# Tensor products and intertwining operators for uniserial representations of
the Lie algebra $\mathfrak{sl}(2)\ltimes V(m)$

Leandro Cagliero  FaMAF-CIEM (CONICET), Universidad Nacional de Córdoba,
Medina Allende s/n, Ciudad Universitaria, 5000 Córdoba, República Argentina.
<EMAIL_ADDRESS>and  Iván Gómez Rivera  FaMAF-CIEM (CONICET),
Universidad Nacional de Córdoba, Medina Allende s/n, Ciudad Universitaria,
5000 Córdoba, República Argentina<EMAIL_ADDRESS>
###### Abstract.

Let $\mathfrak{g}_{m}=\mathfrak{sl}(2)\ltimes V(m)$, $m\geq 1$, where $V(m)$
is the irreducible $\mathfrak{sl}(2)$-module of dimension $m+1$ viewed as an
abelian Lie algebra. It is known that the isomorphism classes of uniserial
$\mathfrak{g}_{m}$-modules consist of a family, say of type $Z$, containing
modules of arbitrary composition length, and some exceptional modules with
composition length $\leq 4$.

Let $V$ and $W$ be two uniserial $\mathfrak{g}_{m}$-modules of type $Z$. In
this paper we obtain the $\mathfrak{sl}(2)$-module decomposition of
$\text{soc}(V\otimes W)$ by giving explicitly the highest weight vectors. It
turns out that $\text{soc}(V\otimes W)$ is multiplicity free. Roughly
speaking, $\text{soc}(V\otimes W)=\text{soc}(V)\otimes\text{soc}(W)$ in half
of the cases, and in these cases we obtain the full socle series of $V\otimes
W$ by proving that $\text{soc}^{t+1}(V\otimes
W)=\sum_{i=0}^{t}\text{soc}^{i+1}(V)\otimes\text{soc}^{t+1-i}(W)$ for all
$t\geq 0$.

As applications of these results, we obtain for which $V$ and $W$, the space
of $\mathfrak{g}_{m}$-module homomorphisms
$\text{Hom}_{\mathfrak{g}_{m}}(V,W)$ is not zero, in which case is
1-dimensional. Finally we prove, for $m\neq 2$, that if $U$ is the tensor
product of two uniserial $\mathfrak{g}_{m}$-modules of type $Z$, then the
factors are determined by $U$. We provide a procedure to identify the factors
from $U$.

###### Key words and phrases:

non-semisimple Lie algebras, uniserial representations, indecomposable, socle,
radical

###### 2010 Mathematics Subject Classification:

17B10, 17B30, 22E27, 16G10

This research was partially supported by an NSERC grant, CONICET PIP
112-2013-01-00511, PIP 112-2012-01-00501, MinCyT Córdoba, FONCYT Pict2013
1391, SeCyT-UNC 33620180100983CB

##  1\. Introduction

We fix throughout a field $\mathbb{F}$ of characteristic zero. All Lie
algebras and representations considered in this paper are assumed to be finite
dimensional over $\mathbb{F}$, unless explicitly stated otherwise.

It is generally acknowledged that the problem of classifying all
indecomposable finite dimensional representations of a Lie algebra is
intractable, one of the most clear manifestation of this is given in [22] for
abelian Lie algebras of dimension greater than or equal to $2$. This is also
discussed in [34] for the 3-dimensional euclidean Lie algebra
$\mathfrak{e}(2)$, and in [27] for virtually any complex Lie algebra other
than semisimple or 1-dimensional.

Rather than attempting to classify all indecomposable modules for a given Lie
algebra, or a family of Lie algebras, it would be very interesting to identify
a class of representations that is sufficiently limited so that we can have a
reasonably comfortable handling of them and, at the same time, large enough to
include many representations that appear naturally in problems of interest.
Just to mention an example, let $A$ be a finite dimensional (associative or
Lie) algebra and let $\text{Der}(A)$ be its Lie algebra of derivations. Except
for very exceptional cases, $\text{Der}(A)$ is not semisimple. We know that
$\text{Der}(A)$ acts in various objects associated to $A$, for instance in its
(Hochschild or Lie) cohomology. There are many results in cohomology obtained
by considering the action of a Levi factor of $\text{Der}(A)$ (and using the
highest weight theory) disregarding the action of its solvable radical. If we
wanted to describe (and/or make use of) the whole $\text{Der}(A)$-module
structure of the cohomology of $A$, there is no a standard way to do this due
to the lack of knowledge we have of an appropriate class of representations of
$\text{Der}(A)$. Moreover, this could be specially useful if we wanted to
describe the whole Gerstenhaber algebra $H\\!H(A)$ of an associative algebra
$A$.

Many authors have considered the idea of describing or classifying a special
class of representations of non-semisimple Lie algebras. For instance, A.
Piard [30] analyzed thoroughly the indecomposable modules $U$, of the complex
Lie algebra $\mathfrak{sl}(2)\ltimes\mathbb{C}^{2}$, such that
$U/\text{rad}(U)$ is irreducible. More recently, various families of
indecomposable modules over various types of non-semisimple Lie algebras have
been constructed and/or classified, see for instance [14, 15, 16, 18, 21, 19,
20, 24].

On the other hand, we have been systematically studying uniserial
representations of Lie algebras. In the articles [10, 11, 12, 13, 8, 7, 9, 17]
we and other authors have classified all uniserial representations for many
different families of Lie algebras. It is worth mentioning that, in the theory
of finite dimensional representations of associative algebras, the class of
uniserial ones is quite relevant, a foundational result here is due to T.
Nakayama [28] (see also [1] or[2]) and it states that every finitely generated
module over a serial ring is a direct sum of uniserial modules. For more
information in the associative case we refer the reader mainly to [1, 2, 31],
see also [5, 23, 29]. We point out that, for Lie algebras, when $\mathfrak{g}$
is 1-dimensional, any representation is a direct sum of uniserial ones. We do
not know if the is a class of Lie algebras, apart from semisimples, for which
this remains true.

If we want to pursue farther the idea of identifying a class of Lie algebras
representations based on the uniserial ones, a natural step forward is to
study morphisms between them and the tensor category that they generate. The
main goal of this article is to start this project with the family of the Lie
algebras $\mathfrak{g}_{m}=\mathfrak{sl}(2)\ltimes V(m)$, $m\geq 1$, where
$V(m)$ is the irreducible $\mathfrak{sl}(2)$-module of dimension $m+1$ viewed
as an abelian Lie algebra. The uniserial $\mathfrak{g}_{m}$-modules where
classified in [10] and the isomorphism classes consist of a general family
$Z(a,\ell)$ and its duals, and some exceptional modules with composition
length $\leq 4$. This is described below in Theorem 2.3. In the family
$Z(a,\ell)$, $a$ and $\ell$ are arbitrary non-negative integers, $\ell+1$ is
the composition length of $Z(a,\ell)$ and the socle of $Z(a,\ell)$ is
isomorphic to the irreducible $\mathfrak{sl}(2)$-module of dimension $a+1$. We
call the uniserial modules $Z(a,\ell)$ and $Z(a,\ell)^{*}$ _uniserials of
type_ $Z$, they constitute the vast majority of uniserial
$\mathfrak{g}_{m}$-modules.

Our main results are the following. First, given two uniserials $V$ and $W$ of
type $Z$, we obtain in Theorem 3.5 the $\mathfrak{sl}(2)$-module decomposition
of $\text{soc}(V\otimes W)$ by explicitly giving its highest weight vectors.
By duality, since the $\mathfrak{sl}(2)$-module decomposition of $V\otimes W$
follows from the Clebsch-Gordan formula, the $\mathfrak{sl}(2)$-module
structure of the radical $\text{rad}(V\otimes W)$ can be derived (see for
instance [1, Chapter V]). It turns out that $\text{soc}(V\otimes W)$ is
multiplicity free ($V\otimes W$ is not at all multiplicity free except for $V$
and $W$ irreducible). In some sense, $\text{soc}(V\otimes
W)=\text{soc}(V)\otimes\text{soc}(W)$ in half of the cases (including when
$V=Z(a,\ell)^{*}$ and $W=Z(b,\ell^{\prime})^{*}$). It turns out that, in these
cases (see Corollary 3.8),

$\text{soc}^{t+1}(V\otimes
W)=\sum_{i=0}^{t}\text{soc}^{i+1}(V)\otimes\text{soc}^{t+1-i}(W)$

for all $t\geq 0$. In other words, this formula holds for all $t$ if and only
if it holds for $t=0$. One of the main steps towards Theorem 3.5 is to deal
with the cases $V$ and $W$ with composition lengths equal to 2. This is done
in Theorem 3.3 and the proof of it required lengthy and precise computations
in which the Clebsch-Gordan coefficients (the 3-$j$ symbols) were a crucial
tool. In order to extend Theorem 3.3 to the exceptional uniserials it is
necessary to work out these computations, but they become harder and so far we
could only arrive to Conjecture 3.4.

Next we study the intertwining operators between uniserials $V$ and $W$ of
type $Z$. From the multiplicity free structure of $\text{soc}(V\otimes W)$ we
obtain that $\text{Hom}_{\mathfrak{g}_{m}}(V,W)$ is either zero or
1-dimensional and we derive from Theorem 3.5 in which cases
$\text{Hom}_{\mathfrak{g}_{m}}(V,W)\neq 0$ (see Theorem 4.1 and Corollary
4.2). A well known result of this flavor is the Bernstein-Gelfand-Gelfand
classification of the intertwining operators among Verma modules and their
generalizations (see [3]).

Finally we prove, for $m\neq 2$, that if $U$ is the tensor product of two
uniserial modules of type $Z$, then the factors are determined by $U$ (see
Theorem 4.3). Moreover, we give explicitly a procedure to identify the
factors. This question about the uniqueness of the factorization of tensor
products is frequently addressed in the literature for irreducible modules. It
is well known that in general, the tensor product of two modules (even when
they are irreducible) do not determine the factors. A very basic example would
be the tensor product of two irreducible modules that is itself irreducible,
and this may happen even if the underlying group or algebra is indecomposable
and none of the factors is 1-dimensional (see for instance [4] or [25, 26] and
the references within them). In contrast, a celebrated result of C. S. Rajan
[32] states that a tensor product of an arbitrary number of irreducible,
finite dimensional representations of a simple Lie algebra over a field of
characteristic zero determines uniquely the factors. This is also true in
other categories of modules, see [36] for a generalization of Rajan’s result
to a natural category of representations of symmetrizable Kac-Moody algebras,
or [33] for a unique factorization result for some special irreducible
representations of Borcherds-Kac-Moody algebras. We are not aware of results
dealing with this problem within a much larger class of modules such as the
class of uniserials. We think that Theorem 4.3 remains valid for $m=2$ and
that our proof only requires a small adjustment that we did not find so far.

We close this introduction with some open questions closely related to this
paper that are of our interest.

  1. $\bullet$

What is the $\mathfrak{sl}(2)$-module structure of $\text{soc}(V\otimes W)$
when $V$ and $W$ are exceptional uniserial $\mathfrak{g}_{m}$-modules? As we
mentioned, we think that the answer for modules of composition length 2 is
given in Conjecture 3.4. The general case should follow without major
difficulties from the result for the case of composition length 2.

  2. $\bullet$

Is it true, for $m=2$, the statement of Theorem 4.3?

  3. $\bullet$

Is it possible to extend Theorem 4.3 to an arbitrary number of uniserial
modules?

  4. $\bullet$

Given two uniserials $\mathfrak{g}_{m}$-modules $V$ and $W$, what are, up to
isomorphism, the extensions of $V$ by $W$? Is it possible to obtain, for any
$m$, results that are similar to those obtained by A. Piard [30] for $m=1$?

  5. $\bullet$

For which uniserials $\mathfrak{g}_{m}$-modules $V$ and $W$ is $V\otimes W$
indecomposable?

##  2\. Preliminaries

###  2.1. The Clebsch-Gordan coefficients

Recall that $\mathbb{F}$ is a field of characteristic zero and that all Lie
algebras and representations are assumed to be finite dimensional over
$\mathbb{F}$, unless explicitly stated otherwise. Let

(2.1) $e=\begin{pmatrix}0&1\\\ 0&0\end{pmatrix},\qquad h=\begin{pmatrix}1&0\\\
0&-1\end{pmatrix},\qquad f=\begin{pmatrix}0&0\\\ 1&0\end{pmatrix}$

be the standard basis of $\mathfrak{sl}(2)$. Let $V(a)$ be the irreducible
$\mathfrak{sl}(2)$-module with highest weight $a\geq 0$. We fix a basis
$\\{v_{0}^{a},\dots,v_{a}^{a}\\}$ of $V(a)$ relative to which the basis
$\\{e,h,f\\}$ acts as follows:

$\displaystyle e\,v_{k}^{a}=$
$\displaystyle\sqrt{\frac{a}{2}\left(\frac{a}{2}+1\right)-\left(\frac{a}{2}-k+1\right)\left(\frac{a}{2}-k\right)}v_{k-1}^{a},$
$\displaystyle h\,v_{k}^{a}=$ $\displaystyle(a-2k)v_{k}^{a},$ $\displaystyle
f\,v_{k}^{a}=$
$\displaystyle\sqrt{\frac{a}{2}\left(\frac{a}{2}+1\right)-\left(\frac{a}{2}-k-1\right)\left(\frac{a}{2}-k\right)}v_{k+1}^{a},$

where $0\leq k\leq a$ and $v_{-1}^{a}=0=v_{a+1}^{a}$. The basis
$\\{v_{0}^{a},\dots,v_{a}^{a}\\}$ has been chosen in a convenient way to
introduce below the Clebsch-Gordan coefficients. Note that, if we denote by
$(x)_{a}$ the matrix of $x\in\mathfrak{sl}(2)$ relative to the basis
$\\{v_{0}^{a},\dots,v_{a}^{a}\\}$, then $\\{(e)_{1},(h)_{1},(f)_{1}\\}$ are as
in (2.1), and

$(e)_{2}=\begin{pmatrix}0&\sqrt{2}&0\\\ 0&0&\sqrt{2}\\\
0&0&0\end{pmatrix},\qquad(h)_{2}=\begin{pmatrix}2&0&0\\\ 0&0&0\\\
0&0&-2\end{pmatrix},\qquad(f)_{2}=\begin{pmatrix}0&0&0\\\ \sqrt{2}&0&0\\\
0&\sqrt{2}&0\end{pmatrix}.$

This means that we may assume that
$\\{v_{0}^{2},v_{1}^{2},v_{2}^{2}\\}=\\{-e,\frac{\sqrt{2}}{2}h,f\\}$.

We know that $V(a)\simeq V(a)^{*}$ as $\mathfrak{sl}(2)$-modules. More
precisely, if $\\{(v_{0}^{a})^{*},\dots,(v_{a}^{a})^{*}\\}$ is the dual basis
of $\\{v_{0}^{a},\dots,v_{a}^{a}\\}$ then the map

(2.2) $\begin{split}V(a)&\rightarrow V(a)^{*}\\\
v_{k}^{a}&\mapsto(-1)^{a-k}(v_{a-k}^{a})^{*}\end{split}$

gives an explicit $\mathfrak{sl}(2)$-isomorphism.

