# Good filtrations for generalized Schur algebras

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>To the memory of James Humphreys

###### Abstract.

Given a quasi-hereditary superalgebra $A$, the first author and R. Muth have
defined generalized Schur bi-superalgebras $T^{A}(n)$ and proved that these
algebras are again quasi-hereditary. In particular, $T^{A}(n)$ comes with a
family of standard modules. Developing the work of Donkin and Mathieu on good
filtrations, we prove that tensor product of standard modules over $T^{A}(n)$
has a standard filtration.

###### 2021 Mathematics Subject Classification:

16G30, 20C20

The first author was supported by the NSF grant DMS-2101791.

##  1\. Introduction

Let $S(n,d)$ be the classical Schur algebra as in [7]. A fundamental fact
going back to [8] is that the algebra $S(n,d)$ is (based) quasi-hereditary,
and so by [1], the module category $S(n,d)\\!\operatorname{-mod}$ is a highest
weight category, cf. [15, (2.5.3)], with standard modules
$\\{\Delta(\lambda)\mid\lambda\in\Lambda_{+}(n,d)\\}$, where
$\Lambda_{+}(n,d)$ is the set of partitions of $d$ with at most $n$ parts.
Moreover, there is a coproduct  ${\tt\Delta}$ on
$S(n)=\bigoplus_{d=0}^{\infty}S(n,d)$, which makes it into a bialgebra. Upon
restriction,

$\reflectbox{\rotatebox[origin={c}]{180.0}{${\tt\Delta}$}}:S(n,d)\to\bigoplus_{d_{1}+d_{2}=d}S(n,d_{1})\otimes
S(n,d_{2}),$

so for $\lambda\in\Lambda_{+}(n,d_{1})$ and $\mu\in\Lambda_{+}(n,d_{2})$, the
tensor product $\Delta(\lambda)\otimes\Delta(\mu)$ becomes an
$S(n,d_{1}+d_{2})$-module. It follows from (the easy type $A$ case of) the
work of Donkin [2] and Mathieu [14] that $\Delta(\lambda)\otimes\Delta(\mu)$
has a standard filtration, i.e. a filtration whose subquotients are standard
modules $\Delta(\nu)$ with $\nu\in\Lambda_{+}(n,d_{1}+d_{2})$.

Let $R$ be a characteristic $0$ domain and $\mathbb{F}$ be a field which is an
$R$-module. For the purposes of this introduction, we can take $R=\mathbb{Z}$
and $\mathbb{F}$ any field. Let $A$ be a (based) quasi-hereditary superalgebra
over $R$, i.e. $A$ is endowed with a heredity data $I,X,Y$, see § 2.3 for
details. We make a technical assumption that the data $I,X,Y$ are
conforming—this assumption is vacuous in the purely even situation and holds
in all the important examples known to us, for example when $A$ is an extended
zigzag superalgebra.

Developing a construction of Turner [16, 17, 18], the first author and Muth
defined generalized Schur (super)algebras

$T^{A}(n)=\bigoplus_{d=0}^{\infty}T^{A}(n,d),$

see [10], cf. also [5]. Importantly, we must first define $R$-forms
$T^{A}(n,d)_{R}$ of the algebras and then extend scalars to get
$T^{A}(n,d)=\mathbb{F}\otimes_{R}T^{A}(n,d)_{R}$, so it is crucial that $A$ is
defined over $R$. When $A$ is the trivial algebra $R$, the generalized Schur
algebra returns the classical Schur algebra $S(n,d)$.

As predicted by Turner, certain generalized Schur algebras play an important
role in modular representation theory, see [16, 6]. For example, it is
conjectured in [11, Conjecture 7.61] that weight $d$ RoCK blocks of the
classical Schur algebras and $q$-Schur algebras are Morita equivalent to the
generalized Schur algebras of the form $T^{{\mathsf{Z}}}(n,d)$, where
${\mathsf{Z}}$ is an extended zigzag algebra.

It is proved in [11] that under the assumption $d\leq n$ the algebras
$T^{A}(n,d)$ are again based quasi-hereditary, with heredity data
$\Lambda^{I}_{+}(n,d),{\mathcal{X}},{\mathcal{Y}}$, where
$\Lambda^{I}_{+}(n,d)$ is the set of $I$-multipartitions of $d$, see §3.4 for
details. In particular, the category of finite dimensional
$T^{A}(n,d)$-modules is a highest weight category with standard and costandard
modules

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

respectively. If $d>n$, the ‘standard’ modules
$\Delta(\text{\boldmath$\lambda$})$ and the ‘costandard’ modules
$\nabla(\text{\boldmath$\lambda$})$ are still defined and play an important
role. For example, if $A$ has a standard anti-involution then $T^{A}(n,d)$ is
cellular with the cell modules $\Delta(\text{\boldmath$\lambda$})$, see [11,
Lemma 6.25].

There is a natural coproduct  ${\tt\Delta}$ on $T^{A}(n)$, with

$\reflectbox{\rotatebox[origin={c}]{180.0}{${\tt\Delta}$}}:T^{A}(n,d)\to\bigoplus_{d_{1}+d_{2}=d}T^{A}(n,d_{1})\otimes
T^{A}(n,d_{2}).$

So for $\text{\boldmath$\lambda$}\in\Lambda^{I}_{+}(n,d_{1})$ and
$\text{\boldmath$\mu$}\in\Lambda^{I}_{+}(n,d_{2})$, the tensor product
$\Delta(\text{\boldmath$\lambda$})\otimes\Delta(\text{\boldmath$\mu$})$
becomes a $T^{A}(n,d_{1}+d_{2})$-module. The main goal of this note is to
prove the following

Main Theorem. The tensor product of two standard (resp. costandard) modules
over $T^{A}(n)$ has a standard (resp. costandard) filtration.

We hope that this theorem might be useful to study Ringel and Koszul duality
for generalized Schur algebras. In particular, we are interested in
generalizing the results of Donkin [3, 4] on Ringel duality for classical (and
$q$-) Schur algebras. In the follow up paper [12], we will use the Main
Theorem to prove that the extended zigzag Schur algebra $T^{\mathsf{Z}}(n,d)$
is Ringel self-dual.

The paper is organized as follows. Section 2 is preliminary. In particular, in
§2.3, we review the theory of based quasi-hereditary algebras, and in
§§2.4,2.5, we introduce the necessary combinatorial notions needed to handle
the formal characters of standard modules over generalized Schur algebras. In
Section 3, we begin to work with generalized Schur algebras. After reviewing
the definition and main properties, in §3.5 we prove a character identity
which is a combinatorial version of the standard filtration of the tensor
product of two standard modules. In Section 4, we prove the Main Theorem. Our
argument, when specialized to the case $A=R$, yields a very easy proof for the
classical Schur algebra case which seems to be new (or at least not well-
known).

##  2\. Preliminaries

###  2.1. General notation

For $m,n\in\mathbb{Z}$, we denote $[m,n]:=\\{k\in\mathbb{Z}\mid m\leq k\leq
n\\}$. If $n\in\mathbb{Z}_{>0}$, we also denote $[n]:=\\{1,2,\dots,n\\}$.

Throughout the paper, $I$ denotes a finite partially ordered set. We always
identify $I$ with the set $\\{0,1,\dots,l\\}$ for $l=|I|-1$, 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:

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

An (arbitrary) ground field is denoted by $\mathbb{F}$. Often we will also
need to work over a characteristic $0$ 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 $R=\mathbb{Z}$). 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 that $V_{\mathbb{F}}:=\mathbb{F}\otimes_{R}V_{R}$ is
a module over $A_{\mathbb{F}}:=\mathbb{F}\otimes_{R}A_{R}$.

###  2.2. Superspaces 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, write
$|v|={\varepsilon}$, and 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 $S$ consists of homogeneous elements then
$S=S_{\bar{0}}\sqcup S_{\bar{1}}$. Let $V$ and $W$ be superspaces. For
$\delta\in\mathbb{Z}/2$, a parity $\delta$ (homogeneous) linear map $f:V\to W$
is a linear map satisfying $f(V_{{\varepsilon}})\subseteq
W_{{\varepsilon}+\delta}$ for all ${\varepsilon}$.

Let $d\in\mathbb{Z}_{>0}$. The group $\mathfrak{S}_{d}$ acts on $V^{\otimes
d}$ on the right with 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$}\rangle:=\sharp\\{(k,l)\in[d]^{2}\mid
k<l,\,\sigma^{-1}k>\sigma^{-1}l,\ \text{and}\ v_{k},v_{l}\in V_{\bar{1}}\\}.$
(2.2)

For $0\leq c\leq d$, denote by ${}^{(c,d-c)}{\mathscr{D}}$ the set of the
shortest coset representatives for
$(\mathfrak{S}_{c}\times\mathfrak{S}_{d-c})\backslash\mathfrak{S}_{d}$. Given
$w_{1}\in V^{\otimes c}$ and $w_{2}\in V^{\otimes(d-c)}$, we define the star
product

$w_{1}*w_{2}:=\sum_{\sigma\in{}^{(c,d-c)}{\mathscr{D}}}(w_{1}\otimes
w_{2})^{\sigma}\in V^{\otimes d}.$ (2.3)

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

$\langle\text{\boldmath$v$},\text{\boldmath$w$}\rangle:=\sharp\\{(k,l)\in[d]^{2}\mid
k>l,\ v_{k}\in V_{\bar{1}},w_{l}\in W_{\bar{1}}\\}.$ (2.4)

Let now $A$ be a unital $\Bbbk$-superalgebra. 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).

For any superspace $V$, we consider the subspace of invariants

$\Gamma^{d}V:=\big{(}V^{\otimes d}\big{)}^{\mathfrak{S}_{d}}=\\{w\in
V^{\otimes d}\mid w^{\sigma}=w\ \text{for all
$\sigma\in\mathfrak{S}_{d}$}\\}.$ (2.5)

If $A$ is a superalgebra, then $\Gamma^{d}A$ is a subsuperalgebra of
$A^{\otimes d}$.

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 linear
map $f:V\to W$ satisfying $f(av)=(-1)^{|f||a|}af(v)$ for all (homogeneous)
$a,v$. Let

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

be the superspace of all $A$-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.

All subspaces, ideals, submodules, etc. are assumed to be homogeneous. For
example, given homogeneous elements $v_{1},\dots,v_{k}$ of an $A$-supermodule
$V$, we have the $A$-submodule $A\langle v_{1},\dots,v_{k}\rangle\subseteq V$
generated by $v_{1},\dots,v_{k}$.

###  2.3. Based quasi-hereditary algebras

The main reference here is [9]. Let $A$ be a $\Bbbk$-superalgebra.

######  Definition 2.6.

[9] 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 initial 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 $A$.

From now on, $A$ is a based quasi-hereditary superalgebra with heredity data
$I,X,Y$. We refer to the idempotents $e_{i}$ as the standard idempotents of
the heredity data. We have $B=B_{\bar{0}}\sqcup B_{\bar{1}}$ and

$B_{\bar{0}}=B_{\mathfrak{a}}\sqcup B_{\mathfrak{c}},$ (2.7)

where

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

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

######  Lemma 2.8.

[9, Lemmas 2.7, 2.8] Let $i,j\in I$ and $x\in X(i)$, $y\in Y(i)$.

  1. (i)

$e_{i}e_{j}=\delta_{i,j}e_{i}$

  2. (ii)

If $j\not\leq i$, then $e_{j}x=ye_{j}=0$.

######  Corollary 2.9.

We have $X\cap Y=\\{e_{i}\mid i\in I\\}$.

###### Proof.

Let $z\in X\cap Y$. As $z\in X$ we have $z\in X(i)$ so $ze_{i}=z$ for some
$i\in I$. As $z\in Y$, we have $z\in Y(j)$ so $e_{j}z=z$ for some $j\in I$. By
Lemma 2.8(ii), $j=i$, and the result follows from Definition 2.6(c). ∎

######  Definition 2.10.

Let $0\neq V\in A\\!\operatorname{-mod}$ and $i\in I$. We call $V$ a highest
weight module (of weight $i$) if there exists a homogeneous $v\in V$ such that
$e_{i}V$ is spanned by $v$, $Av=V$, and $j>i$ implies $e_{j}V=0$. In this case
we refer to $v$ as a highest weight vector of $V$.

######  Lemma 2.11.

Let $i\in I$, $0\neq V\in A\\!\operatorname{-mod}$ and $v\in V$ be a
homogeneous vector. Suppose that $e_{i}v=v$, $Av=V$, and $yv=0$ for all $y\in
Y\setminus\\{e_{i}\\}$. Then $V$ is a highest weight module of weight $i$.

###### Proof.

Since $A$ is based quasi-hereditary, it follows from the assumption $yv=0$ for
all $y\in Y\setminus\\{e_{i}\\}$ that $V$ is spanned by $\\{xv\mid x\in
X(i)\\}$. The result now follows from Definition 2.6(c) and Lemma 2.8(ii). ∎

Fix $i\in I$. Note that $A^{>i}$ is an ideal in $A$ and denote
$\tilde{A}:=A/A^{>i}$, $\tilde{a}:=a+A^{>i}\in\tilde{A}$ for $a\in A$. By
inflation, $\tilde{A}$-(super)modules will be automatically considered as
$A$-(super)modules. In particular, the standard module

$\Delta(i):=\tilde{A}\tilde{e}_{i}$

is considered as an $A$-module. We have that $\Delta(i)$ is a free
$\Bbbk$-module with basis $\\{v_{x}:=\tilde{x}\mid x\in X(i)\\}$ and the
action $av_{x}=\sum_{x^{\prime}\in X(i)}l_{x^{\prime}}^{x}(a)v_{x^{\prime}},$
cf. [9, §2.3]. Denoting

$v_{i}:=v_{e_{i}}\in\Delta(i),$

we have $e_{i}v_{i}=v_{i}$, and $e_{j}\Delta(i)\neq 0$ implies $j\leq i$
thanks to Lemma 2.8. Moreover, for all for all $x\in X(i)$ we have
$xv_{i}=v_{x}$, $e_{i}v_{x}=\delta_{x,e_{i}}v_{x}$. In particular, $\Delta(i)$
is a highest weight module of weight $i$ (with even highest weight vector). If
$V\in A\\!\operatorname{-mod}$ is isomorphic to $\Delta(i)$, then it is easy
to see, using the fact that $e_{i}V$ is free of rank $1$ as a $\Bbbk$-module,
that either $V\simeq\Delta(i)$ or $V\simeq\Pi\Delta(i)$.

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 $\\{w_{y}:=\tilde{y}\mid y\in Y(i)\\}$.

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.

Suppose now until the end of the subsection that $\Bbbk=\mathbb{F}$. Then each
$L(i):=\operatorname{head}\Delta(i)$ is irreducible, and $\\{L(i)\mid i\in
I\\}$ is a complete set of non-isomorphic irreducible $A$-supermodules. We
also have that
$L^{\mathrm{op}}(i):=\operatorname{head}\Delta^{\mathrm{op}}(i)$ is an
irreducible right module, and $\\{L^{\mathrm{op}}(i)\mid i\in I\\}$ is a
complete set of non-isomorphic irreducible right $A$-supermodules.

