)}^{j} \pi(b_{j}^{(2)}).
\end{equation}
Combining $(\ref{e9})$, $(\ref{e10})$ and $(\ref{e11})$ we obtain
\begin{equation}\label{e12}
a\tbinom{a+2}{2}\pi(b_1^{(1)})+\sum_{j=3}^{b-1} {(-1)}^{j-1} \pi(b_{j}^{(2)})=3\pi(b_2^{(2)}).
\end{equation}
But $\sum_{s=1}^{b-1} {(-1)}^{s-1}\pi(\theta_s (T_1^{(2)}))=0$ and $\theta_s(T_1^{(2)})=0$ for every $s\in\{3,\dots,b-1\}$, because of the $3\cdots b$ part of $T_1^{(2)}$ in $\Lambda^{b-2}$, which implies that $\pi(\theta_1(T_1^{(2)}))=\pi(\theta_2(T_1^{(2)}))$ or equivalently $\tbinom{a+2}{2}\pi(b_1^{(1)})=\pi(b_1^{(2)}).$ Equation $(\ref{e12})$ is now equivalent to
\begin{equation*}
a\pi(b_1^{(2)})+\sum_{j=3}^{b-1} {(-1)}^{j-1} \pi(b_{j}^{(2)})=3\pi(b_2^{(2)}),
\end{equation*}
and we have proved the case $i=2$.

The general case for $i\geq 2$ will be proved by induction on $i$. Suppose that for $i\geq 2$ we have
\begin{equation}
\label{inductive}
 a(-1)^i \pi(b_1^{(i)})+ (-1)^{i}\sum_{j=2}^{i-1} (-1)^{j-1} \pi(b_j^{(i)})+ {(-1)}^i\sum_{j=i+1}^{b-1} (-1)^{j-1} \pi(b_j^{(i)}) = 3\pi(b_i^{(i)}).
\end{equation}
Using (\ref{inductive}), Lemma \ref{bl1} yields
\begin{align}\label{1q1q}
a&(-1)^{i+1} \pi(b_1^{(i+1)})+ (-1)^{i+1}\sum_{j=2}^{i} (-1)^{j-1} \pi(b_j^{(i+1)}) \nonumber \\
&=a(-1)^{i+1} \pi(b_1^{(i)})+ (-1)^{i+1}\sum_{j=2}^{i-1} (-1)^{j-1} \pi(b_j^{(i)}) +3\pi(b_i^{(i)}) \nonumber \\
&=-\big(a(-1)^{i} \pi(b_1^{(i)})+ (-1)^{i}\sum_{j=2}^{i-1} (-1)^{j-1} \pi(b_j^{(i)})\big)+3\pi(b_i^{(i)}) \nonumber \\
&={(-1)}^i \sum_{j=i+1}^{b-1}{(-1)}^{j-1}\pi(b_j^{(i)}).
\end{align}
By Lemma \ref{1q1q1} we have
\begin{equation}\label{1q1q1q}
\sum_{j=i+2}^{b-1}{(-1)}^{j-i}\pi(b_j^{(i+1)})=3\pi(b_{i+1}^{(i+1)})+\sum_{j=i+1}^{b-1}{(-1)}^{j-i}\pi(b_j^{(i)}),
\end{equation}
\noindent and adding (\ref{1q1q}) and (\ref{1q1q1q}) we obtain
\begin{equation*}
a(-1)^{i+1} \pi(b_1^{(i+1)})+ (-1)^{i+1}\sum_{j=2}^{i} (-1)^{j-1} \pi(b_j^{(i+1)})+\sum_{j=i+2}^{b-1}{(-1)}^{j-i}\pi(b_j^{(i+1)})=
\end{equation*}
\begin{equation*}
{(-1)}^i\sum_{j=i+1}^{b-1}{(-1)}^{j-1}\pi(b_j^{(i)})+3\pi(b_{i+1}^{(i+1)})+\sum_{j=i+1}^{b-1}{(-1)}^{j-i}\pi(b_j^{(i)})=3\pi(b_{i+1}^{(i+1)}).
\end{equation*} \end{proof}
\subsection{The image of the generator $\pi(\Gamma_{a,b})$}
By straightforward calculations, one verifies that the composition $i_2 \circ \pi_3: D_{a+3}\otimes \Lambda^{b-3}\rightarrow \Delta(h(3))\rightarrow D_{a+2}\otimes\Lambda^{b-2}$ is the map $f:D_{a+3}\otimes \Lambda^{b-3}\xrightarrow{\triangle \otimes 1} D_{a+2}\otimes D_1 \otimes \Lambda^{b-3}\xrightarrow{1 \otimes m}D_{a+2}\otimes \Lambda^{b-2}$.

Let $\phi: Ext^2(\Delta(h),D_{a+3}\otimes\Lambda^{b-3})\rightarrow Ext^2(\Delta(h),D_{a+2}\otimes\Lambda^{b-2})$ be the map induced by $i_2\circ \pi_3: D_{a+3}\otimes \Lambda^{b-3}\rightarrow  D_{a+2}\otimes \Lambda^{b-2}$, i.e. $\phi$ is the composition $i_2^{(2)}\circ \pi_3^{(2)}$,
\begin{align*}
\phi:&Ext^2(\Delta(h),D_{a+3}\otimes\Lambda^{b-3})\xrightarrow{\pi_3^{(2)}}Ext^2(\Delta(h),\Delta(h(3)))\\
&\xrightarrow{i_2^{(2)}}Ext^2(\Delta(h),D_{a+2}\otimes\Lambda^{b-2}).
\end{align*}
In this Subsection we will compute the image $\phi(\pi(\Gamma_{a,b}))$ of the cyclic generator $\pi(\Gamma_{a,b})$ that was defined in Proposition 4.1. More precisely, and using the notation established at the beginning of Subsection 4.2, we will prove the following. \begin{prop}We have \[\phi(\pi(\Gamma_{a,b}))=(a+b)\pi(\gamma_{a,b}),\] where $\gamma_{a,b}=\tbinom{a+2}{3} b_1^{(1)}+\sum_{i=2}^{b-1}{(-1)}^{i-1}b_i^{(i)}$ and $\pi(\gamma_{a,b})$ is a cyclic generator of $Ext^2(\Delta(h),D_{a+2}\otimes \Lambda^{b-2})$.
\end{prop}
 \begin{proof}We recall from \cite{MS}, Lemma 4.4, that $\pi(\gamma_{a,b})$ is a cyclic generator of the group $Ext^2(\Delta(h),D_{a+2}\otimes \Lambda^{b-2}).$