It is well known that the tensor product decomposition of two irreducible
$\mathfrak{sl}(2)$-modules $V(a)$ and $V(b)$ is

(2.3) $V(a)\otimes V(b)\simeq V(a+b)\oplus V(a+b-2)\oplus\cdots\oplus
V(|a-b|).$

This is the well known Clebsch-Gordan formula. The _Clebsch-Gordan
coefficients_ $CG(j_{1},m_{1};j_{2},m_{2}\mid j_{3},m_{3})$ are defined below
and they provide an explicit $\mathfrak{sl}(2)$-embedding $V(c)\rightarrow
V(a)\otimes V(b)$ which is the following

$\displaystyle V(c)$ $\displaystyle\rightarrow V(a)\otimes V(b)$
$\displaystyle v_{k}^{c}$ $\displaystyle\mapsto v_{k}^{a,b,c}$

where, by definition,

(2.4)
$v_{k}^{a,b,c}=\sum_{i,j}CG(\tfrac{a}{2},\tfrac{a}{2}-i;\,\tfrac{b}{2},\tfrac{b}{2}-j\,|\,\tfrac{c}{2},\tfrac{c}{2}-k)\,v_{i}^{a}\otimes
v_{j}^{b},$

where the sum runs over all $i,j$ such that
$\tfrac{a}{2}-i+\tfrac{b}{2}-j=\tfrac{c}{2}-k$ (in fact we could let $i,j$ run
freely since the Clebsch-Gordan coefficient involved is zero if
$\tfrac{a}{2}-i+\tfrac{b}{2}-j\neq\tfrac{c}{2}-k$). Since

(2.5) $\text{Hom}(V(b),V(a))\simeq V(b)^{*}\otimes V(a)\simeq V(a)\otimes
V(b)$

it follows from (2.2) and (2.4) that the map
$V(c)\rightarrow\text{Hom}(V(b),V(a))$ given by

$\displaystyle v_{k}^{c}$
$\displaystyle\mapsto\sum_{i,j}CG(\tfrac{a}{2},\tfrac{a}{2}-i;\,\tfrac{b}{2},\tfrac{b}{2}-j\,|\,\tfrac{c}{2},\tfrac{c}{2}-k)\,v_{i}^{a}\otimes
v_{j}^{b},$
$\displaystyle\mapsto\sum_{i,j}(-1)^{b-j}CG(\tfrac{a}{2},\tfrac{a}{2}-i;\,\tfrac{b}{2},\tfrac{b}{2}-j\,|\,\tfrac{c}{2},\tfrac{c}{2}-k)\,v_{i}^{a}\otimes(v_{b-j}^{b})^{*},$
(2.6)
$\displaystyle\mapsto\sum_{i,j}(-1)^{j}CG(\tfrac{a}{2},\tfrac{a}{2}-i;\,\tfrac{b}{2},-\tfrac{b}{2}+j\,|\,\tfrac{c}{2},\tfrac{c}{2}-k)\,\,(v_{j}^{b})^{*}\otimes
v_{i}^{a}$

is an $\mathfrak{sl}(2)$-module homomorphism.

We now recall briefly the basic definitions and facts about the Clebsch-Gordan
coefficients. We will mainly follow [35].

Given three non-negative integers or half-integers $j_{1},j_{2},j_{3}$, we say
that they _satisfy the triangle condition_ if $j_{1}+j_{2}+j_{3}$ is an
integer and they can be the side lengths of a (possibly degenerate) triangle
(that is $|j_{1}-j_{2}|\leq j_{3}\leq j_{1}+j_{2}$). We now define (see [35,
§8.2, eq.(1)])

$\Delta(j_{1},j_{2},j_{3})=\sqrt{\frac{(j_{1}+j_{2}-j_{3})!(j_{1}-j_{2}+j_{3})!(-j_{1}+j_{2}+j_{3})!}{(j_{1}+j_{2}+j_{3}+1)!}}$

if $j_{1},j_{2},j_{3}$ satisfies the triangle condition; otherwise, we set
$\Delta(j_{1},j_{2},j_{3})=0$.

If in addition $m_{1}$, $m_{2}$ and $m_{3}$ are three integers or half-
integers then the corresponding _Clebsch-Gordan coefficient_

$CG(j_{1},m_{1};j_{2},m_{2}|j_{3},m_{3})$

is zero unless $m_{1}+m_{2}=m_{3}$ and $|m_{i}|\leq j_{i}$ for $i=1,2,3$. In
this case, the following formula is valid for $m_{3}\geq 0$ and $j_{1}\geq
j_{2}$ (see [35, §8.2, eq.(3)])

$CG(j_{1},m_{1};j_{2},m_{2}\mid
j_{3},m_{3})=\Delta(j_{1},j_{2},j_{3})\,\sqrt{(2j_{3}+1)}\\\\[2.84526pt]
\times\sqrt{(j_{1}+m_{1})!(j_{1}-m_{1})!(j_{2}+m_{2})!(j_{2}-m_{2})!(j_{3}+m_{3})!(j_{3}-m_{3})!}\\\\[2.84526pt]
\times\sum_{r}\frac{(-1)^{r}}{r!(j_{1}\\!+\\!j_{2}\\!-\\!j_{3}\\!-\\!r)!(j_{1}\\!-\\!m_{1}\\!-\\!r)!(j_{2}\\!+\\!m_{2}\\!-\\!r)!(j_{3}\\!-\\!j_{2}\\!+\\!m_{1}\\!+\\!r)!(j_{3}\\!-\\!j_{1}\\!-\\!m_{2}\\!+\\!r)!},$

where the sum runs through all integers $r$ for which the argument of every
factorial is non-negative. If either $m_{3}<0$ or $j_{1}<j_{2}$ we have

$\displaystyle CG(j_{1},m_{1};j_{2},m_{2}\mid j_{3},m_{3})$
$\displaystyle=(-1)^{j_{1}+j_{2}-j_{3}}\;CG(j_{1},-m_{1};j_{2},-m_{2}\mid
j_{3},-m_{3})$ (2.7)
$\displaystyle=(-1)^{j_{1}+j_{2}-j_{3}}\;CG(j_{2},m_{2};j_{1},m_{1}\mid
j_{3},m_{3}).$

In addition, it also holds

(2.8) $CG(j_{1},m_{1};j_{2},m_{2}\mid
j_{3},m_{3})=(-1)^{j_{1}-m_{1}}\sqrt{\frac{2j_{3}+1}{2j_{2}+1}}\;CG(j_{1},m_{1};j_{3},-m_{3}\mid
j_{2},-m_{2}).$

In the following sections we will need the following particular values of the
Clebsch-Gordan coefficients. Here $a,b$ are integers and $i=0,\dots,a$,
$j=0,\dots,b$.

(2.9)
$CG(\tfrac{a}{2},\tfrac{a}{2}-i;\,\tfrac{b}{2},\tfrac{b}{2}-j\,|\,\tfrac{a+b}{2},\tfrac{a+b}{2}-i-j)=\sqrt{\frac{a!b!(a+b-i-j)!(i+j)!}{i!j!(a+b)!(a-i)!(b-j)!}},$
(2.10)
$CG(\tfrac{a}{2},\tfrac{a}{2}-i;\,\tfrac{b}{2},j-\tfrac{b}{2}\,|\,\tfrac{a-b}{2},\tfrac{a-b}{2}-i+j)\\\
=(-1)^{j}\sqrt{\frac{(a-i)!\;i!\;b!\;(a-b+1)!}{(a+1)!\;j!\;(b-j)!\;(a-b-i+j)!\;(i-j)!}},$
(2.11)
$CG(\tfrac{a}{2},i-\tfrac{a}{2};\,\tfrac{b}{2},\tfrac{b}{2}-j\,|\,\tfrac{b-a}{2},\tfrac{b-a}{2}+i-j)\\\
=(-1)^{a}CG(\tfrac{b}{2},\tfrac{b}{2}-j;\,\tfrac{a}{2},i-\tfrac{a}{2}\,|\,\tfrac{b-a}{2},\tfrac{b-a}{2}+i-j)\\\
=(-1)^{j}\sqrt{\frac{(b-j)!\;j!\;a!\;(b-a+1)!}{(b+1)!\;i!\;(a-i)!\;(b-a-j+i)!\;(j-i)!}},$
(2.12)
$CG(\tfrac{a}{2},\tfrac{a}{2}-i;\,\tfrac{b}{2},\tfrac{b}{2}-j\,|\,\tfrac{a+b}{2}-i-j,\tfrac{a+b}{2}-i-j)\\\
=(-1)^{i}\sqrt{\frac{(a+b-2i-2j+1)!\;(i+j)!\;(a-i)!\;(b-j)!}{(a+b-i-j+1)!\;(a-i-j)!\;(b-i-j)!\;i!\;j!}}.$

###  2.2. Uniserial representations

Given a Lie algebra $\mathfrak{g}$ and a $\mathfrak{g}$-module $V$, we say
that $V$ is _uniserial_ if it admits a unique composition series. In other
words, $V$ is uniserial if the socle series

$0=\text{soc}^{0}(V)\subset\text{soc}^{1}(V)\subset\cdots\subset\text{soc}^{n}(V)=V$

is a composition series of $V$, that is, the socle factors
$\text{soc}^{i}(V)/\text{soc}^{i-1}(V)$ are irreducible for all $1\leq i\leq
n$. Recall that $\text{soc}^{1}(V)=\text{soc}(V)$ is the sum of all
irreducible $\mathfrak{g}$-submodules of $V$ and
$\text{soc}^{i}(V)/\text{soc}^{i-1}(V)=\text{soc}(V/\text{soc}^{i-1}(V))$.
Note that for uniserial modules, the _composition length_ $n$ of $V$ coincides
with its socle length.

If the Levi decomposition of $\mathfrak{g}$ is
$\mathfrak{g}=\mathfrak{s}\ltimes\mathfrak{r}$, (with $\mathfrak{r}$ the
solvable radical and $\mathfrak{s}$ semisimple) we may choose irreducible
$\mathfrak{s}$-submodules $V_{i}\subset V$, $1\leq i\leq n$, such that

(2.13) $V=V_{1}\oplus\cdots\oplus V_{n}$

with $V_{i}\simeq\text{soc}^{i}(V)/\text{soc}^{i-1}(V)$ as
$\mathfrak{s}$-modules and

$\mathfrak{r}V_{i}\subset V_{1}\oplus\cdots\oplus V_{i}.$

In fact, if $[\mathfrak{s},\mathfrak{r}]=\mathfrak{r}$, then
$\mathfrak{r}V_{i}\subset V_{1}\oplus\cdots\oplus V_{i-1}$, see Lemma 2.1
below. We say that (2.13) is the _socle decomposition_ of $V$, note that the
order of the summands is relevant.

######  Lemma 2.1.

If $\mathfrak{r}=[\mathfrak{s},\mathfrak{r}]$, then
$\text{soc}(U)=U^{\mathfrak{r}}$ for any $\mathfrak{g}$-module $U$.

###### Proof.

On the one hand, $U^{\mathfrak{r}}$ is a completely reducible
$\mathfrak{s}$-submodule of $U$ and hence, a completely reducible
$\mathfrak{g}$-submodule, thus $U^{\mathfrak{r}}\subset\text{soc}(U)$. On the
other hand, if $U_{1}$ is an irreducible $\mathfrak{g}$-submodule of $U$,
since the characteristic of the field $\mathbb{F}$ is 0, we know
$\mathfrak{r}=[\mathfrak{g},\mathfrak{g}]\cap\mathfrak{r}$ acts trivially on
$U_{1}$ (see [6, Chapitre 1, §5.3]). Hence $U_{1}\subset U^{\mathfrak{r}}$ and
therefore $\text{soc}(U)\subset U^{\mathfrak{r}}$. ∎

######  Lemma 2.2.

Assume that $\mathfrak{r}=[\mathfrak{s},\mathfrak{r}]$ and let $V$ be a
$\mathfrak{g}$-module such that it has a vector space decomposition
$V=V_{1}\oplus\cdots\oplus V_{n}$ such that $\mathfrak{r}V_{k}\subset V_{k-1}$
for all $k=2,\dots,n$. If $\text{soc}(V)=V_{1}$ then
$\text{soc}^{k}(V)=V_{1}\oplus\cdots\oplus V_{k}$ for all $k=1,\dots,n$.

###### Proof.

We proceed by induction on $k$. Assume the statement true for all $k$ with
$1\leq k<k_{0}<n$ and let us prove that
$\text{soc}^{k_{0}}(V)=V_{1}\oplus\cdots\oplus V_{k_{0}}$.

Let $U=V/\text{soc}^{k_{0}-1}(V)$ and let $p:V\to U$ the corresponding
projection. We point out that, since

$\text{soc}^{k_{0}-1}(V)=V_{1}\oplus\cdots\oplus V_{k_{0}-1}$

is a $\mathfrak{g}$-submodule, it follows that $p$ is a $\mathfrak{g}$-module
homomorphism and

(2.14) $U=p(V_{k_{0}})\oplus\cdots\oplus p(V_{n})$

as vector spaces.

We know, by the definition of the socle series, that $\text{soc}^{k_{0}}(V)$
is the $\mathfrak{g}$-submodule of $V$ satisfying

$\text{soc}^{k_{0}}(V)/\text{soc}^{k_{0}-1}(V)=\text{soc}(U)$

and it follows from Lemma 2.1 that

$\text{soc}(U)=\big{(}p(V_{k_{0}})\oplus\cdots\oplus
p(V_{n})\big{)}^{\mathfrak{r}}.$

Let $v=v_{k_{0}}+v_{k_{0}+1}+\ldots+v_{n}$ with $v_{k}\in V_{k}$,
$k=k_{0},\dots,n$, such that $Xp(v)=0$ for all $X\in\mathfrak{r}$. Since $p$
is a $\mathfrak{g}$-module homomorphism, we have

$p(Xv_{k_{0}})+p(Xv_{k_{0}+1})+\ldots+p(Xv_{n})=0.$

The hypothesis $\mathfrak{r}V_{k}\subset V_{k-1}$ implies
$X(p(v_{k}))=p(Xv_{k})\in p(V_{k-1})$ and hence $p(Xv_{k_{0}})=0$ and

$p(Xv_{k_{0}+1})+\ldots+p(Xv_{n})=0.$

Since $p|_{V_{i}}$ is injective for all $i\geq k_{0}$, it follows from (2.14)
that

$Xv_{i}=0,\quad\text{for all $i\geq k_{0}+1$ and all $X\in\mathfrak{r}$.}$

Finally, it follows from Lemma 2.1 and the hypothesis $\text{soc}(V)=V_{1}$,
that $V^{\mathfrak{r}}=V_{1}$. This implies that $v_{i}=0$ for all $i\geq
k_{0}+1$. Therefore, we have proved that
$\big{(}p(V_{k_{0}})\oplus\cdots\oplus
p(V_{n})\big{)}^{\mathfrak{r}}=p(V_{k_{0}})$ and thus
$\text{soc}(U)=p(V_{k_{0}})$. This shows that

$\text{soc}^{k_{0}}(V)=V_{1}\oplus\cdots\oplus V_{k_{0}}$

and the induction step is complete. ∎

###  2.3. Uniserial representations of $\mathfrak{sl}(2)\ltimes V(m)$.

In [10] it is obtained the classification, up to isomorphism, of all the
uniserial representations of the Lie algebra $\mathfrak{sl}(2)\ltimes V(m)$,
$m\geq 1$, when the underlying field is $\mathbb{C}$. Nevertheless, the
classification remains true over any field $\mathbb{F}$ of characteristic 0.
The main ingredients of this classification are the modules $E(a,b)$ and
$Z(a,\ell)$ that we present below.

From now on, we fix $m\geq 1$ and set
$\mathfrak{g}_{m}=\mathfrak{sl}(2)\ltimes V(m)$. We have

$\mathfrak{s}=\mathfrak{sl}(2)\quad\text{ and
}\quad\mathfrak{r}=[\mathfrak{s},\mathfrak{r}]=V(m).$

It will be useful to have a special notation for the basis
$\\{v_{0}^{m},\dots,v_{m}^{m}\\}$ of $\mathfrak{r}=V(m)$ as part of the Lie
algebra $\mathfrak{g}_{m}$. Thus, we will denote the basis of $\mathfrak{r}$
by $\\{e_{0},\dots,e_{m}\\}$.

If $a$ and $b$ are non-negative integers such that
$\frac{m}{2},\frac{a}{2},\frac{b}{2}$ satisfy the triangle condition, it
follows from (2.3) and (2.5) that, up to scalar, there is a unique
$\mathfrak{sl}(2)$-module homomorphism

$\mathfrak{r}=V(m)\to\text{Hom}(V(b),V(a)).$

This produces an action of $\mathfrak{r}$ on $V(a)\oplus V(b)$ given by
$\mathfrak{r}V(a)=0$ and

(2.15)
$e_{s}\,v_{j}^{b}=(-1)^{j}\,\sum_{i=0}^{a}CG(\tfrac{a}{2},\tfrac{a}{2}-i;\,\tfrac{b}{2},-\tfrac{b}{2}+j\,|\,\tfrac{m}{2},\tfrac{m}{2}-s)\,v_{i}^{a},\qquad
s=0,\dots,m.$

Note that this is the same as (2.6). Note also that the above sum has, in
fact, at most one summand, that is

(2.16) $e_{s}\,v_{j}^{b}=\begin{cases}0,&\text{if $i\neq
j+s+\frac{a-b-m}{2}$;}\\\\[5.69054pt]
(-1)^{j}CG(\tfrac{a}{2},\tfrac{a}{2}-i;\,\tfrac{b}{2},-\tfrac{b}{2}+j\,|\,\tfrac{m}{2},\tfrac{m}{2}-s)\,\,v_{i}^{a},&\text{if
$i=j+s+\frac{a-b-m}{2}$.}\end{cases}$

This action, combined with the action of $\mathfrak{sl}(2)$ defines a
uniserial $\mathfrak{g}_{m}$-module structure with composition length 2 on

$E(a,b)=V(a)\oplus V(b).$

It is straightforward to see that $E(a,b)^{*}\simeq E(b,a)$. The action given
in (2.15) is the main building block for all other uniserial
$\mathfrak{g}_{m}$-modules as follows.