By [9, Lemma 3.3], $A$ is quasi-hereditary in the sense of Cline, Parshall and
Scott, and $A\\!\operatorname{-mod}$ is a highest weight category with
standard modules $\\{\Delta(i)\mid i\in I\\}$, see [1, Theorem 3.6]. In
particular, the projective cover $P(i)$ of $L(i)$ has a standard filtration
with the top factor $\Delta(i)$ and all other factors of the form $\Delta(j)$
or $\Pi\Delta(j)$ for $j>i$. Moreover, $\Delta(i)$ is the largest quotient of
$P(i)$ such that $[\Delta(i):L(i)]=1$ and $[\Delta(i):L(j)]\neq 0$ implies
$j\leq i$.

######  Proposition 2.12.

(Universality of standard modules) Let $\Bbbk=\mathbb{F}$, $i\in I$, and $V$
be a highest weight module of weight $i$ with highest weight vector $v$. Then
there is an homogeneous surjection $\Delta(i){\twoheadrightarrow}V$ of parity
$|v|$; in particular $e_{j}V\neq 0$ implies $j\leq i$.

###### Proof.

Let $e_{i}V$ be spanned by $v\in V$. There is a homogeneous surjective
$A$-supermodule homomorphism ${\varphi}:Ae_{i}{\twoheadrightarrow}V,\
ae_{i}\mapsto av$ of parity $|{\varphi}|=|v|$. As $e_{i}L(i)$ is
$1$-dimensional and $e_{i}L(j)\neq 0$ implies $i\leq j$, we have that
$Ae_{i}=P(i)\oplus P$, where $P$ is a direct sum of supermodules isomorphic to
$P(j)$ with $j>i$.

Note that $\operatorname{Hom}_{A}(\Delta(j),V)=0$ for any $j>i$, so
$\operatorname{Hom}_{A}(P(j),V)=0$ for all $j>i$, and we deduce
$\operatorname{Hom}_{A}(P,V)=0$. So the map ${\varphi}$ factors to give a
surjection $P(i){\twoheadrightarrow}V$. Moreover, $P(i)$ has a standard
filtration with top factor $\Delta(i)$ and other factors isomorphic to
$\Delta(j)$ with $j>i$, so the map further factors through the surjection
$\Delta(i){\twoheadrightarrow}V$. ∎

The following is a useful criterion for $V$ to have a standard filtration.

######  Corollary 2.13.

Let $\Bbbk=\mathbb{F}$, $V\in A\\!\operatorname{-mod}$, $v_{1}\dots,v_{t}\in
V$ be homogeneous elements, and set $V_{s}:=A\langle v_{1},\dots,v_{s}\rangle$
for $s=1,\dots,t$. Suppose that the following conditions hold:

  1. (1)

$V_{t}=V$;

  2. (2)

for each $s=1,\dots,t$ there exists $i_{s}\in I$ such that
$e_{i_{s}}v_{s}-v_{s}\in V_{s-1}$ and $yv_{s}\in V_{s-1}$ for all $y\in
Y\setminus\\{e_{i_{s}}\\}$;

  3. (3)

$\dim V=\sum_{s=1}^{t}\dim\Delta(i_{s})$.

Then $V_{s}/V_{s-1}\simeq\Pi^{|v_{s}|}\Delta(i_{s})$ for all $s=1,\dots,t$. In
particular, $V$ has a standard filtration.

###### Proof.

By Lemma 2.11 and Proposition 2.12, each $V_{s}/V_{s-1}$ is a quotient of
$\Pi^{|v_{s}|}\Delta(i_{s})$. The result follows by dimensions. ∎

The highest weight category $A\\!\operatorname{-mod}$ comes with costandard
modules $\\{\nabla(i)\mid i\in I\\}$. Let $J(i)$ be the injective hull of
$L(i)$ in $A\\!\operatorname{-mod}$ for $i\in I$. One can define $\nabla(i)$
as the largest submodule of $J(i)$ such that $[\nabla(i):L(i)]=1$ and
$[\nabla(i):L(j)]>0$ implies $j\leq i$. Let $V\in A\\!\operatorname{-mod}$. 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 $i_{r}\in
I$.

Given a right $A$-supermodule $V$, there is a (left) $A$-supermodule structure
on $V^{*}$ with $af(v)=(-1)^{|a||f|+|a||v|}f(va)$ for $a\in A,f\in V^{*},v\in
V$. For example, note that $L^{\mathrm{op}}(i)^{*}$ is irreducible,
$e_{i}L^{\mathrm{op}}(i)^{*}\neq 0$, and $e_{j}L^{\mathrm{op}}(i)^{*}\neq 0$
implies $j\leq i$; therefore $L^{\mathrm{op}}(i)^{*}\simeq L(i)$. Denoting by
$P^{\mathrm{op}}(i)$ the projective cover of $L^{\mathrm{op}}(i)$, we deduce
that $P^{\mathrm{op}}(i)^{*}\simeq J(i)$. This in turn implies easily:

$\nabla(i)\simeq\Delta^{\mathrm{op}}(i)^{*}.$ (2.14)

If $0=W_{0}\subseteq W_{1}\subseteq\dots\subseteq W_{l}=V$ is a standard
filtration of $V$ and $i\in I$ 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 [4, Proposition A2.2],
we have

$(V:\Delta(i))=\dim\operatorname{Hom}_{A}(V,\nabla(i)).$ (2.15)

###  2.4. Partitions and compositions

We denote by $\Lambda_{+}$ the set of all partitions. For
$\lambda\in\Lambda_{+}$, we have the conjugate partition $\lambda^{\prime}$,
see [13, p.2]. 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 $\lambda$. Define the partial
order $\preceq$ on the nodes as follows: $(r,s)\preceq(r^{\prime},s^{\prime})$
if and only if $r\leq r^{\prime}$ and $s\leq s^{\prime}$.

For $\lambda,\mu,\nu\in\Lambda_{+}$, we denote by $c^{\,\lambda}_{\mu,\nu}$
the corresponding Littlewood-Richardson coefficient, see [13, § I.9].

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
$\lambda,\mu\in\Lambda(n)$, we define

$\lambda+\mu:=(\lambda_{1}+\mu_{1},\dots,\lambda_{n}+\mu_{n}).$

For $1\leq r\leq n$, we denote

${\varepsilon}_{r}:=(0,\dots,0,1,0,\dots,0)\in\Lambda(n)$ (2.16)

with $1$ in position $r$. For
$\lambda=(\lambda_{1},\dots,\lambda_{n})\in\Lambda(n)$, set
$|\lambda|:=\lambda_{1}+\dots+\lambda_{n}$.

Denote

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

Sometimes we collect equal parts of $\lambda\in\Lambda_{+}(n)$ to write it as
$\lambda=(l_{1}^{a_{1}},\dots,l_{k}^{a_{k}})$ for $l_{1}>\dots>l_{k}\geq 0$
and $a_{1},\dots,a_{k}>0$ with $a_{1}+\dots+a_{k}=n$. We interpret
$\Lambda_{+}(n)$ as a subset of $\Lambda_{+}$ in the obvious way. For
$d\in\mathbb{Z}_{\geq 0}$, let

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

Let $S$ be a finite set. We will consider the set of $S$-multicompositions and
$S$-multipartitions

$\displaystyle\Lambda^{S}(n)$ $\displaystyle:=$
$\displaystyle\Lambda(n)^{S}=\\{\text{\boldmath$\lambda$}=(\lambda^{(s)})_{s\in
S}\mid\lambda^{(s)}\in\Lambda(n)\ \text{for all $s\in S$}\\},$
$\displaystyle\Lambda_{+}^{S}(n)$ $\displaystyle:=$
$\displaystyle\Lambda_{+}(n)^{S}=\\{\text{\boldmath$\lambda$}=(\lambda^{(s)})_{s\in
S}\mid\lambda^{(s)}\in\Lambda_{+}(n)\ \text{for all $s\in S$}\\}.$

For $\text{\boldmath$\lambda$},\text{\boldmath$\mu$}\in\Lambda^{S}(n)$ we
define $\text{\boldmath$\lambda$}+\text{\boldmath$\mu$}$ to be
$\text{\boldmath$\nu$}\in\Lambda^{S}(n)$ with
$\nu^{(s)}=\lambda^{(s)}+\mu^{(s)}$ for all $s\in S$. For
$\text{\boldmath$\lambda$}\in\Lambda^{S}(n)$, we define its Young diagram to
be $[\text{\boldmath$\lambda$}]:=\bigsqcup_{s\in S}[\lambda^{(s)}]$. We also
set

$\|\text{\boldmath$\lambda$}\|:=(|\lambda^{(s)}|)_{s\in S}\in\mathbb{Z}_{\geq
0}^{S}.$

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

$\displaystyle\Lambda^{S}(n,d)$
$\displaystyle:=\\{\text{\boldmath$\lambda$}\in\Lambda^{S}(n)\mid{\textstyle\sum_{s\in
S}}|\lambda^{(s)}|=d\\},$ $\displaystyle\Lambda_{+}^{S}(n,d)$
$\displaystyle:=\\{\text{\boldmath$\lambda$}\in\Lambda^{S}_{+}(n)\mid{\textstyle\sum_{s\in
S}}|\lambda^{(s)}|=d\\}.$

In the special case $S=I=\\{0,\dots,l\\}$, we also write
$\text{\boldmath$\lambda$}=(\lambda^{(0)},\dots,\lambda^{(l)})$ instead of
$\text{\boldmath$\lambda$}=(\lambda^{(i)})_{i\in I}\in\Lambda^{I}(n)$. For
$i\in I$, and $\lambda\in\Lambda(n,d)$, define

$\text{\boldmath$\iota$}_{i}(\lambda):=(0,\dots,0,\lambda,0,\dots,0)\in\Lambda^{I}(n,d),$
(2.17)

with $\lambda$ in the $i$th position.

Let $\leq$ be a partial order on $S$. We have a partial order $\unlhd_{S}$ on
the set $\mathbb{Z}_{\geq 0}^{S}$ with $(a_{s})_{s\in
S}\unlhd_{S}(b_{s})_{s\in S}$ if and only if $\sum_{t\geq
s}a_{t}\leq\sum_{t\geq s}b_{t}$ for all $s\in S$. 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 have a partial order $\leq_{S}$ on $\Lambda^{S}(n,d)$ defined as follows:
$\text{\boldmath$\lambda$}\leq_{S}\text{\boldmath$\mu$}$ if and only if either
$\|\text{\boldmath$\lambda$}\|\lhd_{S}\|\text{\boldmath$\mu$}\|$, or
$\|\text{\boldmath$\lambda$}\|=\|\text{\boldmath$\mu$}\|$ and
$\lambda^{(s)}\unlhd\mu^{(s)}$ for all $s\in S$.

###  2.5. Tableaux

Let $I,X,Y$ be 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 ${\operatorname{let}}(L):=l$ and
${\operatorname{col}}(L):=x.$ For all $i\in I$, we fix arbitrary total orders
‘$<$’ on the sets ${\mathscr{A}}_{X(i)}$ which satisfy $r^{x}<s^{x}$ if $r<s$
(in the standard order on $[n]$). All definitions of this subsection which
involve $X$ have obvious analogues for $Y$, for example, we have the colored
alphabets ${\mathscr{A}}_{Y}$ and ${\mathscr{A}}_{Y(i)}$.

Let
$\text{\boldmath$\lambda$}=(\lambda^{(0)},\dots,\lambda^{(l)})\in\Lambda^{I}(n,d)$.
Fix $i\in I$. A standard $X(i)$-colored $\lambda^{(i)}$-tableau is a function
$T:[\lambda^{(i)}]\to{\mathscr{A}}_{X(i)}$ such that the following two
conditions are satisfied:

  1. (R)

$T(M)\leq T(N)$ whenever $M\prec N$ are nodes in the same row of
$[\lambda^{(i)}]$, and the equality is allowed only if
${\operatorname{col}}(T(M))\in X(i)_{\bar{0}}$.

  2. (C)

$T(M)\leq T(N)$ whenever $M\prec N$ are nodes in the same column of
$[\lambda^{(i)}]$, and the equality is allowed only if
${\operatorname{col}}(T(M))\in X(i)_{\bar{1}}$.

We denote by $\operatorname{Std}^{X(i)}(\lambda^{(i)})$ the set of all
standard $X(i)$-colored $\lambda^{(i)}$-tableaux. Recalling the idempotents
$e_{i}\in X(i)\cap Y(i)$, the initial $\lambda^{(i)}$-tableau
$T^{\lambda^{(i)}}$ is
$T^{\lambda^{(i)}}:[\lambda^{(i)}]\to{\mathscr{A}}_{X(i)},\ (r,s)\mapsto
r^{e_{i}}.$ Note that $T^{\lambda^{(i)}}$ is in both
$\operatorname{Std}^{X(i)}(\lambda^{(i)})$ and
$\operatorname{Std}^{Y(i)}(\lambda^{(i)})$.

Let $T\in\operatorname{Std}^{X(i)}(\lambda^{(i)})$. Denote
$d_{i}:=|\lambda^{(i)}|$. Reading the entries of $T$ along the rows from left
to right starting from the first row, we get a sequence $l_{1}^{x_{1}}\cdots
l_{d_{i}}^{x_{d_{i}}}\in{\mathscr{A}}_{X(i)}^{d_{i}}$. We denote
$\text{\boldmath$l$}^{T}:=l_{1}\cdots l_{d_{i}}$ and
$\text{\boldmath$x$}^{T}:=x_{1}\cdots x_{d_{i}}$.

For a function
$\text{\boldmath$T$}:[\text{\boldmath$\lambda$}]\to{\mathscr{A}}_{X}$ and
$i\in I$, we set $T^{(i)}:=\text{\boldmath$T$}|_{[\lambda^{(i)}]}$ to be the
restriction of $T$ to $[\lambda^{(i)}]$. We write
$\text{\boldmath$T$}=(T^{(0)},\dots,T^{(l)})$, keeping in mind that the
restrictions $T^{(i)}$ determine $T$ uniquely. A standard $X$-colored
$\lambda$-tableau is a function
$\text{\boldmath$T$}:[\text{\boldmath$\lambda$}]\to{\mathscr{A}}_{X}$ such
that $T^{(i)}\in\operatorname{Std}^{X(i)}(\lambda^{(i)})$ for all $i\in I$. We
denote by $\operatorname{Std}^{X}(\text{\boldmath$\lambda$})$ the set of all
standard $X$-colored $\lambda$-tableaux. For example, we have the initial
$\lambda$-tableau

$\text{\boldmath$T$}^{\text{\boldmath$\lambda$}}=(T^{\lambda^{(0)}},\dots,T^{\lambda^{(l)}})\in\operatorname{Std}^{X}(\text{\boldmath$\lambda$})\cap\operatorname{Std}^{Y}(\text{\boldmath$\lambda$}).$

For $\text{\boldmath$T$}\in\operatorname{Std}^{X}(\text{\boldmath$\lambda$})$,
we denote

$\text{\boldmath$l$}^{\text{\boldmath$T$}}:=\text{\boldmath$l$}^{T^{(0)}}\cdots\,\text{\boldmath$l$}^{T^{(l)}}\in[n]^{d},\quad\text{\boldmath$x$}^{\text{\boldmath$T$}}:=\text{\boldmath$x$}^{T^{(0)}}\cdots\,\text{\boldmath$x$}^{T^{(l)}}\in
X^{d},\quad\text{and}\quad\text{\boldmath$l$}^{\text{\boldmath$\lambda$}}:=\text{\boldmath$l$}^{\text{\boldmath$T$}^{\text{\boldmath$\lambda$}}}.$

The sequence $\text{\boldmath$y$}^{\text{\boldmath$T$}}$ for
$\text{\boldmath$T$}\in\operatorname{Std}^{Y}(\text{\boldmath$\lambda$})$ is
defined similarly to $\text{\boldmath$x$}^{\text{\boldmath$T$}}$.

