# Ringel Duality for Extended ZigZag Schur Algebra

Alexander Kleshchev Department of Mathematics
University of Oregon
Eugene
OR 97403, USA<EMAIL_ADDRESS>and  Ilan Weinschelbaum Department of
Mathematics
University of Oregon
Eugene
OR 97403, USA<EMAIL_ADDRESS>
###### Abstract.

Extended zigzag Schur algebras are quasi-hereditary algebras which are
conjecturally Morita equivalent to RoCK blocks of classical Schur algebras. We
prove that extended zigzag Schur algebras are Ringel self-dual.

##  1\. Introduction

Let $S(m,r)$ be a classical Schur algebra over the ground field $\mathbb{F}$
of characteristic $p>0$, see [10]. A fundamental fact going back to [11, 3] is
that the algebra $S(m,r)$ is (based) quasi-hereditary. The blocks of $S(m,r)$
are classified in [6, (2.12)], [7].

From now on let us assume for simplicity that $r\leq m$. Then the blocks of
$S(m,r)$ and of the symmetric group algebra $\mathbb{F}\mathfrak{S}_{r}$ are
parametrized by pairs $(\rho,d)$ where $\rho$ is a $p$-core and $d$ is a non-
negative integer such that $|\rho|+pd=r$. Let $B_{\rho,d}$ denote the
corresponding block of $S(m,r)$ and $\bar{B}_{\rho,d}$ denote the
corresponding block of $\mathbb{F}\mathfrak{S}_{r}$.

A special role in representation theory of $\mathbb{F}\mathfrak{S}_{r}$ is
played by the so-called RoCK blocks going back to [18, 1]. These are the
blocks $\bar{B}_{\rho,d}$ with $\rho$ satisfying certain combinatorial
genericity condition with respect to $d$. The corresponding block $B_{\rho,d}$
of the Schur algebra $S(m,r)$ is then also called RoCK. The RoCK blocks of
symmetric groups are important because they admit a nice ‘local description.’
Namely, by [9], we have that $\bar{B}_{\rho,d}$ is Morita equivalent to the
zigzag Schur algebra $T^{{\bar{\mathsf{Z}}}}(d,d)$ defined by Turner [20], see
also [8]. In view of [2], this yields a ‘local description’ of all blocks of
symmetric groups up to derived equivalence.

On the other hand, an arbitrary RoCK block $B_{\rho,d}$ of $S(m,r)$ is
conjecturally Morita equivalent to the extended zigzag Schur algebra
$T^{\mathsf{Z}}(d,d)$, see [14, Conjecture 7.60]. As a first evidence for this
conjecture, it is proved in [14, Theorem 1] that $T^{\mathsf{Z}}(d,d)$ is
quasi-hereditary. In this paper we obtain further evidence for this conjecture
in terms of Ringel duality.

In fact, Donkin [4, (3.7),(3.11)], [7, §5(2)], [5, §4.1] proves that $S(m,r)$
is Ringel self-dual. It follows from the results of Donkin that the Ringel
dual of the block $B_{\rho,d}$ is the block $B_{\rho^{\prime},d}$. On the
other hand, if $B_{\rho,d}$ is RoCK then so is $B_{\rho^{\prime},d}$, so we
expect that the extended zigzag Schur algebra $T^{\mathsf{Z}}(d,d)$ must be
Ringel self-dual. This is what we prove in this paper:

Main Theorem. Let $d\leq n$. Then the extended zigzag Schur algebra
$T^{\mathsf{Z}}(n,d)$ is Ringel self-dual.

Here ${\mathsf{Z}}$ stands for the extended zigzag algebra corresponding to
the quiver with $p$ vertices, see §5.1. In fact, $T^{\mathsf{Z}}(n,d)$ is a
special case of the generalized Schur algebras $T^{A}_{\mathfrak{a}}(n,d)$
introduced in [13], with $A={\mathsf{Z}}$ and $\mathfrak{a}$ being the degree
zero component of the graded algebra ${\mathsf{Z}}$. It is well known that
${\mathsf{Z}}$ is quasi-hereditary and Ringel self-dual. So our Main Theorem
is a special case of the following conjecture.

Conjecture. Let $A$ be a based quasi-hereditary algebra and $d\leq n$. If
$A^{\prime}$ is a Ringel dual of $A$, then a Ringel dual of
$T^{A}_{\mathfrak{a}}(n,d)$ is of the form
$T^{A^{\prime}}_{\mathfrak{a}^{\prime}}(n,d)$ for some canonical choice of
$\mathfrak{a}^{\prime}$.

The paper is organized as follows. Section 2 is preliminaries. In particular,
§2.3 details necessary facts on based quasi-hereditary algebras and §2.4 is on
the combinatorics of partitions and tableaux. Section 4 describes the
construction of $T^{A}(n,d)$ and important results about its (co)standard
modules. In Section 3, we define the modified divided power which we will use
extensively to construct a full tilting module for $T^{{\mathsf{Z}}}(n,d)$.
Lastly, in Section 5 prove the Main Theorem. In particular, in §5.2 we
describe a full tilting module for $T^{\mathsf{Z}}(n,d)$; and in §5.3 we
compute its endomorphism algebra.

##  2\. Preliminaries

###  2.1. General notation

For $n\in\mathbb{Z}_{>0}$, we denote $[n]:=\\{1,2,\dots,n\\}$. Throughout the
paper, $I$ denotes a non-empty finite partially ordered set. We always
identify $I$ with the set $\\{0,1,\dots,\ell\\}$ for $\ell=|I|-1$, so that the
standard total order on integers refines the partial order on $I$. For a set
$S$, we often write elements of $S^{d}$ as words
$\text{\boldmath$s$}=s_{1}\cdots s_{d}$ with $s_{1},\dots,s_{d}\in S$.

The symmetric group $\mathfrak{S}_{d}$ acts on the right on $S^{d}$ by place
permutations:

$(s_{1}\cdots s_{d})\sigma=s_{\sigma 1}\cdots s_{\sigma d}.$

For $\text{\boldmath$s$}=s_{1}\cdots s_{d}\in S^{d}$ we have the stabilizer
$\mathfrak{S}_{\text{\boldmath$s$}}:=\\{\sigma\in\mathfrak{S}_{d}\mid\text{\boldmath$s$}\sigma=\text{\boldmath$s$}\\}.$
For $\text{\boldmath$s$},\text{\boldmath$t$}\in S^{d}$, we write
$\text{\boldmath$s$}\sim\text{\boldmath$t$}$ if
$\text{\boldmath$s$}\sigma=\text{\boldmath$t$}$ for some
$\sigma\in\mathfrak{S}_{d}$. If $S_{1},\dots,S_{m}$ are sets, then
$\mathfrak{S}_{d}$ acts on $S_{1}^{d}\times\dots\times S_{m}^{d}$ diagonally.
We write
$(\text{\boldmath$s$}_{1},\dots,\text{\boldmath$s$}_{m})\sim(\text{\boldmath$t$}_{1},\dots,\text{\boldmath$t$}_{m})$
if
$(\text{\boldmath$s$}_{1},\dots,\text{\boldmath$s$}_{m})\sigma=(\text{\boldmath$t$}_{1},\dots,\text{\boldmath$t$}_{m})$
for some $\sigma\in\mathfrak{S}_{d}$. If $U\subseteq
S_{1}^{d}\times\dots\times S_{m}^{d}$ is a $\mathfrak{S}_{d}$-invariant
subset, we denote by $U/\mathfrak{S}_{d}$ a complete set of the
$\mathfrak{S}_{d}$-orbit representatives in $U$ and we identify
$U/\mathfrak{S}_{d}$ with the set of $\mathfrak{S}_{d}$-orbits on $U$.

An (arbitrary) ground field is denoted by $\mathbb{F}$. Often we will also
need to work over a characteristic $0$ principal ideal domain $R$ such that
$\mathbb{F}$ is a $R$-module, so that we can change scalars from $R$ to
$\mathbb{F}$ (in all examples of interest to us, one can use $R=\mathbb{Z}$).
When considering $R$-supermodules (or $\mathbb{F}$-superspaces) below, we
always consider $R$ (and $\mathbb{F}$) as concentrated in degree ${\bar{0}}$.

We use $\Bbbk$ to denote $\mathbb{F}$ or $R$ and use it whenever the nature of
the ground ring is not important. On the other hand, when it is important to
emphasize whether we are working over $R$ or $\mathbb{F}$, we will use lower
indices; for example for an $R$-algebra $A_{R}$ and an $A_{R}$-module $V_{R}$,
after extending scalars we have $V_{\mathbb{F}}:=\mathbb{F}\otimes_{R}V_{R}$
is a module over $A_{\mathbb{F}}:=\mathbb{F}\otimes_{R}A_{R}$.

###  2.2. Superalgebras and supermodules

Let $V=\bigoplus_{{\varepsilon}\in\mathbb{Z}/2}V_{\varepsilon}$ be a
$\Bbbk$-supermodule. If $v\in V_{\varepsilon}\setminus\\{0\\}$ for
${\varepsilon}\in\mathbb{Z}/2$, we say $v$ is homogeneous, we write
$|v|={\varepsilon}$, and we refer to ${\varepsilon}$ as the parity of $v$. If
$S\subseteq V$, we denote $S_{\bar{0}}:=S\cap V_{\bar{0}}$ and
$S_{\bar{1}}:=S\cap V_{\bar{1}}$. If $W$ is another $\Bbbk$-supermodule, the
set of all $\Bbbk$-linear homomorphisms $\operatorname{Hom}_{\Bbbk}(V,W)$ is a
$\Bbbk$-supermodule such that for ${\varepsilon}\in\mathbb{Z}/2$, we have

$\operatorname{Hom}_{\Bbbk}(V,W)_{\varepsilon}=\\{f\in\operatorname{Hom}_{\Bbbk}(V,W)\mid|f(v)|=|v|+{\varepsilon}\
\text{for all homogeneous $v$}\\}.$

The group $\mathfrak{S}_{d}$ acts on $V^{\otimes d}$ on the right by
automorphisms, such that for all homogeneous $v_{1},\dots,v_{d}\in V$ and
$\sigma\in\mathfrak{S}_{d}$, we have

$(v_{1}\otimes\dots\otimes
v_{d})^{\sigma}=(-1)^{\langle\sigma;\text{\boldmath$v$}\rangle}v_{\sigma
1}\otimes\dots\otimes v_{\sigma d},$ (2.1)

where, setting $\text{\boldmath$v$}:=v_{1}\cdots v_{d}\in V^{d}$, we have put:

$\langle\sigma;\text{\boldmath$v$}\rangle:=\sharp\\{(k,l)\in[d]^{2}\mid k<l,\
\sigma^{-1}k>\sigma^{-1}l,\ |v_{k}|=|v_{l}|={\bar{1}}\\}.$ (2.2)

We consider the $d$th divided power $\Gamma^{d}V$, which by definition is the
subspace of invariants

$\Gamma^{d}V:=\\{w\in V^{\otimes d}\mid w^{\sigma}=w\ \text{for all
$\sigma\in\mathfrak{S}_{d}$}\\}.$ (2.3)

Let $0\leq c\leq d$. Given $w_{1}\in V^{\otimes c}$ and $w_{2}\in
V^{\otimes(d-c)}$, we define

$w_{1}*w_{2}:=\sum_{\sigma}(w_{1}\otimes w_{2})^{\sigma}\in V^{\otimes d},$
(2.4)

where the sum is over all shortest coset representatives $\sigma$ for
$(\mathfrak{S}_{c}\times\mathfrak{S}_{d-c})\backslash\mathfrak{S}_{d}$.

Let $V$ and $W$ be $\Bbbk$-supermodules, $d\in\mathbb{Z}_{\geq 0}$, and let
$\text{\boldmath$v$}=v_{1}\cdots v_{d}\in V^{d}$,
$\text{\boldmath$w$}=w_{1}\cdots w_{d}\in W^{d}$ be $d$-tuples of homogeneous
elements. We denote

$\langle\text{\boldmath$v$},\text{\boldmath$w$}\rangle:=\sharp\\{(k,l)\in[d]^{2}\mid
k>l,\ |v_{k}|=|w_{l}|={\bar{1}}\\}.$ (2.5)

Let $A$ be a (unital) $\Bbbk$-superalgebra and $V,W$ be $A$-supermodules. A
homogeneous $A$-supermodule homomorphism $f:V\to W$ is a homogeneous
$\Bbbk$-linear map $f:V\to W$ satisfying $f(av)=(-1)^{|f||a|}af(v)$ for all
(homogeneous) $a,v$. For ${\varepsilon}\in\mathbb{Z}/2$, let
$\operatorname{Hom}_{A}(V,W)_{\varepsilon}$ be the set of all homogeneous
$A$-supermodule homomorphisms of parity ${\varepsilon}$, and let

$\operatorname{Hom}_{A}(V,W):=\operatorname{Hom}_{A}(V,W)_{\bar{0}}\oplus\operatorname{Hom}_{A}(V,W)_{\bar{1}}.$

We refer to the elements of $\operatorname{Hom}_{A}(V,W)$ as the
$A$-supermodule homomorphisms from $V$ to $W$. We denote by
$A\\!\operatorname{-mod}$ the category of all finitely generated (left)
$A$-supermodules and all $A$-supermodule homomorphisms. We denote by ‘$\cong$’
an isomorphism in this category and by ‘$\simeq$’ an even isomorphism in this
category.

We have the parity change functor $\Pi$ on $A\\!\operatorname{-mod}$: for
$V\in A\\!\operatorname{-mod}$ we have $\Pi V\in A\\!\operatorname{-mod}$ with
$(\Pi V)_{\varepsilon}=V_{{\varepsilon}+{\bar{1}}}$ for all
${\varepsilon}\in\mathbb{Z}/2$ and the new action $a\cdot v=(-1)^{|a|}av$ for
$a\in A,v\in V$. We have $V\cong\Pi V$ via the identity map.

Suppose there is an even superalgebra anti-involution $\tau:A\to A$. In
particular, $\tau(ab)=(-1)^{|a||b|}\tau(b)\tau(a)$ for all $a,b\in A$. Then
$\tau$ is an isomorphism $A\to A^{{\mathrm{sop}}}$, where the multiplication
in $A^{{\mathrm{sop}}}$ is defined as $a\cdot b:=(-1)^{|a||b|}ba$. If $V\in
A\\!\operatorname{-mod}$ then the $\tau$-dual $V^{\tau}\in
A\\!\operatorname{-mod}$ is the dual $V^{*}$ as a $\Bbbk$-supermodule
considered as a left $A$-supermodule via $(af)(v):=(-1)^{|a||f|}f(\tau(a)v)$
for $a\in A,f\in V^{*},v\in V$.

As usual, the tensor product $A^{\otimes d}$ is a superalgebra with respect to

$(a_{1}\otimes\dots\otimes a_{d})(b_{1}\otimes\dots
b_{d})=(-1)^{\langle\text{\boldmath$a$},\text{\boldmath$b$}\rangle}a_{1}b_{1}\otimes\dots\otimes
a_{d}b_{d},$

where we have put $\text{\boldmath$a$}:=a_{1}\cdots a_{d}$,
$\text{\boldmath$b$}:=b_{1}\cdots b_{d}$ (here and below, in expressions like
this, we assume that all elements are homogeneous). If $V$ is an
$A$-supermodule then $V^{\otimes d}$ is a supermodule over $A^{\otimes d}$
with respect to

$(a_{1}\otimes\dots\otimes a_{d})(v_{1}\otimes\dots
v_{d})=(-1)^{\langle\text{\boldmath$a$},\text{\boldmath$v$}\rangle}a_{1}v_{1}\otimes\dots\otimes
a_{d}v_{d},$

where we have again put $\text{\boldmath$a$}:=a_{1}\cdots a_{d}$,
$\text{\boldmath$v$}:=v_{1}\cdots v_{d}$.

The divided power $\Gamma^{d}A$ is a subsuperalgebra of $A^{\otimes d}$. If
$V$ is an $A$-supermodule then

$\big{(}(a_{1}\otimes\dots\otimes a_{d})(v_{1}\otimes\dots
v_{d})\big{)}^{\sigma}=(a_{1}\otimes\dots\otimes
a_{d})^{\sigma}(v_{1}\otimes\dots\otimes v_{d})^{\sigma}$

for all $a_{1},\dots,a_{d}\in A,\,v_{1},\dots,v_{d}\in
V,\,\sigma\in\mathfrak{S}_{d}$. So $\Gamma^{d}V$ is a subsupermodule of the
restriction of $V^{\otimes d}$ to $\Gamma^{d}A$. Thus we will always consider
$\Gamma^{d}V$ as a $\Gamma^{d}A$-supermodule.

###  2.3. Based quasi-hereditary algebras

The main reference here is [12]. Let $A$ be a $\Bbbk$-superalgebra.

######  Definition 2.6.

[12] Let $I$ be a finite partially ordered set and let $X=\bigsqcup_{i\in
I}X(i)$ and $Y=\bigsqcup_{i\in I}Y(i)$ be finite sets of homogeneous elements
of $A$ with distinguished elements $e_{i}\in X(i)\cap Y(i)$ for each $i\in I$.
For each $i\in I$, we set

$A^{>i}:=\operatorname{span}\\{xy\mid j>i,\,x\in X(j),\,y\in Y(j)\\}.$

We say that $I,X,Y$ is heredity data if the following axioms hold:

  1. (a)

$B:=\\{xy\mid i\in I,\,x\in X(i),\,y\in Y(i)\\}$ is a basis of $A$;

  2. (b)

For all $i\in I$, $x\in X(i)$, $y\in Y(i)$ and $a\in A$, we have

$ax\equiv\sum_{x^{\prime}\in X(i)}l^{x}_{x^{\prime}}(a)x^{\prime}\text{
}(\text{\rm mod }A^{>i})\,\ \ \text{and}\ \ ya\equiv\sum_{y^{\prime}\in
Y(i)}r^{y}_{y^{\prime}}(a)y^{\prime}\text{ }(\text{\rm mod }A^{>i})\,$

for some $l^{x}_{x^{\prime}}(a),r^{y}_{y^{\prime}}(a)\in\Bbbk$;

  3. (c)

For all $i,j\in I$ and $x\in X(i),\ y\in Y(i)$ we have

$\displaystyle xe_{i}=x,\ e_{i}x=\delta_{x,e_{i}}x,\ e_{i}y=y,\
ye_{i}=\delta_{y,e_{i}}y,$ $\displaystyle e_{j}x=x\ \text{or}\ 0,\ ye_{j}=y\
\text{or}\ 0.$

If $A$ is endowed with heredity data $I,X,Y$, we call $A$ based quasi-
hereditary, and refer to $B$ as a heredity basis of $A$. By (c),
$e_{i}^{2}=e_{i}$ for all $i\in I$, so from now on we call $\\{e_{i}\mid i\in
I\\}$ the standard idempotents of the heredity data. We set

$\displaystyle B_{\mathfrak{a}}$ $\displaystyle:=\\{xy\mid i\in I,x\in
X(i)_{\bar{0}},y\in Y(i)_{\bar{0}}\\},\ $ (2.7) $\displaystyle
B_{\mathfrak{c}}$ $\displaystyle:=\\{xy\mid i\in I,x\in X(i)_{\bar{1}},y\in
Y(i)_{\bar{1}}\\},$ (2.8)

so that

$B_{\bar{0}}=B_{\mathfrak{a}}\sqcup B_{\mathfrak{c}}.$ (2.9)

The heredity data $I,X,Y$ of $A$ is called conforming if $B_{\mathfrak{a}}$
spans a unital subalgebra of $A$.