The above construction can be extended to arbitrary composition length

$V(a_{0})\oplus V(a_{1})\oplus\cdots\oplus V(a_{\ell})$

only when the sequence $a_{i}$ is monotone and $m=|a_{i}-a_{i-1}|$, for all
$i=1,\dots,\ell$. More precisely, for $\alpha$ and $\ell$ non-negative
integers, let $Z(\alpha,\ell)$ be the uniserial $\mathfrak{g}_{m}$-module
defined by

(2.17) $Z(\alpha,\ell)=V(\alpha)\oplus V(\alpha+m)\oplus\cdots\oplus
V(\alpha+\ell m)$

as $\mathfrak{sl}(2)$-module with action of $\mathfrak{r}$ sending

$0\longleftarrow V(\alpha)\longleftarrow
V(\alpha+2m)\longleftarrow\dots\longleftarrow V(\alpha+\ell m)$

as indicated in (2.15) (with $a=\alpha+(i-1)m$, $b=\alpha+im$, for
$i=1,\dots,\ell$).

We notice that $Z(\alpha,0)=V(\alpha)$ ($\mathfrak{r}$ acts trivially) and
$Z(\alpha,1)=E(\alpha,\alpha+m)$. The uniserial modules $Z(\alpha,\ell)$ and
their duals will be called _of type $Z$_, and they are the unique isomorphism
classes of uniserial $\mathfrak{g}_{m}$-modules of composition length $\ell+1$
for $\ell\geq 4$.

For composition lengths $3$ and $4$, very few other ways to “combine” the
modules $E(a,b)$ are possible. For composition length equal to $3$, given
$0\leq c\leq 2m$ and $c\equiv 2m\mod 4$, let

$E_{3}(c)=V(0)\oplus V(m)\oplus V(c)$

as $\mathfrak{sl}(2)$-modules with action of $\mathfrak{r}$ sending

$0$$V(0)$$V(m)$$V(c)$

with the maps $V(c)\to V(m)$ and $V(m)\to V(0)$ given by (2.15).

For composition length equal to $4$, if $m\equiv 0\mod 4$, there is a family
of $\mathfrak{g}_{m}$-modules, parameterized by a non-zero scalar
$t\in\mathbb{F}$, with a fixed socle decomposition. This is defined by

$E_{4}(t)=V(0)\oplus V(m)\oplus V(m)\oplus V(0)$

as $\mathfrak{sl}(2)$-modules with action of $\mathfrak{r}$ sending the
$\mathfrak{sl}(2)$-modules as shown by the arrows

$0$$V(0)$$V(m)$$V(m)$$V(0)$

where the horizontal arrows are given by (2.15) and the bent arrow is $t$
times (2.15). We can now state one of the main results of [10, Thm 10.1].

######  Theorem 2.3.

The following list describes all the isomorphism classes of uniserial
representations of $\mathfrak{g}_{m}=\mathfrak{sl}(2)\ltimes V(m)$.

Length 1. |  $Z(a,0)=V(a)$, $a\geq 0$.
---|---
Length 2. |  $E(a,b)$, with $a+b\equiv m\mod 2$ and $0\leq|a-b|\leq m\leq a+b$.
Length 3. |  $Z(a,2)$, $Z(a,2)^{*}$, $a\geq 0$; and
|  $E_{3}(c)$ with $c\equiv 2m\mod 4$ and $0\leq c\leq 2m$.
Length 4. |  $Z(a,3)$, $Z(a,3)^{*}$, $a\geq 0$; and
|  $E_{4}(t)$, with $0\neq t\in\mathbb{F}$ (this exists only if $m\equiv 0\mod
4$).
Length $\ell\geq 5$. |  $Z(a,\ell-1)$, $Z(a,\ell-1)^{*}$, $a\geq 0$.

##  3\. The socle of the tensor product of two uniserial
$\mathfrak{g}_{m}$-modules

###  3.1. General considerations

Given two $\mathfrak{g}_{m}$-modules $V$ and $W$, it is clear that
$\text{soc}(V)\otimes\,\text{soc}(W)\subset\text{soc}(V\otimes W)$. Therefore,
if $V$ and $W$ are uniserial $\mathfrak{g}_{m}$-modules with socle
decomposition (see (2.13) and the comments below it)

$\displaystyle V$ $\displaystyle=V(a_{0})\oplus V(a_{1})\oplus\ldots\oplus
V(a_{\ell}),$ $\displaystyle W$ $\displaystyle=V(b_{0})\oplus
V(b_{1})\oplus\ldots\oplus V(b_{\ell^{\prime}})$

($V(a_{0})=\text{soc}(V)$, $V(b_{0})=\text{soc}(W)$), we have

$V(a_{0})\otimes V(b_{0})\subset\text{soc}(V\otimes W).$

For convenience we assume $V(a_{i})=V(b_{j})=0$ for $i,j<0$. We know from
Theorem 2.3 that

$\mathfrak{r}V(a_{i})\subset V(a_{i-1})\oplus
V(a_{i-2})\quad\text{and}\quad\mathfrak{r}V(b_{j})\subset V(b_{j-1})\oplus
V(b_{j-2})$

for all $i\leq\ell$, $j\leq\ell^{\prime}$ (in fact we know that
$\mathfrak{r}V(a_{i})\subset V(a_{i-1})$ and $\mathfrak{r}V(b_{j})\subset
V(b_{j-1})$ except for cases of uniserials of type $E_{4}$). This implies that

(3.1) $\mathfrak{r}\;\Big{(}\bigoplus_{i+j=t}V(a_{i})\otimes
V(b_{j})\Big{)}\subset\bigoplus_{i+j<t}V(a_{i})\otimes V(b_{j}).$

Moreover, we point out for future use that if neither $V$ nor $W$ is of type
$E_{4}$ then

(3.2) $\mathfrak{r}\,\big{(}V(a_{i})\otimes V(b_{j})\big{)}\subset
V(a_{i-1})\otimes V(b_{j})\;\oplus\;V(a_{i})\otimes V(b_{j-1})$

Since $\mathfrak{r}=[\mathfrak{s},\mathfrak{r}]$, it follows from Lemma 2.1
that $\text{soc}(U)=U^{\mathfrak{r}}$ for any $\mathfrak{g}_{m}$-module $U$.
Hence, it follows from (3.1) that

$\displaystyle\text{soc}(V\otimes W)$
$\displaystyle=\bigoplus_{t=0}^{\ell+\ell^{\prime}}\Big{(}\text{soc}(V\otimes
W)\cap\bigoplus_{i+j=t}V(a_{i})\otimes V(b_{j})\Big{)}$ (3.3)
$\displaystyle=\bigoplus_{t=0}^{\ell+\ell^{\prime}}\Big{(}\bigoplus_{i+j=t}V(a_{i})\otimes
V(b_{j})\Big{)}^{\mathfrak{r}}.$

For $t=0,\dots,\ell+\ell^{\prime}$, let us define

$S_{t}=S_{t}(V,W)=\Big{(}\bigoplus_{i+j=t}V(a_{i})\otimes
V(b_{j})\Big{)}^{\mathfrak{r}}$

so that $\displaystyle{\text{soc}(V\otimes
W)=\bigoplus_{t=0}^{\ell+\ell^{\prime}}S_{t}}$. It is clear that

$S_{0}=\Big{(}V(a_{0})\otimes V(b_{0})\Big{)}^{\mathfrak{r}}=V(a_{0})\otimes
V(b_{0})=\text{soc}(V)\otimes\text{soc}(W)$

and hence

(3.4) $\text{soc}(V\otimes
W)=\text{soc}(V)\otimes\text{soc}(W)\;\oplus\;\bigoplus_{t=1}^{\ell+\ell^{\prime}}S_{t}$

as $\mathfrak{sl}(2)$-modules. Hence, in order to obtain the
$\mathfrak{sl}(2)$-decomposition of $\text{soc}(V\otimes W)$ we need to find
the highest weight vectors in $S_{t}$ (specially for $t\geq 1$) that are
annihilated by $\mathfrak{r}$.

Given $v\in V(a_{i})\otimes V(b_{j})$ and $e_{s}\in\mathfrak{r}$, let

(3.5) $e_{s}v=(e_{s}v)_{1}+(e_{s}v)_{2}+(e_{s}v)_{3}$

where

$\displaystyle(e_{s}v)_{1}$ $\displaystyle\in V(a_{i-1})\otimes V(b_{j}),$
$\displaystyle(e_{s}v)_{2}$ $\displaystyle\in V(a_{i})\otimes V(b_{j-1}),$
$\displaystyle(e_{s}v)_{3}$ $\displaystyle\in V(a_{i-2})\otimes V(b_{j})\oplus
V(a_{i})\otimes V(b_{j-2})$

(note that $(e_{s}v)_{3}=0$ if neither $V$ nor $W$ is of type $E_{4}$). It is
clear that $(e_{s}v)_{1}=0$ if $i=0$ and $(e_{s}v)_{2}=0$ if $j=0$, the
following lemma states a sort of converse of this for highest weight vectors
in $V(a_{i})\otimes V(b_{j})$.

######  Lemma 3.1.

Let $V=V(a_{0})\oplus\ldots\oplus V(a_{\ell})$ and
$W=V(b_{0})\oplus\ldots\oplus V(b_{\ell^{\prime}})$, with
$\ell,\ell^{\prime}\geq 1$, be the socle decomposition of two uniserial
$\mathfrak{g}_{m}$-modules. If $v_{0}\in V(a_{i_{0}})\otimes V(b_{j_{0}})$ is
a highest weight vector then:

  1. (i)

$(e_{s}v_{0})_{1}=0$ for all $s=0,\dots,m$ if and only if $i_{0}=0$.

  2. (ii)

$(e_{s}v_{0})_{2}=0$ for all $s=0,\dots,m$ if and only if $j_{0}=0$.

###### Proof.

By symmetry it suffices to prove (i). As we already mentioned, it is clear the
“if” part and thus we will prove the “only if” part. So let us assume
$i_{0}>0$ and let us prove $(e_{s}v_{0})_{1}\neq 0$ for some $s=0,\dots,m$.

If $c$ is the weight of $v_{0}$, we may assume that (see (2.4) and (2.12))

$\displaystyle v_{0}=v_{0}^{a,b,c}$
$\displaystyle=\sum_{i+j=\frac{a+b-c}{2}}CG(\tfrac{a}{2},\tfrac{a}{2}-i;\,\tfrac{b}{2},\tfrac{b}{2}-j\,|\,\tfrac{c}{2},\tfrac{c}{2})\,v_{i}^{a}\otimes
v_{j}^{b}$
$\displaystyle=\sum_{i+j=\frac{a+b-c}{2}}(-1)^{i}\sqrt{\frac{(a+b-2i-2j+1)!(i+j)!(a-i)!(b-j)!}{(a+b-i-j+1)!(a-i-j)!(b-i-j)!\,i!\,j!}}\,v_{i}^{a}\otimes
v_{j}^{b}$
$\displaystyle=\sqrt{\frac{(c+1)!\;a!}{(\frac{a+b+c}{2}+1)!\;(\frac{a-b+c}{2})!}}\,\,v_{0}^{a}\otimes
v_{\frac{a+b-c}{2}}^{b}\quad+\quad\sum_{j<\frac{a+b-c}{2}}q_{j}\;v_{\frac{a+b-c}{2}-j}^{a}\otimes
v_{j}^{b}$

where $a=a_{i_{0}}$ and $b=b_{j_{0}}$ and $q_{j}\in\mathbb{F}$ is some scalar.
Now, for $s=0,\dots,m$, we have

$(e_{s}v_{0})_{1}=\sqrt{\frac{(c+1)!\;(a_{i_{0}})!}{(\frac{a_{i_{0}}+b_{j_{0}}+c}{2}+1)!\;(\frac{a_{i_{0}}-b_{j_{0}}+c}{2})!}}\,\,e_{s}v_{0}^{a_{i_{0}}}\otimes
v_{\frac{a_{i_{0}}+b_{j_{0}}-c}{2}}^{b_{j_{0}}}\quad\\\
+\quad\sum_{j<\frac{a_{i_{0}}+b_{j_{0}}-c}{2}}q_{j}\;e_{s}v_{\frac{a_{i_{0}}+b_{j_{0}}-c}{2}-j}^{a_{i_{0}}}\otimes
v_{j}^{b_{j_{0}}}.$

We know from (2.16) (see also (2.8) and (2.12)) that, if $i_{0}>0$ and
$s=\frac{a_{i_{0}}-a_{i_{0}-1}+m}{2}$ then

$\displaystyle e_{s}\,v_{0}^{a}$
$\displaystyle=CG(\tfrac{a_{i_{0}-1}}{2},\tfrac{a_{i_{0}-1}}{2};\,\tfrac{a_{i_{0}}}{2},-\tfrac{a_{i_{0}}}{2}\,|\,\tfrac{m}{2},\tfrac{m}{2}-s)\,\,v_{0}^{a_{i_{0}-1}}$
$\displaystyle=\sqrt{\frac{(m+1)!\;(a_{i_{0}})!\;(a_{i_{0}-1})!}{\big{(}\tfrac{a_{i_{0}}+a_{i_{0}-1}+m}{2}+1\big{)}!\;\big{(}\tfrac{a_{i_{0}}+a_{i_{0}-1}-m}{2}\big{)}!}}\,\,v_{0}^{a_{i_{0}-1}}\neq
0.$

This implies $(e_{s}v_{0})_{1}\neq 0$. ∎

######  Proposition 3.2.

Let $V=V(a_{0})\oplus\ldots\oplus V(a_{\ell})$ and
$W=V(b_{0})\oplus\ldots\oplus V(b_{\ell^{\prime}})$, with
$\ell,\ell^{\prime}\geq 1$, be the socle decomposition of two uniserial
$\mathfrak{g}_{m}$-modules. If $t>\min\\{\ell,\ell^{\prime}\\}$ then

$S_{t}=\Big{(}\bigoplus_{i+j=t}V(a_{i})\otimes
V(b_{j})\Big{)}^{\mathfrak{r}}=0.$

If $0<t\leq\min\\{\ell,\ell^{\prime}\\}$ and $\mu$ is a highest weight in
$S_{t}$ then $\mu$ must be a highest weight in all the summands
$V(a_{i})\otimes V(b_{t-i})$, $i=0,\dots,t$, and, in this case, its weight
space is 1-dimensional and generated by a linear combination

$\sum_{i=0}^{t}q_{i}\,v_{0}^{a_{i},b_{t-i},\mu}$

with $q_{i}\neq 0$ for all $i=0,\dots,t$.

###### Proof.

We fix $t>0$ and we assume that there is a non-zero

$u=\sum_{i+j=t}u_{i,j}\in\Big{(}\bigoplus_{i+j=t}V(a_{i})\otimes
V(b_{j})\Big{)}^{\mathfrak{r}}$

that is a highest weight vector of weight $\mu$. Since $V(a_{i})\otimes
V(b_{j})$ is an $\mathfrak{sl}(2)$-submodule, it follows that $u_{i,j}$ is
either zero or a highest weight vector of weight $\mu$. Let

$I_{t}^{\mu}=\\{(i,j):0\leq i\leq\ell,\;0\leq j\leq\ell^{\prime},\;i+j=t\text{
and }u_{i,j}\neq 0\\}.$

Since $u\neq 0$, it follows that $I_{t}^{\mu}\neq\emptyset$ and

$u=\sum_{(i,j)\in I_{t}^{\mu}}q_{i,j}\,v_{0}^{a_{i},b_{j},\mu}$

for certain non-zero scalars $0\neq q_{i,j}\in\mathbb{F}$. We will show that
$t\leq\min\\{\ell,\ell^{\prime}\\}$ and

(3.6) $I_{t}^{\mu}=\\{(0,t),(1,t-1),\dots,(t,0)\\}.$

Since $u$ is $\mathfrak{r}$-invariant, we have

(3.7) $0=e_{s}u=\sum_{(i,j)\in
I_{t}^{\mu}}q_{i,j}\big{(}(e_{s}v_{0}^{a_{i},b_{j},\mu})_{1}+(e_{s}v_{0}^{a_{i},b_{j},\mu})_{2}+(e_{s}v_{0}^{a_{i},b_{j},\mu})_{3}\big{)}$

for all $0\leq s\leq m$ (see (3.5)).