Let $\text{\boldmath$\lambda$}\in\Lambda_{+}^{I}(n,d)$ and
$\text{\boldmath$T$}\in\operatorname{Std}^{X}(\text{\boldmath$\lambda$})$,
with $\text{\boldmath$l$}^{\text{\boldmath$T$}}=l_{1}\cdots l_{d}$ and
$\text{\boldmath$x$}^{\text{\boldmath$T$}}=x_{1}\cdots x_{d}$. Suppose that
there exist $i_{1},\dots,i_{d}\in I$ such that
$e_{i_{1}}x_{1}=x_{1},\,\dots,\,e_{i_{d}}x_{d}=x_{d}$. Recalling (2.16) and
(2.17), we define the left weight of $T$ to be

$\text{\boldmath$\alpha$}(\text{\boldmath$T$}):=\sum_{c=1}^{d}\text{\boldmath$\iota$}_{i_{c}}({\varepsilon}_{l_{c}})\in\Lambda^{I}(n,d).$

For $\text{\boldmath$\mu$}\in\Lambda^{I}(n,d)$, we denote

$\operatorname{Std}^{X}(\text{\boldmath$\lambda$},\text{\boldmath$\mu$}):=\\{\text{\boldmath$T$}\in\operatorname{Std}^{X}(\text{\boldmath$\lambda$})\mid\text{\boldmath$\alpha$}(\text{\boldmath$T$})=\text{\boldmath$\mu$}\\}.$
(2.18)

###  2.6. Symmetric functions

Let ${\mathrm{Sym}}$ be the ring of symmetric functions over $\mathbb{Z}$ in
infinitely many variables $z_{1},z_{2},\dots$, see [13], with basis consisting
of Schur functions $s_{\lambda}\in{\mathrm{Sym}}$ for $\lambda\in\Lambda_{+}$.
Recall that ${\mathrm{Sym}}$ is a Hopf algebra with coproduct

${\tt\Delta}:{\mathrm{Sym}}\to{\mathrm{Sym}}\otimes{\mathrm{Sym}},\
s_{\lambda}\mapsto\sum_{\mu,\nu\in\Lambda_{+}}c^{\,\lambda}_{\mu,\nu}\,s_{\mu}\otimes
s_{\nu},$

where $c^{\,\lambda}_{\mu,\nu}$ are the Littlewood-Richardson coefficients,
see [13, §I.5].

For $n\in\mathbb{Z}_{>0}$, let
${\mathrm{Sym}}(n)=\mathbb{Z}[z_{1},\dots,z_{n}]^{\mathfrak{S}_{n}}$ be the
ring of symmetric polynomials in $z_{1},\dots,z_{n}$. There is a canonical
homomorphism $\rho_{n}:{\mathrm{Sym}}\to{\mathrm{Sym}}(n)$, see [13, p.18].
For $\lambda\in\Lambda_{+}(n)$, let
$s_{\lambda}(z_{1},\dots,z_{n}):=\rho_{n}(s_{\lambda})\in{\mathrm{Sym}}(n)$.

For a finite set $S$, we introduce $S$-fold tensor products
${\mathrm{Sym}}^{S}:={\mathrm{Sym}}^{\otimes S}$ and
${\mathrm{Sym}}^{S}(n):={\mathrm{Sym}}(n)^{\otimes S}$. We have the canonical
homomorphism

$\rho_{n}^{S}=\rho_{n}^{\otimes
S}:{\mathrm{Sym}}^{S}\to{\mathrm{Sym}}^{S}(n).$ (2.19)

For $d\in\mathbb{Z}_{\geq 0}$, we denote by ${\mathrm{Sym}}^{S}(n,d)$ the
degree $d$ component of ${\mathrm{Sym}}^{S}(n)$.

Given $\text{\boldmath$\nu$}=(\nu^{(s)})_{s\in S}\in\Lambda^{S}_{+}$, we have
an element

$s_{\text{\boldmath$\nu$}}:=\otimes_{t\in
S}s_{\nu^{(t)}}\in{\mathrm{Sym}}^{S}.$

If $\text{\boldmath$\nu$}\in\Lambda^{S}_{+}(n)$, we set

$s_{\text{\boldmath$\nu$}}(z_{1},\dots,z_{n}):=\rho_{n}^{S}(s_{\text{\boldmath$\nu$}})\in{\mathrm{Sym}}^{S}(n).$

If $m=|S|$, iterating the coproduct (and using coassociativity and
cocommutativity) we get the algebra homomorphism

${\tt\Delta}^{m-1}:{\mathrm{Sym}}\to{\mathrm{Sym}}^{S},$ (2.20)

(with ${\tt\Delta}^{0}$ interpreted as the identity map). We introduce the
iterated Littlewood-Richardson coefficients
$c^{\lambda}_{\text{\boldmath$\nu$}}$ from

${\tt\Delta}^{m-1}(s_{\lambda})=\sum_{\text{\boldmath$\nu$}\in\Lambda^{S}_{+}}c^{\lambda}_{\text{\boldmath$\nu$}}s_{\text{\boldmath$\nu$}}.$
(2.21)

##  3\. Generalized Schur algebras

Throughout the section, we fix $n\in\mathbb{Z}_{>0}$. We also fix a based
quasi-hereditary superalgebra $A_{R}$ over $R$ with conforming heredity data
$I,X,Y$.

###  3.1. Definition

Let $S$ be a set and $d\in\mathbb{Z}_{\geq 0}$. Recall that the symmetric
group $\mathfrak{S}_{d}$ acts on $S^{d}$ by place permutations. 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 all $\mathfrak{S}_{d}$-orbits on $U$.

Let $H=H_{\bar{0}}\sqcup H_{\bar{1}}$ be a set of non-zero homogeneous
elements of $A_{R}$. Define $\operatorname{Tri}^{H}(n,d)$ to be the set of all
triples

$(\text{\boldmath$a$},\text{\boldmath$r$},\text{\boldmath$s$})=(a_{1}\cdots
a_{d},\,r_{1}\cdots r_{d},\,s_{1}\cdots s_{d})\in
H^{d}\times[n]^{d}\times[n]^{d}$

such that for all $1\leq k\neq l\leq d$ we have
$(a_{k},r_{k},s_{k})=(a_{l},r_{l},s_{l})$ only if $a_{k}\in H_{\bar{0}}$. Then
$\operatorname{Tri}^{H}(n,d)\subseteq H^{d}\times[n]^{d}\times[n]^{d}$ is a
$\mathfrak{S}_{d}$-invariant subset, so we can choose a set
$\operatorname{Tri}^{H}(n,d)/\mathfrak{S}_{d}$ of $\mathfrak{S}_{d}$-orbit
representatives and identify it with the set of all $\mathfrak{S}_{d}$-orbits
on $\operatorname{Tri}^{H}(n,d)$ as in the previous paragraph.

Sometimes we use a preferred choice of representatives for
$\operatorname{Tri}^{H}(n,d)/\mathfrak{S}_{d}$ defined as follows. Fix a total
order $<$ on $H\times[n]\times[n]$. We have a lexicographic order on
$\operatorname{Tri}^{H}(n,d)$:
$(\text{\boldmath$a$},\text{\boldmath$r$},\text{\boldmath$s$})<(\text{\boldmath$a$}^{\prime},\text{\boldmath$r$}^{\prime},\text{\boldmath$s$}^{\prime})$
if and only if there exists $l\in[d]$ such that
$(a_{k},r_{k},s_{k})=(a_{k}^{\prime},r_{k}^{\prime},s_{k}^{\prime})$ for all
$k<l$ and
$(a_{l},r_{l},s_{l})<(a_{l}^{\prime},r_{l}^{\prime},s_{l}^{\prime})$. Denote

$\operatorname{Tri}^{H}_{0}(n,d)=\\{(\text{\boldmath$a$},\text{\boldmath$r$},\text{\boldmath$s$})\in\operatorname{Tri}^{H}(n,d)\mid(\text{\boldmath$a$},\text{\boldmath$r$},\text{\boldmath$s$})\leq(\text{\boldmath$a$},\text{\boldmath$r$},\text{\boldmath$s$})\sigma\
\text{for all}\ \sigma\in\mathfrak{S}_{d}\\}.$ (3.1)

For
$(\text{\boldmath$a$},\text{\boldmath$r$},\text{\boldmath$s$})\in\operatorname{Tri}^{H}(n,d)$
and $\sigma\in\mathfrak{S}_{d}$, we define

$\displaystyle\langle\text{\boldmath$a$},\text{\boldmath$r$},\text{\boldmath$s$}\rangle$
$\displaystyle:=\sharp\\{(k,l)\in[d]^{2}\mid k<l,\ a_{k},a_{l}\in
H_{\bar{1}},\ (a_{k},r_{k},s_{k})>(a_{l},r_{l},s_{l})\\}.$ (3.2)

Specializing to $H=B$, let
$(\text{\boldmath$b$},\text{\boldmath$r$},\text{\boldmath$s$})\in\operatorname{Tri}^{B}(n,d)$.
For $b\in B$ and $r,s\in[n]$, we denote

$[\text{\boldmath$b$},\text{\boldmath$r$},\text{\boldmath$s$}:b,r,s]:=\sharp\\{k\in[d]\mid(b_{k},r_{k},s_{k})=(b,r,s)\\},$

and, recalling (2.7), we set

$[\text{\boldmath$b$},\text{\boldmath$r$},\text{\boldmath$s$}]^{!}_{\mathfrak{c}}:=\prod_{b\in
B_{\mathfrak{c}},\,r,s\in[n]}[\text{\boldmath$b$},\text{\boldmath$r$},\text{\boldmath$s$}:b,r,s]!.$
(3.3)

Let $M_{n}(A_{R})$ be the superalgebra of $n\times n$ matrices with entries in
$A_{R}$. For $a\in A_{R}$, we denote by $\xi_{r,s}^{a}\in M_{n}(A_{R})$ the
matrix with $a$ in the position $(r,s)$ and zeros elsewhere. By definition,
$|\xi_{r,s}^{a}|=|a|$. For each $d\in\mathbb{Z}_{\geq 0}$ we have a
superalgebra structure on $M_{n}(A_{R})^{\otimes d}$, and then on
$\bigoplus_{d\geq 0}M_{n}(A_{R})^{\otimes d}$.

Recall from [5, §4.1], that $\bigoplus_{d\geq 0}M_{n}(A_{R})^{\otimes d}$ is a
bisuperalgebra with the coproduct  ${\tt\Delta}$ defined by

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

Moreover, recalling (2.3), $\bigoplus_{d\geq 0}M_{n}(A_{R})^{\otimes d}$ is
also a bisuperalgebra with respect to  ${\tt\Delta}$ and $*$, see [5, Lemma
3.12].

According to (2.1) $\mathfrak{S}_{d}$ acts on $M_{n}(A_{R})^{\otimes d}$ with
superalgebra automorphisms, and using the notation (2.5), we have the
subsuperalgebra of invariants $\Gamma^{d}M_{n}(A_{R})\subseteq
M_{n}(A_{R})^{\otimes d}.$ For
$(\text{\boldmath$a$},\text{\boldmath$r$},\text{\boldmath$s$})\in\operatorname{Tri}^{H}(n,d)$,
we have elements

$\xi_{\text{\boldmath$r$},\text{\boldmath$s$}}^{\text{\boldmath$a$}}:=\sum_{(\text{\boldmath$c$},\text{\boldmath$t$},\text{\boldmath$u$})\sim(\text{\boldmath$a$},\text{\boldmath$r$},\text{\boldmath$s$})}(-1)^{\langle\text{\boldmath$a$},\text{\boldmath$r$},\text{\boldmath$s$}\rangle+\langle\text{\boldmath$c$},\text{\boldmath$t$},\text{\boldmath$u$}\rangle}\xi_{t_{1},u_{1}}^{c_{1}}\otimes\dots\otimes\xi_{t_{d},u_{d}}^{c_{d}}\in\Gamma^{d}M_{n}(A_{R}).$

We have the following $R$-basis of $\Gamma^{d}M_{n}(A_{R})$:

$\\{\xi_{\text{\boldmath$r$},\text{\boldmath$s$}}^{\text{\boldmath$b$}}\mid(\text{\boldmath$b$},\text{\boldmath$r$},\text{\boldmath$s$})\in\operatorname{Tri}^{B}(n,d)/\mathfrak{S}_{d}\\}.$
(3.4)

For
$(\text{\boldmath$b$},\text{\boldmath$r$},\text{\boldmath$s$})\in\operatorname{Tri}^{B}(n,d)$,
we also set

$\eta^{\text{\boldmath$b$}}_{\text{\boldmath$r$},\text{\boldmath$s$}}:=[\text{\boldmath$b$},\text{\boldmath$r$},\text{\boldmath$s$}]^{!}_{\mathfrak{c}}\,\xi^{\text{\boldmath$b$}}_{\text{\boldmath$r$},\text{\boldmath$s$}},$

and

$T(n,d)_{R}=T^{A}(n,d)_{R}:=\operatorname{span}_{R}\big{\\{}\,\eta^{\text{\boldmath$b$}}_{\text{\boldmath$r$},\text{\boldmath$s$}}\mid(\text{\boldmath$b$},\text{\boldmath$r$},\text{\boldmath$s$})\in\operatorname{Tri}^{B}(n,d)\,\big{\\}}\subseteq\Gamma^{d}M_{n}(A_{R}).$

Let

$T(n)_{R}:=\bigoplus_{d\geq 0}T(n,d)_{R}.$

By [10, Proposition 3.12, Lemma 3.10], $T(n,d)_{R}$ is a unital
$R$-subsuperalgebra of $M_{n}(A_{R})^{\otimes d}$ with $R$-basis

$\big{\\{}\,\eta^{\text{\boldmath$b$}}_{\text{\boldmath$r$},\text{\boldmath$s$}}\mid(\text{\boldmath$b$},\text{\boldmath$r$},\text{\boldmath$s$})\in\operatorname{Tri}^{B}(n,d)/\mathfrak{S}_{d}\,\big{\\}}.$
(3.5)

Moreover, by [10, Corollary 3.24], $T(n)_{R}$ is a sub-bisuperalgebra of
$\bigoplus_{d\geq 0}M_{n}(A_{R})^{\otimes d}$ (with respect to  ${\tt\Delta}$
and the usual product). Moreover:

######  Lemma 3.6.

[10, Corollary 4.4] $T(n)_{R}$ is a sub-bisuperalgebra of $\bigoplus_{d\geq
0}M_{n}(A_{R})^{\otimes d}$ with respect to the coproduct  ${\tt\Delta}$ and
the product $*$

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

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

We denote
$1_{\mathbb{F}}\otimes\eta^{\text{\boldmath$b$}}_{\text{\boldmath$r$},\text{\boldmath$s$}}\in
T(n,d)_{\mathbb{F}}$ 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 fact, when working over the field, we will
often drop the index and write simply

$T(n,d):=T(n,d)_{\mathbb{F}}.$ (3.7)

If $W_{1}$ is a $T(n,d_{1})$-supermodule and $W_{2}$ is a
$T(n,d_{2})$-supermodule, we consider $W_{1}\otimes W_{2}$ as a
$T(n,d_{1}+d_{2})$-supermodule via the coproduct  ${\tt\Delta}$ .