Let $A$ be a based quasi-hereditary superalgebra with heredity data $I,X,Y$
(not necessarily conforming). By [12, Lemma 3.3], $A$ is quasi-hereditary in
the sense of Cline, Parshall and Scott and $A\\!\operatorname{-mod}$ is a
highest weight category (see [3, Theorem 3.6]). The corresponding standard and
costandard modules are defined as follows. Let $i\in I$. Note that $A^{>i}$
the ideal of $A$ generated by $\\{e_{j}\mid j>i\\}$, and denote
$\tilde{A}:=A/A^{>i}$, $\tilde{a}:=a+A^{>i}\in\tilde{A}$ for $a\in A$. The
standard $A$-module of highest weight $i$ is defined by
$\Delta(i):=\tilde{A}\tilde{e}_{i}$, which is a free $\Bbbk$-module with basis
$\\{v_{x}:=\tilde{x}\mid x\in X(i)\\}$, see [12, §2.3]. We also have the right
standard $A$-module $\Delta^{\mathrm{op}}(i):=\tilde{e}_{i}\tilde{A}$, and by
symmetry every result we have about $\Delta(i)$ has its right analogue for
$\Delta^{\mathrm{op}}(i)$, for example $\Delta^{\mathrm{op}}(i)$ is a free
$\Bbbk$-module with basis $\\{w_{y}:=\tilde{y}\mid y\in Y(i)\\}$. The
costandard $A$-module of highest weight $i$ is defined by
$\nabla(i):=\Delta^{\mathrm{op}}(i)^{*}$ with
$(af)(v)=(-1)^{|a||f|+|a||v|}(va)$ for $a\in
A,f\in\Delta(i)^{*},v\in\Delta(i)$.

Let $V\in A\\!\operatorname{-mod}$. A standard filtration of $V$ is an
$A$-supermodule filtration $0=W_{0}\subseteq W_{1}\subseteq\dots\subseteq
W_{l}=V$ such that for every $r=1,\dots,l$, we have
$W_{r}/W_{r-1}\cong\Delta(i_{r})$ for some $i_{r}\in I$. We refer to
$\Delta(i_{1}),\dots,\Delta(i_{l})$ as the factors of the filtration, and to
$\Delta(i_{1})$ (resp. $\Delta(i_{l})$) as the bottom (resp. top) factor. If
$\Bbbk=\mathbb{F}$, $i\in I$ and $V$ has a standard filtration as above, then
$\sharp\\{1\leq r\leq l\mid W_{r}/W_{r-1}\cong\Delta(i)\\}$ does not depend on
the choice of the standard filtration and is denoted $(V:\Delta(i))$. In fact,
by [5, Proposition A2.2], we have

$(V:\Delta(i))=\dim\operatorname{Hom}_{A}(V,\nabla(i)).$ (2.10)

A costandard filtration of $V$ is an $A$-supermodule filtration
$0=W_{0}\subseteq W_{1}\subseteq\dots\subseteq W_{l}=V$ such that for every
$r=1,\dots,l$, we have $W_{r}/W_{r-1}\cong\nabla(i_{r})$ for some $i_{r}\in
I$.

Let $\operatorname{T}\in A\\!\operatorname{-mod}$. We say that
$\operatorname{T}$ is a tilting supermodule if it has standard and costandard
filtrations. We refer to [19, §4] for the integral version of the tilting
theory. In particular, by [19, Propositions 4.26, 4.27], for every $i\in I$
there exists a (unique up to isomorphism) indecomposable tilting supermodule
$\operatorname{T}(i)$ such that $\Delta(i)\subseteq\operatorname{T}(i)$ and
$\operatorname{T}(i)/\Delta(i)$ has a standard filtration with factors of the
form $\Delta(j)$ for $j<i$; moreover, for every tilting supermodule
$\operatorname{T}$ we have $\operatorname{T}\cong\bigoplus_{i\in
I}\operatorname{T}(i)^{\oplus m_{i}}.$ In this case $\operatorname{T}$ is
called a full tilting supermodule if $m_{i}>0$ for all $i\in I$. If
$\operatorname{T}$ is full tilting, the superalgebra
$A^{\prime}:=\operatorname{End}_{A}(\operatorname{T})^{\mathrm{sop}}$ is
called a Ringel dual of $A$. The algebra $A^{\prime}$ is defined uniquely up
to Morita superequivalence and is quasi-hereditary, see [19, Proposition
4.26].

###  2.4. Multipartitions and tableaux

For a partition $\lambda$, we have the conjugate partition $\lambda^{\prime}$,
see [16, p.2]. For partitions $\lambda,\mu,\nu$, we denote by
$c^{\,\lambda}_{\mu,\nu}$ the corresponding Littlewood-Richardson coefficient,
see [16, § I.9]. The Young diagram of $\lambda$ is
$[\lambda]:=\\{(r,s)\in\mathbb{Z}_{>0}\times\mathbb{Z}_{>0}\mid
s\leq\lambda_{r}\\}$. We refer to $(r,s)\in[\lambda]$ as the nodes of
$\lambda$.

Let $n\in\mathbb{Z}_{>0}$. We denote $\Lambda(n):=\mathbb{Z}_{\geq 0}^{n}$ and
interpret it as the set of compositions
$\lambda=(\lambda_{1},\dots,\lambda_{n})$ with $n$ non-negative parts. For
such $\lambda$, we set $|\lambda|:=\lambda_{1}+\dots+\lambda_{n}$. The
partitions with at most $n$ parts are identified with

$\Lambda_{+}(n):=\\{\lambda=(\lambda_{1},\dots,\lambda_{n})\in\Lambda(n)\mid\lambda_{1}\geq\dots\geq\lambda_{n}\\}.$

For $d\in\mathbb{Z}_{\geq 0}$, we let

$\Lambda(n,d):=\\{\lambda\in\Lambda(n)\mid|\lambda|=d\\}\qquad\text{and}\qquad\Lambda_{+}(n,d):=\\{\lambda\in\Lambda_{+}(n)\mid|\lambda|=d\\}.$

For $\lambda,\mu\in\Lambda(n)$, we define
$\lambda+\mu:=(\lambda_{1}+\mu_{1},\dots,\lambda_{n}+\mu_{n})$.

Recall that $I$ denotes a finite poset. We will consider the set of
$I$-multicompositions

$\Lambda^{I}(n):=\Lambda(n)^{I}=\\{\text{\boldmath$\lambda$}=(\lambda^{(i)})_{i\in
I}\mid\lambda^{(i)}\in\Lambda(n)\ \text{for all $i\in I$}\\}.$ (2.11)

For $\text{\boldmath$\lambda$},\text{\boldmath$\mu$}\in\Lambda^{I}(n)$ we
define $\text{\boldmath$\lambda$}+\text{\boldmath$\mu$}$ to be
$\text{\boldmath$\nu$}\in\Lambda^{I}(n)$ with
$\nu^{(i)}=\lambda^{(i)}+\mu^{(i)}$ for all $i\in I$. For
$d\in\mathbb{Z}_{\geq 0}$, we have the sets of $I$-multicompositions and
$I$-multipartitions of $d$:

$\displaystyle\Lambda^{I}(n,d)$
$\displaystyle:=\\{\text{\boldmath$\lambda$}\in\Lambda^{I}(n)\mid\sum_{i\in
I}|\lambda^{(i)}|=d\\},$ $\displaystyle\Lambda_{+}^{I}(n,d)$
$\displaystyle:=\\{\text{\boldmath$\lambda$}\in\Lambda^{I}(n,d)\mid\lambda^{(i)}\in\Lambda_{+}(n)\
\text{for all $i\in I$}\\}.$

For $\text{\boldmath$\lambda$}\in\Lambda^{I}_{+}(n,d)$, we define
$[\text{\boldmath$\lambda$}]:=\bigsqcup_{i\in I}[\lambda^{(i)}]$ and
$\|\text{\boldmath$\lambda$}\|:=(|\lambda^{(i)}|)_{i\in I}\in\mathbb{Z}_{\geq
0}^{I}$.

Via our identification $I=\\{0,\dots,\ell\\}$, for
$\text{\boldmath$\lambda$}=(\lambda^{(i)})_{i\in I}\in\Lambda^{I}(n)$, we also
write $\text{\boldmath$\lambda$}=(\lambda^{(0)},\dots,\lambda^{(\ell)})$. For
$i\in I$, and $\lambda\in\Lambda(n,d)$, define

$\text{\boldmath$\iota$}_{i}(\lambda):=(0,\dots,0,\lambda,0,\dots,0)\in\Lambda^{I}(n,d),$
(2.12)

with $\lambda$ in the $i$th position. We will slightly abuse this notation
writing $\text{\boldmath$\iota$}_{i}(d)$ for
$\text{\boldmath$\iota$}_{i}\big{(}(d,0,\dots,0)\big{)}$ and
$\text{\boldmath$\iota$}_{i}(1^{d})$ for
$\text{\boldmath$\iota$}_{i}\big{(}(1,\dots,1,0,\dots,0)\big{)}$.

Let $\leq$ be the partial order on $I$. We have a partial order $\unlhd_{I}$
on the set $\mathbb{Z}_{\geq 0}^{I}$ with $(a_{i})_{i\in
I}\unlhd_{I}(b_{i})_{i\in I}$ if and only if $\sum_{j\geq
i}a_{j}\leq\sum_{j\geq i}b_{j}$ for all $i\in I$. Let $\unlhd$ be the usual
dominance partial order on $\Lambda(n,d)$, i.e. $\lambda\unlhd\mu$ if and only
if $\sum_{r=1}^{s}\lambda_{r}\leq\sum_{r=1}^{s}\mu_{r}$ for all $s=1,\dots,n$.
We define a partial order $\leq_{I}$ on $\Lambda^{I}(n,d)$ via:
$\text{\boldmath$\lambda$}\leq_{I}\text{\boldmath$\mu$}$ whenever
$\|\text{\boldmath$\lambda$}\|\lhd_{I}\|\text{\boldmath$\mu$}\|$, or
$\|\text{\boldmath$\lambda$}\|=\|\text{\boldmath$\mu$}\|$ and
$\lambda^{(i)}\unlhd\mu^{(i)}$ for all $i\in I$.

Let $I,X,Y$ be a heredity data on a $\Bbbk$-superalgebra $A$ as in §2.3. We
introduce colored alphabets ${\mathscr{A}}_{X}:=[n]\times
X\quad\text{and}\quad{\mathscr{A}}_{X(i)}:=[n]\times X(i).$ An element
$(l,x)\in{\mathscr{A}}_{X}$ is often written as $l^{x}$. If
$L=l^{x}\in{\mathscr{A}}_{X}$, we denote ${\operatorname{color}}(L):=x.$ For
all $i\in I$, we fix arbitrary total orders ‘$<$’ on the sets
${\mathscr{A}}_{X(i)}$ such that whenever $r<s$ (in the standard order on
$[n]$), $r^{x}<s^{x}$ for all $x\in X(i)$.

Let
$\text{\boldmath$\lambda$}=(\lambda^{(0)},\dots,\lambda^{(l)})\in\Lambda^{I}_{+}(n,d)$.
Let $N_{1},\dots,N_{d}$ be the nodes of
$[\text{\boldmath$\lambda$}]=[\lambda^{(0)}]\sqcup\dots\sqcup[\lambda^{(\ell)}]$
listed along the rows of $[\lambda^{(0)}]$ from left to right starting from
the first row and going down, then along the rows of $[\lambda^{(1)}]$ from
left to right starting with the first row and going down, etc. A function
$\text{\boldmath$T$}:[\text{\boldmath$\lambda$}]\to{\mathscr{A}}_{X}$ is
called a standard $X$-colored $\lambda$-tableau if the following conditions
hold:

  1. $\bullet$

$\text{\boldmath$T$}([\lambda^{(i)}])\subseteq{\mathscr{A}}_{X(i)}$ for all
$i\in I$;

  2. $\bullet$

if $r<s$ and $N_{r},N_{s}$ are in the same row of $[\lambda^{(i)}]$, then
$\text{\boldmath$T$}(N_{r})\leq\text{\boldmath$T$}(N_{s})$ with
$\text{\boldmath$T$}(N_{r})=\text{\boldmath$T$}(N_{s})$ allowed only if
${\operatorname{color}}(\text{\boldmath$T$}(N_{r}))\in X(i)_{\bar{0}}$;

  3. $\bullet$

if $r<s$ and $N_{r},N_{s}$ are in the same column of $[\lambda^{(i)}]$, then
$\text{\boldmath$T$}(N_{r})\leq\text{\boldmath$T$}(N_{s})$ with
$\text{\boldmath$T$}(N_{r})=\text{\boldmath$T$}(N_{s})$ allowed only if
${\operatorname{color}}(\text{\boldmath$T$}(N_{r}))\in X(i)_{\bar{1}}$;

We denote by $\operatorname{Std}^{X}(\text{\boldmath$\lambda$})$ the set of
all standard $X$-colored $\lambda$-tableaux. For
$\text{\boldmath$T$}\in\operatorname{Std}^{X}(\text{\boldmath$\lambda$})$,
letting $\text{\boldmath$T$}(N_{r})=l_{r}^{x_{r}}\in{\mathscr{A}}_{X}$ for
$r=1,\dots,d$, we denote
$\text{\boldmath$l$}^{\text{\boldmath$T$}}:=l_{1}\cdots l_{d}$ and
$\text{\boldmath$x$}^{\text{\boldmath$T$}}:=x_{1}\cdots x_{d}$. For
$\text{\boldmath$\mu$}\in\Lambda^{I}(n,d)$, we say that $T$ has left weight
$\mu$ if there exist $i_{1},\dots,i_{d}\in I$ such that
$e_{i_{1}}x_{1}=x_{1},\,\dots,\,e_{i_{d}}x_{d}=x_{d}$ and
$\text{\boldmath$\mu$}=\sum_{c=1}^{d}\text{\boldmath$\iota$}_{i_{c}}({\varepsilon}_{l_{c}}).$
We have the set of all standard $\lambda$-tableaux of left weight $\mu$:

$\operatorname{Std}^{X}(\text{\boldmath$\lambda$},\text{\boldmath$\mu$}):=\\{\text{\boldmath$T$}\in\operatorname{Std}^{X}(\text{\boldmath$\lambda$})\mid\text{\boldmath$T$}\
\text{has left weight}\ \text{\boldmath$\mu$}\\}.$ (2.13)

##  3\. Modified divided powers

Throughout the section, we fix a non-negative integer $d$.

###  3.1. Calibrated supermodules and their modified divided powers

Let $V_{R}=V_{R,{\bar{0}}}\oplus V_{R,{\bar{1}}}$ be a free $R$-supermodule of
finite rank. We say that $V_{R}$ is calibrated if we are given a decomposition
$V_{R,{\bar{0}}}=V_{R,\mathfrak{a}}\oplus V_{R,\mathfrak{c}}$ into two free
$R$-modules.

Let $V_{R}$ be a calibrated $R$-supermodule. We choose bases
$B^{V}_{\mathfrak{a}}$, $B^{V}_{\mathfrak{c}}$, $B^{V}_{\bar{1}}$ of
$V_{R,\mathfrak{a}},V_{R,\mathfrak{c}},V_{R,{\bar{1}}}$, respectively. Thus
$B^{V}_{\bar{0}}:=B^{V}_{\mathfrak{a}}\sqcup B^{V}_{\mathfrak{c}}$ is a basis
of $V_{R,{\bar{0}}}$ and $B^{V}=B^{V}_{\mathfrak{a}}\sqcup
B^{V}_{\mathfrak{c}}\sqcup B^{V}_{{\bar{1}}}$ is a basis of $V_{R}$. Fix an
arbitrary total order $<$ on $B^{V}$. Let $\text{\boldmath$b$}\in(B^{V})^{d}$.
We define

$\langle\text{\boldmath$b$}\rangle:=\sharp\\{(k,l)\in[d]^{2}\mid k<l,\
b_{k},b_{l}\in B^{V}_{\bar{1}},\ b_{k}>b_{l}\\}.$ (3.1)

For any $b\in B^{V}$, set

$[\text{\boldmath$b$}:b]:=\sharp\\{k\in[d]\mid b_{k}=b\\}$ (3.2)

and let

$[\text{\boldmath$b$}]^{!}_{\mathfrak{c}}:=\prod_{b\in
B^{V}_{\mathfrak{c}}}[\text{\boldmath$b$}:b]!.$ (3.3)

Define $\operatorname{Seq}(B^{V},d)$ to be the set of all $d$-tuples
$\text{\boldmath$b$}=b_{1}\cdots b_{d}\in(B^{V})^{d}$ such that $b_{k}=b_{l}$
for some $1\leq k\neq l\leq d$ only if $b_{k}\in B^{V}_{\bar{0}}$. Then
$\operatorname{Seq}(B^{V},d)\subseteq(B^{V})^{d}$ is a
$\mathfrak{S}_{d}$-invariant subset, so we can choose a corresponding set
$\operatorname{Seq}(B^{V},d)/\mathfrak{S}_{d}$ of $\mathfrak{S}_{d}$-orbit
representatives and identify it with the set of all $\mathfrak{S}_{d}$-orbits
on $\operatorname{Seq}(B^{V},d)$, cf. §2.1.

Recall the divided power $R$-supermodule $\Gamma^{d}V_{R}$ from §2.2. For
$\text{\boldmath$b$}=b_{1}\cdots b_{d}\in\operatorname{Seq}(B^{V},d)$, we have
elements

$x_{\text{\boldmath$b$}}:=\sum_{\text{\boldmath$b$}^{\prime}=b_{1}^{\prime}\cdots
b_{d}^{\prime}\sim\text{\boldmath$b$}}(-1)^{\langle\text{\boldmath$b$}\rangle+\langle\text{\boldmath$b$}^{\prime}\rangle}b_{1}^{\prime}\otimes\dots\otimes
b_{d}^{\prime}\in\Gamma^{d}V_{R}\qquad\text{and}\qquad
y_{\text{\boldmath$b$}}:=[\text{\boldmath$b$}]^{!}_{\mathfrak{c}}\,x_{\text{\boldmath$b$}}\in\Gamma^{d}V_{R}.$

Define the modified divided power

$\tilde{\Gamma}^{d}V_{R}:=\operatorname{span}_{R}\\{y_{\text{\boldmath$b$}}\mid\text{\boldmath$b$}\in\operatorname{Seq}(B^{V},d)\\}\subseteq\Gamma^{d}V_{R}.$

Note that
$\\{x_{\text{\boldmath$b$}}\mid\text{\boldmath$b$}\in\operatorname{Seq}(B^{V},d)/\mathfrak{S}_{d}\\}$
is a basis of $\Gamma^{d}V_{R}$ and
$\\{y_{\text{\boldmath$b$}}\mid\text{\boldmath$b$}\in\operatorname{Seq}(B^{V},d)/\mathfrak{S}_{d}\\}$
is a basis of $\tilde{\Gamma}^{d}V_{R}$, cf. [8, (3.9)].

The $n=1$ case of [13, Proposition 4.11] yields:

######  Lemma 3.4.

$\tilde{\Gamma}^{d}V_{R}$ depends only on $V_{R,\mathfrak{a}}$, and not on
$V_{R,\mathfrak{c}}$ or choice of basis $B^{V}$.

If $V_{R},W_{R}$ are calibrated $R$-supermodules, then $V_{R}\oplus W_{R}$ is
also a calibrated $R$-supermodule with $(V_{R}\oplus
W_{R})_{\mathfrak{a}}:=V_{R,\mathfrak{a}}\oplus W_{R,\mathfrak{a}}$ and
$(V_{R}\oplus W_{R})_{\mathfrak{c}}:=V_{R,\mathfrak{c}}\oplus
W_{R,\mathfrak{c}}$. Moreover, recalling the star product from (2.4).