Since we know that $(e_{s}v_{0}^{a_{i},b_{j},\mu})_{1}\in V(a_{i-1})\otimes
V(b_{i})$, $(e_{s}v_{0}^{a_{i},b_{j},\mu})_{2}\in V(a_{i})\otimes V(b_{i-1})$,
and $(e_{s}v_{0}^{a_{i},b_{j},\mu})_{3}\in V(a_{i-2})\otimes V(b_{j})\oplus
V(a_{i})\otimes V(b_{j-2})$, (3.7) yields a linear system whose unknowns are
the $q_{i,j}$’s and the equations are

(3.8) $\displaystyle
q_{i,j}(e_{s}v_{0}^{a_{i},b_{j},\mu})_{1}+q_{i-1,j+1}(e_{s}v_{0}^{a_{i-1},b_{j+1},\mu})_{2}$
$\displaystyle=0,\quad\text{if $(i,j),(i-1,j+1)\in I_{t}^{\mu}$};$ (3.9)
$\displaystyle q_{i,j}(e_{s}v_{0}^{a_{i},b_{j},\mu})_{1}$
$\displaystyle=0,\quad\text{if $(i,j)\in I_{t}^{\mu}$, $(i-1,j+1)\not\in
I_{t}^{\mu}$};$ (3.10) $\displaystyle
q_{i,j}(e_{s}v_{0}^{a_{i},b_{j},\mu})_{2}$ $\displaystyle=0,\quad\text{if
$(i,j)\in I_{t}^{\mu}$, $(i+1,j-1)\not\in I_{t}^{\mu}$}$

for each $s=0,\dots,m$. Suppose, if possible, that there is $(i_{0},j_{0})\in
I_{t}^{\mu}$, $i_{0}>0$, such that $(i_{0}-1,j_{0}+1)\not\in I_{t}^{\mu}$
(this would happen if $t>\ell^{\prime}$). Then (3.9) implies
$(e_{s}v_{0}^{a_{i},b_{j},\mu})_{1}=0$ for all $s=0,\dots,m$, which
contradicts Lemma 3.1. This proves that $t\leq\ell^{\prime}$ and $(0,t)\in
I_{t}^{\mu}$.

Similarly, it can be shown that if $(i_{0},j_{0})\in I_{t}^{\mu}$, then either
$j_{0}=0$ or $(i_{0}+1,j_{0}-1)\in I_{t}^{\mu}$. This shows that $t\leq\ell$
and (3.6).

Finally, it follows from Lemma 3.1 that the linear system given by (3.8),
(3.9) and (3.10) has at most a 1-dimensional solution space. ∎

###  3.2. The socle of the tensor product of $\mathfrak{g}_{m}$-modules of
type $Z$

Given two uniserial $\mathfrak{g}_{m}$-modules $V_{1}$ and $V_{2}$, of length
2, with socle decomposition $V_{1}=V(a)\oplus V(b)$ and $V_{2}=V(c)\oplus
V(d)$, we know from the previous section that

$\text{soc}(V_{1}\otimes V_{2})=V(a)\otimes V(c)\;\oplus S_{1}$

with

$S_{1}=\text{soc}(V_{1}\otimes V_{2})\cap\Big{(}V(a)\otimes
V(d)\;\oplus\;V(b)\otimes V(d)\Big{)}.$

The following theorem describes $S_{1}$ for uniserials of length 2 of type $Z$
(see §2.3).

######  Theorem 3.3.

Let $V_{1}=V(a)\oplus V(b)$ and $V_{2}=V(c)\oplus V(d)$ be the socle
decomposition of two uniserial $\mathfrak{g}_{m}$-modules of type $Z$. Then
the following table describes $S_{1}$:

${}_{\displaystyle V_{1}}\backslash{\displaystyle V_{2}}$ | $\begin{array}[]{l}Z(c,1)\\\\[2.84526pt] \simeq V(c)\oplus V(c+m)\end{array}$ |  $\begin{array}[]{l}Z(d,1)^{*}\\\\[2.84526pt] \simeq V(d+m)\oplus V(d)\end{array}$
---|---|---
$\begin{array}[]{l}Z(a,1)\\\\[2.84526pt] \simeq V(a)\oplus V(a+m)\end{array}$ | $S_{1}\simeq V(a+d);$ |  $S_{1}\simeq\begin{cases}V(d-a),&\text{if $a\leq d$;}\\\ 0,&\text{if $a>d$;}\end{cases}$
$\begin{array}[]{l}Z(b,1)^{*}\\\\[2.84526pt] \simeq V(b+m)\oplus V(b)\end{array}$ | $S_{1}\simeq\begin{cases}V(b-c),&\text{if $b\geq c$;}\\\ 0,&\text{if $b<c$;}\end{cases}$ |  $S_{1}=0$.

All the isomorphisms in the table are as $\mathfrak{sl}(2)$-modules. A highest
weight vector is

$u_{0}=\sqrt{d+1}\;v_{0}^{a,d,\mu}-\sqrt{b+1}\;v_{0}^{b,c,\mu}$

in the entries (1,1) and (1,2) of the table, with $\mu=a+d=b+c$ and
$\mu=d-a=c-b$ respectively, and

$u_{0}=\sqrt{d+1}\;v_{0}^{a,d,\mu}\;-(-1)^{m}\sqrt{b+1}\;v_{0}^{b,c,\mu}$

in the entries (2,1) of the table with $\mu=a-d=b-c$.

It would be very interesting for us to extend this theorem to any pair of
uniserials of length 2. We have the following conjecture:

######  Conjecture 3.4.

Let $V_{1}=V(a)\oplus V(b)$ and $V_{2}=V(c)\oplus V(d)$ be the socle
decomposition of two uniserial $\mathfrak{g}_{m}$-modules, as in Theorem 2.3,
and assume that $a<c$, or $a=c$ and $b\leq d$. Then $S_{1}=0$ except in the
following cases.

  1. $\bullet$

Case 1: $[a,b]=[0,m]$. Here $S_{1}\simeq V(d)$.

  2. $\bullet$

Cases 2: Here $a>0$.

    1. –

Case 2.1: $a+b=c+d=m$ with $d-a=b-c\geq 0$. Here $S_{1}\simeq V(d-a)$.

    2. –

Case 2.2: $b-a=d-c=m$. Here $S_{1}\simeq V(d+a)$.

    3. –

Case 2.3: $b-a=c-d=m$ with $d-a=c-b\geq 0$. Here $S_{1}\simeq V(d-a)$.

  3. $\bullet$

Case 3: $[c,d]=[b,a]$. Here $S_{1}\simeq V(0)$.

Note that the entries (1,1) and (1,2) in the table of Theorem 3.3 correspond
to Cases 2.2 and 2.3 respectively, while the entry (2,1), with $b\geq c$, is
ruled out in the conjecture by the condition $a\leq c$.

###### Proof of Theorem 3.3.

Let $\mu$ be a possible highest weight in $S_{1}$. We first point out some
general considerations that will be useful for all cases, and next we will
work out the details of each case.

We know from Proposition 3.2 that $\mu$ must be highest weight in both
$V(a)\otimes V(d)$ and $V(b)\otimes V(c)$, that is

(3.11) $|a-d|,|b-c|\leq\mu\leq a+d,b+c$

and $\mu\equiv a+d\equiv b+c\mod 2$. We also know that $\mu$ is indeed highest
weight in $S_{1}$ if and only if there is a linear combination

$u_{0}=q_{1}v_{0}^{a,d,\mu}+q_{2}v_{0}^{b,c,\mu},$

with $q_{1},q_{2}\neq 0$, that is annihilated by $e_{s}$ for all
$s=0,\dots,m$. We now describe $e_{s}v_{0}^{a,d,\mu}$ and
$e_{s}v_{0}^{b,c,\mu}$.

On the one hand we have (see (2.4))

$v_{0}^{a,d,\mu}=\sum_{i,j}CG(\tfrac{a}{2},\tfrac{a}{2}-i;\,\tfrac{d}{2},\tfrac{d}{2}-j\,|\,\tfrac{\mu}{2},\tfrac{\mu}{2})\,v_{i}^{a}\otimes
v_{j}^{d}$

and thus (see (2.15))

$\displaystyle e_{s}v_{0}^{a,d,\mu}$
$\displaystyle=\sum_{i,j}CG(\tfrac{a}{2},\tfrac{a}{2}-i;\,\tfrac{d}{2},\tfrac{d}{2}-j\,|\,\tfrac{\mu}{2},\tfrac{\mu}{2})\,v_{i}^{a}\otimes
e_{s}v_{j}^{d}$
$\displaystyle=\sum_{i,j,k}(-1)^{j}CG(\tfrac{a}{2},\tfrac{a}{2}-i;\,\tfrac{d}{2},\tfrac{d}{2}-j\,|\,\tfrac{\mu}{2},\tfrac{\mu}{2})$
$\displaystyle\hskip 113.81102pt\times
CG(\tfrac{c}{2},\tfrac{c}{2}-k;\,\tfrac{d}{2},-\tfrac{d}{2}+j\,|\,\tfrac{m}{2},\tfrac{m}{2}-s)\,v_{i}^{a}\otimes
v_{k}^{c}$
$\displaystyle=\sum_{i,j,k}(-1)^{k}CG(\tfrac{a}{2},\tfrac{a}{2}-i;\,\tfrac{d}{2},\tfrac{d}{2}-k\,|\,\tfrac{\mu}{2},\tfrac{\mu}{2})$
(3.12) $\displaystyle\hskip 113.81102pt\times
CG(\tfrac{c}{2},\tfrac{c}{2}-j;\,\tfrac{d}{2},-\tfrac{d}{2}+k\,|\,\tfrac{m}{2},\tfrac{m}{2}-s)\,v_{i}^{a}\otimes
v_{j}^{c}.$

In this sum, if the coefficient of $v_{i}^{a}\otimes v_{j}^{c}$ is not zero
then we must have

(3.13) $\begin{split}\frac{a}{2}-i+\frac{d}{2}-k&=\frac{\mu}{2},\\\
\frac{c}{2}-j-\frac{d}{2}+k&=\frac{m}{2}-s.\end{split}$

On the other hand we have (see (2.4))

$v_{0}^{b,c,\mu}=\sum_{i,j}CG(\tfrac{b}{2},\tfrac{b}{2}-i;\,\tfrac{c}{2},\tfrac{c}{2}-j\,|\,\tfrac{\mu}{2},\tfrac{\mu}{2})\,v_{i}^{b}\otimes
v_{j}^{c}.$

and thus (see (2.15))

$\displaystyle e_{s}v_{0}^{b,c,\mu}$
$\displaystyle=\sum_{i,j}CG(\tfrac{b}{2},\tfrac{b}{2}-i;\,\tfrac{c}{2},\tfrac{c}{2}-j\,|\,\tfrac{\mu}{2},\tfrac{\mu}{2})\,e_{s}v_{i}^{b}\otimes
v_{j}^{c}.$
$\displaystyle=\sum_{i,j,k}(-1)^{i}CG(\tfrac{b}{2},\tfrac{b}{2}-i;\,\tfrac{c}{2},\tfrac{c}{2}-j\,|\,\tfrac{\mu}{2},\tfrac{\mu}{2})$
$\displaystyle\hskip 113.81102pt\times
CG(\tfrac{a}{2},\tfrac{a}{2}-k;\,\tfrac{b}{2},-\tfrac{b}{2}+i\,|\,\tfrac{m}{2},\tfrac{m}{2}-s)\,v_{k}^{a}\otimes
v_{j}^{c}$
$\displaystyle=\sum_{i,j,k}(-1)^{k}CG(\tfrac{b}{2},\tfrac{b}{2}-k;\,\tfrac{c}{2},\tfrac{c}{2}-j\,|\,\tfrac{\mu}{2},\tfrac{\mu}{2})$
(3.14) $\displaystyle\hskip 113.81102pt\times
CG(\tfrac{a}{2},\tfrac{a}{2}-i;\,\tfrac{b}{2},-\tfrac{b}{2}+k\,|\,\tfrac{m}{2},\tfrac{m}{2}-s)\,v_{i}^{a}\otimes
v_{j}^{c}.$

In this sum, if the coefficient of $v_{i}^{a}\otimes v_{j}^{c}$ is not zero
then we must have

(3.15) $\begin{split}\frac{b}{2}-k+\frac{c}{2}-j&=\frac{\mu}{2},\\\
\frac{a}{2}-i-\frac{b}{2}+k&=\frac{m}{2}-s.\end{split}$

Either (3.13) or (3.15) imply

(3.16) $i+j=\frac{a+c-m-\mu}{2}+s,$

and recall that $0\leq i\leq a$ and $0\leq j\leq c$.

_The case $Z(a,1)\otimes Z(c,1)$_. Here $b=a+m$, $d=c+m$ and (3.11) implies

(3.17) $\mu=a+c+m-2p,\quad 0\leq p\leq\min\\{a,c+m\\}.$

It follows from (3.16) that

(3.18) $0\leq i+j=p-m+s.$

First we prove that if $p=0$ then $\mu$ is indeed a highest weight in $S_{1}$.
In this case, it follows from (3.18) that

$e_{s}v_{0}^{a,d,\mu}=e_{s}v_{0}^{b,c,\mu}=0$

for all $s=0,\dots,m-1$.

For $s=m$, in the sums describing $e_{m}v_{0}^{a,d,\mu}$ and
$e_{m}v_{0}^{b,c,\mu}$ we must have $i+j=0$, that is $i=j=0$, and thus (see
(3.12) and (3.14))

$\displaystyle e_{m}v_{0}^{a,d,\mu}$
$\displaystyle=CG(\tfrac{a}{2},\tfrac{a}{2};\,\tfrac{c+m}{2},\tfrac{c+m}{2}\,|\,\tfrac{\mu}{2},\tfrac{\mu}{2})\,CG(\tfrac{c}{2},\tfrac{c}{2};\,\tfrac{c+m}{2},-\tfrac{c+m}{2}\,|\,\tfrac{m}{2},-\tfrac{m}{2})\,v_{0}^{a}\otimes
v_{0}^{c}$ $\displaystyle=\sqrt{\frac{m+1}{c+m+1}}\;v_{0}^{a}\otimes
v_{0}^{c}$

and

$\displaystyle e_{m}v_{0}^{b,c,\mu}$
$\displaystyle=CG(\tfrac{a+m}{2},\tfrac{a+m}{2};\,\tfrac{c}{2},\tfrac{c}{2}\,|\,\tfrac{\mu}{2},\tfrac{\mu}{2})\,CG(\tfrac{a}{2},\tfrac{a}{2};\,\tfrac{a+m}{2},-\tfrac{a+m}{2}\,|\,\tfrac{m}{2},-\tfrac{m}{2})\,v_{0}^{a}\otimes
v_{0}^{c}$ $\displaystyle=\sqrt{\frac{m+1}{a+m+1}}\;v_{0}^{a}\otimes
v_{0}^{c}.$

This implies that

$u_{0}=\sqrt{c+m+1}\;v_{0}^{a,d,\mu}-\sqrt{a+m+1}\;v_{0}^{b,c,\mu}$

is, indeed, a highest weight vector, of weight $\mu=a+c+m$, in $S_{1}$.

We now prove that if $p\geq 1$ then $\mu=a+c+m-2p$ is not a highest weight in
$S_{1}$.