###  3.2. Properties of product and coproduct

In this section we work over $R$. Define the structure constants
$\kappa^{b}_{a,c}\in R$ of $A_{R}$ from $ac=\sum_{b\in B}\kappa^{b}_{a,c}b$
for $a,c\in A_{R}$. More generally, for
$\text{\boldmath$b$}=(b_{1},\dots,b_{d})\in
B^{d}\quad\text{and}\quad\text{\boldmath$a$}=(a_{1},\dots,a_{d}),\,\text{\boldmath$c$}=(c_{1},\dots,c_{d})\in
A_{R}^{d},$ we define

$\kappa^{\text{\boldmath$b$}}_{\text{\boldmath$a$},\text{\boldmath$c$}}:=\kappa^{b_{1}}_{a_{1},c_{1}}\cdots\kappa^{b_{d}}_{a_{d},c_{d}}\in
R.$

Recall the notation (3.2), (2.4). The following generalization of [7, (2.3b)]
follows from [5, (6.14)], cf. [10, Proposition 3.6].

######  Proposition 3.8.

Let
$(\text{\boldmath$a$},\text{\boldmath$p$},\text{\boldmath$q$}),\,(\text{\boldmath$c$},\text{\boldmath$u$},\text{\boldmath$v$})\in\operatorname{Tri}^{B}(n,d)$.
Then in $T(n,d)_{R}$ we have

$\eta^{\text{\boldmath$a$}}_{\text{\boldmath$p$},\text{\boldmath$q$}}\,\eta^{\text{\boldmath$c$}}_{\text{\boldmath$u$},\text{\boldmath$v$}}=\sum_{[\text{\boldmath$b$},\text{\boldmath$r$},\text{\boldmath$s$}]\in\operatorname{Tri}^{B}(n,d)/\mathfrak{S}_{d}}g_{\text{\boldmath$a$},\text{\boldmath$p$},\text{\boldmath$q$};\text{\boldmath$c$},\text{\boldmath$u$},\text{\boldmath$v$}}^{\text{\boldmath$b$},\text{\boldmath$r$},\text{\boldmath$s$}}\,\eta^{\text{\boldmath$b$}}_{\text{\boldmath$r$},\text{\boldmath$s$}}$

where

$g_{\text{\boldmath$a$},\text{\boldmath$p$},\text{\boldmath$q$};\text{\boldmath$c$},\text{\boldmath$u$},\text{\boldmath$v$}}^{\text{\boldmath$b$},\text{\boldmath$r$},\text{\boldmath$s$}}=\frac{[\text{\boldmath$a$},\text{\boldmath$p$},\text{\boldmath$q$}]^{!}_{\mathfrak{c}}\cdot[\text{\boldmath$c$},\text{\boldmath$u$},\text{\boldmath$v$}]^{!}_{\mathfrak{c}}}{[\text{\boldmath$b$},\text{\boldmath$r$},\text{\boldmath$s$}]^{!}_{\mathfrak{c}}}\sum_{\text{\boldmath$a$}^{\prime},\text{\boldmath$c$}^{\prime},\text{\boldmath$t$}}(-1)^{\langle\text{\boldmath$a$},\text{\boldmath$p$},\text{\boldmath$q$}\rangle+\langle\text{\boldmath$c$},\text{\boldmath$u$},\text{\boldmath$v$}\rangle+\langle\text{\boldmath$a$}^{\prime},\text{\boldmath$r$},\text{\boldmath$t$}\rangle+\langle\text{\boldmath$c$}^{\prime},\text{\boldmath$t$},\text{\boldmath$s$}\rangle+\langle\text{\boldmath$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$}^{\prime}\in B^{d}$ and
$\text{\boldmath$t$}\in[n]$ such that
$(\text{\boldmath$a$}^{\prime},\text{\boldmath$r$},\text{\boldmath$t$})\sim(\text{\boldmath$a$},\text{\boldmath$p$},\text{\boldmath$q$})$
and
$(\text{\boldmath$c$}^{\prime},\text{\boldmath$t$},\text{\boldmath$s$})\sim(\text{\boldmath$c$},\text{\boldmath$u$},\text{\boldmath$v$})$.

######  Lemma 3.9.

[10, Lemma 4.6] Let $q\in\mathbb{Z}_{>0}$,
$d_{1},\dots,d_{q}\in\mathbb{Z}_{\geq 0}$ with $d_{1}+\dots+d_{q}=d$, and for
$m=1,\dots,q$, we have
$(\text{\boldmath$b$}^{m},\text{\boldmath$r$}^{m},\text{\boldmath$s$}^{m})\in\operatorname{Tri}^{B}(n,d_{m})$
with $\text{\boldmath$b$}^{m}=b^{m}_{1}\cdots b^{m}_{d_{m}},\
\text{\boldmath$r$}^{m}=r^{m}_{1}\cdots r^{m}_{d_{m}},\
\text{\boldmath$s$}^{m}=s^{m}_{1}\cdots s^{m}_{d_{m}}$. If
$(b^{m}_{t},r^{m}_{t},s^{m}_{t})\neq(b^{l}_{u},r^{l}_{u},s^{l}_{u})$ for all
$1\leq m\neq l\leq q$, $1\leq t\leq d_{m}$ and $1\leq u\leq d_{l}$, then

$\eta^{\text{\boldmath$b$}^{1}\cdots\text{\boldmath$b$}^{q}}_{\text{\boldmath$r$}^{1}\cdots\text{\boldmath$r$}^{q},\text{\boldmath$s$}^{1}\cdots\text{\boldmath$s$}^{q}}=\eta_{\text{\boldmath$r$}^{1},\text{\boldmath$s$}^{1}}^{\text{\boldmath$b$}^{1}}*\dots*\eta_{\text{\boldmath$r$}^{q},\text{\boldmath$s$}^{q}}^{\text{\boldmath$b$}^{q}}.$

To describe  ${\tt\Delta}$ on basis elements, let
${\mathcal{T}}=(\text{\boldmath$b$},\text{\boldmath$r$},\text{\boldmath$s$})\in\operatorname{Tri}^{B}_{0}(n,d)$.
We write
$\eta_{\mathcal{T}}:=\eta^{\text{\boldmath$b$}}_{\text{\boldmath$r$},\text{\boldmath$s$}}$
and
${\mathcal{T}}\sigma:=(\text{\boldmath$b$},\text{\boldmath$r$},\text{\boldmath$s$})\sigma$
for $\sigma\in\mathfrak{S}_{d}$. We have that the stabilizer
$\mathfrak{S}_{\mathcal{T}}:=\\{\sigma\in\mathfrak{S}_{d}\mid{\mathcal{T}}\sigma={\mathcal{T}}\\}$
is a standard parabolic subgroup. Let ${}^{\mathcal{T}}{\mathscr{D}}$ be the
set of the shortest coset representatives in
$\mathfrak{S}_{\mathcal{T}}\backslash\mathfrak{S}_{d}$. We also set

$[{\mathcal{T}}]^{!}_{\mathfrak{c}}:=[\text{\boldmath$b$},\text{\boldmath$r$},\text{\boldmath$s$}]^{!}_{\mathfrak{c}}.$
(3.10)

If $d=d_{1}+d_{2}$,
${\mathcal{T}}^{1}=(\text{\boldmath$b$}^{1},\text{\boldmath$r$}^{1},\text{\boldmath$s$}^{1})\in\operatorname{Tri}^{B}(n,d_{1})$
and
${\mathcal{T}}^{2}=(\text{\boldmath$b$}^{2},\text{\boldmath$r$}^{2},\text{\boldmath$s$}^{2})\in\operatorname{Tri}^{B}(n,d_{2})$,
we denote
${\mathcal{T}}^{1}{\mathcal{T}}^{2}:=(\text{\boldmath$b$}^{1}\text{\boldmath$b$}^{2},\text{\boldmath$r$}^{1}\text{\boldmath$r$}^{2},\text{\boldmath$s$}^{1}\text{\boldmath$s$}^{2})\in
B^{d}\times[n]^{d}\times[n]^{d}.$ Recall the notation (3.1). For
${\mathcal{T}}\in\operatorname{Tri}^{B}_{0}(n,d)$ define

$\textup{Spl}({\mathcal{T}}):=\bigsqcup_{0\leq e\leq
d}\big{\\{}({\mathcal{T}}^{1},{\mathcal{T}}^{2})\in\operatorname{Tri}^{B}_{0}(n,e)\times\operatorname{Tri}^{B}_{0}(n,d-e)\mid{\mathcal{T}}^{1}{\mathcal{T}}^{2}\sim{\mathcal{T}}\big{\\}}.$
(3.11)

For $({\mathcal{T}}^{1},{\mathcal{T}}^{2})\in\textup{Spl}({\mathcal{T}})$, let
$\sigma^{{\mathcal{T}}}_{{\mathcal{T}}^{1},{\mathcal{T}}^{2}}$ be the unique
element of ${}^{{\mathcal{T}}}\mathscr{D}$ such that
${\mathcal{T}}\sigma^{{\mathcal{T}}}_{{\mathcal{T}}^{1},{\mathcal{T}}^{2}}={\mathcal{T}}^{1}{\mathcal{T}}^{2}.$
Recalling the notation (2.2), we have:

######  Lemma 3.12.

[10, Corollary 3.24] If
${\mathcal{T}}=(\text{\boldmath$b$},\text{\boldmath$r$},\text{\boldmath$s$})\in\operatorname{Tri}^{B}_{0}(n,d)$
then

$\displaystyle\reflectbox{\rotatebox[origin={c}]{180.0}{${\tt\Delta}$}}(\eta_{\mathcal{T}})=\sum_{({\mathcal{T}}^{1},{\mathcal{T}}^{2})\in\textup{Spl}({\mathcal{T}})}(-1)^{\langle\sigma^{{\mathcal{T}}}_{{\mathcal{T}}^{1},{\mathcal{T}}^{2}};\text{\boldmath$b$}\rangle}{\small\frac{[{\mathcal{T}}]^{!}_{\mathfrak{c}}}{[{\mathcal{T}}^{1}]^{!}_{\mathfrak{c}}[{\mathcal{T}}^{2}]^{!}_{\mathfrak{c}}}}\eta_{{\mathcal{T}}^{1}}\otimes\eta_{{\mathcal{T}}^{2}}.$

###  3.3. Idempotents and characters

Let $\lambda\in\Lambda(n,d)$. Set
$\text{\boldmath$l$}^{\lambda}:=1^{\lambda_{1}}\cdots n^{\lambda_{n}}.$ For an
idempotent $e\in A$ we have an idempotent
$\eta_{\lambda}^{e}:=\eta^{e^{d}}_{\text{\boldmath$l$}^{\lambda},\text{\boldmath$l$}^{\lambda}}\in
T(n,d)$. Let $e_{0},\dots,e_{l}\in A$ be the standard idempotents. For each
$\text{\boldmath$\lambda$}=(\lambda^{(0)},\dots,\lambda^{(l)})\in\Lambda^{I}(n,d)$,
we have the idempotent

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

The idempotents $\eta_{\text{\boldmath$\lambda$}}$ are orthogonal.

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

Following [10, §5A], see especially [10, Lemma 5.9], for a $T(n,d)$-module
$V$, 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$}}\in{\mathrm{Sym}}^{I}(n,d).$

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$}}V\subsetneq
V$.

######  Lemma 3.13.

[10, Lemma 5.10] If $W_{1}\in T(n,d_{1})\\!\operatorname{-mod}$ and $W_{2}\in
T(n,d_{2})\\!\operatorname{-mod}$, then $\operatorname{ch}(W_{1}\otimes
W_{2})=\operatorname{ch}(W_{1})\,\operatorname{ch}(W_{2}).$

The group $\mathfrak{S}_{n}$ acts on $\Lambda(n)$ on the left via

$\sigma\lambda:=(\lambda_{\sigma^{-1}1},\dots,\lambda_{\sigma^{-1}n}).$ (3.14)

The group $\mathfrak{S}_{n}^{I}:=\prod_{i\in I}\mathfrak{S}_{n}$ acts on
$\Lambda^{I}(n)$ via
$\text{\boldmath$\sigma$}\text{\boldmath$\lambda$}:=(\sigma^{(0)}\lambda^{(0)},\dots,\sigma^{(l)}\lambda^{(l)}),$
for
$\text{\boldmath$\sigma$}=(\sigma^{(0)},\dots,\sigma^{(l)})\in\mathfrak{S}_{n}^{I}$
and
$\text{\boldmath$\lambda$}=(\lambda^{(0)},\dots,\lambda^{(l)})\in\Lambda^{I}(n)$.
For $a\in A$ and $\sigma\in\mathfrak{S}_{n}$, let
$\xi_{\sigma}^{a}:=\sum_{r=1}^{n}\xi_{\sigma(r),r}^{a}\in M_{n}(A)$. For
$\text{\boldmath$\sigma$}=(\sigma^{(0)},\dots,\sigma^{(l)})\in\mathfrak{S}^{I}_{n}$,
we set

$\xi_{d}(\text{\boldmath$\sigma$}):=\sum_{d_{0}+\dots+d_{l}=d}(\xi_{\sigma^{(0)}}^{e_{0}})^{\otimes
d_{0}}*\cdots*(\xi_{\sigma^{(l)}}^{e_{l}})^{\otimes d_{l}}\in T^{A}(n,d).$
(3.15)

######  Lemma 3.16.

[10, Lemmas 5.6, 5.7] For all
$\text{\boldmath$\sigma$},\text{\boldmath$\tau$}\in\mathfrak{S}^{I}_{n}$ and
$\text{\boldmath$\lambda$}\in\Lambda^{I}(n,d)$, we have
$\xi_{d}(\text{\boldmath$\sigma$})\xi_{d}(\text{\boldmath$\tau$})=\xi_{d}(\text{\boldmath$\sigma$}\text{\boldmath$\tau$})$
and
$\xi_{d}(\text{\boldmath$\sigma$})\eta_{\text{\boldmath$\lambda$}}\xi_{d}(\text{\boldmath$\sigma$}^{-1})=\eta_{\text{\boldmath$\sigma$}\text{\boldmath$\lambda$}}$.

######  Corollary 3.17.

For $\text{\boldmath$\sigma$}\in\mathfrak{S}^{I}_{n}$,
$\text{\boldmath$\lambda$}\in\Lambda^{I}(n,d)$ and $V\in
T(n,d)\\!\operatorname{-mod}$, we have
$\xi_{d}(\text{\boldmath$\sigma$})\,\eta_{\text{\boldmath$\lambda$}}V=\eta_{\text{\boldmath$\sigma$}\text{\boldmath$\lambda$}}V$.

######  Lemma 3.18.

For $\text{\boldmath$\sigma$}\in\mathfrak{S}^{I}_{n}$, we have
$\reflectbox{\rotatebox[origin={c}]{180.0}{${\tt\Delta}$}}(\xi_{d}(\text{\boldmath$\sigma$}))=\sum_{c=0}^{d}\xi_{c}(\text{\boldmath$\sigma$})\otimes\xi_{d-c}(\text{\boldmath$\sigma$})$.