As $\phi$ is a linear map, we obtain
\begin{align*}
\phi(\pi(\Gamma_{a,b}))=&\sum_{i=3}^{b-2}{(-1)}^{i-1}\Big(a\phi(\pi(B_{i,1}^{a,b})) -\sum_{j=2}^{i-1} {(-1)}^j \phi(\pi(B_{i,j}^{a,b}))+\sum_{j=i+1}^{b-1} {(-1)}^{j} \phi(\pi(B_{i,j}^{a,b}))  \Big)\\
&+\tbinom{a+2} {3}\sum_{j=2}^{b-1}{(-1)}^{j}\phi(\pi(B_{1,j}^{a,b})),
\end{align*}
where empty sums are equal to zero. Using the straightening law we have
\begin{align*}
\phi(\pi(B_{1,j}^{a,b}))&={(-1)}^j\cdot \pi(1^{(a+2)}\otimes 2\dots (b-1)) + \pi(1^{(a+1)}j \otimes 1\dots \hat{j} \dots (b-1))\\&={(-1)}^j \pi(b_1^{(1)})+\pi(b_j^{(1)}).
\end{align*}
This implies that
\begin{align}\label{eqgen1}
&\tbinom{a+2}{3}\sum_{j=2}^{b-1}{(-1)}^{j}\phi(\pi(B_{1,j}^{a,b}))=\tbinom{a+2}{3}\sum_{j=2}^{b-1} \Big(\pi( b_1^{(1)})+ {(-1)}^j\pi( b_j^{(1)})\Big)\nonumber \\
&=\tbinom{a+2}{3}\Big((b-2)\pi(b_1^{(1)}) + \sum_{j=2}^{b-1} {(-1)}^j \pi(b_j^{(1)})\Big)=\tbinom{a+2}{3}(a+b)\pi(b_1^{(1)}),
\end{align}
as $ \sum_{j=2}^{b-1} {(-1)}^j \pi(b_j^{(1)})=(a+2)\pi(b_1^{(1)})$, by part (a) of Proposition \ref{l1}.
Using again the straightening law, we observe that
\begin{itemize}
\item $\phi(\pi(B_{i,j}^{a,b}))={(-1)}^{j-1}\pi(b_i^{(i)})+{(-1)}^i \pi(b_j^{(i)})$ for $1\leq j<i\leq b-1$, and
\item $\phi(\pi(B_{i,j}^{a,b}))={(-1)}^j \pi(b_i^{(i)})+{(-1)}^{i-1} \pi(b_j^{(i)})$ for $2\leq i<j\leq b-1$.
\end{itemize}
Then, by Proposition \ref{l1} for $i\in\{2,\dots,b-1\}$ and (\ref{eqgen1}), we have
\begin{align*}
\phi (\pi(\Gamma_{a,b}))&=\sum_{i=2}^{b-1}{(-1)}^{i-1}\Big(a\phi(\pi(B_{i,1}^{a,b})) -\sum_{j=2}^{i-1} {(-1)}^j \phi(\pi(B_{i,j}^{a,b}))+\sum_{j=i+1}^{b-1} {(-1)}^{j} \phi(\pi(B_{i,j}^{a,b}))  \Big)\\
&\,\,\,\,\,\,+\tbinom{a+2}{3}(a+b)\pi(b_1^{(1)}) \\
&=\sum_{i=2}^{b-1}{(-1)}^{i-1}\Big((a+b-3)\pi(b_i^{(i)}) + a{(-1)}^i\pi(b_1^{(i)})+{(-1)}^i\sum_{j=2}^{i-1}{(-1)}^{j-1}\pi(b_j^{(i)})\\
&\,\,\,\,\,\,+{(-1)}^i\sum_{j=i+1}^{b-1}{(-1)}^{j-1}\pi(b_j^{(i)})\Big)+\tbinom{a+2}{3}(a+b)\pi(b_1^{(1)})\\
&=(a+b)\Big(\tbinom{a+2}{3} \pi(b_1^{(1)})+\sum_{i=2}^{b-1}{(-1)}^{i-1}\pi(b_i^{(i)})\Big)=(a+b)\pi(\gamma_{a,b}),
\end{align*}
\end{proof}

Using Proposition 4.6, we will now prove the case $k=4$ of Theorem 1.1.
\begin{prop}
We have	$Ext^2(\Delta(h),\Delta(h(4)))=\mathbb{Z}_{gcd(3,a+b)}. $
\end{prop}
\begin{proof}We noted at the beginning of Section 4.3 that $\phi$ factors as follows
	\begin{align*}
	\phi:&Ext^2(\Delta(h),D_{a+3}\otimes\Lambda^{b-3})\xrightarrow{\pi_3^{(2)}}Ext^2(\Delta(h),\Delta(h(3)))\\
	&\xrightarrow{i_2^{(2)}}Ext^2(\Delta(h),D_{a+2}\otimes\Lambda^{b-2}).
	\end{align*}
	We also saw in Proposition 4.6 that $$\phi(\pi(\Gamma_{a,b}))=(a+b)\pi(\gamma_{a,b}),$$ where $\pi(\Gamma_{a,b})$ is a generator of $Ext^2(\Delta(h),D_{a+3}\otimes \Lambda^{b-3})=\mathbb{Z}_3$ and $\pi(\gamma_{a,b})$ is a generator of $Ext^2(\Delta(h),D_{a+2}\otimes \Lambda^{b-2})=\mathbb{Z}_3$. Hence, $\phi$ is an isomorphism if and only if $3\notdivides a+b$ if and only if $\ker \phi =0$.
Using (\ref{long2}) for $k=3$, we obtain the exact sequence
\begin{align*}
0&\longrightarrow Ext^2(\Delta(h),\Delta(h(4)))\xrightarrow{i_3^{(2)}}Ext^2(\Delta(h),D_{a+3}\otimes\Lambda^{b-3})\\
&\xrightarrow{\pi_3^{(2)}}Ext^2(\Delta(h),\Delta(h(3))),
\end{align*}
and so $Ext^2(\Delta(h),\Delta(h(4)))=\ker \pi_3^{(2)}$. As $i_2^{(2)}$ is an injection, we obtain $\ker\phi = \ker \pi_3^{(2)}$ and so we have $Ext^2(\Delta(h),\Delta(h(4)))=\ker \phi$. We have $\ker \phi =0$ if and only if $3\notdivides a+b$ and, also, $\ker \phi=\mathbb{Z}_3$ if and only if $3| a+b$. We conclude that $Ext^2(\Delta(h),\Delta(h(4)))=\mathbb{Z}_{gcd(3,a+b)}$.
\end{proof}

\subsection{Modular $Ext^1$}
Let $K$ be an infinite field of characteristic $p > 0$. The corollary below is a statement for $S_K(n,r)$-modules, where $S_K(n,r)$ is the Schur algebra for $GL_n(K)$ corresponding to homogeneous polynomial representations of $GL_n(K)$ of degree $r$ \cite{Gr}. By $\Delta_K(h)$ we denote the Weyl module of highest weight $h$ for $S_K(n,r)$. From \cite{Ma} it follows that $Ext^1(\Delta_K(h),\Delta_K(h(1)))$ is one dimensional if $p|a+b$ and zero otherwise. Using the universal coefficient theorem (\cite{AB}, Thm. 5.3), Theorem 3.5 by \cite{MS}, and Theorem 1.1 we have the following complete picture.