Let us fix $s=m$. It follows from (3.13) and (3.18) that

$k=j=p-i$

and (see 3.17)

$\displaystyle e_{m}v_{0}^{a,d,\mu}$
$\displaystyle=\sum_{i=0}^{\min\\{p,a\\}}(-1)^{p-i}CG(\tfrac{a}{2},\tfrac{a}{2}-i;\,\tfrac{c+m}{2},\tfrac{c+m}{2}+i-p\,|\,\tfrac{a+c+m}{2}-p,\tfrac{a+c+m}{2}-p)\,$
$\displaystyle\hskip 56.9055pt\times
CG(\tfrac{c}{2},\tfrac{c}{2}+i-p;\,\tfrac{c+m}{2},-\tfrac{c+m}{2}+p-i\,|\,\tfrac{m}{2},-\tfrac{m}{2})\,v_{i}^{a}\otimes
v_{p-i}^{c}$
$\displaystyle=\sum_{i=0}^{\min\\{p,a\\}}(-1)^{i}\sqrt{\frac{(a+c+m-2p+1)!\;p!\;(a-i)!\;(c+m-p+i)!}{(a+c+m-p+1)!\;(a-p)!\;(c+m-p)!\;i!\;(p-i)!}}$
$\displaystyle\hskip
113.81102pt\times\sqrt{\frac{(m+1)!\;c!\;(c+m+i-p)!}{(c+m+1)!\;m!\;(c+i-p)!}}\;\;v_{i}^{a}\otimes
v_{p-i}^{c}.$

This is, up to a non-zero scalar, equal to

$w^{a,d,\mu}=\sum_{i=0}^{\min\\{p,a\\}}(-1)^{i}\sqrt{\frac{(a-i)!\;(c+m+i-p)!^{2}}{\;i!\;(p-i)!\;(c+i-p)!}}\;\;v_{i}^{a}\otimes
v_{p-i}^{c}.$

Similarly, it follows from (3.15) and (3.14) that

$k=i,\quad j=p-i$

and (see 3.17)

$\displaystyle e_{m}v_{0}^{b,c,\mu}$
$\displaystyle=\sum_{i=0}^{\min\\{p,a\\}}(-1)^{i}\;CG(\tfrac{a+m}{2},\tfrac{a+m}{2}-i;\,\tfrac{c}{2},\tfrac{c}{2}+i-p\,|\,\tfrac{a+c-m}{2}-p,\tfrac{a+c-m}{2}-p)$
$\displaystyle\hskip 56.9055pt\times
CG(\tfrac{a}{2},\tfrac{a}{2}-i;\,\tfrac{a+m}{2},-\tfrac{a+m}{2}+i\,|\,\tfrac{m}{2},-\tfrac{m}{2})\,v_{i}^{a}\otimes
v_{p-i}^{c}$
$\displaystyle=\sum_{i=0}^{\min\\{p,a\\}}(-1)^{i}\sqrt{\frac{(a+c+m-2p+1)!\;p!\;(a+m-i)!\;(c+i-p)!}{(a+c+m-p+1)!\;(a+m-p)!\;(c-p)!\;i!\;(p-i)!}}$
$\displaystyle\hskip
113.81102pt\times\sqrt{\frac{(m+1)!\;a!\;(a+m-i)!}{(a+m+1)!\;m!\;(a-i)!}}\;\;v_{i}^{a}\otimes
v_{p-i}^{c},$

and this is, up to a scalar, equal to

$w^{b,c,\mu}=\sum_{i=0}^{\min\\{p,a\\}}(-1)^{i}\sqrt{\frac{(a+m-i)!^{2}\;(c+i-p)!}{i!\;(p-i)!\;(a-i)!}}\;\;v_{i}^{a}\otimes
v_{p-i}^{c}.$

We will show that $w^{a,d,\mu}$ and $w^{b,c,\mu}$ are linearly independent. It
suffices to show that the ratio of the first two coefficients (corresponding
to $i=0,1$) of $w^{a,d,\mu}$ and $w^{b,c,\mu}$ differ from each other. (We
recall that $p\geq 1$ and hence $a\geq 1$ (see 3.17). Note also that all the
coefficients in both $w^{a,d,\mu}$ and $w^{b,c,\mu}$ are non-zero.)

The ratio of the first two coefficients of $w^{a,d,\mu}$ is

$-\frac{\sqrt{\dfrac{a!\;(c+m-p)!^{2}}{p!\;(c-p)!}}}{\sqrt{\dfrac{(a-1)!\;(c+m-p+1)!^{2}}{(p-1)!\;(c+1-p)!}}}=-\sqrt{\dfrac{a\;(c+1-p)}{p\;(c+m-p+1)^{2}}},$

and ratio of the first two coefficients of $w^{b,c,\mu}$ is

$-\frac{\sqrt{\dfrac{(a+m)!^{2}\;(c-p)!}{p!\;a!}}}{\sqrt{\dfrac{(a+m-1)!^{2}\;(c+1-p)!}{(p-1)!\;(a-1)!}}}=-\sqrt{\dfrac{(a+m)^{2}}{p\;a\;(c+1-p)}}.$

This two ratios are different since $a^{2}(c+1-p)^{2}<(a+m)^{2}(c+m-p+1)^{2}.$
This completes the proof of this case.

_The case $Z(b,1)^{*}\otimes Z(d,1)^{*}$_. Here $a=b+m$, $c=d+m$, and (3.11)
implies

$\mu=b+d+m-2p,\quad 0\leq p\leq\min\\{b,d\\}.$

We know from (3.16) that

$0\leq i+j=p+s.$

Recall that (3.13) implies $k=p-i\geq 0$ and (3.12) says

$\displaystyle e_{s}v_{0}^{a,d,\mu}$
$\displaystyle=\sum_{i=0}^{\min\\{a,p\\}}(-1)^{p-i}\;CG(\tfrac{b+m}{2},\tfrac{b+m}{2}-i;\,\tfrac{d}{2},\tfrac{d}{2}-p+i\,|\,\tfrac{b+d+m}{2}-p,\tfrac{b+d+m}{2}-p)$
$\displaystyle\hskip 36.98866pt\times
CG(\tfrac{d+m}{2},\tfrac{d+m}{2}-p-s+i;\,\tfrac{d}{2},-\tfrac{d}{2}+p-i\,|\,\tfrac{m}{2},\tfrac{m}{2}-s)\,v_{i}^{a}\otimes
v_{p+s-i}^{c}$
$\displaystyle=\sum_{i=0}^{\min\\{a,p\\}}(-1)^{i}\sqrt{\frac{(b+m+d-2p+1)!\;p!\;(b+m-i)!\;(d-p+i)!}{(b+m+d-p+1)!\;(b+m-p)!\;(d-p)!\;i!\;(p-i)!}}$
$\displaystyle\hskip
56.9055pt\times\sqrt{\frac{(d+m-p-s+i)!\;(p+s-i)!\;(m+1)!\;d!\;}{(d+m+1)!\;(p-i)!\;(d-p+i)!\;(m-s)!\;s!}}\;\;v_{i}^{a}\otimes
v_{p+s-i}^{c}.$

As always, the reader should check that all the numbers under the factorial
sign are non-negative.

On the other hand, (3.15) implies $k=i-s\geq 0$ and (3.14) says

$\displaystyle e_{s}v_{0}^{b,c,\mu}$
$\displaystyle=\sum_{i=s}^{\min\\{a,p+s\\}}(-1)^{i-s}$ $\displaystyle\hskip
42.67912pt\times
CG(\tfrac{b}{2},\tfrac{b}{2}-i+s;\,\tfrac{d+m}{2},\tfrac{d+m}{2}-p-s+i\,|\,\tfrac{b+d+m}{2}-p,\tfrac{b+d+m}{2}-p)\,$
$\displaystyle\hskip 71.13188pt\times
CG(\tfrac{b+m}{2},\tfrac{b+m}{2}-i;\,\tfrac{b}{2},-\tfrac{b}{2}+i-s\,|\,\tfrac{m}{2},\tfrac{m}{2}-s)\;\;v_{i}^{a}\otimes
v_{p+s-i}^{c}$ $\displaystyle=\sum_{i=s}^{\min\\{a,p+s\\}}(-1)^{i-s}$
$\displaystyle\hskip
39.83368pt\times\sqrt{\frac{(b+m+d-2p+1)!\;p!\;(b-i+s)!\;(d+m-p-s+i)!}{(b+m+d-p+1)!\;(b-p)!\;(d+m-p)!\;(i-s)!\;(p+s-i)!}}$
$\displaystyle\hskip
62.59596pt\times\sqrt{\frac{(b+m-i)!\;i!\;(m+1)!\;b!\;}{(b+m+1)!\;(i-s)!\;(b-i+s)!\;(m-s)!\;s!}}\;\;v_{i}^{a}\otimes
v_{p+s-i}^{c}.$

Again, the reader should check that all the numbers under the factorial sign
are non-negative.

Now, for $s=1$, the sum describing $e_{1}v_{0}^{a,d,\mu}$ starts at $i=0$ with
non-zero coefficient, while the sum describing $e_{1}v_{0}^{b,c,\mu}$ starts
at $i=1$. This proves that $\\{e_{1}v_{0}^{a,d,\mu},e_{1}v_{0}^{b,c,\mu}\\}$
is linearly independent and hence there is no possible $\mu$ in $S_{1}$, that
is $S_{1}=0$. This completes the proof in this case.

_The case $Z(a,1)\otimes Z(d,1)^{*}$_. Here $b=a+m$, $c=d+m$ and we first
assume

$a\leq d.$

In this case

$\mu=d-a+2p,\quad 0\leq p\leq a,$

and it follows from (3.16) that

$0\leq i+j=a-p+s.$

Recall that (3.12) says

$e_{s}v_{0}^{a,d,\mu}=\sum_{i,j,k}(-1)^{k}CG(\tfrac{a}{2},\tfrac{a}{2}-i;\,\tfrac{d}{2},\tfrac{d}{2}-k\,|\,\tfrac{d-a}{2}+p,\tfrac{d-a}{2}+p)\,\\\
\times
CG(\tfrac{d+m}{2},\tfrac{d+m}{2}-j;\,\tfrac{d}{2},-\tfrac{d}{2}+k\,|\,\tfrac{m}{2},\tfrac{m}{2}-s)\;\;v_{i}^{a}\otimes
v_{j}^{c}.$

It follows from (3.13) that

$\displaystyle k$ $\displaystyle=a-p-i,$ $\displaystyle j$
$\displaystyle=a-p+s-i,$

and the condition $k\geq 0$ implies $i\leq a-p$. Hence

$\displaystyle e_{s}v_{0}^{a,d,\mu}$
$\displaystyle=\sum_{i=0}^{a-p}(-1)^{a-p-i}CG(\tfrac{a}{2},\tfrac{a}{2}-i;\,\tfrac{d}{2},\tfrac{d}{2}-a+p+i\,|\,\tfrac{d-a}{2}+p,\tfrac{d-a}{2}+p)\,$
$\displaystyle\hskip 36.98866pt\times
CG(\tfrac{d+m}{2},\tfrac{d+m}{2}-a+p-s+i;\,\tfrac{d}{2},-\tfrac{d}{2}+a-p-i\,|\,\tfrac{m}{2},\tfrac{m}{2}-s)$
$\displaystyle\hskip 256.0748pt\times\;v_{i}^{a}\otimes v_{a-p+s-i}^{c}$
$\displaystyle=\sum_{i=0}^{a-p}(-1)^{i}\sqrt{\frac{(d-a+2p+1)!\;(a-p)!\;(a-i)!\;(d-a+p+i)!}{(d+p+1)!\;p!\;(d-a+p)!\;i!\;(a-p-i)!}}$
$\displaystyle\hskip
36.98866pt\times\sqrt{\frac{(d+m-a+p-s+i)!\;(a-p+s-i)!\;(m+1)!\;d!\;}{(d+m+1)!\;(a-p-i)!\;(d-a+p+i)!(m-s)!\;s!}}$
$\displaystyle\hskip 256.0748pt\times\;v_{i}^{a}\otimes v_{a-p+s-i}^{c}.$

At this point, the reader should check that all the numbers under the
factorial sign are non-negative. This last sum is, up to the non-zero scalar

$\sqrt{\frac{1}{s!\;(m-s)!}}\;\sqrt{\frac{(d-a+2p+1)!\;(a-p)!\;(m+1)!\;d!}{(d+p+1)!\;p!\;(d-a+p)!\;(d+m+1)!}},$

equal to

$w^{a,d,\mu}_{s}=\sum_{i=0}^{a-p}(-1)^{i}\sqrt{\frac{(a-i)!\;(d+m-a+p-s+i)!\;(a-p+s-i)!}{i!\;(a-p-i)!^{2}}}\;\;v_{i}^{a}\otimes
v_{a-p-i}^{c}.$

On the other hand, (3.14) says

$e_{s}v_{0}^{b,c,\mu}=\sum_{i,j,k}(-1)^{k}CG(\tfrac{b}{2},\tfrac{b}{2}-k;\,\tfrac{c}{2},\tfrac{c}{2}-j\,|\,\tfrac{\mu}{2},\tfrac{\mu}{2})\,\\\
\times
CG(\tfrac{a}{2},\tfrac{a}{2}-i;\,\tfrac{b}{2},-\tfrac{b}{2}+k\,|\,\tfrac{m}{2},\tfrac{m}{2}-s)\;\;v_{i}^{a}\otimes
v_{j}^{c}$

and it follows from (3.15) that

$\displaystyle j$ $\displaystyle=a-p+s-i,$ $\displaystyle k$
$\displaystyle=m+i-s,$

and the condition $j\geq 0$ implies $i\leq a-p+s$. Thus

$\displaystyle e_{s}v_{0}^{b,c,\mu}$
$\displaystyle=\sum_{i=0}^{\min\\{a,a-p+s\\}}(-1)^{m+i-s}$
$\displaystyle\hskip 28.45274pt\times
CG(\tfrac{a+m}{2},\tfrac{a-m}{2}-i+s;\,\tfrac{d+m}{2},\tfrac{d+m}{2}-a+p-s+i\,|\,\tfrac{d-a}{2}+p,\tfrac{d-a}{2}+p)\,$
$\displaystyle\hskip 56.9055pt\times
CG(\tfrac{a}{2},\tfrac{a}{2}-i;\,\tfrac{a+m}{2},-\tfrac{a-m}{2}+i-s\,|\,\tfrac{m}{2},\tfrac{m}{2}-s)\;\;v_{i}^{a}\otimes
v_{a-p+s-i}^{c}$ $\displaystyle=\sum_{i=0}^{\min\\{a,a-p+s\\}}(-1)^{i}$
$\displaystyle\hskip
34.14322pt\times\sqrt{\frac{(d-a+2p+1)!\,(m+a-p)!\,(a-i+s)!\,(d+m-a+p+i-s)!}{(m+d+p+1)!\,p!\,(d-a+p)!\,(m+i-s)!\,(a-p-i+s)!}}$
$\displaystyle\hskip
56.9055pt\times\sqrt{\frac{(a-i+s)!\;(m+i-s)!\,a!\,(m+1)!\,}{(m+1+a)!\,(a-i)!\,i!\;s!\;(m-s)!}}\;\;v_{i}^{a}\otimes
v_{a-p+s-i}^{c}.$

As above, at this point, the reader should check that all the numbers under
the factorial sign are non-negative. The above sum is, up to the non-zero
scalar

$\sqrt{\frac{1}{s!\;(m-s)!}}\;\sqrt{\frac{(d-a+2p+1)!\,(m+a-p)!\;(m+1)!\,a!}{(m+d+p+1)!\,p!\,(d-a+p)!\;(m+1+a)!}},$

equal to

$w^{b,c,\mu}_{s}=\sum_{i=0}^{\min\\{a,a-p+s\\}}(-1)^{i}\sqrt{\frac{(a-i+s)!^{2}\,(d+m-a+p+i-s)!\;}{(a-p-i+s)!\,(a-i)!\;i!}}\;\;v_{i}^{a}\otimes
v_{a-p+s-i}^{c}.$

If $p=0$ then

$w^{a,d,\mu}_{s}=w^{b,c,\mu}_{s}=\sum_{i=0}^{a}(-1)^{i}\sqrt{\frac{(a-i+s)!\,(d+m-a+i-s)!\;}{(a-i)!\;i!}}\;\;v_{i}^{a}\otimes
v_{a+s-i}^{c}$

for all $s=0,\dots,m$. This shows that

$u_{0}=\sqrt{d+1}\;v_{0}^{a,d,\mu}\;-\sqrt{b+1}\;v_{0}^{b,c,\mu}$

is, indeed, a highest weight vector, of weight $\mu=d-a$, in $S_{1}$.