###### Proof.

By definition,

$\reflectbox{\rotatebox[origin={c}]{180.0}{${\tt\Delta}$}}((\xi_{\sigma^{(i)}}^{e_{i}})^{\otimes
d_{i}})=\sum_{c_{i}=0}^{d_{i}}\big{(}(\xi_{\sigma^{(i)}}^{e_{i}})^{\otimes
c_{i}}\big{)}\otimes\big{(}(\xi_{\sigma^{(i)}}^{e_{i}})^{\otimes
d_{i}-c_{i}}\big{)}.$

By Lemma 3.6, we have

$\reflectbox{\rotatebox[origin={c}]{180.0}{${\tt\Delta}$}}(\xi_{d}(\text{\boldmath$\sigma$}))=\sum_{d_{0}+\dots+d_{l}=d}\reflectbox{\rotatebox[origin={c}]{180.0}{${\tt\Delta}$}}\big{(}(\xi_{\sigma^{(0)}}^{e_{0}})^{\otimes
d_{0}}\big{)}*\cdots*\reflectbox{\rotatebox[origin={c}]{180.0}{${\tt\Delta}$}}\big{(}(\xi_{\sigma^{(l)}}^{e_{l}})^{\otimes
d_{l}}\big{)},$

and the result follows. ∎

For $N\geq n$, we set $E^{N}_{n}:=\sum_{r=1}^{n}\xi_{r,r}^{1}\in M_{N}(A)$ and
consider the idempotent

$\eta^{N}_{n}(d):=(E^{N}_{n})^{\otimes d}\in T^{A}(N,d).$ (3.19)

######  Lemma 3.20.

[10, Lemma 5.15] Let $N\geq n$. Then we have a unital superalgebra isomorphism

$T(n,d)\stackrel{{\scriptstyle\sim}}{{\longrightarrow}}\eta^{N}_{n}(d)T(N,d)\eta^{N}_{n}(d),\
\eta^{\text{\boldmath$b$}}_{\text{\boldmath$r$},\text{\boldmath$s$}}\mapsto\eta^{\text{\boldmath$b$}}_{\text{\boldmath$r$},\text{\boldmath$s$}}$

######  Lemma 3.21.

[10, Proposition 5.19] Let $d_{1},d_{2}\in\mathbb{Z}_{\geq 0}$ with
$d_{1}+d_{2}=d$, $n\leq N$, $V_{1}\in T(N,d_{1})\\!\operatorname{-mod}$ and
$V_{2}\in T(N,d_{2})\\!\operatorname{-mod}$. Then there is a functorial
isomorphism of $T(n,d)$-modules $\eta^{N}_{n}(d)(V_{1}\otimes
V_{2})\simeq(\eta^{N}_{n}(d_{1})V_{1})\otimes(\eta^{N}_{n}(d_{2})V_{2}).$

###  3.4. Quasi-hereditary structure on $T(n,d)$

Recall that $A$ is a based quasi-hereditary superalgebra with conforming
heredity data $I,X,Y$. Throughout this subsection, we assume that $d\leq n$.
Then, by [11, Theorem 6.6], $T(n,d)=T^{A}(n,d)$ is a based quasi-hereditary
algebra.

We now describe the heredity data
$\Lambda_{+}^{I}(n,d),{\mathcal{X}}(n,d),{\mathcal{Y}}(n,d)$ for $T(n,d)$
following [11, §6]. To start with, we have already defined the partially
ordered set $\Lambda_{+}^{I}(n,d)$ of $I$-multipartitions with partial order
$\leq_{I}$, see §2.4. For $\text{\boldmath$\lambda$}\in\Lambda_{+}^{I}(n,d)$
the corresponding sets
${\mathcal{X}}(\text{\boldmath$\lambda$})=\\{{\mathcal{X}}_{\text{\boldmath$S$}}\mid\text{\boldmath$S$}\in\operatorname{Std}^{X}(\text{\boldmath$\lambda$})\\}$
and
${\mathcal{Y}}(\text{\boldmath$\lambda$})=\\{{\mathcal{Y}}_{\text{\boldmath$T$}}\mid\text{\boldmath$T$}\in\operatorname{Std}^{Y}(\text{\boldmath$\lambda$})\\}$
are labeled by the standard $X$-colored and $Y$-colored $\lambda$-tableaux,
respectively. Recalling the notation
$\text{\boldmath$x$}^{\text{\boldmath$S$}}$,
$\text{\boldmath$l$}^{\text{\boldmath$S$}}$, etc. from §2.5, the elements
${\mathcal{X}}_{\text{\boldmath$S$}}$ and
${\mathcal{Y}}_{\text{\boldmath$T$}}$ are defined as follows.

${\mathcal{X}}_{\text{\boldmath$S$}}:=\eta^{\text{\boldmath$x$}^{\text{\boldmath$S$}}}_{\text{\boldmath$l$}^{\text{\boldmath$S$}},\text{\boldmath$l$}^{\text{\boldmath$\lambda$}}},\quad{\mathcal{Y}}_{\text{\boldmath$T$}}:=\eta^{\text{\boldmath$y$}^{\text{\boldmath$T$}}}_{\text{\boldmath$l$}^{\text{\boldmath$\lambda$}},\text{\boldmath$l$}^{\text{\boldmath$T$}}}.$

For any $\text{\boldmath$\lambda$}\in\Lambda_{+}^{I}(n,d)$, we have
${\mathcal{X}}_{\text{\boldmath$T$}^{\text{\boldmath$\lambda$}}}={\mathcal{Y}}_{\text{\boldmath$T$}^{\text{\boldmath$\lambda$}}}=\eta_{\text{\boldmath$\lambda$}}$,
so
${\mathcal{X}}(\text{\boldmath$\lambda$})\cap{\mathcal{Y}}(\text{\boldmath$\lambda$})=\\{\eta_{\text{\boldmath$\lambda$}}\\}$,
and
$\\{\eta_{\text{\boldmath$\lambda$}}\mid\text{\boldmath$\lambda$}\in\Lambda_{+}^{I}(n,d)\\}$
are the standard idempotents of the heredity data.

Let $\text{\boldmath$\lambda$}\in\Lambda_{+}^{I}(n,d)$. The standard module
$\Delta(\text{\boldmath$\lambda$})$ has basis

$\\{v_{\text{\boldmath$T$}}:={\mathcal{X}}_{\text{\boldmath$T$}}v_{\text{\boldmath$\lambda$}}\mid\text{\boldmath$T$}\in\operatorname{Std}^{X}(\text{\boldmath$\lambda$})\\},$
(3.22)

where $v_{\text{\boldmath$\lambda$}}$ is the (unique up to scalar) vector of
weight $\lambda$ in $\Delta(\text{\boldmath$\lambda$})$. 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 (2.18), then

$v_{\text{\boldmath$T$}}\in\eta_{\text{\boldmath$\mu$}}\Delta(\text{\boldmath$\lambda$}).$
(3.23)

Corollary 3.17 immediately implies:

######  Lemma 3.24.

For $\text{\boldmath$\sigma$}\in\mathfrak{S}^{I}_{n}$ and
$\text{\boldmath$\lambda$}\in\Lambda^{I}_{+}(n,d)$ such that
$\text{\boldmath$\sigma$}\text{\boldmath$\lambda$}=\text{\boldmath$\lambda$}$,
we have $\xi_{d}(\text{\boldmath$\sigma$})v_{\text{\boldmath$\lambda$}}=\pm
v_{\text{\boldmath$\lambda$}}$.

If follows from [11, Theorem 6.6] that the formal character of the standard
module $\Delta(\text{\boldmath$\lambda$})$ is of the form

$\operatorname{ch}\Delta(\text{\boldmath$\lambda$})=z^{\text{\boldmath$\lambda$}}+\sum_{\text{\boldmath$\mu$}<_{I}\text{\boldmath$\lambda$}}c_{\text{\boldmath$\mu$}}z^{\text{\boldmath$\mu$}}.$
(3.25)

This implies:

######  Lemma 3.26.

The formal characters
$\\{\operatorname{ch}\Delta(\text{\boldmath$\lambda$})\mid\text{\boldmath$\lambda$}\in\Lambda_{+}^{I}(n,d)\\}$
are linearly independent. In particular:

  1. (i)

if $V\in T(n,d)\\!\operatorname{-mod}$ has a standard filtration and
$\operatorname{ch}V=\sum_{\text{\boldmath$\lambda$}\in\Lambda^{I}(n,d)}m_{\text{\boldmath$\lambda$}}\operatorname{ch}\Delta(\text{\boldmath$\lambda$})$
then every $\Delta(\text{\boldmath$\lambda$})$ appears as a subquotient of the
filtration exactly $m_{\text{\boldmath$\lambda$}}$ times.

  2. (ii)

if $\text{\boldmath$\lambda$}\in\Lambda_{+}^{I}(n,d)$,
$\text{\boldmath$\mu$}\in\Lambda_{+}^{I}(n,c)$ and
$\Delta(\text{\boldmath$\lambda$})\otimes\Delta(\text{\boldmath$\mu$})$ has a
standard filtration, then
$\Delta(\text{\boldmath$\lambda$}+\text{\boldmath$\mu$})$ appears in this
filtration once and all other subquotients $\Delta(\text{\boldmath$\nu$})$ of
the filtration satisfy
$\text{\boldmath$\nu$}<_{I}\text{\boldmath$\lambda$}+\text{\boldmath$\nu$}$.

######  Lemma 3.27.

Let $\text{\boldmath$\lambda$}\in\Lambda^{I}_{+}(n,d)$,
$\text{\boldmath$r$},\text{\boldmath$s$}\in[n]^{d}$ and $y_{1},\dots,y_{d}\in
Y$ with at least one $y_{r}\not\in X$. Suppose that
$v\in\eta_{\text{\boldmath$\nu$}}\Delta(\text{\boldmath$\lambda$})$ for some
$\text{\boldmath$\nu$}\in\Lambda^{I}(n,d)$ with
$\|\text{\boldmath$\nu$}\|=\|\text{\boldmath$\lambda$}\|$. Then
$\eta^{y_{1}\cdots y_{d}}_{\text{\boldmath$r$},\text{\boldmath$s$}}v=0$.

###### Proof.

Suppose $\eta^{y_{1}\cdots
y_{d}}_{\text{\boldmath$r$},\text{\boldmath$s$}}v\neq 0$. Then
$\eta^{y_{1}\cdots
y_{d}}_{\text{\boldmath$r$},\text{\boldmath$s$}}\eta_{\text{\boldmath$\nu$}}=\eta^{y_{1}\cdots
y_{d}}_{\text{\boldmath$r$},\text{\boldmath$s$}}$. So there exist
$i_{1},\dots,i_{d}\in I$ such that
$y_{1}e_{i_{1}}=y_{1},\,\dots,\,y_{d}e_{i_{d}}=y_{d}$ and for all $i\in I$ we
have $\sharp\\{k\mid i_{k}=i\\}=|\nu^{(i)}|=|\lambda^{(i)}|$. On the other
hand, there exist $j_{1},\dots,j_{d}$ such that
$e_{j_{1}}y_{1}=y_{1},\,\dots,\,e_{j_{d}}y_{d}=y_{d}$. By Lemma 2.8,
$j_{1}\geq i_{1},\dots,j_{d}\geq i_{d}$, and by the assumption that at least
one $y_{r}\not\in X$, we have that at least one $j_{r}>i_{r}$. So
$\eta^{y_{1}\cdots
y_{d}}_{\text{\boldmath$r$},\text{\boldmath$s$}}v\in\eta_{\text{\boldmath$\mu$}}\Delta(\lambda)$
for $\mu$ satisfying
$\|\text{\boldmath$\mu$}\|\rhd_{I}\|\text{\boldmath$\lambda$}\|$, hence
$\text{\boldmath$\mu$}>_{I}\text{\boldmath$\lambda$}$, which contradicts
(3.25). ∎

Recalling (2.17), we have

######  Lemma 3.28.

[11, 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$}_{i}(\lambda^{(i)})).$

###  3.5. Character formula

Throughout the subsection we continue to assume that $d\leq n$. Set

${}_{j}X(i):=\\{x\in X(i)\mid e_{j}x=x\\}\qquad(i,j\in I).$

Note that ${}_{j}X(i)\neq\varnothing$ only if $j\leq i$. For
$\text{\boldmath$\nu$}=(\text{\boldmath$\nu$}^{(x)})_{x\in
X(i)}\in\Lambda_{+}^{X(i)}$ and $j\in I$, we define

${}_{j}\text{\boldmath$\nu$}=(\text{\boldmath$\nu$}^{(x)})_{x\in{}_{j}X(i)}\in\Lambda_{+}^{{}_{j}X(i)}.$

Fix $i\in I$ until the end of the subsection. We define an algebra
homomorphism

$\chi:{\mathrm{Sym}}^{X(i)}\to{\mathrm{Sym}}^{I},\ \bigotimes_{x\in
X(i)}f_{x}\mapsto\bigotimes_{j\in I}\prod_{x\in{}_{j}X(i)}f_{x},$

cf. [11, (7.41)]. By the Littlewood-Richardson rule, for
$\text{\boldmath$\nu$}\in\Lambda_{+}^{X(i)}$, we have

$\chi\left(\text{\boldmath$s$}_{\text{\boldmath$\nu$}}\right)=\sum_{\text{\boldmath$\gamma$}\in\Lambda_{+}^{I}}\prod_{j\in
I}c_{{}_{j}\text{\boldmath$\nu$}}^{\gamma^{(j)}}\text{\boldmath$s$}_{\text{\boldmath$\gamma$}}.$
(3.29)

For a multipartition $\text{\boldmath$\nu$}\in\Lambda^{X(i)}_{+}(n,d)$, we
define its superconjugate multipartition

$\text{\boldmath$\nu$}^{\mathsf{con}}:=(\tilde{\nu}^{(x)})_{x\in X(i)},$

where $\tilde{\nu}^{(x)}:=\nu^{(x)}$ if $x$ is even and $\tilde{\nu}^{(x)}$ is
the conjugate partition $(\nu^{(x)})^{\prime}$ if $x$ is odd. Using [13,
(2.7),(3.8)], we have the algebra homomorphism

${\mathsf{con}}:{\mathrm{Sym}}^{X(i)}\to{\mathrm{Sym}}^{X(i)},s_{\text{\boldmath$\nu$}}\mapsto
s_{\text{\boldmath$\nu$}^{\mathsf{con}}}.$

Let $t:=|X(i)|$. By choosing a total order on $X(i)$ we will identify
$\Lambda_{+}^{X(i)}$ with $\Lambda_{+}^{t}$, ${\mathrm{Sym}}^{X(i)}$ with
${\mathrm{Sym}}^{\otimes t}$, etc. In particular, we have a well-defined map
${\mathsf{con}}:{\mathrm{Sym}}^{\otimes t}\to{\mathrm{Sym}}^{\otimes t}$.
Recalling (2.19) and (2.20), we now have:

######  Theorem 3.30.

Let $d\in\mathbb{Z}_{\geq 0}$, $n\in\mathbb{Z}_{>0}$, with $d\leq n$, and
$\lambda\in\Lambda_{+}(n,d)\subseteq\Lambda_{+}$ and $i\in I$. Then