######  Lemma 3.5.

We have an isomorphism of $R$-supermodules

$\bigoplus_{d_{1}+d_{2}=d}(\tilde{\Gamma}^{d_{1}}V_{R})\otimes(\tilde{\Gamma}^{d_{2}}W_{R})\stackrel{{\scriptstyle\sim}}{{\longrightarrow}}\tilde{\Gamma}^{d}(V_{R}\oplus
W_{R}),\ y\otimes y^{\prime}\mapsto y*y^{\prime}.$

###### Proof.

Note that
$y_{\text{\boldmath$b$}^{1}}*y_{\text{\boldmath$b$}^{2}}=y_{\text{\boldmath$b$}^{1}\text{\boldmath$b$}^{2}}$
for all
$\text{\boldmath$b$}^{1}\in\operatorname{Seq}(B^{V},d_{1}),\,\text{\boldmath$b$}^{2}\in\operatorname{Seq}(B^{W},d_{2})$
and compare bases. ∎

###  3.2. Bilinear form on $\tilde{\Gamma}^{d}V$

Let $V_{R}$ be a calibrated $R$-supermodule. Suppose in addition that we are
given an ($R$-valued) even supersymmetric (or superantisymmetric), non-
degenerate bilinear form $(\cdot,\cdot)$ on $V_{R}$, such that
$(V_{R,\mathfrak{a}},V_{R,\mathfrak{a}})=0$ and the $R$-complement
$V_{R,\mathfrak{c}}$ of $V_{R,\mathfrak{a}}$ in $V_{R,{\bar{0}}}$ can be
chosen so that the restriction of $(\cdot,\cdot)$ to $V_{R,\mathfrak{a}}\times
V_{R,\mathfrak{c}}$ is a perfect pairing. Until the end of this subsection, we
always assume that the complement $V_{R,\mathfrak{c}}$ has this property.

Since the form is non-degenerate we may select bases
$B^{V}_{\mathfrak{a}}=\\{a_{1},\ldots,a_{r}\\},B^{V}_{\mathfrak{c}}=\\{c_{1},\ldots,c_{r}\\}$
and $B^{V}_{\bar{1}}$ for $V_{R,\mathfrak{a}},V_{R,\mathfrak{c}}$, and
$V_{R,{\bar{1}}}$ respectively such that $(a_{i},c_{j})=\delta_{i,j}$ for all
$i,j\in[r]$. Set $B^{V}=B^{V}_{\mathfrak{a}}\cup B^{V}_{\mathfrak{c}}\cup
B^{V}_{\bar{1}}$. Let $(B^{V})^{*}=\\{b^{*}\mid b\in B^{V}\\}$ be the dual
basis with respect to $(\cdot,\cdot)$. Note that $c_{i}^{*}=a_{i}$ for all
$i$, but it is not necessarily true that $a_{i}^{*}=c_{i}$. We take
$V_{R,\mathfrak{c}}^{\prime}$ to be the $R$-span of
$a_{1}^{*},\dots,a_{r}^{*}$, so that $V_{R,\mathfrak{c}}^{\prime}$ is another
$R$-complement of $V_{R,\mathfrak{a}}$ in $V_{R,{\bar{0}}}$. We now have
$(B^{V})^{*}=(B^{V})^{*}_{\mathfrak{a}}\cup(B^{V})^{*}_{\mathfrak{c}}\cup(B^{V})^{*}_{\bar{1}}$
where

$(B^{V})^{*}_{\mathfrak{a}}:=B^{V}_{\mathfrak{a}},\
(B^{V})^{*}_{\mathfrak{c}}:=\\{a_{1}^{*},\dots,a_{r}^{*}\\},\
(B^{V})^{*}_{\bar{1}}=\\{b^{*}\mid b\in B^{V}_{\bar{1}}\\}.$

By Lemma 3.4, $\tilde{\Gamma}^{d}V_{R}$ is independent of the choice of a
complement $V_{R,\mathfrak{c}}$ and of the choice of a corresponding basis. So
we now have two $R$-bases of $\tilde{\Gamma}^{d}V_{R}$:

$\\{y_{\text{\boldmath$b$}}\mid\text{\boldmath$b$}\in\operatorname{Seq}(B^{V},d)/\mathfrak{S}_{d}\\}\quad\text{and}\quad\\{y_{\text{\boldmath$b$}}\mid\text{\boldmath$b$}\in\operatorname{Seq}((B^{V})^{*},d)/\mathfrak{S}_{d}\\}.$

The form $(\cdot,\cdot)$ extends to the form $(\cdot,\cdot)_{d}$ on
$V_{R}^{\otimes d}$ such that

$(v_{1}\otimes\dots\otimes v_{d},w_{1}\otimes\dots\otimes
w_{d})_{d}=(-1)^{\langle\text{\boldmath$v$},\text{\boldmath$w$}\rangle}(v_{1},w_{1})\cdots(v_{d},w_{d}).$
(3.6)

for all $\text{\boldmath$v$}=v_{1}\dots v_{d},\text{\boldmath$w$}=w_{1}\cdots
w_{d}\in V^{d}$. Since the form is even, we have for any
$\sigma\in\mathfrak{S}_{d}$:

$((v_{1}\otimes\cdots\otimes v_{d})^{\sigma},(w_{1}\otimes\cdots\otimes
w_{d})^{\sigma})_{d}=(v_{1}\otimes\cdots\otimes
v_{d},w_{1}\otimes\cdots\otimes w_{d})_{d}.$ (3.7)

Moreover, for
$\text{\boldmath$b$},\text{\boldmath$b$}^{\prime}\in(B^{V})^{d}$, we have

$(b_{1}^{\prime}\otimes\dots\otimes
b_{d}^{\prime},b^{*}_{1}\otimes\dots\otimes
b^{*}_{d})_{d}=(-1)^{\langle\text{\boldmath$b$}^{\prime},\text{\boldmath$b$}\rangle}\delta_{\text{\boldmath$b$},\text{\boldmath$b$}^{\prime}}.$
(3.8)

######  Lemma 3.9.

Let
$\text{\boldmath$b$},\text{\boldmath$b$}^{\prime}\in\operatorname{Seq}(B^{V},d)$.
Then $\text{\boldmath$b$}^{*}:=b_{1}^{*}\cdots
b_{d}^{*}\in\operatorname{Seq}((B^{V})^{*},d)$, and
$(y_{\text{\boldmath$b$}^{\prime}},y_{\text{\boldmath$b$}^{*}})_{d}=\pm
d!\,\delta_{\text{\boldmath$b$}\sim\text{\boldmath$b$}^{\prime}}$.

###### Proof.

By (3.8),
$(y_{\text{\boldmath$b$}}^{\prime},y_{\text{\boldmath$b$}^{*}})_{d}\neq 0$
only if $\text{\boldmath$b$}\sim\text{\boldmath$b$}^{\prime}$. So we may
assume that $\text{\boldmath$b$}^{\prime}=\text{\boldmath$b$}$ and that the
stabilizer
$\mathfrak{S}_{\text{\boldmath$b$}}=\mathfrak{S}_{\text{\boldmath$b$}^{*}}$ is
a standard parabolic subgroup. As no odd element repeats in $b$, we have

$|\mathfrak{S}_{\text{\boldmath$b$}}|=\bigg{(}\prod_{b\in
B^{V}_{\mathfrak{a}}}[\text{\boldmath$b$}:b]!\bigg{)}\bigg{(}\prod_{b\in
B^{V}_{\mathfrak{c}}}[\text{\boldmath$b$}:b]!\bigg{)}=\bigg{(}\prod_{b^{*}\in(B^{V})^{*}_{\mathfrak{c}}}[\text{\boldmath$b$}^{*}:b^{*}]!\bigg{)}\bigg{(}\prod_{b\in
B^{V}_{\mathfrak{c}}}[\text{\boldmath$b$}:b]!\bigg{)}.$

So, using (3.7) and (3.8), we have that
$(y_{\text{\boldmath$b$}},y_{\text{\boldmath$b$}^{*}})_{d}$ equals

$\displaystyle\bigg{(}\bigg{(}\prod_{b\in
B^{V}_{\mathfrak{c}}}[\text{\boldmath$b$}:b]!\bigg{)}x_{\text{\boldmath$b$}},\bigg{(}\prod_{b^{*}\in(B^{V})^{*}_{\mathfrak{c}}}[\text{\boldmath$b$}^{*}:b^{*}]!\bigg{)}x_{\text{\boldmath$b$}^{*}}\bigg{)}_{d}$
$\displaystyle=\,$ $\displaystyle\bigg{(}\bigg{(}\prod_{b\in
B^{V}_{\mathfrak{c}}}[\text{\boldmath$b$}:b]!\bigg{)}\sum_{\sigma\in\mathfrak{S}_{d}/\mathfrak{S}_{\text{\boldmath$b$}}}(b_{1}\otimes\cdots\otimes
b_{d})^{\sigma},\bigg{(}\prod_{b^{*}\in(B^{V})^{*}_{\mathfrak{c}}}[\text{\boldmath$b$}^{*}:b^{*}]!\bigg{)}\sum_{\sigma\in\mathfrak{S}_{d}/\mathfrak{S}_{\text{\boldmath$b$}}}(b_{1}^{*}\otimes\cdots\otimes
b_{d}^{*})^{\sigma}\bigg{)}_{d}$ $\displaystyle=\,$
$\displaystyle\bigg{(}\prod_{b\in
B^{V}_{\mathfrak{c}}}[\text{\boldmath$b$}:b]!\bigg{)}\bigg{(}\prod_{b^{*}\in(B^{V})^{*}_{\mathfrak{c}}}[\text{\boldmath$b$}^{*}:b^{*}]!\bigg{)}[\mathfrak{S}_{d}:\mathfrak{S}_{\text{\boldmath$b$}}](b_{1}\otimes\cdots\otimes
b_{d},b_{1}^{*}\otimes\cdots\otimes b_{d}^{*})_{d}$ $\displaystyle=\,$
$\displaystyle\pm d!$

which completes the proof. ∎

In view of the lemma, we have $(z,w)_{d}$ is divisible by $d!$ for all
$z,w\in\tilde{\Gamma}^{d}V_{R}$. So we can define a new form on
$\tilde{\Gamma}^{d}V_{R}$ by setting

$(z,w)_{\sim}:=\frac{1}{d!}(z,w)_{d}$ (3.10)

for all $z,w\in\tilde{\Gamma}^{d}V_{R}$. The following is now clear from the
lemma:

######  Proposition 3.11.

The bilinear form $(\cdot,\cdot)_{\sim}$ on $\tilde{\Gamma}^{d}V_{R}$ is even
and non-degenerate. Moreover, it is supersymmetric or superantisymmetric
depending on the parity of $d$ and whether $(\cdot,\cdot)$ is supersymmetric
or superantisymmetric.

###  3.3. Modified divided powers of algebras

Let $A_{R}=A_{R,{\bar{0}}}\oplus A_{R,{\bar{1}}}$ be an $R$-superalgebra such
that $A_{R,{\bar{0}}}$ and $A_{R,{\bar{1}}}$ are free of finite rank as
$R$-modules. We say that $A_{R}$ is a calibrated superalgebra if we are given
a free $R$-module decomposition
$A_{R,{\bar{0}}}=\mathfrak{a}_{R}\oplus\mathfrak{c}_{R}$ such that
$\mathfrak{a}_{R}$ is a unital subalgebra of $A_{R}$. Choose bases
$B^{A}_{\mathfrak{a}},B^{A}_{\mathfrak{c}},B^{A}_{\bar{1}}$ of
$\mathfrak{a}_{R},\mathfrak{c}_{R},A_{R,{\bar{1}}}$, respectively, and set
$B^{A}=B^{A}_{\mathfrak{a}}\cup B^{A}_{\mathfrak{c}}\cup B^{A}_{\bar{1}}$. Our
main examples come from based quasihereditary algebras over $R$ with
conforming heredity data as in 2.3. In that case we would take
$\mathfrak{a}_{R}$ to be the $R$-span of $B_{\mathfrak{a}}$, cf. (2.7).

Note that a calibrated $R$-superalgebra is in particular a calibrated
$R$-supermodule as in §3.1, so we have modified divided power
$\tilde{\Gamma}^{d}A_{R}$ and elements
$x_{\text{\boldmath$b$}}\in\Gamma^{d}A_{R}$,
$y_{\text{\boldmath$b$}}\in\tilde{\Gamma}^{d}A_{R}$ for
$\text{\boldmath$b$}\in\operatorname{Seq}(B^{A},d)$. But to differentiate
between algebras and module, when working with algebras, it will be convenient
to use another notation:

$\xi^{\text{\boldmath$b$}}:=x_{\text{\boldmath$b$}}\quad\text{and}\quad\eta^{\text{\boldmath$b$}}:=y_{\text{\boldmath$b$}}.$
(3.12)

Recall from §2.2 that $\Gamma^{d}A_{R}$ is a subsuperalgebra of
$A_{R}^{\otimes d}$. Moreover, $\tilde{\Gamma}^{d}A_{R}$ is a (unital)
subsuperalgebra of $\Gamma^{d}A_{R}$, see [13, Proposition 3.12], with basis

$\\{\eta^{\text{\boldmath$b$}}\mid\text{\boldmath$b$}\in\operatorname{Seq}(B^{V},d)/\mathfrak{S}_{d}\\}.$
(3.13)

Recall from [8, §4.1], that $\bigoplus_{d\geq 0}A_{R}^{\otimes d}$ is a
bisuperalgebra with the coproduct  ${\tt\Delta}$ defined so that

$\displaystyle\reflectbox{\rotatebox[origin={c}]{180.0}{${\tt\Delta}$}}\,:\,\,A_{R}^{\otimes
d}\,\,$ $\displaystyle\to\,\,\bigoplus_{c=0}^{d}A_{R}^{\otimes c}\otimes
A_{R}^{\otimes(d-c)},$ $\displaystyle a_{1}\otimes\cdots\otimes a_{d}\,\,$
$\displaystyle\mapsto\,\,\sum_{c=0}^{d}(a_{1}\otimes\cdots\otimes
a_{c})\otimes(a_{c+1}\otimes\cdots\otimes a_{d}).$

Moreover, by the $n=1$ case of [13, Corollary 3.24], $\bigoplus_{d\geq
0}\tilde{\Gamma}^{d}A_{R}$ is a sub-bisuperalgebra of $\bigoplus_{d\geq
0}A_{R}^{\otimes d}$.

There is also another bisuperalgebra structure on $\bigoplus_{d\geq
0}A_{R}^{\otimes d}$ using the star product $*$ of (2.4). In fact:

######  Lemma 3.14.

[13, Corollary 4.4] $\bigoplus_{d\geq 0}\tilde{\Gamma}^{d}A_{R}$ is a sub-
bisuperalgebra of $\bigoplus_{d\geq 0}A_{R}^{\otimes d}$ with respect to the
coproduct  ${\tt\Delta}$ and the product $*$

###  3.4. $\tilde{\Gamma}^{d}V$ as a module over $\tilde{\Gamma}^{d}A$

Let $A_{R}=\mathfrak{a}_{R}\oplus\mathfrak{c}_{R}\oplus A_{R,{\bar{1}}}$ be a
calibrated superalgebra as in the previous subsection. Let
$V_{R}=V_{R,\mathfrak{a}}\oplus V_{R,\mathfrak{c}}\oplus V_{R,{\bar{1}}}$ be a
calibrated $R$-supermodule with the corresponding basis
$B^{V}=B^{V}_{\mathfrak{a}}\sqcup B^{V}_{\mathfrak{c}}\sqcup B^{V}_{\bar{1}}$
as in §3.1, and assume in addition that $V_{R}$ is an $A_{R}$-supermodule. We
say that $V_{R}$ is a calibrated $A_{R}$-supermodule if
$\mathfrak{a}_{R}V_{R,\mathfrak{a}}\subseteq V_{R,\mathfrak{a}}$.

Recall from §2.2 that $\Gamma^{d}V_{R}$ is naturally a
$\Gamma^{d}A_{R}$-supermodule. So upon restriction to the subalgebra
$\tilde{\Gamma}^{d}A_{R}\subseteq\Gamma^{d}A_{R}$, $\Gamma^{d}V_{R}$ is a
$\tilde{\Gamma}^{d}A_{R}$-supermodule. In Lemma 3.16 we will show that
$\tilde{\Gamma}^{d}V_{R}\subseteq\Gamma^{d}V_{R}$ is a
$\tilde{\Gamma}^{d}A_{R}$-subsupermodule if $V_{R}$ is a calibrated
$A_{R}$-supermodule. .

For $a\in A_{R}$ and $b,c\in B^{V}$, we define the structure constants
$\kappa^{b}_{a,c}\in R$ from $ac=\sum_{b\in B^{V}}\kappa^{b}_{a,c}b$. For
$\text{\boldmath$a$}\in\operatorname{Seq}(B^{A},d)$ and
$\text{\boldmath$b$},\text{\boldmath$c$}\in\operatorname{Seq}(B^{V},d)$, we
also set
$\kappa_{\text{\boldmath$a$},\text{\boldmath$c$}}^{\text{\boldmath$b$}}:=\kappa_{a_{1},c_{1}}^{b_{1}}\cdots\kappa_{a_{d},c_{d}}^{b_{d}}$.
We want to describe the structure constants
$f^{\text{\boldmath$b$}}_{\text{\boldmath$a$},\text{\boldmath$c$}}$ defined
from

$\xi^{\text{\boldmath$a$}}x_{\text{\boldmath$c$}}=\sum_{\text{\boldmath$b$}\in\operatorname{Seq}(B^{V},d)/\mathfrak{S}_{d}}f^{\text{\boldmath$b$}}_{\text{\boldmath$a$},\text{\boldmath$c$}}x_{\text{\boldmath$b$}}.$

Recall the stabilizer $\mathfrak{S}_{\text{\boldmath$s$}}$ from §2.1. The
following lemma is an analogue of [13, Corollary 3.7]:

######  Lemma 3.15.

Let $\text{\boldmath$a$}\in\operatorname{Seq}(B^{A},d)$ and
$\text{\boldmath$b$},\text{\boldmath$c$}\in\operatorname{Seq}(B^{V},d)$. Let
$X$ be the set of all pairs
$(\text{\boldmath$a$}^{\prime},\text{\boldmath$c$}^{\prime})\in\operatorname{Seq}(B^{A},d)\times\operatorname{Seq}(B^{V},d)$
such that $\text{\boldmath$a$}^{\prime}\sim\text{\boldmath$a$}$,
$\text{\boldmath$c$}^{\prime}\sim\text{\boldmath$c$}$ and
$|a^{\prime}_{k}|+|c_{k}^{\prime}|=|b_{k}|$ for all $k=1,\dots,d$. Then

$f^{\text{\boldmath$b$}}_{\text{\boldmath$a$},\text{\boldmath$c$}}=\sum_{(\text{\boldmath$a$}^{\prime},\text{\boldmath$c$}^{\prime})\in
X/\mathfrak{S}_{\text{\boldmath$b$}}}(-1)^{\langle\text{\boldmath$a$}\rangle+\langle\text{\boldmath$a$}^{\prime}\rangle+\langle\text{\boldmath$c$}\rangle+\langle\text{\boldmath$c$}^{\prime}\rangle+\langle\text{\boldmath$a$}^{\prime},\text{\boldmath$c$}^{\prime}\rangle}[\mathfrak{S}_{\text{\boldmath$b$}}:(\mathfrak{S}_{\text{\boldmath$b$}}\cap\mathfrak{S}_{\text{\boldmath$a$}^{\prime}}\cap\mathfrak{S}_{\text{\boldmath$c$}^{\prime}})]\kappa_{\text{\boldmath$a$}^{\prime},\text{\boldmath$c$}^{\prime}}^{\text{\boldmath$b$}}$

###### Proof.

Clearly, we have

$f^{\text{\boldmath$b$}}_{\text{\boldmath$a$},\text{\boldmath$c$}}=\sum(-1)^{\langle\text{\boldmath$a$}\rangle+\langle\text{\boldmath$a$}^{\prime}\rangle+\langle\text{\boldmath$c$}\rangle+\langle\text{\boldmath$c$}^{\prime}\rangle+\langle\text{\boldmath$a$}^{\prime},\text{\boldmath$c$}^{\prime}\rangle}\kappa_{\text{\boldmath$a$}^{\prime},\text{\boldmath$c$}^{\prime}}^{\text{\boldmath$b$}},$

the sum being over all
$(\text{\boldmath$a$}^{\prime},\text{\boldmath$c$}^{\prime})\in\operatorname{Seq}(B^{A},d)\times\operatorname{Seq}(B^{V},d)$
such that $\text{\boldmath$a$}^{\prime}\sim\text{\boldmath$a$}$ and
$\text{\boldmath$c$}^{\prime}\sim\text{\boldmath$c$}$, cf. [8, (3.14)]. It
remains to note that
$\kappa_{\text{\boldmath$a$}^{\prime},\text{\boldmath$c$}^{\prime}}^{\text{\boldmath$b$}}=0$
unless $(\text{\boldmath$a$}^{\prime},\text{\boldmath$c$}^{\prime})\in X$, and
for
$(\text{\boldmath$a$}^{\prime},\text{\boldmath$c$}^{\prime}),(\text{\boldmath$a$}^{\prime\prime},\text{\boldmath$c$}^{\prime\prime})\in
X$ in the same $\mathfrak{S}_{\text{\boldmath$b$}}$-orbit the corresponding
summands are equal to each other, cf. the proof of [13, Corollary 3.7]. ∎

######  Lemma 3.16.

If $V_{R}$ is a calibrated $A_{R}$-supermodule, then
$\tilde{\Gamma}^{d}V_{R}\subseteq\Gamma^{d}V_{R}$ is a
$\tilde{\Gamma}^{d}A_{R}$-submodule.