\begin{cor}
	Let $h=(a,1^b)$ and  $h(k)=(a+k,1^{b-k})$, where $ 1 \le k \le b$. Suppose $n \ge b+1$. Then $Ext^1(\Delta_K (h),\Delta_K (h(k)))$ is 1-dimensional in each of the following cases:
		\begin{itemize}
			\item $k=1$ and $p$ divides $a+b$,
			\item $k=2$ and $p$ divides $ 2(a+b)/{gcd(2,a+b)}^2$,
	 		\item $k=3$ and $p$ divides $ 6/{gcd(2,a+b+1)gcd(3,a+b)}$,
			\item $k=4$ and $p$ divides $ {2gcd(3,a+b)}/{gcd(2,a+b)}$,
			\item $k\geq 5$ and $p=2$, $a+b+k$ is odd.
		\end{itemize}
		In all other cases, $Ext^1(\Delta_K (h),\Delta_K (h(k)))=0.$
\end{cor}


TITLE: The generalized Axiom of Dependent Choice (DC) - is this a valid generalization?
QUESTION [1 upvotes]: After studying the axiom of dependent choice, I tried to think of a possible generalization of the axiom that would work in a similar way on infinite uncountable sets: by replacing the binary relation with a general set membership relation over ordinal-indexed elements, I finally obtained the axiom below.
Remark: in order to make the axiom easier to read, I use the notation $(\underbrace{y_0, y_1, \dots}_{\alpha})$ to denote a map of elements indexed over an ordinal $\alpha$.
Generalized Axiom of Dependent Choice: given a set $X$, an element $x_0\in X$, a limit ordinal $\beta$, and a set $R$ containing elements of $X$ indexed over any $\alpha\in\beta$, suppose that the following two relations hold:
$
(a)\quad \forall \alpha\in\beta\,:\,(\underbrace{y_0, y_1, \dots}_{\alpha})\in R\,\,\Rightarrow\,\,\exists z\in X : (\underbrace{y_0, y_1, \dots, z}_{\alpha+1})\in R
$
$(b)\quad$ if $\sigma$ is a sequence of elements of $X$ indexed by a limit ordinal $\gamma\in\beta$ such that each initial segment of $\sigma$ is an element of $R$, then $\sigma$ is an element of $R$.
Then there exists a function $f\,:\,\beta\,\mapsto\,X$ s.t. $f(0)=x_0$ and
$$
\forall\alpha\in\beta\,:\,(\underbrace{f(0), f(1), \dots}_{\alpha})\in R
$$
Do you think this is a valid and sound generalization? I found simple proofs of the following facts:

if $\beta=\omega$ (i.e. the first infinite ordinal), then the generalized version is equivalent to the axiom of dependent choice with fixed first element.
the generalized axiom of dependent choice implies the axiom of choice on the ordinals (although it is not clear if they are also equivalent)
although inherently non-constructive, the generalized axiom hints to a 'method' for building the function $f\,:\,\beta\,\mapsto\,X$: choose $x_0$ as the first element, then randomly select the following elements $x_1, x_2,\dots$ in such a way that the sequence $(x_0,x_1,x_2,\dots)$ is still in $R$. This seems to be an intuitive process
if we replace the condition $\cdots\Rightarrow\,\,\exists z\in X\cdots$ with the condition $\cdots\Rightarrow\,\,\exists ! z\in X\cdots$ (where the existing element $z\in X$ is also unique) then the function $f\,:\,\beta\,\mapsto\,X$ is provably unique. Although the changed version may seem to be provable in ZF, it is not a consequence of the axiom schema of replacement, nor it is definable via transfinite recursion - in other words, I suspect it is independent of ZF.

I would like to receive some feedback from other logicians: does this look like a valid generalization? Do you think it may be worth of a publication, and in case, in which journal?
Please share your thoughts about this, thank you in advance and have a great day!
Kind regards,
Gianluca Calcagni

REPLY [0 votes]: Noah gave a nice example why this generalization does not work. But there is in fact a generalization of $\sf DC$ to larger [$\aleph$] cardinals, due to Azriel Levy.
$\sf DC_\kappa$ is the statement that, whenever $S$ is a non-empty set, and $R$ is a relation on $S^{<\kappa}\times S$ with domain $S^{<\kappa}$, then there is a function $f\colon\kappa\to S$ such that $f\restriction\alpha\mathrel{R}f(\alpha)$ for all $\alpha<\kappa$.
Namely, if we have a relation $R$ such that for every $\alpha<\kappa$ and every $f\colon\alpha\to S$, there is some $z\in S$ such that $f\mathrel{R}z$, then there is a function from $\kappa$ which satisfies the wanted conclusion.
As a side note, this is equivalent to saying that every $\kappa$-closed tree without leaves has a branch of length $\kappa$; and that Zorn's lemma holds for chains of order type $<\kappa$ (namely, if $P$ is a partial where every chain of order type $<\kappa$ has an upper bound, then $P$ has a maximal element).
Let me finish by saying that this sort of gap is the reason that these two proofs cannot be generalized immediately to accommodate $\sf DC_\kappa$ for uncountable $\kappa$:

If for every ordinal $\alpha$, $\sf AC_\alpha$ holds, then $\sf DC$ holds.
If $M$ is a structure for a finite language, then $M$ has an elementary submodel of size $\leq\kappa$ if and only if $\sf DC+AC_\kappa$.

Both proofs work because for $\kappa=\aleph_0$ we can readily approximate things with finite steps, but we cannot bridge the obvious gap of limit ordinals.


text:
$Q$-learning~\citep{watkins1992q} is one of the most popular classes of methods for solving reinforcement learning (RL) problems.
$Q$-learning tries to estimate the optimal state-action value function ($Q$-function).
With a $Q$-function, at every state, one can just greedily choose the action with the largest $Q$ value to interact with the RL environment.
Compared to another popular class of methods, model-based learning, $Q$-learning algorithms (or more generally, model-free algorithms) often enjoy better memory and time efficiency\footnote{See Section~\ref{sec:pre} for the precise definitions of model-free and model-based algorithms in the tabular setting.}.
These are the main reasons why $Q$-learning is applied in solving a wide range of RL problems~\citep{mnih2015human}.

While model-free methods are widely applied in practice, most theoretical works study model-based RL.
In one of the most fundamental RL frameworks, tabular RL, which is the focus of this paper, the majority of works study model-based algorithms~\citep{kearns1999finite, kakade2003sample, singh1994upper, azar2013minimax,azar2017minimax,dann2015sample,dann2017unifying,agarwal2019optimality,max2019nonasymptotic} with a few exceptions \citep{strehl2006pac,jin2018qlearning,dong2019qlearning,zhang2020optimal}.
From a regret minimization point of view, the state-of-the-art analysis demonstrates that one can achieve a $\sqrt{T}$-type regret bound where $T$ is the number of episodes.
Although these bounds are sharp in the worst-case scenario, they do not reveal the favorable
structures of the environment, which can significantly decrease the regret.