$\operatorname{ch}\Delta(\text{\boldmath$\iota$}_{i}(\lambda))=\rho_{n}^{I}\circ\chi\circ{\mathsf{con}}\circ{\tt\Delta}^{t-1}(s_{\lambda}).$

###### Proof.

By [11, Proposition 7.45], we have

$\displaystyle\operatorname{ch}\Delta(\text{\boldmath$\iota$}_{i}(\lambda))$
$\displaystyle=\sum_{\text{\boldmath$\gamma$}\in\Lambda^{I}_{+}(n)}\,\sum_{\text{\boldmath$\nu$}\in\Lambda_{+}(n)^{t}}c^{\lambda}_{\text{\boldmath$\nu$}^{{\mathsf{con}}}}\bigg{(}\prod_{j\in
I}c^{\gamma^{(j)}}_{\,{}_{j}\text{\boldmath$\nu$}}\bigg{)}s_{\text{\boldmath$\gamma$}}(z_{1},\dots,z_{n})$
$\displaystyle=\rho_{n}^{I}\Bigg{(}\sum_{\text{\boldmath$\gamma$}\in\Lambda^{I}_{+}}\,\sum_{\text{\boldmath$\nu$}\in\Lambda_{+}^{t}}c^{\lambda}_{\text{\boldmath$\nu$}^{{\mathsf{con}}}}\bigg{(}\prod_{j\in
I}c^{\gamma^{(j)}}_{\,{}_{j}\text{\boldmath$\nu$}}\bigg{)}s_{\text{\boldmath$\gamma$}}\Bigg{)}$
$\displaystyle=\rho_{n}^{I}\circ\chi\Bigg{(}\sum_{\text{\boldmath$\nu$}\in\Lambda_{+}^{t}}c^{\lambda}_{\text{\boldmath$\nu$}^{{\mathsf{con}}}}s_{\text{\boldmath$\nu$}}\Bigg{)}$
$\displaystyle=\rho_{n}^{I}\circ\chi\circ{\mathsf{con}}\Bigg{(}\sum_{\text{\boldmath$\nu$}\in\Lambda_{+}^{t}}c^{\lambda}_{\text{\boldmath$\nu$}}s_{\text{\boldmath$\nu$}}\Bigg{)}$
$\displaystyle=\rho_{n}^{I}\circ\chi\circ{\mathsf{con}}\circ{\tt\Delta}^{t-1}(s_{\lambda}),$

where we have used (3.29) for the third equality and (2.21) for the last
equality. ∎

######  Theorem 3.31.

Let $\lambda\in\Lambda_{+}(n,d)$, $\mu\in\Lambda_{+}(n,e)$ and $i\in I$. If
$d+e\leq n$ then

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

###### Proof.

By Lemma 3.13, Theorem 3.30 and the Littlewood-Richardson rule, we have

$\displaystyle\operatorname{ch}\big{(}\Delta(\text{\boldmath$\iota$}_{i}(\lambda))\otimes\Delta(\text{\boldmath$\iota$}_{i}(\mu))\big{)}$
$\displaystyle=\big{(}\operatorname{ch}\Delta(\text{\boldmath$\iota$}_{i}(\lambda))\big{)}\big{(}\operatorname{ch}\Delta(\text{\boldmath$\iota$}_{i}(\mu))\big{)}$
$\displaystyle=\big{(}\rho_{n}^{I}\circ\chi\circ{\mathsf{con}}\circ{\tt\Delta}^{t-1}(s_{\lambda})\big{)}\,\big{(}\rho_{n}^{I}\circ\chi\circ{\mathsf{con}}\circ{\tt\Delta}^{t-1}(s_{\mu})\big{)}$
$\displaystyle=\rho_{n}^{I}\circ\chi\circ{\mathsf{con}}\circ{\tt\Delta}^{t-1}(s_{\lambda}s_{\mu})$
$\displaystyle=\rho_{n}^{I}\circ\chi\circ{\mathsf{con}}\circ{\tt\Delta}^{t-1}\Bigg{(}\sum_{\nu\in\Lambda_{+}(n,d+e)}c^{\,\nu}_{\lambda,\mu}s_{\nu}\Bigg{)}$
$\displaystyle=\sum_{\nu\in\Lambda_{+}(n,d+e)}c^{\,\nu}_{\lambda,\mu}\operatorname{ch}\Delta(\text{\boldmath$\iota$}_{i}(\nu)),$

as required. ∎

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$}}:=\prod_{j\in
I}c^{\,\nu^{(j)}}_{\lambda^{(j)},\mu^{(j)}}.$ (3.32)

######  Corollary 3.33.

Let $\text{\boldmath$\lambda$}\in\Lambda^{I}_{+}(n,d)$ and
$\text{\boldmath$\mu$}\in\Lambda^{I}_{+}(n,e)$. If $d+e\leq n$ then

$\operatorname{ch}(\Delta(\text{\boldmath$\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$}).$

###### Proof.

This follows from Theorem 3.31 and Lemma 3.28. ∎

##  4\. Tensor products of standard modules have standard filtrations

We again fix a based quasi-hereditary superalgebra $A_{R}$ over $R$ with
conforming heredity data $I,X,Y$. Recalling the convention (3.7), we have the
($\mathbb{F}$-)superalgebra $T(n,d)=T^{A}(n,d)$. Under the assumption $d\leq
n$, this superalgebra is based quasi-hereditary with heredity data
$\Lambda_{+}^{I}(n,d),{\mathcal{X}}(n,d),{\mathcal{Y}}(n,d)$, see §3.4.

###  4.1. Reduction

We begin to prove Main Theorem by reducing to the case of ‘one color’ and
‘fundamental dominant weights’, cf. [19, (3.5)], [2, Proposition 3.5.4(i)].
For integer $0\leq c\leq n$, recalling (2.16), we have

$\omega_{c}:={\varepsilon}_{1}+\dots+{\varepsilon}_{c}\in\Lambda_{+}(n,c).$

######  Proposition 4.1.

Suppose that for all $n\in\mathbb{Z}_{>0}$, $d,c\in\mathbb{Z}_{\geq 0}$ with
$d+c\leq n$, $\lambda\in\Lambda_{+}(n,d)$, and $i\in I$, the tensor product
$\Delta(\text{\boldmath$\iota$}_{i}(\lambda))\otimes\Delta(\text{\boldmath$\iota$}_{i}(\omega_{c}))$
has a standard filtration. Then for all $n\in\mathbb{Z}_{>0}$,
$d,c\in\mathbb{Z}_{\geq 0}$ with $d+c\leq n$,
$\text{\boldmath$\lambda$}\in\Lambda^{I}_{+}(n,d)$,
$\text{\boldmath$\mu$}\in\Lambda_{+}^{I}(n,c)$, the tensor product
$\Delta(\text{\boldmath$\lambda$})\otimes\Delta(\text{\boldmath$\mu$})$ has a
standard filtration.

###### Proof.

We apply induction on the total degree $d+c$, the base case $d+c=0$ being
trivial, since $T(n,0)\cong\mathbb{F}$. Let $d+c>0$. Take
$\text{\boldmath$\lambda$}=(\lambda^{(0)},\dots,\lambda^{(l)})\in\Lambda^{I}_{+}(n,d)$
and
$\text{\boldmath$\mu$}=(\mu^{(0)},\dots,\mu^{(l)})\in\Lambda^{I}_{+}(n,c)$.
For all $i\in I$, set $d_{i}=|\lambda^{(i)}|$ and $c_{i}:=|\mu^{(i)}|$. By
Theorem 3.28, we have

$\Delta(\text{\boldmath$\lambda$})\otimes\Delta(\text{\boldmath$\mu$})\simeq\bigotimes_{i\in
I}\Big{(}\Delta(\text{\boldmath$\iota$}_{i}(\lambda^{(i)}))\otimes\Delta(\text{\boldmath$\iota$}_{i}(\mu^{(i)}))\Big{)}.$
(4.2)

Suppose there exist distinct $j,k\in I$ with $d_{j},d_{k}>0$. Then $d_{i}<d$
for all $i\in I$. By the inductive assumption, for all $i\in I$, we then have
that
$\Delta(\text{\boldmath$\iota$}_{i}(\lambda^{(i)}))\otimes\Delta(\text{\boldmath$\iota$}_{i}(\mu^{(i)}))$
has a standard filtration. It follows from Lemma 3.26 and Theorem 3.31 that in
this filtration only subquotients of the form
$\Delta(\text{\boldmath$\iota$}_{i}(\nu^{(i)}))$ with
$\nu^{(i)}\in\Lambda_{+}(n,d_{i}+c_{i})$ appear. Hence by Theorem 3.28, the
right hand side of (4.2) has a filtration with subquotients of the form
$\bigotimes_{i\in
I}\Delta(\text{\boldmath$\iota$}_{i}(\nu^{(i)}))\simeq\Delta(\text{\boldmath$\nu$})$.
Thus we may assume that there exists a unique $i$ with $d_{i}=d$ and $d_{k}=0$
for all $k\neq i$, i.e.
$\text{\boldmath$\lambda$}=\text{\boldmath$\iota$}_{i}(\lambda)$ for some
$i\in I$ and $\lambda\in\Lambda_{+}(n,d)$. Similarly we may assume that
$\text{\boldmath$\mu$}=\text{\boldmath$\iota$}_{j}(\mu)$ for some $j\in I$ and
$\mu\in\Lambda_{+}(n,c)$. Moreover, we may assume that $j=i$ since otherwise
$\Delta(\text{\boldmath$\lambda$})\otimes\Delta(\text{\boldmath$\mu$})=\Delta(\text{\boldmath$\iota$}_{i}(\lambda))\otimes\Delta(\text{\boldmath$\iota$}_{j}(\mu))$
is a standard module, thanks to Theorem 3.28.

We now also apply induction on the dominance order on $\mu$. If $\mu$ is
minimal in the dominance order, then $\mu=\omega_{c}$ and we are done by
assumption. Otherwise, we can write $\mu=\gamma+\omega_{r}$ for
$\gamma\in\Lambda_{+}(n,s)$ with $0<s,r<c$. By the inductive assumption on the
degree, we have that
$\Delta(\text{\boldmath$\iota$}_{i}(\gamma))\otimes\Delta(\text{\boldmath$\iota$}_{i}(\omega_{r}))$
has a standard filtration. By Lemma 3.26 and Theorem 3.31, in this filtration
$\Delta(\text{\boldmath$\iota$}_{i}(\mu))$ appears once and other standard
subquotients are of the form $\Delta(\text{\boldmath$\iota$}_{i}(\nu))$ with
$\nu\lhd\mu$. By [4, Proposition A2.2(i)], there is a short exact sequence

$0\to\Delta(\text{\boldmath$\iota$}_{i}(\mu))\to\Delta(\text{\boldmath$\iota$}_{i}(\gamma))\otimes\Delta(\text{\boldmath$\iota$}_{i}(\omega_{r}))\to
Q\to 0,$

where $Q$ has a standard filtration with subquotients of the form
$\Delta(\text{\boldmath$\iota$}_{i}(\nu))$ with $\nu\lhd\mu$. Tensoring with
$\Delta(\text{\boldmath$\iota$}_{i}(\lambda))$ we get a short exact sequence

$0\to\Delta(\text{\boldmath$\iota$}_{i}(\lambda))\otimes\Delta(\text{\boldmath$\iota$}_{i}(\mu))\to\Delta(\text{\boldmath$\iota$}_{i}(\lambda))\otimes\Delta(\text{\boldmath$\iota$}_{i}(\gamma))\otimes\Delta(\text{\boldmath$\iota$}_{i}(\omega_{r}))\to\Delta(\text{\boldmath$\iota$}_{i}(\lambda))\otimes
Q\to 0.$

By induction on the dominance order,
$\Delta(\text{\boldmath$\iota$}_{i}(\lambda))\otimes Q$ has a standard
filtration. By induction on the degree, using Lemma 3.26 and Theorem 3.31, we
have that
$\Delta(\text{\boldmath$\iota$}_{i}(\lambda))\otimes\Delta(\text{\boldmath$\iota$}_{i}(\gamma))$
has a standard filtration with subquotients of the form
$\Delta(\text{\boldmath$\iota$}_{i}(\kappa))$, hence by assumption, the middle
term has a standard filtration. By [4, Proposition A2.2(v)], the first term
has a standard filtration. ∎

###  4.2. The filtration

In view of Proposition 4.1, we now fix $i\in I$, $\lambda\in\Lambda_{+}(n,d)$,
$c\in\mathbb{Z}_{>0}$ such that $d+c\leq n$, and set

$\text{\boldmath$\lambda$}:=\text{\boldmath$\iota$}_{i}(\lambda),\quad\text{\boldmath$\mu$}:=\text{\boldmath$\iota$}_{i}(\omega_{c}).$

We have highest weight vectors
$v_{\text{\boldmath$\lambda$}}\in\Delta(\text{\boldmath$\lambda$})$ and
$v_{\text{\boldmath$\mu$}}\in\Delta(\text{\boldmath$\mu$})$.

Recalling the action of $\mathfrak{S}_{n}$ on $\Lambda(n)$ from (3.14), we
denote

$\mathfrak{S}_{\lambda}:=\\{\sigma\in\mathfrak{S}_{n}\mid\sigma\lambda=\lambda\\}$

If $\lambda=(l_{1}^{a_{1}},\dots,l_{k}^{a_{k}})$ for $l_{1}>\dots>l_{k}\geq 0$
and $a_{1},\dots,a_{k}>0$ with $a_{1}+\dots+a_{k}=n$ then
$\mathfrak{S}_{\lambda}=\mathfrak{S}_{a_{1}}\times\dots\times\mathfrak{S}_{a_{k}}$.

Let $\Omega:=\\{P\subseteq[n]\mid|P|=c\\}$. The group $\mathfrak{S}_{n}$ acts
on $\Omega$ via $\sigma P=\\{\sigma p_{1},\dots,\sigma p_{c}\\}$ for
$P=\\{p_{1},\dots,p_{c}\\}\in\Omega$ and $\sigma\in\mathfrak{S}_{n}$. Denote

${\varepsilon}_{P}:={\varepsilon}_{p_{1}}+\dots+{\varepsilon}_{p_{c}}\in\Lambda(n,c).$

Note that $\sigma({\varepsilon}_{P})={\varepsilon}_{\sigma P}$ for all
$\sigma\in\mathfrak{S}_{n}$ and $P\in\Omega$. We denote

$\Omega_{\lambda}:=\\{P\in\Omega\mid\lambda+{\varepsilon}_{P}\in\Lambda_{+}(n,d+c)\\}.$

Given $P=\\{p_{1},\dots,p_{c}\\}$ and $Q=\\{q_{1},\dots,q_{c}\\}$ in $\Omega$,
with $1\leq p_{1}<\dots<p_{c}\leq n$ and $1\leq q_{1}<\dots<q_{c}\leq n$, we
write $P<Q$ if and only if $(p_{1},\dots,p_{c})<(q_{1},\dots,q_{c})$
lexicographically. This yields the total order on $\Omega$. Let
$\Omega_{\lambda}=\\{P_{1},P_{2},\dots,P_{t}\\}$ with

$P_{1}=\\{1,2,\dots,c\\}<P_{2}<\dots<P_{t}.$

The following is easy to see:

######  Lemma 4.3.

Let $1\leq r\leq t$. Then

  1. (i)

$P_{r}$ is the minimal element of the orbit $\mathfrak{S}_{\lambda}\cdot
P_{r}$;

  2. (ii)

$\Omega=\bigsqcup_{r=1}^{t}\mathfrak{S}_{\lambda}\cdot P_{r}$.