###### Proof.

For each $\text{\boldmath$a$}\in\operatorname{Seq}(B^{A},d)$ and
$\text{\boldmath$c$}\in\operatorname{Seq}(B^{V},d)$ we have (working over the
field of quotients of $R$):

$\displaystyle\eta^{\text{\boldmath$a$}}y_{\text{\boldmath$c$}}=([\text{\boldmath$a$}]_{\mathfrak{c}}^{!}\xi^{\text{\boldmath$a$}})([\text{\boldmath$c$}]_{\mathfrak{c}}^{!}x_{\text{\boldmath$c$}})=\sum_{\text{\boldmath$b$}\in\operatorname{Seq}(B^{V},d)/\mathfrak{S}_{d}}[\text{\boldmath$a$}]_{\mathfrak{c}}^{!}[\text{\boldmath$c$}]_{\mathfrak{c}}^{!}f^{\text{\boldmath$b$}}_{\text{\boldmath$a$},\text{\boldmath$c$}}x_{\text{\boldmath$b$}}=\sum_{\text{\boldmath$b$}\in\operatorname{Seq}(B^{V},d)/\mathfrak{S}_{d}}\frac{[\text{\boldmath$a$}]_{\mathfrak{c}}^{!}[\text{\boldmath$c$}]_{\mathfrak{c}}^{!}f^{\text{\boldmath$b$}}_{\text{\boldmath$a$},\text{\boldmath$c$}}}{[\text{\boldmath$b$}]_{\mathfrak{c}}^{!}}y_{\text{\boldmath$b$}}.$

So in view of Lemma 3.15, it suffices to prove that for fixed
$\text{\boldmath$a$}\in\operatorname{Seq}(B^{A},d)$,
$\text{\boldmath$b$},\text{\boldmath$c$}\in\operatorname{Seq}(B^{V},d)$ and
$(\text{\boldmath$a$}^{\prime},\text{\boldmath$c$}^{\prime})\in X$ satisfying
$\kappa^{\text{\boldmath$b$}}_{\text{\boldmath$a$}^{\prime},\text{\boldmath$c$}^{\prime}}\neq
0$, the integer

$M_{\text{\boldmath$a$},\text{\boldmath$c$}}^{\text{\boldmath$b$}}:=[\text{\boldmath$a$}]_{\mathfrak{c}}^{!}[\text{\boldmath$c$}]_{\mathfrak{c}}^{!}[\mathfrak{S}_{\text{\boldmath$b$}}:(\mathfrak{S}_{\text{\boldmath$b$}}\cap\mathfrak{S}_{\text{\boldmath$a$}^{\prime}}\cap\mathfrak{S}_{\text{\boldmath$c$}^{\prime}})]$

is divisible by $[\text{\boldmath$b$}]_{\mathfrak{c}}^{!}$. Here, as in Lemma
3.15, $X$ consists of all pairs
$(\text{\boldmath$a$}^{\prime},\text{\boldmath$c$}^{\prime})\in\operatorname{Seq}(B^{A},d)\times\operatorname{Seq}(B^{V},d)$
such that $\text{\boldmath$a$}^{\prime}\sim\text{\boldmath$a$}$,
$\text{\boldmath$c$}^{\prime}\sim\text{\boldmath$c$}$, and
$|a_{k}^{\prime}|+|c_{k}^{\prime}|=|b_{k}|$ for all $k=1,\dots,d$.

For $a\in B^{A}$ and $b,c\in B^{V}$, let

$m_{a,c}^{b}=\\#\\{k\in[d]\mid a^{\prime}_{k}=a,c^{\prime}_{k}=c,b_{k}=b\\}.$

Then, recalling the notation (3.2), we have

$\displaystyle|\mathfrak{S}_{\text{\boldmath$b$}}\cap\mathfrak{S}_{\text{\boldmath$a$}^{\prime}}\cap\mathfrak{S}_{\text{\boldmath$c$}^{\prime}}|$
$\displaystyle=\prod_{a\in B^{A},b,c\in B^{V}}m_{a,c}^{b}!,$
$\displaystyle[\text{\boldmath$a$}:a]$
$\displaystyle=[\text{\boldmath$a$}^{\prime}:a]=\sum_{b,c\in
B^{V}}m_{a,c}^{b},$ $\displaystyle[\text{\boldmath$c$}:c]$
$\displaystyle=[\text{\boldmath$c$}^{\prime}:c]=\sum_{a\in B^{A},b\in
B^{V}}m_{a,c}^{b},$ $\displaystyle[\text{\boldmath$b$}:b]$
$\displaystyle=\sum_{a\in B^{A},c\in B^{V}}m_{a,c}^{b}.$

In particular, for all $b,c\in B^{V}$ and $a\in B^{A}$, we have integers

$z_{b}:=\frac{[\text{\boldmath$b$}:b]!}{\prod_{a\in B^{A},c\in
B^{V}}m_{a,c}^{b}!},\ Z_{c}:=\frac{[\text{\boldmath$c$}:c]!}{\prod_{a\in
B^{A}_{\mathfrak{a}},b\in B_{\mathfrak{c}}^{V}}m_{a,c}^{b}!},\
Z_{a}:=\frac{[\text{\boldmath$a$}:a]!}{\prod_{b\in B_{\mathfrak{c}}^{V},c\in
B_{\bar{0}}^{V}}m_{a,c}^{b}!}.$

Denoting $C=\prod_{b\in B_{\mathfrak{a}}^{V}\sqcup B_{\bar{1}}^{V}}z_{b}$, we
have

$[\mathfrak{S}_{\text{\boldmath$b$}}:\mathfrak{S}_{\text{\boldmath$b$}}\cap\mathfrak{S}_{\text{\boldmath$a$}}^{\prime}\cap\mathfrak{S}_{\text{\boldmath$c$}}^{\prime}]=\frac{\prod_{b\in
B^{V}}[\text{\boldmath$b$}:b]!}{\prod_{a\in B^{A},b,c\in
B^{V}}m^{b}_{a,c}!}=\prod_{b\in B^{V}}z_{b}=C\prod_{b\in
B_{\mathfrak{c}}^{V}}z_{b}.$

Let $b\in B_{\mathfrak{c}}^{V}$. If $a\in B_{{\bar{1}}}^{A}$ or $c\in
B^{V}_{\bar{1}}$, then $m_{a,c}^{b}\leq 1$ because there are no repeated odd
elements in tuples in $\operatorname{Seq}(B^{A},d)$ or
$\operatorname{Seq}(B^{V},d)$. Also observe that if $a\in
B_{\mathfrak{a}}^{A}$ and $c\in B_{\mathfrak{a}}^{V}$, then $ac\in
V_{R,\mathfrak{a}}$ by assumption, so, since $b\in B_{\mathfrak{c}}^{V}$, we
have $\kappa^{b}_{a,c}=0$, hence $m_{a,c}^{b}=0$. So

$z_{b}=\frac{[\text{\boldmath$b$}:b]!}{\left(\prod_{a\in
B^{A}_{\mathfrak{c}},c\in
B_{{\bar{0}}}^{V}}m_{a,c}^{b}!\right)\left(\prod_{a\in
B^{A}_{\mathfrak{a}},c\in B_{\mathfrak{c}}^{V}}m_{a,c}^{b}!\right)}.$

Thus we have

$\displaystyle
M_{\text{\boldmath$a$},\text{\boldmath$c$}}^{\text{\boldmath$b$}}$
$\displaystyle=\Bigg{(}\prod_{a\in
B_{\mathfrak{c}}^{A}}[\text{\boldmath$a$}:a]!\Bigg{)}\Bigg{(}\prod_{c\in
B_{\mathfrak{c}}^{V}}[\text{\boldmath$c$}:c]!\Bigg{)}\cdot C\prod_{b\in
B_{\mathfrak{c}}^{V}}\frac{[\text{\boldmath$b$}:b]!}{\left(\prod_{a\in
B^{A}_{\mathfrak{c}},c\in
B_{{\bar{0}}}^{V}}m_{a,c}^{b}!\right)\left(\prod_{a\in
B^{A}_{\mathfrak{a}},c\in B_{\mathfrak{c}}^{V}}m_{a,c}^{b}!\right)}$
$\displaystyle=C\Bigg{(}\prod_{a\in
B^{A}_{\mathfrak{c}}}\frac{[\text{\boldmath$a$}:a]!}{\prod_{b\in
B_{\mathfrak{c}}^{V},c\in
B_{\bar{0}}^{V}}m_{a,c}^{b}!}\Bigg{)}\Bigg{(}\prod_{c\in
B_{\mathfrak{c}}^{V}}\frac{[\text{\boldmath$c$}:c]!}{\prod_{a\in
B^{A}_{\mathfrak{a}},b\in
B_{\mathfrak{c}}^{V}}m_{a,c}^{b}!}\Bigg{)}\Bigg{(}\prod_{b\in
B^{V}_{\mathfrak{c}}}[\text{\boldmath$b$}:b]!\Bigg{)}$
$\displaystyle=C\Bigg{(}\prod_{a\in
B^{A}_{\mathfrak{c}}}Z_{a}\Bigg{)}\Bigg{(}\prod_{c\in
B_{\mathfrak{c}}^{V}}Z_{c}\Bigg{)}[\text{\boldmath$b$}]_{\mathfrak{c}}^{!},$

which completes the proof. ∎

###  3.5. More on $\tilde{\Gamma}^{d}A_{R}$-module $\tilde{\Gamma}^{d}V_{R}$

Throughout the subsection $A_{R}$ is a calibrated superalgebra and $V_{R}$ is
a calibrated $A_{R}$-supermodule.

Let $e\in\mathfrak{a}_{R}$ be an idempotent such that $be=b$ or $0$ for all
$b\in B^{A}$, cf. [13, §5]. Let $B^{A}e=\\{b\in B^{A}\mid be=b\\}$,
$B^{A}_{\mathfrak{a}}e:=\\{b\in B^{A}_{\mathfrak{a}}\mid be=b\\}$ and
$B^{A}_{\mathfrak{c}}e:=\\{b\in B^{A}_{\mathfrak{c}}\mid be=b\\}$. We have an
idempotent $\eta^{e^{d}}=e^{\otimes d}\in\tilde{\Gamma}^{d}A_{R}$. In the
special case where $V_{R}=A_{R}e$, we always take
$V_{R,\mathfrak{a}}:=\mathfrak{a}_{R}e$ with basis $B^{A}_{\mathfrak{a}}e$ and
$V_{R,\mathfrak{c}}:=\mathfrak{c}_{R}e$ with basis $B^{A}_{\mathfrak{c}}$. In
this case we can describe the $\tilde{\Gamma}^{d}A_{R}$-module
$\tilde{\Gamma}^{d}V_{R}$ explicitly as follows:

######  Lemma 3.17.

Let $e\in\mathfrak{a}_{R}$ be an idempotent such that $be=b$ or $0$ for all
$b\in B^{A}$. Then
$\tilde{\Gamma}^{d}(A_{R}e)\cong(\tilde{\Gamma}^{d}A_{R})\eta^{e^{d}}$.

###### Proof.

Note that $\tilde{\Gamma}^{d}(A_{R}e)$ has basis
$\\{y_{\text{\boldmath$b$}}\mid\text{\boldmath$b$}\in\operatorname{Seq}(B^{A}e,d)/\mathfrak{S}_{d}\\}$
and $(\tilde{\Gamma}^{d}A_{R})\eta^{e^{d}}$ has basis
$\\{\eta^{\text{\boldmath$b$}}\mid\text{\boldmath$b$}\in\operatorname{Seq}(B^{A}e,d)/\mathfrak{S}_{d}\\}$.
There is a $\tilde{\Gamma}^{d}A_{R}$-module map
${\varphi}:(\tilde{\Gamma}^{d}A_{R})\eta^{e^{d}}\to\tilde{\Gamma}^{d}(A_{R}e)$
with ${\varphi}(\eta^{e^{d}})=y_{e^{d}}$. It remains to notice that
${\varphi}(\eta^{\text{\boldmath$b$}})=y_{\text{\boldmath$b$}}$ for all
$\text{\boldmath$b$}\in\operatorname{Seq}(B^{A}e,d)/\mathfrak{S}_{d}$. ∎

Suppose there is an even superalgebra anti-involution $\tau:A_{R}\to A_{R}$.
We make the additional assumption that $\tau(\mathfrak{a})=\mathfrak{a}$, in
which case $\tau^{\otimes d}$ restricts to a superalgebra anti-involution
$\tau_{d}$ on $\tilde{\Gamma}^{d}A_{R}$, see [13, (4.12)].

Given $W\in\tilde{\Gamma}^{d}A_{R}\\!\operatorname{-mod}$, its $\tau_{d}$-dual
$W^{\tau_{d}}$ is defined as $W^{*}:=\operatorname{Hom}_{R}(W,R)$ with the
action $(xf)(w)=(-1)^{|x||f|}f(\tau_{d}(x)w)$ for all $f\in W^{*},w\in
W,x\in\tilde{\Gamma}^{d}A_{R}$. Note that $W\simeq W^{\tau_{d}}$ if and only
if there is a non-degenerate $\tau_{d}$-contravariant form $(\cdot,\cdot)$ on
$W$, where $\tau_{d}$-contravariance means
$(xv,w)=(-1)^{|x||v|}(v,\tau_{d}(x)w)$ for all $x\in\tilde{\Gamma}^{d}A$ and
$v,w\in W$.

######  Lemma 3.18.

Let $(\cdot,\cdot)$ be an even supersymmetric or superantisymmetric, non-
degenerate bilinear form on $V_{R}$, such that
$(V_{R,\mathfrak{a}},V_{R,\mathfrak{a}})=0$ and the $R$-complement
$V_{R,\mathfrak{c}}$ of $V_{R,\mathfrak{a}}$ in $V_{R,{\bar{0}}}$ can be
chosen so that the restriction of $(\cdot,\cdot)$ to $V_{R,\mathfrak{a}}\times
V_{R,\mathfrak{c}}$ is a perfect pairing. Then $(\cdot,\cdot)_{\sim}$ is a
non-degenerate $\tau_{d}$-contravariant form on the
$\tilde{\Gamma}^{d}A_{R}$-module $\tilde{\Gamma}^{d}V_{R}$.

###### Proof.

For $\text{\boldmath$a$}=a_{1}\cdots a_{d}\in A^{d}$ and
$\text{\boldmath$v$}=v_{1}\cdots v_{d}\in V^{d}$, we set
$\text{\boldmath$a$}\cdot\text{\boldmath$v$}:=(a_{1}v_{1})\cdots(a_{d}v_{d})\in
V^{d}$. Then
$(-1)^{\langle\text{\boldmath$a$}\cdot\text{\boldmath$v$},\text{\boldmath$w$}\rangle}=(-1)^{\langle\text{\boldmath$a$},\text{\boldmath$w$}\rangle+\langle\text{\boldmath$v$},\text{\boldmath$w$}\rangle}.$
Using this, it is easy to establish that $(\cdot,\cdot)_{d}$ is a
$\tau^{\otimes d}$-contravariant form on $V^{\otimes d}$, cf. (3.6). The lemma
now follows from Proposition 3.11. ∎

Recall the coproduct from from §3.3, for supermodules
$X\in\tilde{\Gamma}^{d}A_{R}\\!\operatorname{-mod}$ and
$Y\in\tilde{\Gamma}^{e}A_{R}\\!\operatorname{-mod}$, we have a structure of a
$\tilde{\Gamma}^{d+e}A_{R}$-module on $X\otimes Y$. Recalling Lemma 3.14 and
the isomorphism from Lemma 3.5, we now obtain:

######  Lemma 3.19.

Let $V_{R}$ and $W_{R}$ be calibrated $A_{R}$-supermodules. Then we have an
isomorphism of $\tilde{\Gamma}^{d}A_{R}$-modules

$\bigoplus_{d_{1}+d_{2}=d}(\tilde{\Gamma}^{d_{1}}V_{R})\otimes(\tilde{\Gamma}^{d_{2}}W_{R})\stackrel{{\scriptstyle\sim}}{{\longrightarrow}}\tilde{\Gamma}^{d}(V_{R}\oplus
W_{R}),\ y\otimes y^{\prime}\mapsto y*y^{\prime}.$

##  4\. Generalized Schur algebras

Throughout the section we fix $n\in\mathbb{Z}_{>0}$ and a calibrated
superalgebra $A_{R}$ over $R$ as in §3.3. In particular we have fixed
$\mathfrak{a}_{R},\mathfrak{c}_{R},A_{R,{\bar{1}}}$ with bases
$B_{\mathfrak{a}},B_{\mathfrak{c}},B_{\bar{1}}$, and $B=B_{\mathfrak{a}}\sqcup
B_{\mathfrak{c}}\sqcup B_{\bar{1}}$ is a basis of $A_{R}$.

###  4.1. The algebra $T^{A}(n,d)$

We consider the matrix superalgebra $M_{n}(A_{R})$ with
$M_{n}(A_{R})_{\bar{0}}=M_{n}(A_{R,{\bar{0}}})$ and
$M_{n}(A_{R})_{\bar{1}}=M_{n}(A_{R,{\bar{1}}})$. For $b\in B$, we denote by
$\xi_{r,s}^{b}\in M_{n}(A_{R})$ so that

$B^{M_{n}(A)}:=\\{\xi_{r,s}^{b}\mid b\in B,\,1\leq r,s\leq n\\}$ (4.1)

is a basis of $M_{n}(A_{R})$.