One such structure is the existence of a strictly positive sub-optimality gap, i.e., for every state, there is a strictly positive value gap between the optimal action(s) and the rest (cf. Definition~\ref{defn:gap}).
In practice, arguably, nearly all environments with finite action sets satisfy some sub-optimality gap conditions.

In Atari-games, e.g., Freeway, the optimal action has a value that is usually very distinctive from the rest of actions.
In many other environments with finite number of actions, e.g. those control environments in OpenAI gym \citep{1606.01540},  the gap condition usually holds.
Similar gap conditions can be observed in other environments (see e.g. \cite{kakade2003sample}).

Theoretically, the sub-optimality gap is extensively investigated in the bandit problems, which can be viewed as RL problems with the planning horizon being $1$.
With this structure, one can drastically decrease the $\sqrt{T}$-type regret to $\log T$-type regret~\citep{bubeck2012regret,lattimore2018bandit,slivkins2019introduction}.
For RL, most existing works that can leverage this structure require additional assumptions about the environment, such as finite hitting time and ergodicity~\citep{jaksch2010near,tewari2007reinforcement,ok2018exploration} or access to a generator~\citep{zanette2019almost}.\footnote{The simulator allows the user to query any state-action pair.}
Recently, \citet{max2019nonasymptotic} presented a systematic study of episodic tabular RL with the gap structure.
They presented a novel algorithm which achieves the near-optimal $\sqrt{T}$-type regret in the worst scenario and $\log T$-type regret if there exists a strictly positive sub-optimality gap.
Furthermore, they also provided instance-dependent lower bounds for a class of reasonable algorithms.
See Section~\ref{sec:rel} for more detailed discussions.

However, to our knowledge, all existing works that obtain $\log T$-type regret bounds are about model-based algorithms.
It remains open whether model-free algorithms such as $Q$-learning can achieve $\log T$-type regret bounds.
Indeed, this is a challenging task.
As discussed in \cite{max2019nonasymptotic}, their analysis framework cannot be applied to model-free algorithms directly.
Later in this section, we also provide some technical explanations on why their approach is difficult to adopt.

\paragraph{Our Contributions}
We answer the aforementioned open problem by proving that the optimistic $Q$-learning algorithm studied in \cite{jin2018qlearning} enjoys $\cO\!\left(\frac{SA H^6}{\gapmin}\log \left(SAT\right)\right)$ cumulative regret where $S$ is the number states, $A$ is the number of actions, $H$ is the planning horizon and $\gapmin$ is the minimum sub-optimality gap.
To our knowledge, this is the first result showing model-free algorithms can achieve $\log T$-type regret.
Furthermore, our bound matches the lower bound by~\cite{max2019nonasymptotic} in terms of $S$, $A$ and $T$ up to a $\log\left(SA\right)$ factor.
Importantly, the algorithm does not need to know $\gapmin$.

Second, we extend our analysis to the infinite-horizon discounted setting with the regret defined in \cite{liu2020regret}, for which we show the optimistic $Q$-learning achieves $\cO\!\left(\frac{SA}{\gapmin\left(1-\gamma\right)^6}\log\left(\frac{SAT}{\gapmin \left(1-\gamma\right)}\right)\right)$ regret where $0 \!<\! \gamma\! <\! 1$ is the discount factor.

\paragraph{Main Challenges}
\label{sec:challenge}

Here we explain the main challenges of using existing analyses and give an  overview of our main techniques at a high level.
The existing proof in \cite{jin2018qlearning} bounds the regret in terms of a weighted sum of the estimation error of $Q$-function.
Note the estimation error scales

$1/\sqrt{T}$ which in turn gives a $\sqrt{T}$-type regret, but cannot give a $\log T$-type regret bound.

For model-based algorithms, \citet{max2019nonasymptotic} introduced a novel notion, \emph{optimistic surplus} (cf. Equation~\eqref{eqn:opt_surplus}),
 which can be bounded by the estimation error of the transition probability.
The logarithmic regret bound can be proved via a clipping trick on top of the optimistic surplus.

Unfortunately, as acknowledged by \citet{max2019nonasymptotic}, their analysis is highly tailored to model-based algorithms.
First, model-free algorithms do not estimate the probability transition, so we cannot bound the optimistic surplus via this approach.
Secondly, although we can also obtain a formula for the optimistic surplus in each episode using the update rules of the $Q$-learning algorithm, the formula depends on the estimation error of $Q$-function in previous episodes.
This dependency makes it difficult to bound the optimistic surplus.
See Section~\ref{sec:opt_surplus} for more technical details.

\paragraph{Technique Overview}
In this paper, we adopt an entirely different \emph{counting} approach. We first write the total regret as expected sum over sub-optimality gaps appearing in the whole learning process, then use the estimation error of $Q$-function and the definition of sub-optimality gap to upper bound the number of times the algorithm takes suboptimal actions.

To obtain a sharp dependency on $\gapmin$, we divide the interval $[\gapmin,H]$ (the range of all gaps) into multiple subintervals.
We then bound the sum of learning error in each subinterval by its maximum value times the number of steps falling into this subinterval.
The number of steps in each layer is bounded through computing the weighted sum of  learning error across all the episodes $k\in[K]$.
See detailed discussion in Lemma~\ref{lemma:weighed-sum-learning-error} and Lemma~\ref{lemma:count-in-each-layer}.

\paragraph{Organization}
This paper is organized as follows.
In Section~\ref{sec:rel} we discuss related works.
In Section~\ref{sec:pre}, we introduce necessary definitions and backgrounds.
In Section~\ref{sec:main_results}, we present our main results and discussions.
In Section~\ref{sec:proof_sketch}, we give the proof of our theorem on the episodic setting.
We conclude in Section~\ref{sec:conclusion} and leave remaining proofs to the appendix.

config: arxiv


text:
\subsection{Relation with linear independence}
\label{sec:rel.w.aff.indep}

In Euclidean simplex is non-degenerate if and only if for any point $x$ in Euclidean space we can find $n$ vertices
such that the vectors from $x$ to the vertices are linearly independent. Linear independence likewise plays an
important role in the definition of a non-degenerate Riemannian
simplex. We remind ourselves, see Section \ref{sec:bary.map}, that a Riemannian simplex $\sigma_M$ is
\defn{non-degenerate} if the barycentric coordinate map
$\stdsplxn \to \sigma_M$ is a smooth embedding.

\begin{lem}
\label{ConsequencesLinearIndependence} If for any  $x$ in the image
of the map given in Definition \ref{def:riem.splx} ($\sigma_M$)
there are $n$ tangents to geodesics connecting this point $x$ to some
subset of the vertices $v_0 , \ldots , v_{j-1}, v_{j+1}, \ldots, v_n$
(this choice does depend on $x$) that are linearly independent then
\begin{itemize}
\item[$\bullet$] The map $\stdsplxn \to  \sigma_{M}$ is
bijective
\item[$\bullet$] The inverse of $\stdsplxn \to
\sigma_{M}$ is smooth \end{itemize}
\end{lem}