On the other hand, assume $p\geq 1$. Then, for $s=1$, the sum defining
$w^{b,c,\mu}_{1}$ has the index $i$ running up to $i=a+1-p$ while in the sum
defining $w^{a,d,\mu}_{1}$ the index $i$ only runs up to $i=a-p$. In both
cases, all the coefficients are non-zero, and thus
$\\{w^{a,d,\mu}_{1},w^{b,c,\mu}_{1}\\}$ is linearly independent. This
completes the proof in the case $d\geq a$

We now assume

$a>d.$

In this case

$\mu=a-d+2p,\quad 0\leq p\leq d,$

and it follows from (3.16) that

$0\leq i+j=d-p+s.$

From (3.12) we have

$e_{s}v_{0}^{a,d,\mu}=\sum_{i,j,k}(-1)^{k}CG(\tfrac{a}{2},\tfrac{a}{2}-i;\,\tfrac{d}{2},\tfrac{d}{2}-k\,|\,\tfrac{a-d}{2}+p,\tfrac{a-d}{2}+p)\,\\\
\times
CG(\tfrac{d+m}{2},\tfrac{d+m}{2}-j;\,\tfrac{d}{2},-\tfrac{d}{2}+k\,|\,\tfrac{m}{2},\tfrac{m}{2}-s)\;\;v_{i}^{a}\otimes
v_{j}^{c}.$

It follows from (3.13) that

$\displaystyle j$ $\displaystyle=d-p+s-i,$ $\displaystyle k$
$\displaystyle=d-p-i.$

and the condition $k\geq 0$ implies $i\leq d-p$. Hence

$\displaystyle e_{s}v_{0}^{a,d,\mu}$
$\displaystyle=\sum_{i=0}^{d-p}(-1)^{d-p-i}CG(\tfrac{a}{2},\tfrac{a}{2}-i;\,\tfrac{d}{2},\tfrac{d}{2}-d+p+i\,|\,\tfrac{a-d}{2}+p,\tfrac{a-d}{2}+p)\,$
$\displaystyle\hskip 36.98866pt\times
CG(\tfrac{d+m}{2},\tfrac{d+m}{2}-d+p-s+i;\,\tfrac{d}{2},-\tfrac{d}{2}+d-p-i\,|\,\tfrac{m}{2},\tfrac{m}{2}-s)$
$\displaystyle\hskip 250.38434pt\times\;v_{i}^{a}\otimes v_{d-p+s-i}^{c}$
$\displaystyle=\sum_{i=0}^{d-p}(-1)^{i}\sqrt{\frac{(a-d+2p+1)!\;(d-p)!\;(a-i)!\;(p+i)!}{(a+p+1)!\;p!\;(a-d+p)!\;i!\;(d-p-i)!}}$
$\displaystyle\hskip
48.36958pt\times\sqrt{\frac{(m+p-s+i)!\;(d-p+s-i)!\;(m+1)!\;d!\;}{(d+m+1)!\;(d-p-i)!\;(p+i)!(m-s)!\;s!}}\;\;v_{i}^{a}\otimes
v_{d-p+s-i}^{c}.$

At this point, the reader should check that all the numbers under the
factorial sign are non-negative. This last sum is, up to the non-zero scalar

$\sqrt{\frac{(a-d+2p+1)!\;(d-p)!\;(m+1)!\;d!}{(a+p+1)!\;p!\;(a-d+p)!\;(d+m+1)!\;(m-s)!\;s!}},$

equal to

$w^{a,d,\mu}_{s}=\sum_{i=0}^{d-p}(-1)^{i}\sqrt{\frac{(a-i)!\;(m+p-s+i)!\;(d-p+s-i)!}{i!\;(d-p-i)!^{2}}}\;\;v_{i}^{a}\otimes
v_{d-p+s-i}^{c}.$

On the other hand, recall that (3.14) is

$e_{s}v_{0}^{b,c,\mu}=\sum_{i,j,k}(-1)^{k}CG(\tfrac{b}{2},\tfrac{b}{2}-k;\,\tfrac{c}{2},\tfrac{c}{2}-j\,|\,\tfrac{\mu}{2},\tfrac{\mu}{2})\,\\\
\times
CG(\tfrac{a}{2},\tfrac{a}{2}-i;\,\tfrac{b}{2},-\tfrac{b}{2}+k\,|\,\tfrac{m}{2},\tfrac{m}{2}-s)\;\;v_{i}^{a}\otimes
v_{j}^{c}$

and it follows from (3.15) that

$\displaystyle j$ $\displaystyle=d-p+s-i,$ $\displaystyle k$
$\displaystyle=m+i-s,$

and the condition $j\geq 0$ implies $i\leq d-p+s$. Thus

$\displaystyle e_{s}v_{0}^{b,c,\mu}$
$\displaystyle=\sum_{i=0}^{\min\\{a,d-p+s\\}}(-1)^{m+i-s}$
$\displaystyle\hskip 39.83368pt\times
CG(\tfrac{a+m}{2},\tfrac{a-m}{2}-i+s;\,\tfrac{d+m}{2},\tfrac{m-d}{2}+p-s+i\,|\,\tfrac{a-d}{2}+p,\tfrac{a-d}{2}+p)\,$
$\displaystyle\hskip 56.9055pt\times
CG(\tfrac{a}{2},\tfrac{a}{2}-i;\,\tfrac{a+m}{2},-\tfrac{a-m}{2}+i-s\,|\,\tfrac{m}{2},\tfrac{m}{2}-s)\;\;v_{i}^{a}\otimes
v_{d-p+s-i}^{c}$ $\displaystyle=\sum_{i=0}^{\min\\{a,d-p+s\\}}(-1)^{i}$
$\displaystyle\hskip
39.83368pt\times\sqrt{\frac{(a-d+2p+1)!\,(m+d-p)!\,(a-i+s)!\,(m+p+i-s)!}{(m+a+p+1)!\,p!\,(a-d+p)!\,(m+i-s)!\,(d-p-i+s)!}}$
$\displaystyle\hskip
56.9055pt\times\sqrt{\frac{(a-i+s)!\;(m+i-s)!\,a!\,(m+1)!\,}{(m+1+a)!\,(a-i)!\,i!\;s!\;(m-s)!}}\;\;v_{i}^{a}\otimes
v_{d-p+s-i}^{c}.$

As above, at this point, the reader should check that all the numbers under
the factorial sign are non-negative. The above sum is, up to the non-zero
scalar

$\sqrt{\frac{(a-d+2p+1)!\,(m+d-p)!\;(m+1)!\,a!}{(m+a+p+1)!\,p!\,(a-d+p)!\;(m+1+a)!\,s!\;(m-s)!}},$

equal to

$w^{b,c,\mu}_{s}=\sum_{i=0}^{\min\\{a,d-p+s\\}}(-1)^{i}\sqrt{\frac{(a-i+s)!^{2}\,(m+p+i-s)!}{(d-p-i+s)!\,(a-i)!\;i!}}\;\;v_{i}^{a}\otimes
v_{d-p+s-i}^{c}.$

Since $a>d$, for $s=1$, the sum defining $w^{b,c,\mu}_{1}$ has the index $i$
running up to $i=d+1-p$ while the sum defining $w^{a,d,\mu}_{1}$ has the index
$i$ running only up to $i=d-p$. In both cases, all the coefficients are non-
zero, and thus $\\{w^{a,d,\mu}_{1},w^{b,c,\mu}_{1}\\}$ is linearly
independent. This shows that there is no possible $\mu$ in $S_{1}$ and thus
$S_{1}=0$. This completes the proof in this case.

Since the case $Z(b,1)^{*}\otimes Z(c,1)$ is derived from the case
$Z(a,1)\otimes Z(d,1)^{*}$, we have completed the proof of the theorem. We
warn the reader that in order to obtain the highest weight vector in this case
from the case $Z(a,1)\otimes Z(d,1)^{*}$, it is needed to swap the tensor
factors to go from $v_{0}^{d,a,\mu}$ and $v_{0}^{c,b,\mu}$ to
$v_{0}^{a,d,\mu}$ and $v_{0}^{b,c,\mu}$ respectively. To do this it is needed
(2.7). ∎

######  Theorem 3.5.

Let $V=V(a_{0})\oplus\ldots\oplus V(a_{\ell})$ and
$W=V(b_{0})\oplus\ldots\oplus V(b_{\ell^{\prime}})$ be socle decomposition of
two uniserial $\mathfrak{g}_{m}$-modules of type $Z$. Then

$\text{soc}(V\otimes
W)=\text{soc}(V)\otimes\text{soc}(W)\;\oplus\;\bigoplus_{t=1}^{\min\\{\ell,\ell^{\prime}\\}}S_{t}$

where each $S_{t}$ is as follows:

  1. (i)

For $V=Z(a_{0},\ell)$ and $W=Z(b_{0},\ell^{\prime})$ we have $S_{t}\simeq
V(a_{0}+b_{0}+mt)$ and thus

$\text{soc}(V\otimes
W)\simeq\bigoplus_{k=0}^{\min\\{a_{0},b_{0}\\}}V(a_{0}+b_{0}-2k)\;\;\oplus\;\;\bigoplus_{t=1}^{\min\\{\ell,\ell^{\prime}\\}}V(a_{0}+b_{0}+mt).$

  2. (ii)

For $V=Z(a_{0},\ell)$ and $W=Z(b_{\ell^{\prime}},\ell^{\prime})^{*}$ we have

$S_{t}\simeq\begin{cases}0,&\text{if $a_{0}>b_{t}$};\\\
V(b_{0}-a_{0}-tm),&\text{if $a_{0}\leq b_{t}$};\end{cases}$

(note that $b_{t}=b_{0}-tm$) and thus

$\text{soc}(V\otimes
W)\simeq\begin{cases}\displaystyle\bigoplus_{k=0}^{b_{0}}V(a_{0}-b_{0}+2k),&\text{if
$a_{0}>b_{0}$};\\\\[14.22636pt]
\displaystyle\bigoplus_{k=0}^{a_{0}}V(b_{0}-a_{0}+2k)\;\;\oplus\;\;\bigoplus_{t=1}^{T}V(b_{0}-a_{0}-tm),&\text{if
$a_{0}\leq b_{0}$},\end{cases}$

with
$T=\min\left\\{\ell,\ell^{\prime},\left\lfloor\frac{b_{0}-a_{0}}{m}\right\rfloor\right\\}$.

  3. (iii)

For $V=Z(a_{\ell},\ell)^{*}$ and $W=Z(b_{\ell^{\prime}},\ell^{\prime})^{*}$ we
have $S_{t}=0$ for all $t\geq 1$, and thus

$\text{soc}(V\otimes
W)\simeq\bigoplus_{k=0}^{\min\\{a_{0},b_{0}\\}}V(|a_{0}-b_{0}|+2k).$

In particular, $\text{soc}(V\otimes W)$ is multiplicity free as a
representation of $\mathfrak{sl}(2)$.

######  Remark 3.6.

Items (i), (ii) and (iii) are not mutually exclusive since they have
intersection when $\ell=0$ or $\ell^{\prime}=0$. We can make them exclusive by
requiring $\ell,\ell^{\prime}\geq 1$ in (ii). In that case, the second sum in
$\text{soc}(V\otimes W)$ in (ii) is non-empty if and only if $a_{0}+m\leq
b_{0}$.

######  Remark 3.7.

If $U\simeq V\otimes W$ with $V$ and $W$ as in Theorem 3.5, then the list of
highest weights appearing in the $\mathfrak{sl}(2)$-decomposition of
$\text{soc}(U)$ consists of the union of two sets $A_{2}(U)$ and $A_{m}(U)$
whose elements are in arithmetic progressions with common differences 2 and
$m$ respectively. In all cases

$A_{2}(U)=\\{|a_{0}-b_{0}|,|a_{0}-b_{0}|+2,\dots,a_{0}+b_{0}\\}$

($A_{2}(U)$ consists of all the highest weights of
$\text{soc}(V)\otimes\text{soc}(W)$) and

$A_{m}(U)=\begin{cases}\left\\{a_{0}+b_{0}+m,\dots,a_{0}+b_{0}+\min\\{\ell,\ell^{\prime}\\}m\right\\},&\text{in
case (i) with $\ell,\ell^{\prime}\geq 1$};\\\\[5.69054pt]
\left\\{b_{0}\\!-\\!a_{0}\\!-\\!\min\\{\ell,\ell^{\prime},\lfloor\frac{b_{0}-a_{0}}{m}\rfloor\\}m,\dots,b_{0}\\!-\\!a_{0}\\!-\\!m\right\\},&\text{in
case (ii) with $\ell,\ell^{\prime}\geq 1$}\\\\[-2.84526pt] &\text{and
$b_{0}\geq a_{0}+m$};\\\\[5.69054pt] \emptyset,&\text{ otherwise.}\end{cases}$

Note that $A_{2}(U)$ and $A_{m}(U)$ are disjoint sets and given $A_{2}(U)\cup
A_{m}(U)$, if $m\neq 2$, it is clear how to identify $A_{2}(U)$ and
$A_{m}(U)$.

###### Proof of Theorem 3.5.

It follows from (3.4) and the Clebsch-Gordan formula for the decomposition of
the tensor product of irreducible $\mathfrak{sl}(2)$-modules, that in order to
prove this theorem we only need to study $S_{t}$.

Let us fix $t_{0}>1$ and assume $S_{t_{0}}\neq 0$. We know from Proposition
3.2 that $t_{0}\leq\min\\{\ell,\ell^{\prime}\\}$. We will first show that if
$\mu$ is the weight of a highest weight vector

$u\in S_{t_{0}}=S_{t_{0}}(V,W)=\Big{(}\bigoplus_{i+j=t_{0}}V(a_{i})\otimes
V(b_{j})\Big{)}^{\mathfrak{r}},$

then $\mu$ is as claimed in the theorem.

We have

$u=\bigoplus_{i+j=t_{0}}u_{i,j}$

with $u_{i,j}\in V(a_{i})\otimes V(b_{j})$, then each $u_{i,j}$ must be a
highest weight vector of weight $\mu$. Moreover, since $\mathfrak{r}u=0$ it
follows from (3.2) that

(3.19) $(X\,u_{i,j})_{2}+(X\,u_{i+1,j-1})_{1}=0$

for all $X\in\mathfrak{r}$ and for all $i,j$ such that $i+j=t_{0}$ and
$i+1\leq\ell$ and $j\geq 1$. In other words, let us fix $i_{0},j_{0}$ so that
$i_{0}+j_{0}=t_{0}-1$. In particular $i_{0}<\ell$ and $j_{0}<\ell^{\prime}$.
Now we may consider uniserial subquotients $\tilde{V}$ and $\tilde{W}$ whose
socle decompositions are

$\tilde{V}=V(a_{i_{0}})\oplus V(a_{i_{0}+1})\qquad\tilde{W}=V(b_{j_{0}})\oplus
V(b_{j_{0}+1}).$

Then, (3.19) is equivalent to say that

$u_{i_{0},j_{0}+1}+u_{i_{0}+1,j_{0}}\in S_{1}(\tilde{V},\tilde{W})$

and we know that it is a highest weight vector of weight $\mu$.

We now apply Theorem 3.3 and we obtain that

  1. (i)

If $V=Z(a_{0},\ell)$ and $W=Z(b_{0},\ell^{\prime})$, then $a_{i}=a_{0}+im$ and
$b_{j}=b_{0}+jm$. This implies $V=Z(a_{i_{0}},1)$, $W=Z(b_{j_{0}},1)$ and
hence $S_{1}(\tilde{V},\tilde{W})\simeq V(a_{i_{0}}+b_{j_{0}}+m)$. that is

$\mu=a_{i_{0}}+b_{j_{0}}+m=a_{0}+i_{0}m+b_{0}+j_{0}m+m=a_{0}+b_{0}+t_{0}m.$

  2. (ii)

If $V=Z(a_{0},\ell)$ and $W=Z(b_{0},\ell^{\prime})^{*}$, then $a_{i}=a_{0}+im$
and $b_{j}=b_{0}-jm$. This implies $V=Z(a_{i_{0}},1)$,
$W=Z(b_{j_{0}+1},1)^{*}$ and hence

$S_{1}(\tilde{V},\tilde{W})\simeq\begin{cases}V(b_{j_{0}+1}-a_{i_{0}}),&\text{if
$a_{i_{0}}\leq b_{j_{0}+1}$;}\\\ 0,&\text{if
$a_{i_{0}}>b_{j_{0}+1}$;}\end{cases}$

which is equivalent to

$S_{1}(\tilde{V},\tilde{W})\simeq\begin{cases}V(b_{0}-a_{0}-t_{0}m),&\text{if
$a_{0}+i_{0}m\leq b_{0}-j_{0}m-m$;}\\\ 0,&\text{if
$a_{0}+i_{0}m>b_{0}-j_{0}m-m$.}\end{cases}$

Thus, if $S_{1}(\tilde{V},\tilde{W})\neq 0$ then $\mu=b_{0}-a_{0}-t_{0}m$.