This superalgebra is calibrated via
$M_{n}(A_{R})_{\mathfrak{a}}:=M_{n}(\mathfrak{a}_{R})$ and
$M_{n}(A_{R})_{\mathfrak{c}}:=M_{n}(\mathfrak{c}_{R})$. So for
$d\in\mathbb{Z}_{\geq 0}$, we may define

$T^{A}(n,d)_{R}:=\tilde{\Gamma}^{d}M_{n}(A_{R}).$

The theory developed in Section 3 applies with $A_{R}$ replaced by
$M_{n}(A_{R})$. To facilitate the transition from $A_{R}$ to $M_{n}(A_{R})$,
it is convenient to adopt a special notation. First of all, taking into
account that the basis of $M_{n}(A_{R})$ is of the form (4.1), we have element
$\eta^{\text{\boldmath$x$}}$ from (3.12) labeled by $d$-tuples
$\text{\boldmath$x$}=\xi^{b_{1}}_{r_{1},s_{1}}\,\dots\,\xi^{b_{d}}_{r_{d},s_{d}}$
of basis elements from (4.1) such that
$(b_{i},r_{i},s_{i})=(b_{j},r_{j},s_{j})$ for $i\neq j$ only if $b_{i}$ is
even. It is convenient to denote such $\eta^{\text{\boldmath$x$}}$ rather by
$\eta^{\text{\boldmath$b$}}_{\text{\boldmath$r$},\text{\boldmath$s$}}$ where
$\text{\boldmath$b$}:=b_{1}\dots b_{d}$, $\text{\boldmath$r$}:=r_{1}\cdots
r_{d}$ and $\text{\boldmath$s$}:=s_{1}\dots s_{d}$. Thus,

$\eta_{\text{\boldmath$r$},\text{\boldmath$s$}}^{\text{\boldmath$b$}}:=[\text{\boldmath$b$},\text{\boldmath$r$},\text{\boldmath$s$}]^{!}_{\mathfrak{c}}\sum_{(\text{\boldmath$c$},\text{\boldmath$t$},\text{\boldmath$u$})\sim(\text{\boldmath$b$},\text{\boldmath$r$},\text{\boldmath$s$})}(-1)^{\langle\text{\boldmath$b$},\text{\boldmath$r$},\text{\boldmath$s$}\rangle+\langle\text{\boldmath$c$},\text{\boldmath$t$},\text{\boldmath$u$}\rangle}\xi_{t_{1},u_{1}}^{c_{1}}\otimes\dots\otimes\xi_{t_{d},u_{d}}^{c_{d}}\in
T^{A}(n,d)_{R},$

where

$\displaystyle\langle\text{\boldmath$b$},\text{\boldmath$r$},\text{\boldmath$s$}\rangle$
$\displaystyle:=|\\{(k,l)\in[d]^{2}\mid k<l,\ b_{k},b_{l}\in B_{\bar{1}},\
(b_{k},r_{k},s_{k})>(b_{l},r_{l},s_{l})\\}|$ (4.2)

is the analogue of (3.1) (where ‘$>$’ is a fixed total order on
$B\times[n]\times[n]$) and

$[\text{\boldmath$b$},\text{\boldmath$r$},\text{\boldmath$s$}]^{!}_{\mathfrak{c}}:=\prod_{b\in
B_{\mathfrak{c}},\,r,s\in[n]}|\\{k\in[d]\mid(b_{k},r_{k},s_{k})=(b,r,s)\\}|!$
(4.3)

is the analogue of (3.3).

Define $\operatorname{Tri}^{B}(n,d)$ to be the set of all triples

$(\text{\boldmath$b$},\text{\boldmath$r$},\text{\boldmath$s$})=(b_{1}\cdots
b_{d},\,r_{1}\cdots r_{d},\,s_{1}\cdots s_{d})\in
B^{d}\times[n]^{d}\times[n]^{d}$

such that for all $1\leq k\neq l\leq d$ we have
$(b_{k},r_{k},s_{k})=(b_{l},r_{l},s_{l})$ only if $b_{k}\in B_{\bar{0}}$. To
connect with the theory developed in §3, we identify
$\operatorname{Tri}^{B}(n,d)$ with $\operatorname{Seq}(B^{M_{n}(A)},d)$ via
the map

$\operatorname{Tri}^{B}(n,d)\stackrel{{\scriptstyle\sim}}{{\longrightarrow}}\operatorname{Seq}(B^{M_{n}(A)},d),\
(\text{\boldmath$b$},\text{\boldmath$r$},\text{\boldmath$s$})\mapsto\xi^{b_{1}}_{r_{1},s_{1}}\,\cdots\,\xi^{b_{d}}_{r_{d},s_{d}}.$

Since $\operatorname{Tri}^{B}(n,d)\subseteq B^{d}\times[n]^{d}\times[n]^{d}$
is a $\mathfrak{S}_{d}$-invariant subset, we can choose a corresponding set
$\operatorname{Tri}^{B}(n,d)/\mathfrak{S}_{d}$ of $\mathfrak{S}_{d}$-orbit
representatives and identify it with the set of all $\mathfrak{S}_{d}$-orbits
on $\operatorname{Tri}^{B}(n,d)$ as in §2.1. Then we have a basis

$\\{\eta_{\text{\boldmath$r$},\text{\boldmath$s$}}^{\text{\boldmath$b$}}\mid(\text{\boldmath$b$},\text{\boldmath$r$},\text{\boldmath$s$})\in\operatorname{Tri}^{B}(n,d)/\mathfrak{S}_{d}\\}$

of $T^{A}(n,d)_{R}$ which is the analogue of the basis (3.13).

Applying the theory developed in §3.3 replacing $A_{R}$ with $M_{n}(A_{R})$,
we also have a coproduct  ${\tt\Delta}$ on $\bigoplus_{d\geq
0}T^{A}(n,d)_{R}$, which allows us to consider the tensor product $V\otimes W$
of $V\in T^{A}(n,d)_{R}\\!\operatorname{-mod}$ and $W\in
T^{A}(n,e)_{R}\\!\operatorname{-mod}$ as a supermodule over
$T^{A}(n,d+e)_{R}$.

Extending scalars from $R$ to $\mathbb{F}$, we now define the
$\mathbb{F}$-superalgebra

$T^{A}(n,d):=\mathbb{F}\otimes_{R}T^{A}(n,d)_{R}.$

We denote
$1_{\mathbb{F}}\otimes\eta^{\text{\boldmath$b$}}_{\text{\boldmath$r$},\text{\boldmath$s$}}\in
T^{A}(n,d)$ again by
$\eta^{\text{\boldmath$b$}}_{\text{\boldmath$r$},\text{\boldmath$s$}}$, the
map
$\mathop{\mathrm{id}}\nolimits_{\mathbb{F}}\otimes\,\reflectbox{\rotatebox[origin={c}]{180.0}{${\tt\Delta}$}}$
again by  ${\tt\Delta}$ , etc. In particular, given $V\in
T^{A}(n,d)\\!\operatorname{-mod}$ and $W\in T^{A}(n,e)\\!\operatorname{-mod}$,
we consider $V\otimes W$ as a $T^{A}(n,d+e)$-supermodule via  ${\tt\Delta}$ .

###  4.2. Quasi-hereditary structure on $T^{A}(n,d)$

Throughout the section, we assume that $d\leq n$.

Let $A_{R}$ be a based quasi-hereditary superalgebra with conforming heredity
data $I,X,Y$, see §2.3. In particular, $A_{R}$ comes with the standard
idempotents $\\{e_{i}\mid i\in I\\}$ in $\mathfrak{a}_{R}$.

For $\lambda\in\Lambda(n,d)$, set
$\text{\boldmath$l$}^{\lambda}:=1^{\lambda_{1}}\cdots
n^{\lambda_{n}}\in[n]^{d}.$ For an idempotent $e\in\mathfrak{a}_{R}$ we have
an idempotent
$\eta_{\lambda}^{e}:=\eta^{e^{d}}_{\text{\boldmath$l$}^{\lambda},\text{\boldmath$l$}^{\lambda}}\in
T^{A}(n,d)$. For all
$\text{\boldmath$\lambda$}=(\lambda^{(0)},\dots,\lambda^{(\ell)})\in\Lambda^{I}(n,d)$,
we have the orthogonal idempotents

$\eta_{\text{\boldmath$\lambda$}}:=\eta_{\lambda^{(0)}}^{e_{0}}*\cdots*\eta_{\lambda^{(\ell)}}^{e_{\ell}}\in
T^{A}(n,d).$

Given a $T^{A}(n,d)$-supermodule $V$ we refer to the vectors of
$\eta_{\text{\boldmath$\lambda$}}V$ as vectors of weight $\lambda$.

For $\lambda=(\lambda_{1},\dots,\lambda_{n})\in\Lambda(n)$, define the
monomial $z^{\lambda}:=z_{1}^{\lambda_{1}}\cdots
z_{n}^{\lambda_{n}}\in\mathbb{Z}[z_{1},\dots,z_{n}].$ For
$\text{\boldmath$\lambda$}\in\Lambda^{I}(n)$, we now set

$z^{\text{\boldmath$\lambda$}}:=z^{\lambda^{(0)}}\otimes
z^{\lambda^{(1)}}\otimes\dots\otimes
z^{\lambda^{(\ell)}}\in\mathbb{Z}[z_{1},\dots,z_{n}]^{\otimes I}.$

Following [13, §5A], see especially [13, Lemma 5.9], for a
$T^{A}(n,d)$-supermodule $V$, we define its formal character

$\operatorname{ch}V:=\sum_{\text{\boldmath$\mu$}\in\Lambda^{I}(n,d)}(\dim\,\eta_{\text{\boldmath$\mu$}}V)z^{\text{\boldmath$\mu$}}\in\mathbb{Z}[z_{1},\dots,z_{n}]^{\otimes
I}.$

If $\sum_{i\in I}e_{i}=1_{A}$, then
$1_{T^{A}(n,d)}=\sum_{\text{\boldmath$\lambda$}\in\Lambda^{I}(n,d)}\eta_{\text{\boldmath$\lambda$}},$
but we do not need to assume this. So in general we might have
$\sum_{\text{\boldmath$\mu$}\in\Lambda^{I}(n,d)}\eta_{\text{\boldmath$\mu$}}V\subsetneq
V$ for $V\in T^{A}(n,d)\\!\operatorname{-mod}$. (Another fact that is not
going to be used directly is that $\operatorname{ch}V$ is a symmetric
function.)

######  Lemma 4.4.

[13, Lemma 5.10] If $V\in T^{A}(n,d)\\!\operatorname{-mod}$ and $W\in
T^{A}(n,e)\\!\operatorname{-mod}$, then $\operatorname{ch}(V\otimes
W)=\operatorname{ch}(V)\,\operatorname{ch}(W).$

By [14, Theorem 6.6], $T^{A}(n,d)$ is a based quasi-hereditary algebra with
standard idempotents
$\\{\eta_{\text{\boldmath$\lambda$}}\mid\text{\boldmath$\lambda$}\in\Lambda_{+}^{I}(n,d)\\}$,
see [14, §6]. The corresponding simple, standard and costandard modules are
denoted

$\\{L(\text{\boldmath$\lambda$})\mid\text{\boldmath$\lambda$}\in\Lambda_{+}^{I}(n,d)\\},\quad\\{\Delta(\text{\boldmath$\lambda$})\mid\text{\boldmath$\lambda$}\in\Lambda_{+}^{I}(n,d)\\},\quad\\{\nabla(\text{\boldmath$\lambda$})\mid\text{\boldmath$\lambda$}\in\Lambda_{+}^{I}(n,d)\\},$

respectively.

Recalling the $X$-colored standard tableaux from §2.4, for
$\text{\boldmath$\lambda$}\in\Lambda_{+}^{I}(n,d)$, the standard module
$\Delta(\text{\boldmath$\lambda$})$ has basis

$\\{v_{\text{\boldmath$T$}}\mid\text{\boldmath$T$}\in\operatorname{Std}^{X}(\text{\boldmath$\lambda$})\\}.$
(4.5)

Moreover, if
$\text{\boldmath$T$}\in\operatorname{Std}^{X}(\text{\boldmath$\lambda$},\text{\boldmath$\mu$})$
for some $\text{\boldmath$\mu$}\in\Lambda^{I}(n,d)$, see (2.13), then

$v_{\text{\boldmath$T$}}\in\eta_{\text{\boldmath$\mu$}}\Delta(\text{\boldmath$\lambda$}).$
(4.6)

######  Lemma 4.7.

[15, Lemma 3.23] The formal characters
$\\{\operatorname{ch}\Delta(\text{\boldmath$\lambda$})\mid\text{\boldmath$\lambda$}\in\Lambda_{+}^{I}(n,d)\\}$
are linearly independent.

Recalling the notation (2.12), we have:

######  Lemma 4.8.

[14, Theorem 6.17(i)] Let $\text{\boldmath$\lambda$}=(\lambda^{(i)})_{i\in
I}\in\Lambda_{+}^{I}(n,d)$. Then

$\Delta(\text{\boldmath$\lambda$})\simeq\bigotimes_{i\in
I}\Delta(\text{\boldmath$\iota$}_{i}(\lambda^{(i)})).$

For $\text{\boldmath$\lambda$}=(\lambda^{(j)})_{j\in
I}\in\Lambda^{I}_{+}(n,d)$, $\text{\boldmath$\mu$}=(\mu^{(j)})_{j\in
I}\in\Lambda^{I}_{+}(n,e)$ and $\text{\boldmath$\nu$}=(\nu^{(j)})_{j\in
I}\in\Lambda_{+}^{I}(n,d+e)$ we define

$c^{\,\text{\boldmath$\nu$}}_{\text{\boldmath$\lambda$},\text{\boldmath$\mu$}}:=\prod_{j\in
I}c^{\,\nu^{(j)}}_{\lambda^{(j)},\mu^{(j)}}.$ (4.9)

We will crucially use the following

######  Theorem 4.10.

[15, Main Theorem, Corollary 3.30] Let
$\text{\boldmath$\lambda$}\in\Lambda^{I}_{+}(n,d)$ and
$\text{\boldmath$\mu$}\in\Lambda^{I}_{+}(n,e)$. If $d+e\leq n$ then
$\Delta(\text{\boldmath$\lambda$})\otimes\Delta(\text{\boldmath$\mu$})$ (resp.
$\nabla(\text{\boldmath$\lambda$})\otimes\nabla(\text{\boldmath$\mu$})$) has a
standard (resp. costandard) filtration, and

$\operatorname{ch}(\Delta(\text{\boldmath$\lambda$})\otimes\Delta(\text{\boldmath$\mu$}))=\sum_{\text{\boldmath$\nu$}\in\Lambda_{+}^{I}(n,d+e)}c^{\,\text{\boldmath$\nu$}}_{\text{\boldmath$\lambda$},\text{\boldmath$\mu$}}\operatorname{ch}\Delta(\text{\boldmath$\nu$}).$

An immediate corollary of Theorem 4.10 is

######  Corollary 4.11.

Let $\operatorname{T}\in T^{A}(n,d)\\!\operatorname{-mod}$ and
$\operatorname{T}^{\prime}\in T^{A}(n,e)\\!\operatorname{-mod}$ be tilting
supermodules. If $d+e\leq n$ then the tensor product
$\operatorname{T}\otimes\operatorname{T}^{\prime}$ is a tilting supermodule
over $T^{A}(n,d+e)$.

###  4.3. Modified divided powers of modules over $M_{n}(A_{R})$

In this subsection we work over $R$.

Let $A_{R}=\mathfrak{a}_{R}\oplus\mathfrak{c}_{R}\oplus A_{R,{\bar{1}}}$ be a
calibrated superalgebra as in §3.3 with basis $B^{A}=B^{A}_{\mathfrak{a}}\cup
B^{A}_{\mathfrak{c}}\cup B^{A}_{\bar{1}}$, and $V_{R}=V_{R,\mathfrak{a}}\oplus
V_{R,\mathfrak{c}}\oplus V_{R,{\bar{1}}}$ be a calibrated $A_{R}$-supermodule
as in §3.4. Recall that by definition $T^{A}(n,d)_{R}$ is
$\tilde{\Gamma}^{d}M_{n}(A_{R})$ for the calibration on $M_{n}(A_{R})$ induced
by that on $A_{R}$.

The $R$-supermodule of column vectors
$\operatorname{Col}_{n}(V_{R})=V_{R}^{\oplus n}$ is a left supermodule over
$M_{n}(A_{R})$ in a natural way. In fact, it is a calibrated
$M_{n}(A_{R})$-supermodule with
$\operatorname{Col}_{n}(V_{R})_{\mathfrak{a}}:=\operatorname{Col}_{n}(V_{R,\mathfrak{a}})$
and
$\operatorname{Col}_{n}(V_{R})_{\mathfrak{c}}:=\operatorname{Col}_{n}(V_{R,\mathfrak{c}})$.
Then by Lemma 3.16, we have $\tilde{\Gamma}^{d}\operatorname{Col}_{n}(V_{R})$
is a left supermodule over $T^{A}(n,d)_{R}=\tilde{\Gamma}^{d}M_{n}(A_{R})$.

Extending scalars to $\mathbb{F}$ we get the left module

$\tilde{\Gamma}^{d}\operatorname{Col}_{n}(V):=\mathbb{F}\otimes_{R}\tilde{\Gamma}^{d}\operatorname{Col}_{n}(V_{R})$

over $T^{A}(n,d)=\mathbb{F}\otimes_{R}T^{A}(n,d)_{R}$.

######  Lemma 4.12.

Let $e\in\mathfrak{a}_{R}$ be an idempotent such that $be=b$ or $0$ for all
$b\in B^{A}$. Then $\tilde{\Gamma}^{d}(\operatorname{Col}_{n}(A_{R}e))\simeq
T^{A}(n,d)_{R}\eta_{1^{d},1^{d}}^{e^{d}}$.

###### Proof.

First notice that $\operatorname{Col}_{n}(A_{R}e)\simeq
M_{n}(A_{R})\xi_{1,1}^{e}$ as $M_{n}(A_{R})$-supermodules. The result now
follows by applying Lemma 3.17. ∎

We will also need the right module versions of the results of this subsection.
If $V_{R}$ is a calibrated right $A_{R}$-supermodule, we consider the row
vectors $\operatorname{Row}_{n}(V_{R})=V_{R}^{\oplus n}$ as a calibrated right
$M_{n}(A_{R})$-supermodule. Then we obtain a right
$T^{A}(n,d)_{R}$-supermodule structure on
$\tilde{\Gamma}^{d}\operatorname{Row}_{n}(V_{R})$ and a right
$T^{A}(n,d)$-module
$\tilde{\Gamma}^{d}\operatorname{Row}_{n}(V):=\mathbb{F}\otimes_{R}\tilde{\Gamma}^{d}\operatorname{Row}_{n}(V_{R})$.
The right module analogue of Lemma 4.12 is then clear.

##  5\. The extended zigzag Schur algebra

###  5.1. The extended zigzag algebra

Fix $\ell\geq 1$ and set

$I:=\\{0,1,\dots,\ell\\},\quad J:=\\{0,1,\dots,\ell-1\\}.$

Let $\Gamma$ be the quiver with vertex set $I$ and arrows
$\\{{\mathsf{a}}_{j,j+1},{\mathsf{a}}_{j+1,j}\mid j\in J\\}$ as in the
picture:

The extended zigzag algebra ${\mathsf{Z}}$ is the path algebra $\Bbbk\Gamma$
modulo the following relations:

  1. (i)

All paths of length three or greater are zero.

  2. (ii)

All paths of length two that are not cycles are zero.

  3. (iii)

All length-two cycles based at the same vertex are equivalent.

  4. (iv)

${\mathsf{a}}_{\ell,\ell-1}{\mathsf{a}}_{\ell-1,\ell}=0$.

Length zero paths yield the standard idempotents $\\{{\mathsf{e}}_{i}\mid i\in
I\\}$ with
${\mathsf{e}}_{i}{\mathsf{a}}_{i,j}{\mathsf{e}}_{j}={\mathsf{a}}_{i,j}$ for
all admissible $i,j$. The algebra ${\mathsf{Z}}$ is graded by the path length:
${\mathsf{Z}}={\mathsf{Z}}^{0}\oplus{\mathsf{Z}}^{1}\oplus{\mathsf{Z}}^{2}.$
We consider ${\mathsf{Z}}$ as a superalgebra with
${\mathsf{Z}}_{\bar{0}}={\mathsf{Z}}^{0}\oplus{\mathsf{Z}}^{2}\quad\text{and}\quad{\mathsf{Z}}_{\bar{1}}={\mathsf{Z}}^{1}.$
Define ${\mathsf{c}}_{j}:={\mathsf{a}}_{j,j+1}{\mathsf{a}}_{j+1,j}$ for all
$j\in J$.