In the proof we shall need the following observation: Within any
ball smaller than the injectivity radius containing $v_i$, the vector field $v_i(x) =\exp_{x} ^{-1}(v_i)$
depends smoothly on the point $x$ for all $x \neq v_i$.  This is obvious if we
consider Riemannian normal coordinates at $v_i$. The geodesic
between $x$ and origin ($v_i$) is a straight line, that depends
smoothly on $x$. The same holds for the tangent to the geodesic at $x$, this is precisely
$v_i(x)=\exp_{x} ^{-1}(v_i)$.

\begin{proof}
We now prove the first of our claims by contradiction. Let us assume
that \begin{align} \sum \lambda_i v_i(x) =  \sum
\tilde{\lambda}_i v_i(x)  =0 \nonumber  \end{align} for
some $\lambda, \tilde{\lambda} \in \stdsplxn $, $\lambda
\neq \tilde{\lambda} $.  Because $v_0(x), \ldots
,v_{j-1}(x) ,v_{j+1} (x), \ldots ,
v_n(x) $ are assumed to be linearly independent we have $\lambda_j
\neq 0, \tilde{\lambda}_j \neq 0$. This mean that we can solve for
$v_j$ in both cases, so
\begin{align} \frac{ \lambda_0}{\lambda_j} v_0(x)+ \ldots+\frac{
\lambda_{j-1}}{\lambda_j} v_{j-1}(x)+
\frac{\lambda_{j+1}}{\lambda_j} v_{j+1}(x)+\ldots  + \frac{
\lambda_{n}}{\lambda_j} v_{n}(x)  = \nonumber \\
\frac{ \tilde{\lambda} _0}{\tilde{\lambda}_j} v_0(x)+ \ldots+\frac{
\tilde{\lambda}_{j-1}}{\tilde{\lambda}_j} v_{j-1}(x) +
\frac{\tilde{\lambda}_{j+1}}{\tilde{\lambda}_j} v_{j+1}(x) +\ldots  +
\frac{ \tilde{\lambda}_{n}}{\tilde{\lambda}_j} v_{n} (x) . \nonumber
\end{align}
This contradicts the assumption of linear independence. This
establishes injectivity.

We can use a similar argument to show that the inverse of
$\stdsplxn \to  \sigma_{M}$ is smooth. As we have seen
linear independence implies that $\lambda_j \neq 0$, which means
that we have
\begin{align} \lambda_0 v_0(x)+ \ldots +
\lambda_{j-1} v_{j-1}(x) + \lambda_{j+1} v_{j+1}(x) + \ldots + \lambda_{n} v_{n}(x) & =-\lambda_j
v_{j}(x). \nonumber
\end{align} We can now regard the left hand side as the product of
the matrix with columns $ (v_{i}(x))_{i \neq j}$ with the
vector $(\lambda_i)_{i\neq j}$. We can divide by $-\lambda_j$ and
bring the matrix to the right hand side by inverting, because
$\{v_{i}(x)\}_{i \neq j}$  is a linear independent set
this is possible. We now find \begin{align} (v_0(x)  ,
\ldots ,v_{j-1}(x)  ,v_{j+1}(x)  ,\ldots
v_{n}(x) ) ^{-1}& v_{j}(x)   =
\frac{1}{\lambda_j} (\lambda_0 , \ldots
,\lambda_{j-1},\lambda_{j+1} ,\ldots , \lambda_n)^{t} , \nonumber
\end{align} which is smooth because $v_{i}(x)$ is smooth
and $\{v_{i}(x)\}_{i \neq j}$ are linear independent by
assumption.
\end{proof}

In Lemma \ref{ConsequencesLinearIndependence} we refer to points
lying in $\sigma_M$, because $\sigma_M$ is not so easy to determine
a priori, we will need to determine a neighbourhood that contains
$\sigma_M$ where we can determine linear independence. To this end
we observe the following:
\begin{remark}
\label{RemarkNeighbourhood} $\sigma_M$ lies within a ball
centred at any of the vertices $v_r$ of radius $\distmax$, where $\distmax= \max d_M(v_i,v_r)$, provided $\distmax$ is smaller than the injectivity radius and the ball is convex.
\end{remark}
Karcher \cite{Karcher} noted that the centre of mass of any mass distribution is contained in any convex set that contains the support of the mass distribution, so in particular this ball.

\subsection{Determining linear independence}
\label{SubsectionLinearIndependence}

In the previous subsection we established that if for any point $x
\in \sigma_M$ there are $n$ tangents to geodesics connecting this
point to some subset of the vertices $v_0 , \ldots , v_{j-1},
v_{j+1}, \ldots, v_n$ (depending on $x$) are linearly independent,
then the simplex $\sigma_M$ is well defined. In this subsection we shall
formulate conditions on the vertex set $v_0 , \ldots,  v_n$ such
that we can guarantee linear independence. These conditions are
simple for surfaces. For higher dimensional manifolds we shall need
bounds on the quality of the simplex found by taking the convex hull
of the image of the inverse exponential map at one of the vertices. The
quality of the simplex is considered good if the ratio between the
volume of the simplex and the $n^{\textrm{th}}$ power of the largest edge length is large, which we shall make precise in Theorem \ref{ConditionSimplices}, see also \cite{whitney1957, boissonnat2013manmesh.inria}.

As mentioned, linear dependence for surfaces is easy to determine.
Let us suppose $v_0(x)=\exp_{x}^{-1}(v_0)$, $v_1(x)$,
$ v_2(x) $ do not span $T_x M$. Because $M$ is a surface
it follows that $v_0(x)$, $v_1(x)$,
$v_2(x) $ are co-linear. This is in turn equivalent to
$v_0,v_1$ and $v_2$ lying on a geodesic. Using lemma
\ref{ConsequencesLinearIndependence} we find that $\sigma_M$ is
diffeomorphic to the standard simplex if all three vertices do not
lie on a geodesic. In the two dimensional setting bijection has been argued previously
by Rustamov \cite{Rustamov}.