  3. (iii)

If $V=Z(a_{0},\ell)^{*}$ and $W=Z(b_{0},\ell^{\prime})^{*}$, then
$S_{1}(\tilde{V},\tilde{W})=0$.

This completes the first part of the proof, that is if $S_{t_{0}}(V,W)\neq 0$
then the have proved that only highest weight $\mu$ appearing in
$S_{t_{0}}(V,W)$ are as claimed.

Conversely, assume that $\mu$ is a weight claimed to appear in
$S_{t_{0}}(V,W)$. The above analysis shows, in each case, that
$S_{1}(\tilde{V},\tilde{W})$ is isomorphic to $V(\mu)$ for all
$i_{0}+j_{0}=t_{0}-1$. Now we can choose a highest weight vector

$u_{i_{0},j_{0}+1}+u_{i_{0}+1,j_{0}}\in V(a_{i_{0}})\otimes
V(a_{j_{0}+1})\;\oplus\;V(a_{i_{0}+1})\otimes V(b_{j_{0}})\subset
S_{1}(\tilde{V},\tilde{W})$

in a recursive way so that

$\sum_{i=0}^{t_{0}-1}u_{i,t_{0}-i}+u_{i+1,t_{0}-i-1}$

is a highest weight vector of weight $\mu$ in $S_{t_{0}}(V,W)$. This completes
the proof of the theorem. ∎

######  Corollary 3.8.

Let $V=V(a_{0})\oplus\ldots\oplus V(a_{\ell})$ and
$W=V(b_{0})\oplus\ldots\oplus V(b_{\ell^{\prime}})$ be socle decomposition of
two uniserial $\mathfrak{g}_{m}$-modules of type $Z$. Then:

  1. (i)

$\text{soc}(V\otimes W)=\text{soc}(V)\otimes\text{soc}(W)$ if and only if

    1. (a)

$\ell=0$ or $\ell^{\prime}=0$,

    2. (b)

$V=Z(a_{\ell},\ell)^{*}$ and $W=Z(b_{\ell^{\prime}},\ell^{\prime})^{*}$,
$\ell,\ell^{\prime}\geq 0$,

    3. (c)

$V=Z(a_{0},\ell)$ and $W=Z(b_{\ell^{\prime}},\ell^{\prime})^{*}$ with
$b_{0}<a_{0}+m$ and $\ell,\ell^{\prime}\geq 1$,

    4. (d)

$V=Z(a_{\ell},\ell)^{*}$ and $W=Z(b_{0},\ell^{\prime})$ with $a_{0}<b_{0}+m$
and $\ell,\ell^{\prime}\geq 1$.

  2. (ii)

In any of the cases described in (i), the socle length of $V\otimes W$ is
$\ell+\ell^{\prime}+1$ and

$\text{soc}^{t+1}(V\otimes
W)=\sum_{i=0}^{t}\text{soc}^{i+1}(V)\otimes\text{soc}^{t+1-i}(W)=\bigoplus_{0\leq
i+j\leq t}V(a_{i})\otimes V(b_{j})$

as $\mathfrak{sl}(2)$-modules for all $0\leq t\leq\ell+\ell^{\prime}$.

###### Proof.

Part (i) follows at once from Theorem 3.5. Part (ii) is a consequence of Lemma
2.2 applied to the decomposition

$V\otimes W=\bigoplus_{k=1}^{\ell+\ell^{\prime}+1}(V\otimes W)_{k}$

with $(V\otimes W)_{k}=\bigoplus_{i+j=k-1}V(a_{i})\otimes V(b_{j})$. The
hypothesis $\mathfrak{r}V_{k}\subset V_{k-1}$ required by the lemma follows
from (3.2) and since we are in the cases described in (i) we have
$\text{soc}(V\otimes W)=(V\otimes W)_{1}$ as required by the lemma. ∎

##  4\. Applications

Recall from §2.3 that a uniserial $\mathfrak{g}_{m}$-module is of type $Z$ if
it is isomorphic to $Z(a,\ell)$ or $Z(a,\ell)^{*}$ for some non-negative
integers $a$ and $\ell$.

###  4.1. Invariants and intertwining operators

The main goal of this subsection is to obtain the intertwining operators
between two uniserial $\mathfrak{g}_{m}$-modules $V$ and $W$ of type $Z$.

Since $\text{Hom}(V,W)\simeq V^{*}\otimes W$ as $\mathfrak{g}_{m}$-modules, we
have

$\text{Hom}_{\mathfrak{g}_{m}}(V,W)\simeq(V^{*}\otimes W)^{\mathfrak{g}_{m}}.$

In turn, since $\text{soc}(V^{*}\otimes W)=(V^{*}\otimes W)^{\mathfrak{r}}$
(see Lemma 2.1 and (3.3)), it follows that $(V^{*}\otimes
W)^{\mathfrak{g}_{m}}$ is the subspace of $\mathfrak{sl}(2)$-invariant vectors
in $\text{soc}(V^{*}\otimes W)$. Since Theorem 3.5 shows that
$\mathfrak{sl}(2)$-decomposition is multiplicity free, it follows immediately
that $\dim\text{Hom}_{\mathfrak{g}_{m}}(V,W)$ (or $\dim(V^{*}\otimes
W)^{\mathfrak{g}_{m}}$) is either 0 or 1.

The following two corollaries describe exactly in which cases these dimensions
are 1.

######  Corollary 4.1.

Let $V=V(a_{0})\oplus\ldots\oplus V(a_{\ell})$ and
$W=V(b_{0})\oplus\ldots\oplus V(b_{\ell^{\prime}})$ be socle decomposition of
two uniserial $\mathfrak{g}_{m}$-modules of type $Z$. Then

$(V\otimes W)^{\mathfrak{g}_{m}}\neq 0$

if and only if $b_{0}\in\\{a_{0},\ldots,a_{\ell}\\}$ and
$a_{0}\in\\{b_{0},\ldots,b_{\ell^{\prime}}\\}$ and in this case $(V\otimes
W)^{\mathfrak{g}_{m}}$ is 1-dimensional.

###### Proof.

We have to consider the possibilities $V$ and $W$ isomorphic to $Z(c,k)$ or
$Z(c,k)^{*}$.

If either $V=Z(a_{0},\ell)$ and $W=Z(b_{0},\ell^{\prime})$, or
$V=Z(a_{\ell},\ell)^{*}$ and $W=Z(b_{\ell},\ell^{\prime})^{*}$, then Theorem
3.5 implies that the trivial $\mathfrak{sl}(2)$-module appears in
$\text{soc}(V\otimes W)$ if and only if $b_{0}=a_{0}$. In the first and second
cases the sequences $\\{a_{i}\\}$ and $\\{b_{j}\\}$ look like:

$\begin{matrix}a_{0}&a_{0}+m&a_{0}+2m&\ldots\\\\[5.69054pt]
\rotatebox{90.0}{$\,=$}&\rotatebox{90.0}{$\,=$}&\rotatebox{90.0}{$\,=$}&\\\
b_{0}&b_{0}+m&b_{0}+2m&\ldots.\end{matrix}\qquad\text{ or
}\qquad\begin{matrix}\ldots&a_{0}-2m&a_{0}-m&a_{0}\\\\[5.69054pt]
&\rotatebox{90.0}{$\,=$}&\rotatebox{90.0}{$\,=$}&\rotatebox{90.0}{$\,=$}\\\
\ldots&b_{0}-2m&b_{0}-m&b_{0}\end{matrix}$

respectively. In these cases, this precisely coincides with the condition
$b_{0}\in\\{a_{i}:i=0,\dots\ell\\}$ and
$a_{0}\in\\{b_{j}:j=0,\dots\ell^{\prime}\\}$.

If $V=Z(a_{0},\ell)$ and $W=Z(b_{\ell^{\prime}},\ell^{\prime})^{*}$, it
follows from Theorem 3.5 that $\text{soc}(V\otimes W)$ contains the trivial
representation of $\mathfrak{sl}(2)$ if and only if $b_{0}=a_{0}+tm$ with
$0\leq t\leq\min\\{\ell,\ell^{\prime}\\}$. Thus, the sequences $\\{a_{i}\\}$
and $\\{b_{j}\\}$ look like:

$\begin{matrix}&&a_{0}&\ldots&a_{0}+tm&\ldots&a_{0}+\ell m\\\\[5.69054pt]
&&\rotatebox{90.0}{$\,=$}&&\rotatebox{90.0}{$\,=$}&&&&\\\
b_{0}-{\ell^{\prime}}m&\ldots&b_{0}-tm&\ldots&b_{0}.\end{matrix}$

Again, in this case, this precisely coincides with the condition
$b_{0}\in\\{a_{i}:i=0,\dots\ell\\}$ and
$a_{0}\in\\{b_{j}:j=0,\dots\ell^{\prime}\\}$.

The case $V=Z(a_{\ell},\ell)^{*}$ and $W=Z(b_{0},\ell^{\prime})$ is symmetric
to the previous one. ∎

######  Corollary 4.2.

Let $V=V(a_{0})\oplus\ldots\oplus V(a_{\ell})$ and
$W=V(b_{0})\oplus\ldots\oplus V(b_{\ell^{\prime}})$ be socle decomposition of
two uniserial $\mathfrak{g}_{m}$-modules of type $Z$. Then
$\dim\text{Hom}_{\mathfrak{g}_{m}}(V,W)$ is either 0 or 1 and

$\dim\text{Hom}_{\mathfrak{g}_{m}}(V,W)=1$

if and only if $b_{0}\in\\{a_{0},\ldots,a_{\ell}\\}$ and
$a_{\ell}\in\\{b_{0},\ldots,b_{\ell^{\prime}}\\}$, that is, these sets look
like

$\begin{matrix}a_{0}&\ldots&a_{i}&\ldots&a_{\ell}&&\\\\[5.69054pt]
&&\rotatebox{90.0}{$\,=$}&&\rotatebox{90.0}{$\,=$}&&\\\
&&b_{0}&\ldots&b_{j}&\ldots&b_{\ell^{\prime}}.\end{matrix}$

###### Proof.

This is basically a direct consequence of Corollary 4.1. Since

$\text{Hom}_{\mathfrak{g}_{m}}(V,W)\simeq(V^{*}\otimes W)^{\mathfrak{g}_{m}}$

we need to apply Corollary 4.1 to the $\mathfrak{g}_{m}$-modules $V^{*}$ and
$W$ whose socle decomposition are $V^{*}=V(a_{\ell})\oplus\ldots\oplus
V(a_{0})$ and $W=V(b_{0})\oplus\ldots\oplus V(b_{\ell^{\prime}})$
respectively. Therefore $(V^{*}\otimes W)^{\mathfrak{g}_{m}}\neq 0$ if and
only if $b_{0}\in\\{a_{0},\ldots,a_{\ell}\\}$ and
$a_{\ell}\in\\{b_{0},\ldots,b_{\ell^{\prime}}\\}$. ∎

###  4.2. Isomorphisms between tensor products

In this section we use Theorem 3.5 to prove, for $m\neq 2$, that if $U$ is the
tensor product of two uniserial $\mathfrak{g}_{m}$-modules of type $Z$, then
the factors are determined by $U$.

If $U$ is the tensor product of two uniserial $\mathfrak{g}_{m}$-modules of
type $Z$, then so is $U^{*}$. Recall also that, in this case, Theorem 3.5
implies that the list of highest weights appearing in the
$\mathfrak{sl}(2)$-decomposition of $\text{soc}(U)$ consists of the union of
two disjoint sets $A_{2}(U)$ and $A_{m}(U)$ whose elements are in arithmetic
progressions with common differences 2 and $m$ respectively (see Remark 3.7).
Thus we have

$\text{soc}(U)\simeq\bigoplus_{k\in A_{2}(U)}V(k)\;\oplus\;\bigoplus_{k\in
A_{m}(U)}V(k).$

We know that always $A_{2}(U)\neq\emptyset$ but $A_{m}(U)$ might be empty and,
as pointed out in Remark 3.7, both sets can be obtained from $U$ when $m\neq
2$. If $m=2$, we do not know whether it is possible to read off $A_{2}(U)$ and
$A_{m}(U)$ from $U$.

######  Theorem 4.3.

Let $m\neq 2$ and let $U$ be the tensor product of two uniserial
$\mathfrak{g}_{m}$-modules of type $Z$. Then the factors are determined by
$U$. More precisely, let $\Lambda$ be the greatest (highest) weight in $U$ and
let $A_{2}=A_{2}(U)$, $A_{m}=A_{m}(U)$, $A_{2}^{*}=A_{2}(U^{*})$ and
$A_{m}^{*}=A_{m}(U^{*})$. Then:

  1. (i)

Assume that $\max A_{2}^{*}=\Lambda$. Set

$\ell^{\prime}=\begin{cases}\dfrac{\max A_{m}-\max A_{2}}{m},&\text{if
$A_{m}\neq\emptyset$;}\\\ 0,&\text{if
$A_{m}=\emptyset$;}\end{cases}\qquad\ell=\frac{\max A_{2}^{*}-\max
A_{2}}{m}-\ell^{\prime}.$

If $(\ell-\ell^{\prime})m=\min A_{2}^{*}-\min A_{2}$ then set

$a=\frac{\max A_{2}+\min A_{2}}{2},\quad b=\frac{\max A_{2}-\min A_{2}}{2},$

else set

$a=\frac{\max A_{2}-\min A_{2}}{2},\quad b=\frac{\max A_{2}+\min A_{2}}{2}.$

We have $U\simeq Z(a,\ell)\otimes Z(b,\ell^{\prime})$.

  2. (ii)

Assume $\max A_{2}=\Lambda$. Set

$\ell^{\prime}=\begin{cases}\dfrac{\max A_{m}^{*}-\max A_{2}^{*}}{m},&\text{if
$A_{m}^{*}\neq\emptyset$;}\\\ 0,&\text{if
$A_{m}^{*}=\emptyset$;}\end{cases}\qquad\ell=\frac{\max A_{2}-\max
A_{2}^{*}}{m}-\ell^{\prime}.$

If $(\ell-\ell^{\prime})m=\min A_{2}-\min A_{2}^{*}$ then set

$a=\frac{\max A_{2}^{*}+\min A_{2}^{*}}{2},\quad b=\frac{\max A_{2}^{*}-\min
A_{2}^{*}}{2},$

else set

$a=\frac{\max A_{2}^{*}-\min A_{2}^{*}}{2},\quad b=\frac{\max A_{2}^{*}+\min
A_{2}^{*}}{2}.$

We have $U\simeq Z(a,\ell)^{*}\otimes Z(b,\ell^{\prime})^{*}$.

  3. (iii)

Assume that neither $\max A_{2}$ nor $\max A_{2}^{*}$ is $\Lambda$. Set

$\ell^{\prime}=\frac{\Lambda-\max A_{2}^{*}}{m},\qquad\ell=\frac{\Lambda-\max
A_{2}}{m}.$

If $(\ell+\ell^{\prime})m=\min A_{2}^{*}-\min A_{2}$ then set

$a=\frac{\max A_{2}+\min A_{2}}{2},\quad b=\frac{\max A_{2}-\min
A_{2}}{2}-\ell^{\prime}m,$

else set

$a=\frac{\max A_{2}-\min A_{2}}{2},\quad b=\frac{\max A_{2}+\min
A_{2}}{2}-\ell^{\prime}m.$

We have $U\simeq Z(a,\ell)\otimes Z(b,\ell^{\prime})^{*}$.

###### Proof.

We know that $U$ is one of the following possibilities:

$Z(a_{0},\ell_{0})\otimes Z(b_{0},\ell^{\prime}_{0}),\quad
Z(a_{0},\ell_{0})^{*}\otimes Z(b_{0},\ell^{\prime}_{0})^{*},\quad
Z(a_{0},\ell_{0})\otimes Z(b_{0},\ell^{\prime}_{0})^{*}.$

In any case, $\Lambda=a_{0}+b_{0}+(\ell_{0}+\ell^{\prime}_{0})m$.