The superalgebra ${\mathsf{Z}}$ has an anti-involution $\tau$ with

$\tau({\mathsf{e}}_{i})={\mathsf{e}}_{i},\
\tau({\mathsf{a}}_{ij})={\mathsf{a}}_{ji},\
\tau({\mathsf{c}}_{j})=-{\mathsf{c}}_{j}.$

We consider the total order on $I$ given by $0<1<\dots<\ell$. For $i\in I$, we
set

${\mathsf{X}}(i):=\left\\{\begin{array}[]{ll}\\{{\mathsf{e}}_{i},{\mathsf{a}}_{i-1,i}\\}&\hbox{if
$i>0$,}\\\ \\{{\mathsf{e}}_{0}\\}&\hbox{if
$i=0$,}\end{array}\right.\quad{\mathsf{Y}}(i):=\left\\{\begin{array}[]{ll}\\{{\mathsf{e}}_{i},{\mathsf{a}}_{i,i-1}\\}&\hbox{if
$i>0$,}\\\ \\{{\mathsf{e}}_{0}\\}&\hbox{if $i=0$.}\end{array}\right.$ (5.1)

With respect to this data we have:

######  Lemma 5.2.

[12, Lemma 4.14] The graded superalgebra ${\mathsf{Z}}$ is based quasi-
hereditary with conforming heredity data $I,{\mathsf{X}},{\mathsf{Y}}$. For
the corresponding heredity basis ${\mathsf{B}}$ we have
${\mathsf{B}}_{\mathfrak{a}}=\\{{\mathsf{e}}_{i}\mid i\in I\\},\
{\mathsf{B}}_{\mathfrak{c}}=\\{{\mathsf{c}}_{j}\mid j\in J\\},\
{\mathsf{B}}_{\bar{1}}=\\{{\mathsf{a}}_{j,j+1},{\mathsf{a}}_{j+1,j}\mid j\in
J\\}.$

For $i\in I$, let ${\mathsf{L}}(i)=\Bbbk\cdot{\mathsf{v}}_{i}$ with
$|{\mathsf{v}}_{i}|={\bar{0}}$ and the action
${\mathsf{e}}_{i}{\mathsf{v}}_{i}={\mathsf{v}}_{i}$,
${\mathsf{b}}{\mathsf{v}}_{i}=0$ for all
${\mathsf{b}}\in{\mathsf{B}}\setminus\\{{\mathsf{e}}_{i}\\}$. This makes
${\mathsf{L}}(i)$ a ${\mathsf{Z}}$-supermodule, and, up to isomorphism,
$\\{{\mathsf{L}}(i)\mid i\in I\\}$ is a complete set of irreducible
${\mathsf{Z}}$-supermodules. Note that $L(i)^{\tau}\simeq L(i)$.

The standard modules $\Delta(i)$ have similarly explicit description:
$\Delta(0)\simeq{\mathsf{L}}(0)$, and for $i>0$, $\Delta(i)$ has basis
$\\{{\mathsf{v}}_{i},{\mathsf{w}}_{i}\\}$ with $|{\mathsf{v}}_{i}|={\bar{0}}$,
$|{\mathsf{w}}_{i}|={\bar{1}}$ and the only non-trivial action is:
${\mathsf{e}}_{i}{\mathsf{v}}_{i}={\mathsf{v}}_{i}$,
${\mathsf{e}}_{i-1}{\mathsf{w}}_{i}={\mathsf{w}}_{i}$,
${\mathsf{a}}_{i-1,i}{\mathsf{v}}_{i}={\mathsf{w}}_{i}$. For the costandard
modules we have $\nabla(0)\simeq\Delta(0)^{\tau}\simeq{\mathsf{L}}(0)$, and
for $i>0$, $\nabla(i)\simeq\Delta(i)^{\tau}$ has basis
$\\{{\mathsf{v}}_{i}^{*},{\mathsf{w}}_{i}^{*}\\}$ with
$|{\mathsf{v}}_{i}^{*}|={\bar{0}}$, $|{\mathsf{w}}_{i}^{*}|={\bar{1}}$ and the
only non-trivial action is:
${\mathsf{e}}_{i}{\mathsf{v}}_{i}^{*}={\mathsf{v}}_{i}^{*}$,
${\mathsf{e}}_{i-1}{\mathsf{w}}_{i}^{*}={\mathsf{w}}_{i}^{*}$,
${\mathsf{a}}_{i,i-1}{\mathsf{w}}_{i}^{*}=-{\mathsf{v}}_{i}^{*}$.

The indecomposable tilting supermodules over ${\mathsf{Z}}$ are as follows:
$\mathsf{T}(0)\simeq{\mathsf{L}}(0)$ and
$\mathsf{T}(i)=\Pi{\mathsf{Z}}{\mathsf{e}}_{i-1}$ for $i>0$. It will actually
be more convenient for us to work with the tilting modules $\Pi\mathsf{T}(i)$.
Thus $\Pi\mathsf{T}(0)=\Pi{\mathsf{L}}(0)$ is $1$-dimensional with basis
$\\{{\mathsf{v}}_{0}\\}$ where ${\mathsf{v}}_{0}$ is odd,
$\Pi\mathsf{T}(1)={\mathsf{Z}}{\mathsf{e}}_{0}$ has basis
$\\{{\mathsf{e}}_{0},{\mathsf{a}}_{1,0},{\mathsf{c}}_{0}\\}$, and, for $i>1$,
$\Pi\mathsf{T}(i)={\mathsf{Z}}{\mathsf{e}}_{i-1}$ has basis
$\\{{\mathsf{e}}_{i-1},{\mathsf{a}}_{i-2,i-1},{\mathsf{a}}_{i,i-1},{\mathsf{c}}_{i-1}\\}$.
For $i>0$, $\Pi\mathsf{T}(i)$ has a standard filtration
$\Pi\Delta(i)|\Delta(i-1)$ (this means $\Pi\Delta(i)\subseteq\mathsf{T}(i)$
and $\mathsf{T}(i)/\Pi\Delta(i)\simeq\Delta(i-1)$) and a costandard filtration
$\nabla(i-1)|\Pi\nabla(i)$.

We have a full tilting module

$\mathsf{T}:=\bigoplus_{i\in I}\Pi\mathsf{T}(i)=\Pi
L(0)\oplus\bigoplus_{i=0}^{\ell-1}{\mathsf{Z}}{\mathsf{e}}_{i}.$

and the Ringel dual algebra
${\mathsf{Z}}^{\prime}=\operatorname{End}_{\mathsf{Z}}(\mathsf{T})^{\mathrm{sop}}$.
In fact, ${\mathsf{Z}}$ is Ringel self-dual, i.e. there is an isomorphism of
superalgebras ${\mathsf{Z}}^{\prime}\cong{\mathsf{Z}}$ which we now proceed to
describe.

For any $i\in I$, let $\iota_{i}:\Pi\mathsf{T}(i)\to\mathsf{T}$ and
$\pi_{i}:\mathsf{T}\to\Pi\mathsf{T}(i)$ be the natural embedding and
projection. We have the right multiplication maps
$\rho_{{\mathsf{c}}_{i}}:{\mathsf{Z}}{\mathsf{e}}_{i}\to{\mathsf{Z}}{\mathsf{e}}_{i},{\mathsf{v}}\mapsto{\mathsf{v}}{\mathsf{c}}_{i}$
and
$\rho_{{\mathsf{a}}_{ij}}:{\mathsf{Z}}{\mathsf{e}}_{i}\to{\mathsf{Z}}{\mathsf{e}}_{j},{\mathsf{v}}\mapsto(-1)^{|{\mathsf{v}}|}{\mathsf{v}}{\mathsf{a}}_{ij}$.
Let
$f:\Pi\mathsf{T}(0)=\Pi{\mathsf{L}}(0)\,{\hookrightarrow}\,\Pi\mathsf{T}(1)={\mathsf{Z}}{\mathsf{e}}_{0}$
be the embedding given by ${\mathsf{v}}_{0}\mapsto{\mathsf{c}}_{0}$, and let
$g:\Pi\mathsf{T}(1)={\mathsf{Z}}{\mathsf{e}}_{0}\,{\twoheadrightarrow}\,\Pi\mathsf{T}(0)=\Pi{\mathsf{L}}(0)$
be the surjection such that ${\mathsf{e}}_{0}\mapsto{\mathsf{v}}_{0}$. Note
that $f$ and $g$ are odd. Define the following elements of
${\mathsf{Z}}^{\prime}$:

  1. $\bullet$

${\mathsf{e}}_{i}^{\prime}:=\pi_{\ell-i}$ for all $i\in I$;

  2. $\bullet$

${\mathsf{c}}_{i}^{\prime}:=\iota_{\ell-i}\circ\rho_{{\mathsf{c}}_{\ell-i-1}}\circ\pi_{\ell-i}$
for all $i\in J$;

  3. $\bullet$

${\mathsf{a}}_{i+1,i}^{\prime}:=\iota_{\ell-i}\circ\rho_{{\mathsf{a}}_{\ell-i-2,\ell-i-1}}\circ\pi_{\ell-i-1}$
and
${\mathsf{a}}_{i,i+1}^{\prime}:=\iota_{\ell-i-1}\circ\rho_{{\mathsf{a}}_{\ell-i-1,\ell-i-2}}\circ\pi_{\ell-i}$,
for all $i=0,\dots,\ell-2$;

  4. $\bullet$

${\mathsf{a}}_{\ell,\ell-1}^{\prime}:=\iota_{1}\circ f\circ\pi_{0}$ and
${\mathsf{a}}_{\ell-1,\ell}^{\prime}:=\iota_{0}\circ g\circ\pi_{1}$.

######  Lemma 5.3.

Mapping ${\mathsf{e}}_{i}\mapsto{\mathsf{e}}_{i}^{\prime},\
{\mathsf{a}}_{i,j}\mapsto{\mathsf{a}}_{i,j}^{\prime},\
{\mathsf{c}}_{i}\mapsto{\mathsf{c}}_{i}^{\prime}$ is an isomorphism of
superalgebras
${\mathsf{Z}}\stackrel{{\scriptstyle\sim}}{{\longrightarrow}}{\mathsf{Z}}^{\prime}$.
In particular, ${\mathsf{Z}}$ is Ringel self-dual.

###### Proof.

Using the fact that
$\dim\operatorname{Hom}_{{\mathsf{Z}}}(\Delta(i),\nabla(j))=0$ if $i>j$, and
is equal to $1$ if $i=j$, it is easy to see that

$\\{{\mathsf{e}}_{i}^{\prime}\mid i\in
I\\}\sqcup\\{{\mathsf{a}}_{j,j+1}^{\prime},{\mathsf{a}}_{j+1,j}^{\prime},{\mathsf{c}}_{j}^{\prime}\mid
j\in J\\}$

is a basis for ${\mathsf{Z}}^{\prime}$. Hence the given map is a linear
isomorphism.

To check that the map is an algebra homomorphism, one proceeds by cases. For
example, for $i\in J$, in
${\mathsf{Z}}^{\prime}=\operatorname{End}_{{\mathsf{Z}}}(\mathsf{T})^{\mathrm{sop}}$
we have

$\displaystyle{\mathsf{a}}_{i,i+1}^{\prime}\cdot{\mathsf{a}}_{i+1,i}^{\prime}$
$\displaystyle=-{\mathsf{a}}_{i+1,i}^{\prime}\circ{\mathsf{a}}_{i,i+1}^{\prime}$
$\displaystyle=-\iota_{\ell-i}\circ\rho_{{\mathsf{a}}_{\ell-i-2,\ell-i-1}}\circ\pi_{\ell-i-1}\circ\iota_{\ell-i-1}\circ\rho_{{\mathsf{a}}_{\ell-i-1,\ell-i-2}}\circ\pi_{\ell-i}$
$\displaystyle=-\iota_{\ell-i}\circ\rho_{{\mathsf{a}}_{\ell-i-2,\ell-i-1}}\circ\rho_{{\mathsf{a}}_{\ell-i-1,\ell-i-2}}\circ\pi_{\ell-i}$
$\displaystyle=\iota_{\ell-i}\circ\rho_{c_{\ell-i-1}}\circ\pi_{\ell-i}$
$\displaystyle={\mathsf{c}}_{i}^{\prime}.$

The other cases are checked similarly. ∎

We use the isomorphism
${\mathsf{Z}}\stackrel{{\scriptstyle\sim}}{{\longrightarrow}}{\mathsf{Z}}^{\prime}$
of Lemma 5.3 to transport the heredity data $I,{\mathsf{X}},{\mathsf{Y}}$ from
${\mathsf{Z}}$ onto a heredity data
$I^{\prime},{\mathsf{X}}^{\prime},{\mathsf{Y}}^{\prime}$ for
${\mathsf{Z}}^{\prime}$ so that $I^{\prime}=I$ with the same order, and

${\mathsf{X}}^{\prime}(i):=\left\\{\begin{array}[]{ll}\\{{\mathsf{e}}_{i}^{\prime},{\mathsf{a}}_{i-1,i}^{\prime}\\}&\hbox{if
$i>0$,}\\\ \\{{\mathsf{e}}_{0}^{\prime}\\}&\hbox{if
$i=0$,}\end{array}\right.\quad{\mathsf{Y}}^{\prime}(i):=\left\\{\begin{array}[]{ll}\\{{\mathsf{e}}_{i}^{\prime},{\mathsf{a}}_{i,i-1}^{\prime}\\}&\hbox{if
$i>0$,}\\\ \\{{\mathsf{e}}_{0}^{\prime}\\}&\hbox{if $i=0$.}\end{array}\right.$

(Usually, one gets the opposite partial order on $I$ in this place but we have
built in a relabeling $i\mapsto\ell-i$ into the construction). With this
hereditary data, we have the right modules
${\mathsf{L}}^{\prime}(i),\Delta^{\prime}(i),\nabla^{\prime}(i)$ and
$\mathsf{T}^{\prime}(i)$. For example,
$\mathsf{T}^{\prime}(0)\simeq{\mathsf{L}}^{\prime}(0)$ (with
${\mathsf{e}}_{0}^{\prime}$ acting as identity and all the other standard
generators acting as $0$) and
$\mathsf{T}^{\prime}(i)\simeq\Pi{\mathsf{e}}^{\prime}_{i-1}{\mathsf{Z}}^{\prime}$
for $i>0$. A routine check shows that, as a right
${\mathsf{Z}}^{\prime}$-module, $\mathsf{T}$ decomposes as follows:

$\mathsf{T}=\bigoplus_{i\in I}\Pi\mathsf{T}^{\prime}(i),$

where the summands are defined explicitly as follows:

  1. $\bullet$

$\Pi\mathsf{T}^{\prime}(0)=\Bbbk\cdot{\mathsf{a}}_{\ell,\ell-1}\subseteq{\mathsf{Z}}{\mathsf{e}}_{\ell-1}=\Pi\mathsf{T}(\ell)\subseteq\mathsf{T}$;

  2. $\bullet$

$\Pi\mathsf{T}^{\prime}(\ell)=\operatorname{span}_{\Bbbk}({\mathsf{v}}_{0},{\mathsf{e}}_{0},{\mathsf{c}}_{0},{\mathsf{a}}_{0,1})\subseteq\Pi\mathsf{T}(0)\oplus\Pi\mathsf{T}(1)\oplus\Pi\mathsf{T}(2)\subseteq\mathsf{T}$
(dropping ${\mathsf{a}}_{0,1}$ if $\ell=1$);

  3. $\bullet$

for for $i\neq 0,\ell$, we set

$\displaystyle\Pi\mathsf{T}^{\prime}(i)$
$\displaystyle=\operatorname{span}_{\Bbbk}({\mathsf{e}}_{\ell-i},{\mathsf{a}}_{\ell-i,\ell-i-1},{\mathsf{a}}_{\ell-i,\ell-i+1},{\mathsf{c}}_{\ell-i})$
$\displaystyle\subseteq\Pi\mathsf{T}(\ell-i+1)\oplus\Pi\mathsf{T}(\ell-i)\oplus\Pi\mathsf{T}(\ell-i+2)\subseteq\mathsf{T}$

(dropping ${\mathsf{a}}_{\ell-i,\ell-i+1}$ if $i=1$).

In this last paragraph we suppose that $\Bbbk=R$ and recall the theory of
§§3.3,3.4. We have the subalgebra
$\mathfrak{a}_{R}:=\operatorname{span}(B_{\mathfrak{a}})=\operatorname{span}_{R}({\mathsf{e}}_{i}\mid
i\in I)\subseteq{\mathsf{Z}}_{R,{\bar{0}}}$ and the analogous subalgebra
$\mathfrak{a}^{\prime}_{R}=\operatorname{span}_{R}({\mathsf{e}}_{i}^{\prime}\mid
i\in I)\subseteq{\mathsf{Z}}^{\prime}_{R,{\bar{0}}}$. We have $R$-module
decomposition
$\mathsf{T}_{R,{\bar{0}}}=\mathsf{T}_{R,\mathfrak{a}}\oplus\mathsf{T}_{R,\mathfrak{c}}$
where
$\mathsf{T}_{R,\mathfrak{a}}=\operatorname{span}_{R}({\mathsf{e}}_{j}\mid j\in
J)$ and
$\mathsf{T}_{R,\mathfrak{c}}=\operatorname{span}_{R}({\mathsf{c}}_{j}\mid j\in
J)$. Then it is clear from the explicit construction above that

$\mathfrak{a}_{R}\cdot(\mathsf{T}_{R,\mathfrak{a}})\cdot\mathfrak{a}^{\prime}_{R}\subseteq\mathsf{T}_{R,\mathfrak{a}}.$
(5.4)

For any $i\in I$,
$\Pi\mathsf{T}(i)_{R,\mathfrak{a}}:=\Pi\mathsf{T}(i)_{R}\cap\mathsf{T}_{R,\mathfrak{a}}$,
$\Pi\mathsf{T}(i)_{R,\mathfrak{c}}:=\Pi\mathsf{T}(i)_{R}\cap\mathsf{T}_{R,\mathfrak{c}}$,
$\Pi\mathsf{T}^{\prime}(i)_{R,\mathfrak{a}}:=\Pi\mathsf{T}^{\prime}(i)_{R}\cap\mathsf{T}_{R,\mathfrak{a}}$,
$\Pi\mathsf{T}^{\prime}(i)_{R,\mathfrak{c}}:=\Pi\mathsf{T}^{\prime}(i)_{R}\cap\mathsf{T}_{R,\mathfrak{c}}$.
These define calibrations on each $\Pi\mathsf{T}(i)_{R}$ and $\mathsf{T}_{R}$
as left ${\mathsf{Z}}_{R}$-supermodules, as well as on each
$\Pi\mathsf{T}^{\prime}(i)_{R}$ and $\mathsf{T}_{R}$ as right
${\mathsf{Z}}^{\prime}_{R}$-supermodules.

###  5.2. A full tilting module for $T^{Z}(n,d)$

From now on until the end of the paper we fix $n\in\mathbb{Z}_{>0}$ and
$d\in\mathbb{Z}_{\geq 0}$ such that $d\leq n$.

To construct a full tilting module for based quasihereditary superalgebra
$T^{\mathsf{Z}}(n,d)$, we need to first work integrally. Recall from the
previous subsection that $\mathsf{T}_{R}$ is a
$({\mathsf{Z}}_{R},{\mathsf{Z}}^{\prime}_{R})$-bisupermodule.