###### Proof.

Write $\lambda=(l_{1}^{a_{1}},\dots,l_{k}^{a_{k}})$ for $l_{1}>\dots>l_{k}\geq
0$ and $a_{1},\dots,a_{k}>0$ with $a_{1}+\dots+a_{k}=n$, so that
$\mathfrak{S}_{\lambda}=\mathfrak{S}_{a_{1}}\times\dots\times\mathfrak{S}_{a_{k}}$
and $A_{1}:=[1,a_{1}],A_{2}:=[a_{1}+1,a_{1}+a_{2}],\dots,A_{k}:=[n-a_{k}+1,n]$
are the orbits of $\mathfrak{S}_{\lambda}$ on $[n]$. Now $P,Q\in\Omega$ are in
the same $\mathfrak{S}_{\lambda}$-orbit if and only if $|P\cap A_{s}|=|Q\cap
A_{s}|$ for all $s=1,\dots,k$, and it is clear that each orbit has a unique
element from $\Omega_{\lambda}$ which is the lexicographically minimal element
of the orbit. ∎

Let $P=\\{p_{1},\dots,p_{c}\\}\in\Omega$ with $p_{1}<\dots<p_{c}$. There is a
unique tableau
$\text{\boldmath$T$}^{P}\in\operatorname{Std}^{X}(\text{\boldmath$\mu$})$ with
$\text{\boldmath$l$}^{\text{\boldmath$T$}^{P}}=p_{1}\cdots p_{c}$ and
$\text{\boldmath$x$}^{\text{\boldmath$T$}^{P}}=e_{i}^{c}$. We denote the
corresponding standard basis vector

$w_{P}:=v_{\text{\boldmath$T$}^{P}}=\eta^{e_{i}^{c}}_{p_{1}\cdots
p_{c},12\cdots c}\,v_{\text{\boldmath$\mu$}}\in\Delta(\text{\boldmath$\mu$}),$

see (3.22). Note that the vectors $w_{P}$ do not exhaust the standard basis of
$\Delta(\text{\boldmath$\mu$})$.

######  Lemma 4.4.

Let $\nu\in\Lambda(n,c)$. If
$\eta_{\text{\boldmath$\iota$}_{i}(\nu)}\Delta(\text{\boldmath$\mu$})\neq 0$,
then $\nu$ is of the form ${\varepsilon}_{P}$ and $w_{P}$ spans
$\eta_{\text{\boldmath$\iota$}_{i}(\nu)}\Delta(\text{\boldmath$\mu$})$.

###### Proof.

By (3.22), (3.23), the weight space
$\eta_{\text{\boldmath$\iota$}_{i}(\nu)}\Delta(\text{\boldmath$\mu$})\neq 0$
is spanned by the basis elements $v_{\text{\boldmath$T$}}$ such that
$\text{\boldmath$T$}\in\operatorname{Std}^{X}(\text{\boldmath$\mu$},\text{\boldmath$\iota$}_{i}(\nu))$.
As $\text{\boldmath$\mu$}=\text{\boldmath$\iota$}_{i}(\omega_{c})$, we deduce,
using the property (c) of Definition 2.6, that
$\text{\boldmath$T$}=\text{\boldmath$T$}_{P}$ for some $P\in\Omega$, i.e.
$v_{\text{\boldmath$T$}}=w_{P}$. ∎

For $\sigma\in\mathfrak{S}_{n}$ denote

$\text{\boldmath$\iota$}_{i}(\sigma):=(1,\dots,1,\sigma,1,\dots,1)\in\mathfrak{S}_{n}^{I},$

with $\sigma$ in the $i$th position. Recalling (3.15), we have an element

$\xi_{c}(\text{\boldmath$\iota$}_{i}(\sigma))=(\xi_{\sigma}^{e_{i}})^{\otimes
c}\in T^{A}(n,c).$

######  Lemma 4.5.

Let $P=\\{p_{1},\dots,p_{c}\\}\in\Omega$ with $p_{1}<\dots<p_{c}$, and
$\sigma\in\mathfrak{S}_{n}$ such that $\sigma p_{1}<\dots<\sigma p_{c}$. Then
$\xi_{c}(\text{\boldmath$\iota$}_{i}(\sigma))w_{P}=w_{\sigma P}$.

###### Proof.

By definition of $\xi_{\sigma}^{e_{i}}$, we have in $T(n,c)$:

$(\xi_{\sigma}^{e_{i}})^{\otimes c}\,\eta^{e_{i}^{c}}_{p_{1}\cdots
p_{c},12\cdots c}=\eta^{e_{i}^{c}}_{\sigma p_{1}\cdots\sigma p_{c},12\cdots
c}.$

So

$\displaystyle\xi_{c}(\text{\boldmath$\iota$}_{i}(\sigma))w_{P}=(\xi_{\sigma}^{e_{i}})^{\otimes
c}\,\eta^{e_{i}^{c}}_{p_{1}\cdots p_{c},12\cdots
c}\,v_{\text{\boldmath$\mu$}}=\eta^{e_{i}^{c}}_{\sigma p_{1}\cdots\sigma
p_{c},12\cdots c}\,v_{\text{\boldmath$\mu$}}=w_{\sigma P},$

as required. ∎

######  Corollary 4.6.

Let $P\in\Omega$. Then $T(n,c)w_{P}=\Delta(\text{\boldmath$\mu$})$.

###### Proof.

Take $\sigma\in\mathfrak{S}_{n}$ with $\sigma(p_{a})=a$ for $a=1,\dots,c$. By
Lemma 4.5, we have
$\xi_{c}(\text{\boldmath$\iota$}_{i}(\sigma))w_{P}=w_{\\{1,\dots,c\\}}=v_{\text{\boldmath$\mu$}}$,
and the result follows since
$T(n,c)v_{\text{\boldmath$\mu$}}=\Delta(\text{\boldmath$\mu$})$. ∎

######  Lemma 4.7.

We have $T(n,d+c)(v_{\text{\boldmath$\lambda$}}\otimes
w_{P_{t}})=\Delta(\text{\boldmath$\lambda$})\otimes\Delta(\text{\boldmath$\mu$})$.

###### Proof.

Let $h$ be maximal with $\lambda_{h}>0$, so
$\text{\boldmath$l$}^{\text{\boldmath$\lambda$}}=1^{\lambda_{1}}\cdots
h^{\lambda_{h}}$. Since $n\geq d+c$, we have $P_{t}=\\{h+1,\dots,h+c\\}$.

By (3.22), $\Delta(\text{\boldmath$\lambda$})$ is spanned by elements of the
form
$\eta_{\text{\boldmath$r$},\text{\boldmath$l$}^{\text{\boldmath$\lambda$}}}^{\text{\boldmath$x$}}v_{\text{\boldmath$\lambda$}}$
for
$(\text{\boldmath$x$},\text{\boldmath$r$},\text{\boldmath$l$}^{\text{\boldmath$\lambda$}})\in\operatorname{Tri}^{X}(n,d)$.
Let ${\mathcal{T}}^{\prime}\in\operatorname{Tri}_{0}^{X}(n,d)$ with
${\mathcal{T}}^{\prime}\sim(\text{\boldmath$x$},\text{\boldmath$r$},\text{\boldmath$l$}^{\text{\boldmath$\lambda$}})$,
see (3.1). Then

$[{\mathcal{T}}^{\prime}]^{!}_{\mathfrak{c}}=1,$ (4.8)

since $x=xe_{i}\in B_{\mathfrak{a}}$ for all $x\in X(i)$, see (2.7) and
(3.10).

On the other hand, for
$(\text{\boldmath$b$},\text{\boldmath$t$},\text{\boldmath$u$})\in\operatorname{Tri}^{B}(n,c)$,
we have that
$\eta_{\text{\boldmath$t$},\text{\boldmath$u$}}^{\text{\boldmath$b$}}w_{P_{t}}=0$
unless
$\eta_{\text{\boldmath$t$},\text{\boldmath$u$}}^{\text{\boldmath$b$}}\eta_{\text{\boldmath$\iota$}_{i}({\varepsilon}_{P_{t}})}=\eta_{\text{\boldmath$t$},\text{\boldmath$u$}}^{\text{\boldmath$b$}}$,
and, in view of Corollary 4.6, $\Delta(\text{\boldmath$\mu$})$ is spanned by
all
$\eta_{\text{\boldmath$t$},\text{\boldmath$u$}}^{\text{\boldmath$b$}}w_{P_{t}}$
with $\text{\boldmath$u$}\sim(h+1)\cdots(h+c)$.

Let
$(\text{\boldmath$x$},\text{\boldmath$r$},\text{\boldmath$l$}^{\text{\boldmath$\lambda$}})\in\operatorname{Tri}^{X}(n,d)$
and
$(\text{\boldmath$b$},\text{\boldmath$t$},\text{\boldmath$u$})\in\operatorname{Tri}^{B}(n,c)$
satisfy $\text{\boldmath$u$}\sim(h+1)\cdots(h+c)$, and
${\mathcal{T}}\in\operatorname{Tri}^{B}_{0}(n,d+c)$ be the initial triple with
${\mathcal{T}}\sim(\text{\boldmath$x$}\text{\boldmath$b$},\text{\boldmath$r$}\text{\boldmath$t$},\text{\boldmath$l$}^{\text{\boldmath$\lambda$}}\text{\boldmath$u$})$.
Let $({\mathcal{T}}^{1},{\mathcal{T}}^{2})\in\textup{Spl}({\mathcal{T}})$ with
${\mathcal{T}}^{1}\in\operatorname{Tri}^{B}_{0}(n,d)$ and
${\mathcal{T}}^{2}\in\operatorname{Tri}^{B}_{0}(n,c)$. Suppose
${\mathcal{T}}^{1}=(\text{\boldmath$a$},\text{\boldmath$v$},\text{\boldmath$s$})\not\sim(\text{\boldmath$x$},\text{\boldmath$r$},\text{\boldmath$l$}^{\text{\boldmath$\lambda$}})$.
Since $\text{\boldmath$l$}^{\text{\boldmath$\lambda$}}=1^{\lambda_{1}}\cdots
h^{\lambda_{h}}$ and $\text{\boldmath$u$}\sim(h+1)\cdots(h+c)$, we necessarily
have that $s_{k}\in\\{h+1,\dots,h+c\\}$ for some $1\leq k\leq d$. Hence
$\eta_{{\mathcal{T}}^{1}}v_{\text{\boldmath$\lambda$}}=0$. Now, by Lemma 3.12
and (4.8),

$\eta_{\text{\boldmath$r$}\text{\boldmath$t$},\text{\boldmath$l$}^{\text{\boldmath$\lambda$}}\text{\boldmath$u$}}^{\text{\boldmath$x$}\text{\boldmath$b$}}(v_{\text{\boldmath$\lambda$}}\otimes
w_{P_{t}})=(\eta_{\text{\boldmath$r$},\text{\boldmath$l$}^{\text{\boldmath$\lambda$}}}^{\text{\boldmath$x$}}v_{\text{\boldmath$\lambda$}})\otimes(\eta^{\text{\boldmath$b$}}_{\text{\boldmath$t$},\text{\boldmath$u$}}w_{P_{t}}),$

which implies the lemma. ∎

For $r=0,1,\dots,t$, we denote

$M_{r}:=T(n,d)\langle v_{\text{\boldmath$\lambda$}}\otimes w_{P_{s}}\mid 1\leq
s\leq
r\rangle\subseteq\Delta(\text{\boldmath$\lambda$})\otimes\Delta(\text{\boldmath$\mu$}).$

In view of Lemma 4.7, we have a filtration

$0=M_{0}\subseteq M_{1}\subseteq\dots\subseteq
M_{t}=\Delta(\text{\boldmath$\lambda$})\otimes\Delta(\text{\boldmath$\mu$}).$
(4.9)

Our goal is to show that
$M_{r}/M_{r-1}\simeq\Delta(\text{\boldmath$\iota$}_{i}(\lambda+{\varepsilon}_{P_{r}}))$
for all $r=1,\dots,t$, to get the required standard filtration of
$\Delta(\text{\boldmath$\lambda$})\otimes\Delta(\text{\boldmath$\mu$})$.

######  Lemma 4.10.

If $1\leq r\leq t$ and $P\in\mathfrak{S}_{\lambda}\cdot P_{r}$, then
$v_{\text{\boldmath$\lambda$}}\otimes w_{P}\in M_{r}$.

###### Proof.

Write $P=\\{p_{1},\dots,p_{c}\\}$ with $p_{1}<\dots<p_{c}$. Let
$\sigma\in\mathfrak{S}_{\lambda}$ be such that $\sigma P_{r}=P$ and $\sigma
p_{1}<\dots<\sigma p_{c}$. Note using Lemmas 3.24, 4.5 and 3.18 that

$v_{\text{\boldmath$\lambda$}}\otimes
w_{P}=\pm(\xi_{d}(\text{\boldmath$\iota$}_{i}(\sigma))v_{\text{\boldmath$\lambda$}})\otimes(\xi_{c}(\text{\boldmath$\iota$}_{i}(\sigma))w_{P_{r}})=\pm\xi_{d+c}(\text{\boldmath$\iota$}_{i}(\sigma))(v_{\text{\boldmath$\lambda$}}\otimes
w_{P_{r}})\in M_{r},$

as required. ∎

######  Lemma 4.11.

Let $1\leq r\leq t$ and $E\in{\mathcal{Y}}(n,d+c)$. If
$E(v_{\text{\boldmath$\lambda$}}\otimes w_{P_{r}})\not\in M_{r-1}$ then
$E=\eta_{\text{\boldmath$\iota$}_{i}(\lambda+{\varepsilon}_{P_{r}})}$. In
particular, $M_{r}/M_{r-1}$ is a highest weight module of weight
$\text{\boldmath$\iota$}_{i}(\lambda+{\varepsilon}_{P_{r}})$.