\begin{figure}[!htb]
    \centerline{
\mbox{
\begin{tikzpicture}
  \coordinate [label={below left:$v_r$}] (A) at (0, 0);
  \coordinate [label={above:$v_0$}] (B) at (1, 4);
  \coordinate [label={below right:$v_{j}$}] (C) at (4, 1);
  \coordinate [label={above right:$v_n$}] (D) at (3.5, 3);
  \coordinate [label={left:$x$}] (E) at (1.7, 2.3);
  \draw [color=blue] (A) -- (B);
  \draw [color=blue] (A) -- (C);
  \draw [color=blue] (A) -- (D);
 	\draw [very thick] (A) -- (E);
	\draw [color=red,dashed] (E) -- (B);
  \draw [color=red,dashed] (E) -- (C);
  \draw [color=red,dashed] (E) -- (D);
  \draw [color=red,dashed] (C) -- (B);
  \draw [color=red,dashed] (D) -- (B);
  \draw [color=red,dashed] (D) -- (C);
	\draw[fill] (A) circle (0.02);
	\draw[fill] (B) circle (0.02);
	\draw[fill] (C) circle (0.02);
	\draw[fill] (D) circle (0.02);
	\draw[fill] (E) circle (0.02);
\end{tikzpicture}
}
    }
\caption{A schematic depiction of $\sigma^{\mathbb{E}} (v_r )$, where we use red dotted
lines to indicate that these lengths of these edges are not equal to
the lengths of the corresponding edges in $\sigma_M$.  }
\label{EuclideanSimplex}
\end{figure}

Returning to manifolds of arbitraty dimension, we discuss conditions such that for any point $x$ in a ball of
radius $\distmax$ centred at the vertex $v_r$, the vectors $v_i(x)$
in the tangent space at $x$ form an affinely independent set. Because of Remark \ref{RemarkNeighbourhood} this is the neighbourhood of interest, because it suffices to show independence here. Assume that the sectional curvatures $K$ of $M$ are bounded in
absolute value: $|K| \leq \curvabsbnd$.
Define $\sigma^{\mathbb{E}} (v_r )$ to be the convex hull of
$(v_i(v_r))_{i=0}^{n}$ in $T_{v_r}M$.  It will be on $\sigma^{\mathbb{E}}(v_r)$ that we impose condition to ensure that the Riemannian simplex $\sigma_M$ in non-degenerate. Note that given $\sigma^{\mathbb{E}}(v_r)$ we in particular have the lengths of all geodesics from $v_r$ to $v_i$ and the angles between their tangents. Using the Toponogov
comparison theorem, we bound $d_\man(x,v_i)$ for each $i$ by means of
Rauch hinges in $\mathbb{H}^n(\curvabsbnd)$ and
$\mathbb{H}^n(-\curvabsbnd)$ the lengths of the closing edges of the hinges
are denoted by
$d_{ \mathbb{H}^n(\curvabsbnd) } (x,v_i) $ and $d_{ \mathbb{H}^n(-\curvabsbnd) } (x,v_i) $. Lemma \ref{lemSpaceConstantCurvature1} implies that
\begin{align}  ( d_{ \mathbb{H}^n(\pm \curvabsbnd) }  (x,v_i))^2 & = | x(v_r) - v_i(v_r) |^2 + E_{(x,v_i)
, \pm \curvabsbnd}  , \nonumber
\end{align}
with $\exp_{v_r}^{-1}(x)= x(v_r)$ as usual, $E_{(x,v_i) , \pm \curvabsbnd}$ an error term satisfying the
bound $|E_{(x,v_i) , \pm \curvabsbnd}| < 5 \curvabsbnd  (2 \distmax
)^4 $, provided
\begin{align}
|v_i(v_r) | , |x(v_r) |  & \leq \distmax & \text{ and }
& & \sqrt{\curvabsbnd}  \distmax  < \frac{1}{2}.
\nonumber
\end{align} Here the radius of the geodesic ball $\distmax$ is the maximum distance $d_{\textrm{max}}$ in the spaces of constant curvature $\mathbb{H}^n(\pm \curvabsbnd)$, introduced in Lemmas \ref{lemSpaceConstantCurvature1} and \ref{lemSpaceConstantCurvature2}.
Because $|E_{(x,v_i) , \pm \curvabsbnd}| < 5 \curvabsbnd  (2 \distmax
)^4 $ we conclude that
\begin{align}
 d_\man(x,v_i)^2
 = | x(v_r) - v_i(v_r) |^2
+ E_{(x,v_i)}. \nonumber \\
d_\man(v_l,v_k)^2= | v_l(v_r) - v_k(v_r) |^2 + E_{(v_l,v_k)} ,
\label{distanceBounds1}
\end{align}
with $|E_{(v_l,v_k)}| ,|E_{(0,v_i)}|< 5 \curvabsbnd ( 2 \distmax )^4 $.

\begin{figure}[!htb]
    \centerline{
\mbox{\includegraphics[height=2.2in]{pictures/ToponogovPictureWithGeodesicsAndTangentPlane2}}
    }
\caption{A symbolic sketch of the procedure: the lengths of edges
and angles between geodesics in a manifold of arbitrary curvature
(symbolized by the ellipsiod in the centre) are approximated by those
in the spaces of constant curvature (the two spheres). Here in turn
the triangles are approximated by the Euclidean simplex in `the
tangent space'.  } \label{Toponogov2a}
\end{figure}

At this point we know all the lengths of the geodesics between the
points $x, v_0,\ldots, v_n$ in the manifold up to a small and
explicit deviation term, where the deviation is from the Euclidean space or $T_{v_r}M$ in which $x(v_r)$ and $\sigma^{\mathbb{E}}(v_r)$ lie. Any three points from the set $\{x, v_0,\ldots, v_n\}$ together with the
geodesics connecting them can be regarded as a geodesic triangle.
For a geodesic triangle of which we know all edge lengths the
Toponogov comparison theorem gives bounds on the angles in terms of
the Alexandrov triangles in the spaces $\mathbb{H}^n (\curvabsbnd)$
and $\mathbb{H}^n (-\curvabsbnd)$. Let us denote by $\theta_{il}^{\mathbb{H}^n}$ the angle
$\angle v_i x v_l$ between the geodesics in $\mathbb{H}^n(\pm
\curvabsbnd) $ and let $\theta_{il}^{\mathbb{E}}$ denote the angle
$\angle v_i(v_r) x(v_r) v_l(v_r) $ in $T_{v_r}M$, which we may regard as
Euclidean space. If we have a lower bound on the geodesic edge lengths in the simplex as well as on the distance between $x$ and the vertices under consideration, Lemma \ref{lemSpaceConstantCurvature1} in turn gives us bounds on the angles in $\mathbb{H}^n (\curvabsbnd)$
and $\mathbb{H}^n (-\curvabsbnd)$ compared to the corresponding angle in Euclidean space.
To be precise
\begin{align}
d_{\mathbb{E}} (p,q) > \curvabsbnd^{1/4} (d_{\textrm{max}})^{3/2}, \label{LowerBoundDist2}
\end{align}
with $p,q \in \{ x(v_r), v_i(v_r),v_l (v_r) \mid i \neq j , l \neq j \}$, $p \neq q$, then the distance bounds \eqref{distanceBounds1} in $\mathbb{H}^d (\curvabsbnd)$ imply
\begin{align}
  |\cos \theta_{il}^{\mathbb{H}^n} - \cos \theta_{il}^{\mathbb{E}} |
  \leq 80  \sqrt{\curvabsbnd} d_{\textrm{max}} .  \label{angleBoundsS}
\end{align}