Recall the constructions of §4.3. In particular, we have the left
$M_{n}({\mathsf{Z}}_{R})$-module structure on
$\operatorname{Col}_{n}(\mathsf{T}_{R})$ and a right
$M_{n}({\mathsf{Z}}^{\prime}_{R})$-module structure on
$\operatorname{Row}_{n}(\mathsf{T}_{R})$. Setting as in §4.3,
$\operatorname{Col}_{n}(\mathsf{T}_{R})_{\mathfrak{a}}:=\operatorname{Col}_{n}(\mathsf{T}_{R,\mathfrak{a}})$
and
$\operatorname{Row}_{n}(\mathsf{T}_{R})_{\mathfrak{a}}:=\operatorname{Row}_{n}(\mathsf{T}_{R,\mathfrak{a}})$,
we have by (5.4) that
$M_{n}(\mathfrak{a})\operatorname{Col}_{n}(\mathsf{T}_{R})_{\mathfrak{a}}\subseteq\operatorname{Col}_{n}(\mathsf{T}_{R})_{\mathfrak{a}}$
and
$\operatorname{Row}_{n}(\mathsf{T}_{R})_{\mathfrak{a}}M_{n}(\mathfrak{a})\subseteq\operatorname{Row}_{n}(\mathsf{T}_{R})_{\mathfrak{a}}$.
So by Lemma 3.16 (and its right module analogue),
$\tilde{\Gamma}^{d}\operatorname{Col}_{n}(\mathsf{T}_{R})$ is a left module
over $T^{\mathsf{Z}}(n,d)_{R}$ and
$\tilde{\Gamma}^{d}\operatorname{Row}_{n}(\mathsf{T}_{R})$ is a right module
over $T^{{\mathsf{Z}}^{\prime}}(n,d)_{R}$. Similarly, for every $i\in I$, we
have left $T^{\mathsf{Z}}(n,d)_{R}$-modules
$\tilde{\Gamma}^{d}\operatorname{Col}_{n}(\Pi\mathsf{T}(i)_{R})$ and right
$T^{{\mathsf{Z}}^{\prime}}(n,d)_{R}$-modules
$\operatorname{Row}_{n}(\Pi\mathsf{T}^{\prime}(i)_{R})$. Extending scalars, we
have a left supermodule

${\mathscr{T}}^{d}_{i}:=\mathbb{F}\otimes_{R}\tilde{\Gamma}^{d}\operatorname{Col}_{n}(\Pi\mathsf{T}(i)_{R})$

over $T^{\mathsf{Z}}(n,d)=\mathbb{F}\otimes_{R}T^{\mathsf{Z}}(n,d)_{R}$.

Recall from §4.2 that for $d\leq n$, the algebra $T^{\mathsf{Z}}(n,d)$ is
quasi-hereditary with respect to the poset $\Lambda^{I}_{+}(n,d)$ with partial
order $\leq_{I}$, so it has its own standard modules
$\\{\Delta(\text{\boldmath$\lambda$})\mid\text{\boldmath$\lambda$}\in\Lambda^{I}_{+}(n,d)\\}$,
costandard modules
$\\{\nabla(\text{\boldmath$\lambda$})\mid\text{\boldmath$\lambda$}\in\Lambda^{I}_{+}(n,d)\\}$
and indecomposable tilting modules
$\\{\operatorname{T}(\text{\boldmath$\lambda$})\mid\text{\boldmath$\lambda$}\in\Lambda^{I}_{+}(n,d)\\}$.
Moreover, by [14, Proposition 6.20], the anti-involution $\tau$ on
${\mathsf{Z}}$ extends to the anti-involution

$\tau_{n,d}:T^{\mathsf{Z}}(n,d)\to T^{\mathsf{Z}}(n,d),\
\eta_{\text{\boldmath$r$},\text{\boldmath$s$}}^{\text{\boldmath$b$}}\mapsto\eta_{\text{\boldmath$s$},\text{\boldmath$r$}}^{\tau(\text{\boldmath$b$})}$

where for a $d$-tuple
$\text{\boldmath$b$}={\mathsf{b}}_{1}\cdots{\mathsf{b}}_{d}\in{\mathsf{B}}^{d}$
we denote
$\tau(\text{\boldmath$b$}):=\tau({\mathsf{b}}_{1})\cdots\tau({\mathsf{b}}_{d})$.
As in [15, (2.14)], we then have for all
$\text{\boldmath$\lambda$}\in\Lambda^{I}_{+}(n,d)$:

$\Delta(\text{\boldmath$\lambda$})^{\tau_{n,d}}\simeq\nabla(\text{\boldmath$\lambda$}).$
(5.5)

Since $\tau_{n,d}(\eta_{\text{\boldmath$\mu$}})=\eta_{\text{\boldmath$\mu$}}$
for all $\text{\boldmath$\mu$}\in\Lambda^{I}(n,d)$, we deduce:

######  Lemma 5.6.

For all $\text{\boldmath$\lambda$}\in\Lambda^{I}_{+}(n,d)$, we have
$\operatorname{ch}\Delta(\text{\boldmath$\lambda$})=\operatorname{ch}\nabla(\text{\boldmath$\lambda$})$.

######  Proposition 5.7.

For all $i\in I$, the left $T^{\mathsf{Z}}(n,d)$-module
${\mathscr{T}}^{d}_{i}$ is tilting and has a unique maximal weight
$\text{\boldmath$\iota$}_{i}(1^{d})$ (with respect to $\leq_{I}$).

###### Proof.

Suppose first that $i=0$. Since $\text{\boldmath$\iota$}_{0}(1^{d})$ is
minimal in $\Lambda^{+}_{I}(n,d)$ it follows that
$\operatorname{T}(\text{\boldmath$\iota$}_{0}(1^{d}))\simeq\Delta(\text{\boldmath$\iota$}_{0}(1^{d}))\simeq
L(\text{\boldmath$\iota$}_{0}(1^{d}))$. Using the assumption $d\leq n$, we
note that $\text{\boldmath$\iota$}_{0}(1^{d})$ is the only weight of
${\mathscr{T}}^{d}_{0}$ that lies in $\Lambda^{+}_{I}(n,d)$, and the
corresponding weight multiplicity is $1$, so

${\mathscr{T}}^{d}_{0}\cong
L(\text{\boldmath$\iota$}_{0}(1^{d}))\simeq\operatorname{T}(\text{\boldmath$\iota$}_{0}((1^{d})))\simeq\Delta(\text{\boldmath$\iota$}_{0}((1^{d}))).$
(5.8)

Let now $i>0$. Since
$\Pi\mathsf{T}(i)_{R}={\mathsf{Z}}_{R}{\mathsf{e}}_{i-1}$, we have
$\operatorname{Col}_{n}(\Pi\mathsf{T}(i)_{R})\simeq
M_{n}({\mathsf{Z}}_{R})\xi_{1,1}^{{\mathsf{e}}_{i-1}}$, and so by Lemma 4.12
and extension of scalars, we have

${\mathscr{T}}^{d}_{i}\simeq
T^{\mathsf{Z}}(n,d)\eta^{{\mathsf{e}}_{i-1}^{d}}_{1^{d},1^{d}}=T^{\mathsf{Z}}(n,d)\eta_{\text{\boldmath$\iota$}_{i-1}(d)}.$
(5.9)

In particular, ${\mathscr{T}}^{d}_{i}$ is projective, and thus has a standard
filtration. To prove that ${\mathscr{T}}^{d}_{i}$ also has a costandard
filtration, it suffices to show that it is $\tau_{n,d}$-self-dual, or
equivalently possesses a non-degenerate $\tau_{n,d}$-contravariant bilinear
form.

To construct this form we work over $R$. Recall that for $i>1$, we have
$\Pi\mathsf{T}(i)_{R}={\mathsf{Z}}_{R}{\mathsf{e}}_{i-1}$ has basis
${\mathsf{B}}^{\Pi\mathsf{T}(i)}:=\\{{\mathsf{e}}_{i-1},{\mathsf{c}}_{i-1},{\mathsf{a}}_{i-2,i-1},{\mathsf{a}}_{i,i-1}\\}$.
Consider the bilinear form $(\cdot,\cdot)$ on $\Pi\mathsf{T}(i)_{R}$ such that

$({\mathsf{e}}_{i-1},{\mathsf{c}}_{i-1})=-({\mathsf{c}}_{i-1},{\mathsf{e}}_{i-1})=({\mathsf{a}}_{i-2,i-1},{\mathsf{a}}_{i-2,i-1})=({\mathsf{a}}_{i,i-1},{\mathsf{a}}_{i,i-1})=1,$

and all the other pairings of basis elements are $0$. Note that this form is
even, non-degenerate, $\tau$-contravariant, and superantisymmetric. Extending
this form in the obvious way to $\operatorname{Col}_{n}(\Pi\mathsf{T}(i)_{R})$
results in a non-degenerate, superantisymmetric, $\tau_{n,1}$-contravariant
form again denoted $(\cdot,\cdot)$. Note that this form satisfies the
assumptions of Lemma 3.18. Applying Lemma 3.18, we obtain an even, non-
degenerate, $\tau_{n,d}$-contravariant form on
$\tilde{\Gamma}^{d}\operatorname{Col}_{n}(\Pi\mathsf{T}(i)_{R})$. Extending
scalars, we deduce that, ${\mathscr{T}}^{d}_{i}$ is $\tau_{n,d}$-self-dual.

Denote by $v_{r}^{\mathsf{b}}\in\operatorname{Col}_{n}(\Pi\mathsf{T}(i)_{R})$
the column vector with ${\mathsf{b}}\in\Pi\mathsf{T}(i)_{R}$ in the $r$th
position and $0$s elsewhere. To see that the unique maximal weight of
${\mathscr{T}}^{d}_{i}$ is $\text{\boldmath$\iota$}_{i}(1^{d})$, observe that
the unique maximal weight vector of $\Pi\mathsf{T}(i)_{R}$ is
${\mathsf{a}}_{i,i-1}$, which is odd, and so the vector
$v_{1}^{{\mathsf{a}}_{i,i-1}}*\cdots*v_{d}^{{\mathsf{a}}_{i,i-1}}\in\tilde{\Gamma}^{d}\operatorname{Col}_{n}(\Pi\mathsf{T}(i)_{R})$
has weight $\text{\boldmath$\iota$}_{i}(1^{d})$ and all other weight vectors
appearing in $\tilde{\Gamma}^{d}\operatorname{Col}_{n}(\Pi\mathsf{T}(i)_{R})$
have smaller weight.

The case $i=1$ is similar to the case $i>1$ but
$\Pi\mathsf{T}(1)_{R}={\mathsf{Z}}_{R}{\mathsf{e}}_{0}$ has basis
${\mathsf{B}}^{\Pi\mathsf{T}(1)}:=\\{{\mathsf{e}}_{0},{\mathsf{c}}_{0},{\mathsf{a}}_{1,0}\\}$,
and we use the form such that

$({\mathsf{e}}_{0},{\mathsf{c}}_{0})=-({\mathsf{c}}_{0},{\mathsf{e}}_{0})=({\mathsf{a}}_{1,0},{\mathsf{a}}_{1,0})=1$

are the only non-trivial pairings of basis elements. ∎

Recalling that $\mathsf{T}_{R}$ is a
$({\mathsf{Z}}_{R},{\mathsf{Z}}^{\prime}_{R})$-bisupermodule, we can now
consider $M_{n}(\mathsf{T}_{R})$ as an
$(M_{n}({\mathsf{Z}}_{R}),M_{n}({\mathsf{Z}}^{\prime}_{R}))$-bisupermodule in
the obvious way. Take
$M_{n}(\mathsf{T}_{R})_{\mathfrak{a}}:=M_{n}(\mathsf{T}_{R,\mathfrak{a}})$ and
$M_{n}(\mathsf{T}_{R})_{\mathfrak{c}}:=M_{n}(\mathsf{T}_{R,\mathfrak{c}})$. In
view of Lemma 3.16 (and its right module analogue),
$\tilde{\Gamma}^{d}M_{n}(\mathsf{T}_{R})$ is a
$(T^{\mathsf{Z}}(n,d)_{R},T^{{\mathsf{Z}}^{\prime}}(n,d)_{R})$-bisupermodule.
We now extend the scalars from $R$ to $\mathbb{F}$ to get the
$(T^{\mathsf{Z}}(n,d),T^{{\mathsf{Z}}^{\prime}}(n,d))$-bisupermodule

${\mathscr{T}}:=\mathbb{F}\otimes_{R}\tilde{\Gamma}^{d}M_{n}(\mathsf{T}_{R}).$

For each composition $\mu\in\Lambda(n,d)$ define
${\mathscr{T}}_{i}^{\mu}:={\mathscr{T}}_{i}^{\mu_{1}}\otimes\cdots\otimes{\mathscr{T}}_{i}^{\mu_{n}}$.
Furthermore, for each multicomposition
$\text{\boldmath$\mu$}=(\mu^{(0)},\dots,\mu^{(\ell)})\in\Lambda^{I}(n,d)$
define
${\mathscr{T}}^{\text{\boldmath$\mu$}}:={\mathscr{T}}_{0}^{\mu^{(0)}}\otimes\cdots\otimes{\mathscr{T}}_{\ell}^{\mu^{(\ell)}}$.

Since $\mathsf{T}_{R}=\bigoplus_{i\in I}\Pi\mathsf{T}(i)_{R}$, as left
supermodules over $M_{n}({\mathsf{Z}}_{R})$, we have

$M_{n}(\mathsf{T}_{R})\simeq\operatorname{Col}_{n}(\mathsf{T}_{R})^{\oplus
n}\simeq\bigoplus_{i\in I}\operatorname{Col}_{n}(\Pi\mathsf{T}(i)_{R})^{\oplus
n}.$

Now, using Lemmas 3.19 and extending scalars, we have as left
$T^{\mathsf{Z}}(n,d)$-supermodules:

${\mathscr{T}}\simeq\bigoplus_{\text{\boldmath$\mu$}\in\Lambda^{I}(n,d)}{\mathscr{T}}^{\text{\boldmath$\mu$}}.$
(5.10)

For
$\text{\boldmath$\lambda$}=(\lambda^{(0)},\dots,\lambda^{(\ell)})\in\Lambda^{I}_{+}(n,d)$,
we define the conjugate multipartition

$\text{\boldmath$\lambda$}^{\prime}:=((\lambda^{(0)})^{\prime},\dots,(\lambda^{(\ell)})^{\prime})\in\Lambda^{I}_{+}(n,d).$
(5.11)

######  Proposition 5.12.

As a left $T^{\mathsf{Z}}(n,d)$-supermodule, ${\mathscr{T}}$ is a full tilting
supermodule.

###### Proof.

Note that each ${\mathscr{T}}^{\text{\boldmath$\mu$}}$ is tilting by
Proposition 5.7 and Corollary 4.11. So ${\mathscr{T}}$ is tilting by (5.10).
To show that ${\mathscr{T}}$ is full tilting, it suffices for each
$\text{\boldmath$\lambda$}\in\Lambda^{+}_{I}(n,d)$ to find a summand
${\mathscr{T}}^{\text{\boldmath$\mu$}}$ in (5.10) which has maximal weight
$\lambda$. Fix
$\text{\boldmath$\lambda$}=(\lambda^{(0)},\dots,\lambda^{(\ell)})\in\Lambda^{+}_{I}(n,d)$
and take $\text{\boldmath$\mu$}=\text{\boldmath$\lambda$}^{\prime}$. By
Proposition 5.7 again, ${\mathscr{T}}_{i}^{s}$ has unique maximal weight
$\text{\boldmath$\iota$}_{i}(1^{s})$ for each $s\in\mathbb{Z}_{>0}$. So

$\sum_{i=0}^{\ell}\sum_{r=1}^{n}\text{\boldmath$\iota$}_{i}(1^{\mu^{(i)}_{r}})=\sum_{i=0}^{\ell}\text{\boldmath$\iota$}_{i}(\lambda^{(i)})=\text{\boldmath$\lambda$}$

is the unique maximal weight of ${\mathscr{T}}^{\text{\boldmath$\mu$}}$. ∎

######  Corollary 5.13.

As a left $T^{\mathsf{Z}}(n,d)$-supermodule and as a right
$T^{{\mathsf{Z}}^{\prime}}(n,d)$-supermodule, ${\mathscr{T}}$ is faithful.

###### Proof.

As a left $T^{\mathsf{Z}}(n,d)$-supermodule, ${\mathscr{T}}$ is faithful since
it is full tilting by Proposition 5.12 (a full tilting supermodule is faithful
for example by [17, Lemma 6]). The second statement follows similarly from the
right module analogue of that proposition. ∎

###  5.3. $T^{Z}(n,d)$ is Ringel self-dual

In view of Corollary 5.13, we have an embedding of
$T^{{\mathsf{Z}}^{\prime}}(n,d)$ into
$\operatorname{End}_{T^{{\mathsf{Z}}}(n,d)}({\mathscr{T}})^{\mathrm{sop}}$. To
prove that this embedding is an isomorphism, we now compute the dimension of
$\operatorname{End}_{T^{{\mathsf{Z}}}(n,d)}({\mathscr{T}})$.

Recalling (2.13), for $\text{\boldmath$\lambda$}\in\Lambda^{I}_{+}(n,d)$ and
$\text{\boldmath$\mu$}\in\Lambda^{I}(n,d)$, let

$k_{\text{\boldmath$\lambda$},\text{\boldmath$\mu$}}:=|\operatorname{Std}^{\mathsf{X}}(\text{\boldmath$\lambda$},\text{\boldmath$\mu$})|.$

By (4.5), (4.6) and Lemma 5.6, we have

$k_{\text{\boldmath$\lambda$},\text{\boldmath$\mu$}}=\dim\eta_{\text{\boldmath$\mu$}}\Delta(\text{\boldmath$\lambda$})=\dim\eta_{\text{\boldmath$\mu$}}\nabla(\text{\boldmath$\lambda$}).$
(5.14)

Let $i\in I$. If $i\neq 0$, we define

$\text{\boldmath$\beta$}_{i}(d,s)=\text{\boldmath$\iota$}_{i-1}((s))+\text{\boldmath$\iota$}_{i}((1^{d-s}))\in\Lambda^{I}_{+}(n,d)$

for all $0\leq s\leq d$ We also define

$\text{\boldmath$\beta$}_{0}(d,0):=\text{\boldmath$\iota$}_{0}((1^{d})).$

We define by $\Xi_{d,i}$ to be the set of all
$\text{\boldmath$\beta$}_{i}(d,s)$’s, i.e.