###### Proof.

Suppose $E(v_{\text{\boldmath$\lambda$}}\otimes w_{P_{r}})\not\in M_{r-1}$.
Write
$E={\mathcal{Y}}_{\text{\boldmath$T$}}:=\eta^{\text{\boldmath$y$}^{\text{\boldmath$T$}}}_{\text{\boldmath$l$}^{\text{\boldmath$\nu$}},\text{\boldmath$l$}^{\text{\boldmath$T$}}}$,
with $\text{\boldmath$T$}\in\operatorname{Std}^{Y}(\text{\boldmath$\nu$})$ for
some $\text{\boldmath$\nu$}\in\Lambda^{I}_{+}(n,d+c)$. By Lemma 3.12,
$\reflectbox{\rotatebox[origin={c}]{180.0}{${\tt\Delta}$}}(\eta^{\text{\boldmath$y$}^{\text{\boldmath$T$}}}_{\text{\boldmath$l$}^{\text{\boldmath$\nu$}},\text{\boldmath$l$}^{\text{\boldmath$T$}}})$
is a linear combination of elements of the form
$\eta_{\text{\boldmath$r$},\text{\boldmath$s$}}^{\text{\boldmath$y$}}\otimes\eta_{\text{\boldmath$r$}^{\prime},\text{\boldmath$s$}^{\prime}}^{\text{\boldmath$y$}^{\prime}}$
such that
$\text{\boldmath$y$}\text{\boldmath$y$}^{\prime}\sim\text{\boldmath$y$}^{T}$.
By Lemma 3.27,
$\eta_{\text{\boldmath$r$},\text{\boldmath$s$}}^{\text{\boldmath$y$}}v_{\text{\boldmath$\lambda$}}\neq
0$ only if $\text{\boldmath$y$}=e_{i}^{d}$, and
$\eta_{\text{\boldmath$r$}^{\prime},\text{\boldmath$s$}^{\prime}}^{\text{\boldmath$y$}^{\prime}}w_{P_{r}}\neq
0$ only if $\text{\boldmath$y$}^{\prime}=e_{i}^{c}$. We conclude that
$\text{\boldmath$y$}^{\text{\boldmath$T$}}=e_{i}^{d+c}$, and so $E$ can be
written in the form
$E=\eta^{e_{i}^{d+c}}_{\text{\boldmath$r$},\text{\boldmath$s$}}$ with
$r_{k}\leq s_{k}$ for all $k$.

If $r_{k}<s_{k}$ for some $k$, then in view of Lemma 3.12 and Lemma 4.4,
$E(v_{\text{\boldmath$\lambda$}}\otimes w_{P_{r}})$ is a multiple of
$v_{\text{\boldmath$\lambda$}}\otimes w_{P}$ for some $P<P_{r}$. By Lemma 4.3,
$P\in\mathfrak{S}_{\lambda}\cdot P_{s}$ for some $s<r$, hence
$v_{\text{\boldmath$\lambda$}}\otimes w_{P}\in M_{s}\subseteq M_{r-1}$ by
Lemma 4.10, giving a contradiction. So $r_{k}=s_{k}$ for all $k$. Then $E$ is
of the form $\eta_{\text{\boldmath$\nu$}}$, and
$E(v_{\text{\boldmath$\lambda$}}\otimes w_{P_{r}})\neq 0$ implies
$\text{\boldmath$\nu$}=\text{\boldmath$\iota$}_{i}(\lambda+{\varepsilon}_{P_{r}})$.

The second statement now follows from Lemma 2.11. ∎

######  Theorem 4.12.

We have a filtration of $T(n,d+c)$-modules

$0=M_{0}\subseteq M_{1}\subseteq\dots\subseteq
M_{t}=\Delta(\text{\boldmath$\lambda$})\otimes\Delta(\text{\boldmath$\mu$})$

such that
$M_{r}/M_{r-1}\simeq\Delta(\text{\boldmath$\iota$}_{i}(\lambda+{\varepsilon}_{P_{r}}))$
for all $r=1,\dots,t$.

###### Proof.

We consider the filtration (4.9). By Lemma 4.11, each $M_{r}/M_{r-1}$ is a
highest weight module of weight
$\text{\boldmath$\iota$}_{i}(\lambda+{\varepsilon}_{P_{r}})$. Moreover,
recalling that
$\text{\boldmath$\lambda$}=\text{\boldmath$\iota$}_{i}(\lambda)$ and
$\text{\boldmath$\mu$}=\text{\boldmath$\iota$}_{i}(\omega_{c})$, by Theorem
3.31, we have

$\displaystyle\operatorname{ch}\big{(}\Delta(\text{\boldmath$\lambda$})\otimes\Delta(\text{\boldmath$\mu$})\big{)}=\sum_{\nu\in\Lambda_{+}(n,d+e)}c^{\,\nu}_{\lambda,\omega_{c}}\operatorname{ch}\Delta(\text{\boldmath$\iota$}_{i}(\nu))=\sum_{r=1}^{t}\operatorname{ch}\Delta(\text{\boldmath$\iota$}_{i}(\lambda+{\varepsilon}_{P_{r}})),$

where we have used Pieri’s rule for the last equality. Therefore, using linear
independence of characters, we get

$\dim\big{(}\Delta(\text{\boldmath$\lambda$})\otimes\Delta(\text{\boldmath$\mu$})\big{)}=\sum_{r=1}^{t}\dim\Delta(\text{\boldmath$\iota$}_{i}(\lambda+{\varepsilon}_{P_{r}})).$

An application of Corollary 2.13 yields that each $M_{r}/M_{r-1}$ must be
isomorphic to
$\Delta(\text{\boldmath$\iota$}_{i}(\lambda+{\varepsilon}_{P_{r}}))$. ∎

Recall (2.15) and (3.32).

######  Corollary 4.13.

Let $n\in\mathbb{Z}_{>0}$, $d,c\in\mathbb{Z}_{\geq 0}$ with $d+c\leq n$,
$\text{\boldmath$\lambda$}\in\Lambda^{I}_{+}(n,d)$,
$\text{\boldmath$\mu$}\in\Lambda_{+}^{I}(n,c)$ and
$\text{\boldmath$\nu$}\in\Lambda^{I}_{+}(n,d+c)$. Then the tensor product
$\Delta(\text{\boldmath$\lambda$})\otimes\Delta(\text{\boldmath$\mu$})$ has a
standard filtration, and

$(\Delta(\text{\boldmath$\lambda$})\otimes\Delta(\text{\boldmath$\mu$}):\Delta(\text{\boldmath$\nu$}))=c_{\text{\boldmath$\lambda$},\text{\boldmath$\mu$}}^{\text{\boldmath$\nu$}}.$

###### Proof.

The first statement follows from Proposition 4.1 and Theorem 4.12. The second
statement now follows from Corollary 3.33 using linear independence of formal
characters. ∎

By a symmetric argument (switching the roles of ${\mathcal{X}}$ and
${\mathcal{Y}}$ everywhere), we also have the right module version of
Corollary 4.13, which claims that the right $T(n,d+c)$-module
$\Delta^{\mathrm{op}}(\text{\boldmath$\lambda$})\otimes\Delta^{\mathrm{op}}(\text{\boldmath$\mu$})$
has a $\Delta^{\mathrm{op}}$-filtration. In view of (2.14), by dualizing, we
now get:

######  Corollary 4.14.

Let $n\in\mathbb{Z}_{>0}$, $d,c\in\mathbb{Z}_{\geq 0}$ with $d+c\leq n$,
$\text{\boldmath$\lambda$}\in\Lambda^{I}_{+}(n,d)$,
$\text{\boldmath$\mu$}\in\Lambda_{+}^{I}(n,c)$ and
$\text{\boldmath$\nu$}\in\Lambda^{I}_{+}(n,d+c)$. Then the tensor product
$\nabla(\text{\boldmath$\lambda$})\otimes\nabla(\text{\boldmath$\mu$})$ has a
costandard filtration.

######  Remark 4.15.

Note that in Theorem 4.12, the factors of the standard filtration are
isomorphic to standard modules via even isomorphisms. Using this fact and (an
appropriate strengthening of) Proposition 4.1, one can similarly strengthen
Corollaries 4.13 and 4.14.

###  4.3. The case of small $n$

Let $d\in\mathbb{Z}_{\geq 0}$ and $n\in\mathbb{Z}_{>0}$. If $n<d$, the algebra
$T(n,d)$ does not have to be quasihereditary, but it still has a natural
family of ‘standard’ and ‘costandard’ modules which play an important role.
For example, if $A$ has a standard anti-involution then $T(n,d)$ is cellular
with ‘standard’ modules being the cell modules, see [11, Lemma 6.25]. These
‘standard’ (resp. ‘costandard’) modules are obtained by an idempotent
truncation from the standard modules $\Delta(\text{\boldmath$\lambda$})$
(resp. costandard modules $\nabla(\text{\boldmath$\lambda$})$) over $T(N,d)$
for any $N\geq d$. We now provide details on that.

Throughout the subsection we assume that $N\geq n$. In view of Lemma 3.20, we
now always identify the algebras $T(n,d)$ and
$\eta^{N}_{n}(d)T(N,d)\eta^{N}_{n}(d)$. For any $T(N,d)$-module $V$, we
consider $\eta^{N}_{n}(d)V$ as a module over
$T(n,d)=\eta^{N}_{n}(d)T(N,d)\eta^{N}_{n}(d)$.

We always consider $\Lambda_{+}^{I}(n,d)$ as a subset of
$\Lambda_{+}^{I}(N,d)$ by adding $N-n$ zeroes to every component
$\lambda^{(i)}$ of
$\text{\boldmath$\lambda$}=(\lambda^{(0)},\dots,\lambda^{(l)})\in\Lambda_{+}^{I}(n,d)$.
Note that this embedding is a bijection if $n\geq d$. However, when we
consider $\text{\boldmath$\lambda$}\in\Lambda_{+}^{I}(n,d)$ as an element of
$\Lambda_{+}^{I}(N,d)$ the set
$\operatorname{Std}^{X}(\text{\boldmath$\lambda$})$ of standard $X$-colored
$\lambda$-tableaux changes, so in this subsection we will use the more
detailed notation $\operatorname{Std}^{X}_{n}(\text{\boldmath$\lambda$})$ to
indicate that the entries of the tableaux are of the form $r^{x}$ with
$r\in[n]$. We will also use the more detailed notation
$\Delta_{n}(\text{\boldmath$\lambda$})$ for the standard $T(n,d)$-module
$\Delta(\text{\boldmath$\lambda$})$ which so far has only been defined for all
$\text{\boldmath$\lambda$}\in\Lambda_{+}^{I}(n,d)$ when $n\geq d$. Recall from
(3.22) that for $n\geq d$ we have that $\Delta_{n}(\text{\boldmath$\lambda$})$
has basis
$\\{v_{\text{\boldmath$T$}}:={\mathcal{X}}_{\text{\boldmath$T$}}v_{\text{\boldmath$\lambda$}}\mid\text{\boldmath$T$}\in\operatorname{Std}^{X}_{n}(\text{\boldmath$\lambda$})\\}$.

Let $N\geq d$. Fix $\text{\boldmath$\lambda$}\in\Lambda_{+}^{I}(N,d)$. Recall
the idempotent $\eta^{N}_{n}(d)$ of (3.19). It is easy to see that for
$\text{\boldmath$T$}\in\operatorname{Std}_{N}^{X}(\text{\boldmath$\lambda$})$,
we have

$\eta^{N}_{n}(d)v_{\text{\boldmath$T$}}=\left\\{\begin{array}[]{ll}v_{\text{\boldmath$T$}}&\hbox{if
$\text{\boldmath$T$}\in\operatorname{Std}_{n}^{X}(\text{\boldmath$\lambda$})$},\\\
0&\hbox{otherwise.}\end{array}\right.$ (4.16)

If $n\geq d$ it follows from (4.16) that
$\dim\Delta_{n}(\text{\boldmath$\lambda$})=\dim\eta^{N}_{n}(d)\Delta_{N}(\text{\boldmath$\lambda$})$.
Since the $T(n,d)$-module
$\eta^{N}_{n}(d)\Delta_{N}(\text{\boldmath$\lambda$})$ is easily seen to be a
highest weight module of weight $\lambda$, Proposition 2.12 now yields an
isomorphism of $T(n,d)$-modules

$\Delta_{n}(\text{\boldmath$\lambda$})\simeq\eta^{N}_{n}(d)\Delta_{N}(\text{\boldmath$\lambda$}).$
(4.17)

Now for $n<d$ and $\text{\boldmath$\lambda$}\in\Lambda_{+}^{I}(N,d)$, we
define the ‘standard’ module

$\Delta_{n}(\text{\boldmath$\lambda$}):=\eta^{N}_{n}(d)\Delta_{N}(\text{\boldmath$\lambda$}).$

By (4.17), this definition does not depend on the choice of $N\geq d$.
However, note that some of the $\Delta_{n}(\text{\boldmath$\lambda$})$’s might
be zero. Define

${\mathscr{P}}_{+}^{X}(n,d):=\\{\text{\boldmath$\lambda$}\in\Lambda_{+}^{I}(N,d)\mid\operatorname{Std}^{X}_{n}(\text{\boldmath$\lambda$})\neq\varnothing\\}.$

Note that ${\mathscr{P}}_{+}^{X}(n,d)$ does not depend on the choice of $N\geq
d$. Moreover,

$\Lambda_{+}^{I}(n,d)\subseteq{\mathscr{P}}_{+}^{X}(n,d)\subseteq\Lambda_{+}^{I}(N,d),$

with containments being equalities when $n\geq d$. By (4.16), we have:

######  Lemma 4.18.

Let $N\geq d>n$ and $\text{\boldmath$\lambda$}\in\Lambda_{+}^{I}(N,d)$. Then
$\Delta_{n}(\text{\boldmath$\lambda$})\neq 0$ if and only if
$\text{\boldmath$\lambda$}\in{\mathscr{P}}_{+}^{X}(n,d)$.

The story for the costandard modules
$\nabla_{n}(\text{\boldmath$\lambda$}):=\eta^{N}_{n}(d)\nabla_{N}(\text{\boldmath$\lambda$})$
is entirely similar, the non-zero ones being labeled by
${\mathscr{P}}_{+}^{Y}(n,d):=\\{\text{\boldmath$\lambda$}\in\Lambda_{+}^{I}(N,d)\mid\operatorname{Std}^{Y}_{n}(\text{\boldmath$\lambda$})\neq\varnothing\\}.$

######  Theorem 4.19.

Let $\text{\boldmath$\lambda$}\in{\mathscr{P}}_{+}^{X}(n,d)$ and
$\text{\boldmath$\mu$}\in{\mathscr{P}}_{+}^{X}(n,c)$. Then the
$T(n,d+c)$-module
$\Delta_{n}(\text{\boldmath$\lambda$})\otimes\Delta_{n}(\text{\boldmath$\mu$})$
has a filtration with factors of the form $\Delta_{n}(\text{\boldmath$\nu$})$
with $\text{\boldmath$\nu$}\in{\mathscr{P}}_{+}^{X}(n,d+c)$. Similarly for
$\text{\boldmath$\lambda$}\in{\mathscr{P}}_{+}^{Y}(n,d)$ and
$\text{\boldmath$\mu$}\in{\mathscr{P}}_{+}^{Y}(n,c)$, the $T(n,d+c)$-module
$\nabla_{n}(\text{\boldmath$\lambda$})\otimes\nabla_{n}(\text{\boldmath$\mu$})$
has a filtration with factors of the form $\nabla_{n}(\text{\boldmath$\nu$})$
with $\text{\boldmath$\nu$}\in{\mathscr{P}}_{+}^{Y}(n,d+c)$.

###### Proof.

We prove the result for the $\Delta$’s, the proof for $\nabla$’s being
similar. Choose $N\geq d+c$. By Corollary 4.13,
$\Delta_{N}(\text{\boldmath$\lambda$})\otimes\Delta_{N}(\text{\boldmath$\mu$})$
has a filtration with factors of the form $\Delta_{N}(\text{\boldmath$\nu$})$
with $\text{\boldmath$\nu$}\in\Lambda_{+}^{I}(N,d+c)$. Applying the exact
functor

$T(N,d+c)\\!\operatorname{-mod}\to T(n,d+c)\\!\operatorname{-mod},\
V\mapsto\eta^{N}_{n}(d)V$

to this filtration and using Lemma 3.21, we get the required result. ∎

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