Formula \eqref{angleBoundsS} holds for the
upper and lower bounds that appear in the Toponogov comparison
theorem. Thus
\begin{align}
  |\cos \theta_{il} - \cos
  \theta_{il}^{\mathbb{E}} | \leq 160  \sqrt{\curvabsbnd} \distmax  ,
\nonumber
\end{align}
where $\theta_{il}$ denotes the angle
$\angle v_i x v_l$ between the geodesics in the manifold, assuming
that the conditions above are satisfied. A sufficient condition on the simplex $\sigma^{\mathbb{E}}(v_r)$ for \eqref{LowerBoundDist2} to be satisfied (for some choice of $j$) is that for all $j$ the altitude $\textrm{Alt}_j$
\begin{align}
\textrm{Alt}_j (\sigma^{\mathbb{E}}(v_r))> n  \curvabsbnd^{1/4} (\distmax )^{3/2} , \nonumber
\end{align}
which can be weakened\footnote{by which we mean that following inequality implies the previous. } to
\begin{align}
\frac{\textrm{Alt}_j (\sigma^{\mathbb{E}}(v_r))}{L(\sigma^{\mathbb{E}}(v_r) )} > n  \curvabsbnd^{1/4} ( \distmax )^{1/2}, \label{altcondition}
\end{align}
with $L(\sigma^{\mathbb{E}}(v_r) )$ the longest edge length of $\sigma^{\mathbb{E}}(v_r) $.

Using reduced Gram matrices and the estimates by Friedland we now see:
\begin{align}
|\det (\cos \theta_{il} )_j| \geq |\det (\cos \theta_{il}^{\mathbb{E}} )_j| - 160 n   \sqrt{\curvabsbnd}  \distmax ,
\nonumber
\end{align}
with $(\cos \theta_{il} )_j$ and $(\cos \theta_{il}^{\mathbb{E}} )_j$ the matrix cosines of angles between the tangents of geodesics emanating from $x$ to $v_0, \ldots, v_{j-1}, v_{j+1}, \ldots ,v_n$ and corresponding cosines for $\sigma^{\mathbb{E}}(v_r)$,
which is equivalent to, using \eqref{ToConvexHull},
\begin{align}
|\det (\cos \theta_{il} )_j| & \geq \frac{\det(v_0(v_r)-x(v_r), \ldots ,v_{j-1}(v_r)-x(v_r), v_{j+1}(v_r)-x(v_r), \ldots, v_{n}(v_r)-x(v_r)  )^2 }{ |v_0(v_r)-x(v_r)|^2 \cdot  \ldots  \cdot |v_{j-1}(v_r)-x(v_r)|^2 \cdot | v_{j+1}(v_r)-x(v_r)|^2 \cdot  \ldots \cdot |v_{n}(v_r)-x(v_r) |^2 }   \nonumber\\ & -160 n    \sqrt{\curvabsbnd} \distmax , \nonumber
\end{align}
Lemma \ref{ConsequencesLinearIndependence} states that we have non-degeneracy of the simplex if for any $x$ in $B(v_r, \distmax) $ we have that $|\det (\cos \theta_{il} )_j| >0$ for some $j$, this means that if
\begin{align}
\min_{x \in B(v_r,  \distmax )  } \max_{j \in \{ 0 , \ldots ,n\} }
\frac{\det(v_0(v_r)-x(v_r), \ldots ,v_{j-1}(v_r)-x(v_r), v_{j+1}(v_r)-x(v_r), \ldots, v_{n}(v_r)-x(v_r)  )^2 }{ |v_0(v_r)-x(v_r)|^2 \cdot  \ldots  \cdot |v_{j-1}(v_r)-x(v_r)|^2 \cdot | v_{j+1}(v_r)-x(v_r)|^2 \cdot  \ldots\cdot |v_{n}(v_r)-x(v_r) |^2 }   \nonumber\\  >160 n   \sqrt{\curvabsbnd}  \distmax  , \nonumber
\end{align}
non-degeneracy is established. This can be simplified using that $| v_{j+1}(v_r)-x(v_r)| \leq 2 \distmax$ and remarking that the mimimum of
\begin{align}
\max_{j \in \{ 0 , \ldots ,n\} }
\det(v_0(v_r)-x(v_r), \ldots ,v_{j-1}(v_r)-x(v_r), v_{j+1}(v_r)-x(v_r), \ldots, v_{n}(v_r)-x(v_r)  )
\nonumber
\end{align}
is attained in the barycenter and equals $(n-1 )! \textrm{vol} (\sigma^{\mathbb{E}}(v_r) )$.

This means that we now have the condition for non-degeneracy
\begin{align}
  \left(\frac{ (n-1)! \textrm{vol} (\sigma^{\mathbb{E}} (v_r ) ) }{
(2\distmax)^n } \right)^2  > 160 n \sqrt{\curvabsbnd} \distmax, \tag{\ref{QualityCrit}}
\end{align}
assuming that also \eqref{altcondition} is satisfied. Using lemma \ref{prop:alt.non.degen} we can prove that Equation \eqref{QualityCrit} cannot be
satisfied when \eqref{altcondition} is violated. Firstly note that
\begin{align}
\frac{\textrm{vol} (\sigma^{\mathbb{E}} (v_r ) ) }{ (2\distmax)^n }
\leq
\frac{\textrm{vol} (\sigma^{\mathbb{E}} (v_r ) ) }{ (L(\sigma^{\mathbb{E}} (v_r ) ))^n }
= \Theta (\sigma^{\mathbb{E}} (v_r ) )
\leq \frac{t}{(n-1)!}
= \frac{\min_j \textrm{Alt}_j (\sigma^{\mathbb{E}}(v_r))}{ n! L(\sigma^{\mathbb{E}}(v_r) )}, \nonumber
\end{align}
so that \eqref{QualityCrit} yields
\begin{align}
\frac{\min_j \textrm{Alt}_j (\sigma^{\mathbb{E}}(v_r))}{  L(\sigma^{\mathbb{E}}(v_r) )}>  \sqrt{160} n^{3/2} \curvabsbnd^{1/4} \distmax^{1/2}, \nonumber
\end{align} which implies \eqref{altcondition}.