$\Xi_{d,i}:=\left\\{\begin{array}[]{ll}\\{\text{\boldmath$\beta$}_{i}(d,s)\mid
0\leq s\leq d\\}&\hbox{if $i\neq 0$,}\\\
\\{\text{\boldmath$\beta$}_{0}(d,0)\\}&\hbox{if $i=0$.}\end{array}\right.$

######  Lemma 5.15.

Let $\text{\boldmath$\beta$}\in\Lambda^{I}_{+}(n,d)$ and $i\in I$. Then

$({\mathscr{T}}_{i}^{d}:\Delta(\text{\boldmath$\beta$}))=\begin{cases}1&\text{
if }\text{\boldmath$\beta$}\in\Xi_{d,i},\\\ 0&\text{ otherwise}.\\\
\end{cases}$

###### Proof.

By (5.8), we have
${\mathscr{T}}_{0}^{d}\cong\Delta(\text{\boldmath$\iota$}_{0}((1^{d})))$, so
we may assume that $i\neq 0$. Then by (5.9), we have
${\mathscr{T}}_{i}^{d}\simeq
T^{\mathsf{Z}}(n,d)\eta_{\text{\boldmath$\iota$}_{i-1}(d)}$. Now, using (2.10)
and (5.14), we get

$\displaystyle({\mathscr{T}}_{i}^{d}:\Delta(\text{\boldmath$\beta$}))$
$\displaystyle=\dim\operatorname{Hom}_{T^{\mathsf{Z}}(n,d)}({\mathscr{T}}_{i}^{d},\nabla(\text{\boldmath$\beta$}))$
$\displaystyle=\dim\operatorname{Hom}_{T^{\mathsf{Z}}(n,d)}(T^{\mathsf{Z}}(n,d)\eta_{\text{\boldmath$\iota$}_{i-1}(d)},\nabla(\text{\boldmath$\beta$}))$
$\displaystyle=\dim\eta_{\text{\boldmath$\iota$}_{i-1}(d)}\nabla(\text{\boldmath$\beta$})$
$\displaystyle=k_{\text{\boldmath$\beta$},\text{\boldmath$\iota$}_{i-1}(d)}.$

It remains to observe that
$k_{\text{\boldmath$\beta$},\text{\boldmath$\iota$}_{i-1}(d)}=1$ if
$\text{\boldmath$\beta$}=\text{\boldmath$\beta$}_{i}(d,s)$ for some $0\leq
s\leq d$ and $k_{\text{\boldmath$\beta$},\text{\boldmath$\iota$}_{i-1}(d)}=0$
otherwise.∎

Let $0\leq r\leq d$. Recalling (4.9), our next goal is to compute the
Littlewood-Richardson coefficient
$c^{\text{\boldmath$\lambda$}}_{\text{\boldmath$\alpha$},\text{\boldmath$\beta$}}$
for all $\text{\boldmath$\lambda$}\in\Lambda^{I}_{+}(n,d)$,
$\text{\boldmath$\alpha$}\in\Lambda^{I}_{+}(n,d-r)$ and
$\text{\boldmath$\beta$}\in\Xi_{r,i}$. Let $i\in I$ and
$\text{\boldmath$\beta$}=\text{\boldmath$\beta$}_{i}(r,s)\in\Xi_{r,i}$, in
particular, $0\leq s\leq r$, and $s=0$ if $i=0$. We define
$\Omega^{\text{\boldmath$\lambda$}}_{\text{\boldmath$\beta$}}$ to be the set
of all
$\text{\boldmath$\alpha$}=(\alpha^{(0)},\ldots,\alpha^{(\ell)})\in\Lambda^{I}_{+}(n,d-r)$
such that $\alpha^{(j)}=\lambda^{(j)}$ for all $j\notin\\{i-1,i\\}$,
$[\alpha^{(i-1)}]$ is obtained from $[\lambda^{(i-1)}]$ by removing $s$ boxes
from distinct columns, and $[\alpha^{(i)}]$ is obtained from $[\lambda^{(i)}]$
by removing $r-s$ boxes from distinct rows (if $i=0$, then the condition on
$[\alpha^{(i-1)}]$ should be dropped).

######  Lemma 5.16.

Let $0\leq r\leq d$, $i\in I$,
$\text{\boldmath$\lambda$}\in\Lambda^{I}_{+}(n,d)$,
$\text{\boldmath$\alpha$}\in\Lambda^{I}_{+}(n,d-r)$ and
$\text{\boldmath$\beta$}=\Xi_{r,i}$. Then

$c_{\text{\boldmath$\alpha$},\text{\boldmath$\beta$}}^{\text{\boldmath$\lambda$}}=\begin{cases}1&\text{
if
}\text{\boldmath$\alpha$}\in\Omega^{\text{\boldmath$\lambda$}}_{\text{\boldmath$\beta$}}\\\
0&\text{ otherwise.}\\\ \end{cases}$

###### Proof.

The result is an immediate consequence of the Littlewood-Richardson rule. ∎

For each $\text{\boldmath$\mu$}\in\Lambda^{I}(n,d)$, define
$\overleftarrow{\text{\boldmath$\mu$}}=(\overleftarrow{\mu}^{(0)},\dots,\overleftarrow{\mu}^{(\ell)})\in\Lambda^{I}(n,d)$
by setting $\overleftarrow{\mu}^{(i)}:=\mu^{(\ell-i)}$ for all $i\in I$.
Recall (5.11).

######  Proposition 5.17.

Let $\text{\boldmath$\lambda$}\in\Lambda^{I}_{+}(n,d)$ and
$\text{\boldmath$\mu$}\in\Lambda^{I}(n,d)$. Then
$({\mathscr{T}}^{\text{\boldmath$\mu$}}:\Delta(\text{\boldmath$\lambda$}))=k_{\overleftarrow{\text{\boldmath$\lambda$}}^{\prime},\overleftarrow{\text{\boldmath$\mu$}}}\,.$

###### Proof.

We proceed by induction on the number of non-zero parts of $\mu$. To start the
induction, we suppose that $\mu$ has only one row, in which case
${\mathscr{T}}^{\text{\boldmath$\mu$}}\cong{\mathscr{T}}_{i}^{d}$ for some
$i$, and the result follows from Lemma 5.15. So we may assume that $\mu$ has
at least two rows.

Let now $i$ be maximal such that $\mu^{(i)}\neq\varnothing$ and pick the
largest $t$ such that $\mu^{(i)}_{t}\neq 0$. Denote $r:=\mu^{(i)}_{t}$. Let
$\nu^{(i)}$ be $\mu^{(i)}$ with last non-zero row removed:

$\nu^{(i)}:=(\mu^{(i)}_{1},\dots,\mu^{(i)}_{t-1},0,\dots,0)\in\Lambda(n,|\mu^{(i)}|-r),$

and $\nu^{(j)}:=\mu^{(j)}$ for all $j\neq i$. Set

$\text{\boldmath$\nu$}:=(\nu^{(0)},\dots,\nu^{(\ell)})\in\Lambda^{I}(n,d-r).$

Then
${\mathscr{T}}^{\text{\boldmath$\mu$}}\cong{\mathscr{T}}^{\text{\boldmath$\nu$}}\otimes{\mathscr{T}}_{i}^{r}$.

By (5.1), we have ${\mathsf{X}}(0)=\\{{\mathsf{e}}_{0}\\}$ and
${\mathsf{X}}(i)=\\{{\mathsf{e}}_{i},{\mathsf{a}}_{i-1,i}\\}$ for $i\neq 0$.
Recalling §2.4, for $i\neq 0$, we put the total order on
${\mathscr{A}}_{{\mathsf{X}}(i)}$ given by:

$1^{{\mathsf{e}}_{i}}<\cdots<n^{{\mathsf{e}}_{i}}<1^{{\mathsf{a}}_{i-1,i}}<\cdots<n^{{\mathsf{a}}_{i-1,i}}.$

And we endow ${\mathscr{A}}_{{\mathsf{X}}(0)}$ with the order
$1^{{\mathsf{e}}_{0}}<\cdots<n^{{\mathsf{e}}_{0}}$.

By [15, Corollary (4.13)], inductive hypothesis, Lemma 5.15, and Lemma 5.16 we
have

$\displaystyle({\mathscr{T}}^{\text{\boldmath$\mu$}}:\Delta(\text{\boldmath$\lambda$}))$
$\displaystyle=({\mathscr{T}}^{\text{\boldmath$\nu$}}\otimes{\mathscr{T}}_{i}^{r}:\Delta(\text{\boldmath$\lambda$}))$
$\displaystyle=\sum_{\text{\boldmath$\alpha$}\in\Lambda^{I}_{+}(n,d-r)}\sum_{\text{\boldmath$\beta$}\in\Lambda^{I}_{+}(n,r)}c_{\text{\boldmath$\alpha$},\text{\boldmath$\beta$}}^{\text{\boldmath$\lambda$}}({\mathscr{T}}^{\text{\boldmath$\nu$}}:\Delta(\text{\boldmath$\alpha$}))({\mathscr{T}}_{i}^{r}:\Delta(\text{\boldmath$\beta$}))$
$\displaystyle=\sum_{\text{\boldmath$\alpha$}\in\Lambda^{I}_{+}(n,d-r)}\sum_{\text{\boldmath$\beta$}\in\Xi_{r,i}}c_{\text{\boldmath$\alpha$},\text{\boldmath$\beta$}}^{\text{\boldmath$\lambda$}}k_{\overleftarrow{\text{\boldmath$\alpha$}}^{\prime},\overleftarrow{\text{\boldmath$\nu$}}}$
$\displaystyle=\sum_{\text{\boldmath$\beta$}\in\Xi_{r,i}}\sum_{\text{\boldmath$\alpha$}\in\Omega^{\text{\boldmath$\lambda$}}_{\text{\boldmath$\beta$}}}k_{\overleftarrow{\text{\boldmath$\alpha$}}^{\prime},\overleftarrow{\text{\boldmath$\nu$}}}$
$\displaystyle=\sum_{\text{\boldmath$\beta$}\in\Xi_{r,i}}\sum_{\text{\boldmath$\alpha$}\in\Omega^{\text{\boldmath$\lambda$}}_{\text{\boldmath$\beta$}}}|\operatorname{Std}^{\mathsf{X}}(\overleftarrow{\text{\boldmath$\alpha$}}^{\prime},\overleftarrow{\text{\boldmath$\nu$}})|.$

Since
$k_{\overleftarrow{\text{\boldmath$\lambda$}}^{\prime},\overleftarrow{\text{\boldmath$\mu$}}}=|\operatorname{Std}^{\mathsf{X}}(\overleftarrow{\text{\boldmath$\lambda$}}^{\prime},\overleftarrow{\text{\boldmath$\mu$}})|$
it remains to prove that there is a bijection

$\bigsqcup_{\text{\boldmath$\beta$}\in\Xi_{r,i}}\bigsqcup_{\text{\boldmath$\alpha$}\in\Omega^{\text{\boldmath$\lambda$}}_{\text{\boldmath$\beta$}}}\operatorname{Std}^{\mathsf{X}}(\overleftarrow{\text{\boldmath$\alpha$}}^{\prime},\overleftarrow{\text{\boldmath$\nu$}})\stackrel{{\scriptstyle\sim}}{{\longrightarrow}}\operatorname{Std}^{\mathsf{X}}(\overleftarrow{\text{\boldmath$\lambda$}}^{\prime},\overleftarrow{\text{\boldmath$\mu$}}).$

Let $\text{\boldmath$\beta$}\in\Xi_{r,i}$,
$\text{\boldmath$\alpha$}\in\Omega^{\text{\boldmath$\lambda$}}_{\text{\boldmath$\beta$}}$
and
$\text{\boldmath$T$}\in\operatorname{Std}^{\mathsf{X}}(\overleftarrow{\text{\boldmath$\alpha$}}^{\prime},\overleftarrow{\text{\boldmath$\nu$}})$.
By definition, $\beta$ is of the form $\text{\boldmath$\beta$}_{i}(r,s)$.
Moreover, the Young diagram $[\text{\boldmath$\alpha$}]$ is obtained by
removing $s$ nodes from distinct columns of the $(i-1)$st component
$[\lambda^{(i-1)}]$ and $r-s$ nodes from distinct rows of the $i$th component
$[\lambda^{(i)}]$ of $[\text{\boldmath$\lambda$}]$. Therefore
$[\overleftarrow{\text{\boldmath$\alpha$}}^{\prime}]$ is obtained by removing
$s$ nodes $N_{1},\dots,N_{s}$ from distinct rows of the $(\ell-i+1)$st
component of $[\overleftarrow{\text{\boldmath$\lambda$}}^{\prime}]$ and $r-s$
nodes $M_{1},\dots,M_{r-s}$ from distinct columns of the $(\ell-i)$th
component of $[\overleftarrow{\text{\boldmath$\lambda$}}^{\prime}]$. Now
extend $T$ to the tableau
$\hat{\text{\boldmath$T$}}\in\operatorname{Std}^{\mathsf{X}}(\overleftarrow{\text{\boldmath$\lambda$}}^{\prime},\overleftarrow{\text{\boldmath$\mu$}})$
by setting

$\hat{\text{\boldmath$T$}}(N_{1})=\dots=\hat{\text{\boldmath$T$}}(N_{s})=t^{{\mathsf{a}}_{\ell-i,\ell-i+1}}\qquad\text{and}\qquad\hat{\text{\boldmath$T$}}(M_{1})=\dots=\hat{\text{\boldmath$T$}}(M_{r-s})=t^{{\mathsf{e}}_{\ell-i}}.$

The tableaux $\hat{\text{\boldmath$T$}}$ is indeed standard since, by
maximality of $i$ and $t$, we have
$\text{\boldmath$T$}(N)<t^{{\mathsf{a}}_{\ell-i,\ell-i+1}}$ for all $N$ in the
$(\ell-i+1)$st component of
$[\overleftarrow{\text{\boldmath$\alpha$}}^{\prime}]$ and
$\text{\boldmath$T$}(N)<t^{{\mathsf{e}}_{\ell-i}}$ for all $N$ in the
$(\ell-i)$th component of
$[\overleftarrow{\text{\boldmath$\alpha$}}^{\prime}]$. The map
$\text{\boldmath$T$}\mapsto\hat{\text{\boldmath$T$}}$ is clearly injective. To
see that it is surjective, it suffices to show that for any
$\text{\boldmath$S$}\in\operatorname{Std}^{\mathsf{X}}(\overleftarrow{\text{\boldmath$\lambda$}}^{\prime},\overleftarrow{\text{\boldmath$\mu$}})$
there exists $\text{\boldmath$\beta$}\in\Xi_{r,i}$ and
$\text{\boldmath$\alpha$}\in\Omega^{\text{\boldmath$\lambda$}}_{\text{\boldmath$\beta$}}$
with

$[\overleftarrow{\text{\boldmath$\lambda$}}^{\prime}]\setminus\\{N\mid\text{\boldmath$S$}(N)\in\\{t^{{\mathsf{e}}_{\ell-i}},t^{{\mathsf{a}}_{\ell-i,\ell-i+1}}\\}\\}=[\overleftarrow{\text{\boldmath$\alpha$}}^{\prime}].$

Indeed, there are exactly $\overleftarrow{\mu}^{(\ell-i)}_{t}=\mu^{(i)}_{t}=r$
nodes $N$ in the Young diagram
$[\overleftarrow{\text{\boldmath$\lambda$}}^{\prime}]$ such that
$\text{\boldmath$S$}(N)\in\\{t^{{\mathsf{e}}_{\ell-i}},t^{{\mathsf{a}}_{\ell-i,\ell-i+1}}\\}$.
So for some $0\leq s\leq r$, we can write

$\displaystyle\\{N\in[\overleftarrow{\text{\boldmath$\lambda$}}^{\prime}]\mid\text{\boldmath$S$}(N)=t^{{\mathsf{a}}_{\ell-i,\ell-i+1}}\\}$
$\displaystyle=\\{N_{1},\dots,N_{s}\\},$
$\displaystyle\\{N\in[\overleftarrow{\text{\boldmath$\lambda$}}^{\prime}]\mid\text{\boldmath$S$}(N)=t^{{\mathsf{e}}_{\ell-i}}\\}$
$\displaystyle=\\{M_{1},\dots,M_{r-s}\\}.$

By maximality of $i$ and $t$, we have that the nodes $N_{1},\dots,N_{s}$ are
in the ends of distinct rows of the $(\ell-i+1)$st component of
$[\overleftarrow{\text{\boldmath$\lambda$}}^{\prime}]$ and the nodes
$M_{1},\dots,M_{r-s}$ are in the ends of distinct columns of the $(\ell-i)$th
component of $[\overleftarrow{\text{\boldmath$\lambda$}}^{\prime}]$. It
remains to note that removing these nodes produces a shape
$[\overleftarrow{\text{\boldmath$\alpha$}}^{\prime}]$ for
$\text{\boldmath$\alpha$}\in\Omega^{\text{\boldmath$\lambda$}}_{\text{\boldmath$\beta$}_{i}(r,s)}$.
∎

######  Theorem 5.18.

Let $d\leq n$. We have
$\operatorname{End}_{T^{{\mathsf{Z}}}(n,d)}({\mathscr{T}})^{\mathrm{sop}}\cong
T^{{\mathsf{Z}}^{\prime}}(n,d)$. In particular, $T^{{\mathsf{Z}}}(n,d)$ is
Ringel self-dual.

###### Proof.

In view of Lemma 5.3, we have that $T^{\mathsf{Z}}(n,d)\cong
T^{{\mathsf{Z}}^{\prime}}(n,d)$. In particular the second statement of the
theorem follows from the first one.

By Corollary 5.13, $T^{{\mathsf{Z}}^{\prime}}(n,d)$ embeds into
$\operatorname{End}_{T^{{\mathsf{Z}}}(n,d)}({\mathscr{T}})^{\mathrm{sop}}$. So
it suffices to show that
$\dim\operatorname{End}_{T^{{\mathsf{Z}}}(n,d)}({\mathscr{T}})=\dim
T^{{\mathsf{Z}}}(n,d)$.

In view of (5.5), we have that each ${\mathscr{T}}^{\text{\boldmath$\mu$}}$ is
$\tau_{n,d}$-self-dual and
$({\mathscr{T}}^{\text{\boldmath$\mu$}}:\Delta(\text{\boldmath$\lambda$}))=({\mathscr{T}}^{\text{\boldmath$\mu$}}:\nabla(\text{\boldmath$\lambda$}))$
for all $\text{\boldmath$\lambda$}\in\Lambda^{I}_{+}(n,d)$. We now have:

$\displaystyle\dim\operatorname{End}_{T^{{\mathsf{Z}}}(n,d)}({\mathscr{T}})$
$\displaystyle=\sum_{\text{\boldmath$\mu$},\text{\boldmath$\nu$}\in\Lambda^{I}(n,d)}\dim\operatorname{Hom}_{T^{\mathsf{Z}}(n,d)}({\mathscr{T}}^{\text{\boldmath$\mu$}},{\mathscr{T}}^{\text{\boldmath$\nu$}})$
$\displaystyle=\sum_{\text{\boldmath$\lambda$}\in\Lambda^{I}_{+}(n,d)}\sum_{\text{\boldmath$\mu$},\text{\boldmath$\nu$}\in\Lambda^{I}(n,d)}({\mathscr{T}}^{\text{\boldmath$\mu$}}:\Delta(\text{\boldmath$\lambda$}))({\mathscr{T}}^{\text{\boldmath$\nu$}}:\nabla(\text{\boldmath$\lambda$}))$
$\displaystyle=\sum_{\text{\boldmath$\lambda$}\in\Lambda^{I}_{+}(n,d)}\sum_{\text{\boldmath$\mu$},\text{\boldmath$\nu$}\in\Lambda^{I}(n,d)}({\mathscr{T}}^{\text{\boldmath$\mu$}}:\Delta(\text{\boldmath$\lambda$}))({\mathscr{T}}^{\text{\boldmath$\nu$}}:\Delta(\text{\boldmath$\lambda$}))$
$\displaystyle=\sum_{\text{\boldmath$\lambda$}\in\Lambda^{I}_{+}(n,d)}\sum_{\text{\boldmath$\mu$},\text{\boldmath$\nu$}\in\Lambda^{I}(n,d)}k_{\overleftarrow{\text{\boldmath$\lambda$}}^{\prime},\overleftarrow{\text{\boldmath$\mu$}}}k_{\overleftarrow{\text{\boldmath$\lambda$}}^{\prime},\overleftarrow{\text{\boldmath$\nu$}}}$
$\displaystyle=\sum_{\text{\boldmath$\lambda$}\in\Lambda^{I}_{+}(n,d)}\sum_{\text{\boldmath$\mu$},\text{\boldmath$\nu$}\in\Lambda^{I}(n,d)}k_{\text{\boldmath$\lambda$},\text{\boldmath$\mu$}}k_{\text{\boldmath$\lambda$},\text{\boldmath$\nu$}}$
$\displaystyle=\dim T^{{\mathsf{Z}}}(n,d),$

where we have used (5.10) for the first equality, [5, Proposition A2.2] for
the second equality, Proposition 5.17 for the fourth equality and [14, Theorem
5.17] for the last equality. ∎

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