In order to apply Theorem 3.5, it is convenient to recall that the socle
decompositions of the modules $Z(c,t)$ and $Z(c,t)^{*}$ are

$\displaystyle Z(c,t)$ $\displaystyle=V(c)\oplus V(c+m)\oplus\dots\oplus
V(c+tm),$ $\displaystyle Z(c,t)^{*}$ $\displaystyle=V(c+tm)\oplus
V(c+(t-1)m)\oplus\dots\oplus V(c).$

Thus (see Remark 3.7)

If $U=Z(a_{0},\ell_{0})\otimes Z(b_{0},\ell^{\prime}_{0})$ then

$\displaystyle A_{2}$
$\displaystyle=\\{|a_{0}-b_{0}|,|a_{0}-b_{0}|+2,\dots,a_{0}+b_{0}\\},$
$\displaystyle A_{2}^{*}$
$\displaystyle=\\{|a_{0}\\!+\\!\ell_{0}m-b_{0}\\!-\\!\ell^{\prime}_{0}m|,|a_{0}\\!+\\!\ell_{0}m-b_{0}\\!-\\!\ell^{\prime}_{0}m|+2,\dots,a_{0}\\!+\\!\ell_{0}m+b_{0}\\!+\\!\ell^{\prime}_{0}m\\},$
$\displaystyle A_{m}$
$\displaystyle=\\{a_{0}+b_{0}+m,\dots,a_{0}+b_{0}+\min\\{\ell_{0},\ell^{\prime}_{0}\\}m\\}\text{
if $\ell_{0},\ell^{\prime}_{0}>0$, else $A_{m}=\emptyset$},$ $\displaystyle
A_{m}^{*}$ $\displaystyle=\emptyset.$

If $U=Z(a_{0},\ell_{0})^{*}\otimes Z(b_{0},\ell^{\prime}_{0})^{*}$ then

$\displaystyle A_{2}^{*}$
$\displaystyle=\\{|a_{0}-b_{0}|,|a_{0}-b_{0}|+2,\dots,a_{0}+b_{0}\\},$
$\displaystyle A_{2}$
$\displaystyle=\\{|a_{0}\\!+\\!\ell_{0}m-b_{0}\\!-\\!\ell^{\prime}_{0}m|,|a_{0}\\!+\\!\ell_{0}m-b_{0}\\!-\\!\ell^{\prime}_{0}m|+2,\dots,a_{0}\\!+\\!\ell_{0}m+b_{0}\\!+\\!\ell^{\prime}_{0}m\\},$
$\displaystyle A_{m}^{*}$
$\displaystyle=\\{a_{0}+b_{0}+m,\dots,a_{0}+b_{0}+\min\\{\ell_{0},\ell^{\prime}_{0}\\}m\\}\text{
if $\ell_{0},\ell^{\prime}_{0}>0$, else $A_{m}=\emptyset$},$ $\displaystyle
A_{m}$ $\displaystyle=\emptyset.$

If $U=Z(a_{0},\ell_{0})\otimes Z(b_{0},\ell^{\prime}_{0})^{*}$ and
$\ell_{0},\ell^{\prime}_{0}>0$ then

$\displaystyle A_{2}$
$\displaystyle=\\{|a_{0}-b_{0}-\ell^{\prime}_{0}m|,|a_{0}-b_{0}-\ell^{\prime}_{0}m|+2,\dots,a_{0}+b_{0}+\ell^{\prime}_{0}m\\},$
$\displaystyle A_{2}^{*}$
$\displaystyle=\\{|a_{0}+\ell_{0}m-b_{0}|,|a_{0}+\ell_{0}m-b_{0}|+2,\dots,a_{0}+\ell_{0}m+b_{0}\\},$
$\displaystyle A_{m}$
$\displaystyle=\left\\{b_{0}+\ell^{\prime}_{0}m-a_{0}-tm:t=1,\dots,T\right\\}\text{
if $b_{0}+\ell^{\prime}_{0}m-a_{0}\geq m$, else $A_{m}=\emptyset$},$
$\displaystyle A_{m}^{*}$
$\displaystyle=\left\\{a_{0}+\ell_{0}m-b_{0}-tm:t=1,\dots,T^{*}\right\\}\text{
if $a_{0}+\ell_{0}m-b_{0}\geq m$, else $A_{m}=\emptyset$},$

with

$T=\min\left\\{\ell_{0},\ell^{\prime}_{0},\left\lfloor\tfrac{b_{0}+\ell^{\prime}_{0}m-a_{0}}{m}\right\rfloor\right\\},\qquad
T^{*}=\min\left\\{\ell_{0},\ell^{\prime}_{0},\left\lfloor\tfrac{a_{0}+\ell_{0}m-b_{0}}{m}\right\rfloor\right\\}.$

In what follows we need the following fact: given $x,y\geq 0$ and $z>0$ then

(4.1) $|x-y+z|-|x-y|=z\quad\text{if and only if}\quad x\geq y.$

Now we begin the proof. Suppose first that $\ell_{0}=\ell^{\prime}_{0}=0$.
Then $U\simeq Z(a_{0},0)\otimes Z(b_{0},0)$ and $U$ falls into cases (i) and
(ii). Either (i) or (ii) yield $\ell=\ell^{\prime}=0$ and
$a=\max\\{a_{0},b_{0}\\}$, $b=\min\\{a_{0},b_{0}\\}$, which is correct.

Suppose now that $\ell^{\prime}_{0}=0$ and $\ell_{0}>0$. Then either $U\simeq
Z(a_{0},\ell)\otimes Z(b_{0},0)$ or $U\simeq Z(a_{0},\ell)^{*}\otimes
Z(b_{0},0)$.

If $U\simeq Z(a_{0},\ell)\otimes Z(b_{0},0)$ then $U$ falls only in case (i)
(since $\max A_{2}<\Lambda$) and this yield $\ell^{\prime}=0$,
$\ell=\ell_{0}$. Since

$\min A_{2}=|a_{0}-b_{0}|\quad\text{ and }\quad\min A_{2}^{*}=|a_{0}+\ell
m-b_{0}|$

it follows that $\min A_{2}^{*}-\min A_{2}=\ell m$ if and only if $a_{0}\geq
b_{0}$. In any case, we obtain $a=a_{0}$ and $b=b_{0}$.

Similarly, $U\simeq Z(a_{0},\ell)^{*}\otimes Z(b_{0},0)$ then $U$ falls only
in case (ii) (since $\max A_{2}^{*}<\Lambda$) and this yield
$\ell^{\prime}=0$, $\ell=\ell_{0}$. Now

$\min A_{2}=|a_{0}+\ell m-b_{0}|\quad\text{ and }\quad\min
A_{2}^{*}=|a_{0}-b_{0}|$

and hence $\min A_{2}-\min A_{2}^{*}=\ell m$ if and only if $a_{0}\geq b_{0}$.
In any case, we obtain $a=a_{0}$ and $b=b_{0}$.

We finally suppose that $\ell_{0},\ell^{\prime}_{0}>0$. Then cases (i), (ii),
(iii) correspond exactly (without superposition) to the cases when $U$ is
isomorphic to $Z(a_{0},\ell_{0})\otimes Z(b_{0},\ell^{\prime}_{0})$, or
$Z(a_{0},\ell_{0})^{*}\otimes Z(b_{0},\ell^{\prime}_{0})^{*}$, or
$Z(a_{0},\ell_{0})\otimes Z(b_{0},\ell^{\prime}_{0})^{*}$ respectively.

In cases (i) and (ii) we obtain $\ell=\max\\{\ell_{0},\ell^{\prime}_{0}\\}$,
$\ell^{\prime}=\min\\{\ell_{0},\ell^{\prime}_{0}\\}$. We need to show that
$a_{0}$ and $b_{0}$ are obtained correctly.

In case (i) we have to distinguish four cases:

$\begin{array}[]{l}\ell_{0}\geq\ell^{\prime}_{0}\\\ a_{0}\geq
b_{0}\end{array},\qquad\begin{array}[]{l}\ell_{0}\geq\ell^{\prime}_{0}\\\
a_{0}\leq
b_{0}\end{array},\qquad\begin{array}[]{l}\ell_{0}\leq\ell^{\prime}_{0}\\\
a_{0}\geq
b_{0}\end{array},\qquad\begin{array}[]{l}\ell_{0}\leq\ell^{\prime}_{0}\\\
a_{0}\leq b_{0}\end{array}.$

Let us consider, for instance, the third case. We obtain
$\ell=\ell^{\prime}_{0}$ and $\ell^{\prime}=\ell_{0}$. If $a_{0}=b_{0}$ it is
clear that the result will be correct. Otherwise, $b_{0}<a_{0}$ and since

$\min A_{2}=|b_{0}-a_{0}|\quad\text{ and }\quad\min
A_{2}^{*}=|b_{0}-a_{0}+(\ell^{\prime}_{0}-\ell_{0})m|$

it follows from (4.1) that $(\ell-\ell^{\prime})m\neq\min A_{2}^{*}-\min
A_{2}$ and hence $a=b_{0}$ and $b=b_{0}$, which is correct. All the other
three cases work similarly.

The case (ii) is analogous to the case (i).

In case (iii) we obtain $\ell=\ell_{0}$ and $\ell^{\prime}=\ell^{\prime}_{0}$.
Now, since

$\min A_{2}=|a_{0}-b_{0}-\ell^{\prime}m|\quad\text{ and }\quad\min
A_{2}^{*}=|a_{0}-b_{0}+\ell
m|=|a_{0}-b_{0}-\ell^{\prime}m+(\ell+\ell^{\prime})m|$

it follows from (4.1) that $(\ell+\ell^{\prime})m=\min A_{2}^{*}-\min A_{2}$
if and only if $a_{0}\geq b_{0}+\ell^{\prime}m$. In either case we obtain
$a=a_{0}$ and $b=b_{0}$, which is correct.

This completes the proof. ∎

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# Human-Machine Co-Creativity with Older Adults – A Learning Community to
Study Explainable Dialogues

Marianne Bossema<EMAIL_ADDRESS>0000-0002-0622-0552 Amsterdam University of
Applied Sciences, Leiden UniversityThe Netherlands ,  Rob Saunders  Leiden
UniversityThe Netherlands and  Somaya Ben Allouch  Amsterdam University of
Applied Sciences, University of AmsterdamThe Netherlands

(2023; 2 May 2023; 25 May 2023)

###### Abstract.

This position paper is part of a long-term research project on human-machine
co-creativity with older adults. The goal is to investigate how robots and AI-
generated content can contribute to older adults’ creative experiences, with a
focus on collaborative drawing and painting. The research has recently
started, and current activities are centred around literature studies,
interviews with seniors and artists, and developing initial prototypes. In
addition, a course “Drawing with Robots”, is being developed to establish
collaboration between human and machine learners: older adults, artists,
students, researchers, and artificial agents. We present this course as a
learning community and as an opportunity for studying how explainable AI and
creative dialogues can be intertwined in human-machine co-creativity with
older adults.

Computational Creativity, XAI, Gerontechnology

††copyright: acmcopyright††journalyear: 2023††doi:
XXXXXXX.XXXXXXX††conference:  ACM Creativity & Cognition; June 19-21, 2023;
††ccs: Human-centered computing††ccs: Human-centered computing Interaction
design theory, concepts and paradigms††ccs: Human-centered computing
Participatory design

Reference:
Marianne Bossema, Rob Saunders, and Somaya Ben Allouch. 2023. In the 1st
International Workshop on Explainable AI for the Arts (XAIxArts), ACM
Creativity and Cognition (C&C) 2023. Online, 3 pages.

##  1\. Introduction

The global population is rapidly aging. According to the United Nations
(United Nations, 2022), the number of people aged 60 years or older is
expected to more than double by 2050, reaching approximately 2.1 billion. This
has significant social, economic, and health implications. Based on a review
of 900 studies, the World Health Organization concluded that engaging in
creative activities can promote health and well-being, and help prevent and
slow age-related physical and cognitive decline (Fancourt and Finn, 2019).
Social robots offer unique opportunities to support creativity through
assistance and social interaction. In addition, advancements in generative AI
bring new possibilities for creative collaborations between humans and
machines. There are many open questions, however, about how to design human-
machine co-creativity (HMCC) for and with the target group of older adults.

When studying the design of technology, there are several motivators for
involving users throughout the research (Fischer et al., 2020). In general,
this allows for gaining insights into user needs and perspectives, and for
collecting feedback on ideas and prototypes. In our case, it helps to derive
grounded user-centered design principles for HMCC with older adults. It can
also contribute to their empowerment by eliciting curiosity and positive
feelings, increasing familiarity and trust with technology, and enriching
creative experiences.

HMCC systems are “mixed-initiative creative interfaces”, involving humans and
computers in an interactive loop where each produces, evaluates, modifies, and
selects creative outputs in response to the other (Deterding et al., 2017). An
example is “Drawing Apprentice”, a co-creative system that collaborates with
users in real-time abstract drawing. HMCC emerges through dialogic interaction
across multiple modalities including language, drawing, music, etc. Ethical
considerations, such as transparency, bias, and power relations, must be taken
into account from the beginning of the design research process, when
investigating these dialogues.

##  2\. Human-centered XAI for co-creativity

Current methods for XAI may not always offer explanations that are easily
understandable to humans (Angelov et al., 2021). Instead, they often provide
only superficial insights into the “black box” of AI models, through post hoc
hints about specific features or local information. This approach differs
significantly from human reasoning, which involves holistic comparisons of
items, rather than focusing on individual features (Angelov et al., 2021).
Visualizations can contribute to XAI. For example, saliency maps highlight the
important features or inputs that contributed to a particular decision,
helping users understand the decision-making process of an AI system for that
specific case (Samek et al., 2017). Interactive visualizations allow users to
manipulate the inputs or parameters of the model and visualize the resulting
changes, helping them to gain insights into what can be expected, and build
trust. The effectiveness of explanation, however, lies in the perception of
the person receiving the explanation (Liao and Varshney, 2021). In HMCC with
visual artifacts, both visualizations and interactivity can contribute to XAI,
and can be designed to meet the needs for explainability of non-experts, such
as older adults.

In creative collaborations, dialogues are essential to develop understanding
and common ground, either language-based or through creative artifacts. A
dialogic approach for computational creativity, as suggested by Bown et al.
(Bown et al., 2020), enables both human and artificial agents to actively
influence the creative process and products, and adapt to the other’s
behavior. HMCC studies with robots sometimes have used speech, e.g. for
scaffolding creative tasks, and prompting creative reflection (Ali et al.,
2021; Hubbard et al., 2021). Both interactive visualisations and language-
based (speech) dialogues may contribute to explainable HMCC for the target
group of older adults.

Llano et al. (Llano et al., 2022) propose Explainable Computational Creativity
(XCC) to develop systems that can effectively communicate and explain
processes, decisions, and ideas throughout the creative process. They suggest
two-way communication channels that promote co-creation, understandable to
both humans and machines. For example, an artificial agent engaged in
collaborative drawing, may ask questions about a human-made sketch and use the
answers to adapt and make new suggestions, while the human interaction partner
may interact with, and select suggested artifacts and thereby manipulate the
parameters of the model. This type of bi-directional dialogue can promote
reflexivity as being an important part of XAI and at the same time contribute
to the creative process.

##  3\. A learning community exploring dialogues for explainable co-creativity

We present a “Drawing with Robots” course for older adults as a learning
community. The goal is to support a mutual, collaborative learning process for
both human and artificial agents. In human-machine dialogues for co-
creativity, elements of XAI can be explored, implemented, evaluated, and
improved. Research in psychology showed that when children view a task as a
performance situation, they tend to be more risk-averse and less persistent
than children who view the same task as a learning situation. Those with a
learning frame engage in more experimentation and are more likely to try out
new strategies (Edmondson, 2003). This framing of a learning situation may
also be beneficial for older adults. Research has shown that compared to
younger people, seniors are more likely to avoid trial-and-error strategies,
and rely on existing knowledge(Sakaki et al., 2018; Romero et al., 2012). This
depends on conditions, however, such as the efforts required to set goals,
take action, and evaluate results (Oudeyer et al., 2016). The setting of a
learning community and the framing of an artificial agent as one of the
learners can engage older adults to adopt new strategies, ask for
explanations, and share perspectives.

In this learning community, agile design and development processes can help to
improve the deployed technology iteratively. Older adults can develop an
understanding of technological possibilities, allowing them to better think
along and provide feedback. This will also offer insights into shifting needs
for explainability, when users become more familiar with the technology over
time. Overall, this project will contribute to a deeper understanding of how
dialogues, either language-based or through visual artifacts, can support
explainable HMCC with older adults. It also informs how to implement XAI
strategies for this often overlooked target group. This will allow the
research field to gain further insights in how dialogues can be designed to
support XAI in creative processes.

###### Acknowledgements.

This publication is part of the project ‘Social robotics and generative AI to
support and enhance creative experiences for older adults’, with project
number 023.019.021 of the research program Doctoral Grant for Teachers
financed by the Dutch Research Council (NWO).

## References

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