We can now summarize
\begin{thm} \label{ConditionSimplices}
Let $v_0, \ldots ,v_n$ be a set of vertices lying in a Riemannian manifold $M$,
whose sectional curvatures are bounded in absolute value by
$\curvabsbnd$, within a convex geodesic ball of radius $\distmax$ centred at one of the vertices ($v_r$) and such that $\sqrt{\curvabsbnd} \distmax <1/2$. If $\sigma^{\mathbb{E}} (v_r )$, the convex hull of $(\exp^{-1}_{v_r}(v_i))_{i=0}^{n}= (v_i(v_r))_{i=0}^{n}$,
satisfies
\begin{align}
\left(\frac{ (n-1)! \textrm{vol} (\sigma^{\mathbb{E}} (v_r ) ) }{
(2\distmax)^n } \right)^2  > 160 n \sqrt{\curvabsbnd} \distmax, \label{QualityCrit}
\end{align}
then the Riemannian simplex with vertices $v_0, \ldots , v_n$ is non-degenerate, that is diffeomorphic to the standard $n$-simplex.
\end{thm}

\begin{remark}
Equation \eqref{QualityCrit} can be weakened to
\begin{align} \left(\frac{ (n-1)! \Theta (\sigma^{\mathbb{E}} (v_r ) ) }{
2  ^n } \right)^2  =
\left(\frac{ (n-1)! \textrm{vol} (\sigma^{\mathbb{E}} (v_r ) ) }{
(2 L( \sigma^{\mathbb{E}} (v_r ) ) )^n } \right)^2  > 160 n \sqrt{\curvabsbnd} \distmax, \nonumber
\end{align}
or
\begin{align}
\left(\frac{ (n-1)! \textrm{vol} (\sigma^{\mathbb{E}} (v_r ) ) }{
(2 L( \sigma^{M} ) )^n } \right)^2  > 160 n \sqrt{\curvabsbnd} \distmax, \nonumber
\end{align}
with $L( \sigma^{\mathbb{E}} (v_r ) )=\max_{i,j} d_{\mathbb{E} } (v_i (v_r) , v_j(v_r) )$, $L( \sigma^{M} )=\max_{ij} d_M (v_i,v_j)$ and $\Theta$ the fatness.
\end{remark}


TITLE: whats wrong with my singular value decomposition
QUESTION [0 upvotes]: I am trying to calculate one by hand from Strang's "Linear Algebra and Its Applications...
$$
A = [-1, 2, 2]
$$
I can see the subspaces are...
$$
C(A) = [1]
\quad C(A^T) = \begin{bmatrix} -1 \\ 2 \\ 2 \end{bmatrix} \quad
N(A) = \begin{bmatrix} 2 \\ 1 \\ 0 \end{bmatrix},\begin{bmatrix} 2 \\ 0 \\ 1 \end{bmatrix} \quad
N(A^T) = \text{zero vector}
$$
from the rules

first r columns of $U = C(A)$
las m-r coluns of $U = N(A^T)$
first r columns of $V = C(A^T)$
last n-r columns of $V = N(A)$
$\Sigma$ is the square root of the eigenvectors of $A^T A$

I get the SVD
$$
A =
\begin{bmatrix}
1
\end{bmatrix}
\begin{bmatrix}
3 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{bmatrix}
\begin{bmatrix}
\frac{-1}{\sqrt5} & \frac{2}{\sqrt3} & \frac{2}{\sqrt3} \\
\frac{2}{\sqrt5} & \frac{1}{\sqrt3} & 0 \\
\frac{2}{\sqrt5} & 0 & \frac{-2}{\sqrt3} \\
\end{bmatrix}
$$
but numpy is telling me the $V^T$ matrix is
$$
\begin{bmatrix}
-\frac{1}{3} & \frac{2}{3} & \frac{2}{3} \\
\frac{2}{3} & \frac{2}{3} & -\frac{1}{3} \\
-\frac{2}{3} & -\frac{1}{3} & \frac{2}{3} \\
\end{bmatrix}
$$
So I must have done something totally wrong somewhere, but I can't see where

REPLY [2 votes]: For a full SVD, a $1 \times 3$ matrix $A$ will decompose into $U \Sigma V^T$ with $U \in \mathbb{R}^{1 \times 1},\Sigma \in \mathbb{R}^{1 \times 3},V^T \in \mathbb{R}^{3 \times 3}$. (In general, for a full SVD, $\Sigma$ has the same shape as $A$, which means that $U$ and $V$ are square.) This means that you should not have the rows of zeros in $\Sigma$ even for a full SVD. Besides that, your $V$ is not orthogonal: you need to orthogonalize the subspace corresponding to the singular value of zero.
For a reduced SVD, a $1 \times 3$ matrix $A$ other than the zero matrix will decompose with $U \in \mathbb{R}^{1 \times 1},\Sigma \in \mathbb{R}^{1 \times 1},V^T \in \mathbb{R}^{1 \times 3}$. (In general, for a reduced SVD, $\Sigma$ is $r \times r$, which forces $U$ to be $m \times r$ and $V^T$ to be $r \times n$.) In your case this will just be $\begin{bmatrix} 1 \end{bmatrix} \cdot \begin{bmatrix} 3 \end{bmatrix} \cdot (A/3)$.
Note that you can, if you want, make a "bloated SVD", which in this situation would use the shapes $U \in \mathbb{R}^{1 \times 3},\Sigma \in \mathbb{R}^{3 \times 3},V^T \in \mathbb{R}^{3 \times 3}$. Then you would have your $\Sigma$ (including the rows of zeros), but your $U$ would need to be padded with zeros, and your $V^T$ would still need to be orthogonalized. There is really no reason, practical or theoretical, to use such a "bloated SVD", however.


TITLE: Probability with permutations?
QUESTION [0 upvotes]: Suppose we have a bag with 5 different colored balls. We pick a ball, note its color, then put the ball back inside the bag. We do this 3 times. I want to know the probability of getting 3 balls of the same color.
P(3 balls of same color)=$({1\over 5}\cdot{1\over 5}\cdot{1\over 5})\cdot5={1\over 5}\cdot{1\over 5}$
I see this problem in 2 ways:

Pick  a ball (100% chance to pick a color here because you need a color to start with) and now we need P(2 balls of same color as first one)=${1\over 5}\cdot{1\over 5}$
Or the first ball had $1\over 5$ probability of being picked, so it's ${1\over 5}\cdot{1\over 5}$ to get 3 balls of the same color, however we could've picked any other ball so multiply the whole probability by 5

Both approaches lead me to the same result $1\over 25$ but I am not sure if both are correct in their reasoning.

REPLY [1 votes]: Your question comes down to two different ways of working out how to draw the first ball without colour constraint: either the probability of dawing a ball is 1; or the probability of drawing a ball of each colour is $\frac{1}{5}$, since there are 5 colours, the total chance is $\frac{1}{5}*5$, which is 1.
Your calculation is correct in both cases.

REPLY [1 votes]: If I get it correctly, ir doesn't matter to you what the color of the first ball is, as long as the two following balls are the same, ending with a probability of $1\over 25$. If it is the case, then it is correct. I would stress that just to be sure.
That aspect is more compound, taking all the sub-cases and then adding their probabilities. (Or multiplying the probability of a random color by 5, as the number of every ball in a specific color is the same for all colors, $x=1$. This is why the probability won't be affected if $x$ is different).

REPLY [1 votes]: Yes.   The reasoning is sound.
There is an identical probability of selecting any particular colour on any particular draw.   That is $1/5$.
So: (1) It doesn't matter what colour is drawn first, the probability of drawing that colour in the next two draws is $1/5^2$.
Also: (2) The probability of drawing any of the $5$ colours thrice is $5/5^3$.
As you had.   The answer is $1/25$.