\documentclass[ALCO, Unicode]{cedram} \usepackage{makecell,amscd,url,bm} \usepackage{float,blkarray} \usepackage{tikz} \usepackage{tikz-cd} \newcommand{\mysharp}{\#} \makeindex \newcommand{\conjeq}{\stackrel{?}{=}} \begin{DefTralics} \newcommand{\GL}{\mathrm{GL}} \newcommand{\calU}{{\mathcal{U}}} \newcommand{\F}{\mathbb{F}} \end{DefTralics} \newcommand{\GL}{\mathrm{GL}} \newcommand{\SL}{\mathrm{SL}} \newcommand{\PGL}{\mathrm{PGL}} \newcommand{\calG}{\mathcal{G}} \newcommand{\calT}{\mathcal{T}} \newcommand{\calF}{\mathcal{F}} \newcommand{\bF}{\bm{\mathcal{F}}} \newcommand{\B}{\mathrm{B}} \newcommand{\calP}{\mathcal{P}} \newcommand{\bv}{{\mathbf{v}}} \newcommand{\bu}{\mathbf{u}} \newcommand{\be}{{\mathbf{e}}} \newcommand{\C}{{\bf C}} \newcommand{\Z}{{\mathbb{Z}}} \newcommand{\N}{{\mathbb{N}}} \newcommand{\muhat}{{\boldsymbol \mu}} \newcommand{\alphahat}{{\boldsymbol \alpha}} \newcommand{\tauhat}{{\boldsymbol \tau}} \newcommand{\omegahat}{{\boldsymbol \omega}} \newcommand{\calU}{{\mathcal{U}}} \newcommand{\x}{{\bf x}} \DeclareMathOperator{\Tr}{Tr} \DeclareMathOperator{\Ind}{Ind} \DeclareMathOperator*{\bigboxtimes}{\mathlarger{\mathlarger{\mathlarger\boxtimes}}} \newcommand{\calX}{\mathcal{X}} \newcommand{\F}{\mathbb{F}} \newcommand{\Q}{\overline{\mathbb{Q}}_\ell} \newcommand{\calC}{\mathcal{C}} \newcommand{\gl}{\mathrm{gl}} %% User definitions. %% User definitions MUST use \newcommand. %% The commands \def, \renewcommand and \let are not allowed, as they may overwrite definitions in ALCO's style file. %% Note that definitions are forbidden inside titles, abstracts or the bibliography. %\newcommand{\la}{\longrightarrow} % ... %% The title of the paper: amsart's syntax. \title [An analogue of the Saxl conjecture] {The Saxl conjecture and the tensor square of unipotent characters of \texorpdfstring{$\mathrm{GL}_n(q)$}{GL\_n(q)}} %% Authors according to amsart's syntax + distinction between Given %% and Proper names: \author{ \firstname{Emmanuel} \lastname{Letellier}} \address{Universit\'e Paris Cit\'e\\ IMJ-PRG\\ CNRS\\ 45 Rue des Saints-Pères\\ Paris 75006 (France)} \email{emmanuel.letellier@imj-prg.fr} \author{\firstname{GyeongHyeon} \lastname{Nam}} \address{Ajou University\\ Department of Mathematics\\ 206 World cup-ro\\ Suwon 16499 (Republic of Korea)} \email{ghnam@ajou.ac.kr} % Keywords and phrases: \keywords{tensor product, unipotent character, Kronecker coefficient, symmetric function} %% Mathematical classification (2020) %% See https://mathscinet.ams.org/mathscinet/msc/msc2020.html %% This will not be printed but can be useful for database search \subjclass{20C33, 20C30, 05E05} \begin{document} %% Abstracts must be placed before \maketitle \begin{abstract} We know from \cite{L1} that if for some triple of partitions $(\lambda,\mu,\nu)$ of $n$ the Kronecker coefficient $\langle \chi^\lambda\otimes\chi^\mu,\chi^\nu\rangle$ is non-zero then the corresponding multiplicity $\langle\calU^\lambda\otimes\calU^\mu,\calU^\nu\rangle$ for the unipotent characters of $\GL_n(\F_q)$ is also non-zero. A conjecture of Saxl says that if $\mu$ is a staircase partition, then all irreducible characters of $S_{|\mu|}$ appear non-trivially in the tensor square $\chi^\mu\otimes\chi^\mu$. Therefore the Saxl conjecture implies its analogue for unipotent characters, i.e. all unipotent characters of $\GL_{|\mu|}(\F_q)$ appear non-trivially in the tensor square $\calU^\mu\otimes\calU^\mu$ when $\mu$ is a staircase partition. In this paper we prove the analogue of the Saxl conjecture for unipotent characters. In a second part we describe conjecturally the set of all partitions $\mu$ for which the tensor square $\calU^\mu\otimes\calU^\mu$ contains non-trivially all the unipotent characters of $\GL_{|\mu|}(\F_q)$. \end{abstract} \maketitle %% Theorems should be defined using amsthm syntax %% The commented out theorem styles are already defined, please do not redefine them. \theoremstyle{plain} %\newtheorem{theo}{Theorem}[section] %\newtheorem{conj}[theo]{Conjecture} %\newtheorem{coro}[theo]{Corollary} %\newtheorem{lemm}[theo]{Lemma} %\newtheorem{prop}[theo]{Proposition} \newtheorem{axio}[cdrthm]{Axiom} \theoremstyle{definition} %\newtheorem{defi}{Definition}[section] \theoremstyle{remark} %\newtheorem{rema}{Remark}[section] %\newtheorem*{nota*}{Notation} \section{Introduction}\label{intro} For a partition $\mu$ of $n$, we let $\chi^\mu$ denote the corresponding irreducible character of the symmetric group $S_n$ and we let $\calU^\mu$ denote the corresponding unipotent character of $\GL_n(\F_q)$ (see \S \ref{unipotent}). If $\mu=(n)$ then $\chi^\mu$ and $\calU^\mu$ are the trivial characters and if $\mu=(1^n)$, then $\chi^\mu$ is the sign character and $\calU^\mu$ is the Steinberg character ${\rm St}$. Given a triple of partitions $\muhat=(\mu^1,\mu^2,\mu^3)$ of $n$ we consider $$ g_\muhat:=\langle\chi^{\mu^1}\otimes\chi^{\mu^2}\otimes\chi^{\mu^3},1\rangle_{S_n},\hspace{.5cm}U_\muhat(q):=\left\langle\calU^{\mu^1}\otimes\calU^{\mu^2}\otimes\calU^{\mu^3},1\right\rangle_{\GL_n(\F_q)}.$$ The first one is a non-negative integer known as a \emph{Kronecker coefficient} and the second one is a polynomial in $q$. One of the most challenging problem in algebraic combinatorics (going back to Murnaghan in 1938, cf. \cite{Mu}) is to describe combinatorially the set $$ \{\muhat=(\mu^1,\mu^2,\mu^3)\,|\, g_\muhat\neq 0\}.$$ The analogous problem for $U_\muhat(q)$ is relatively new. As far as we know, it was first investigated in \cite{HLM}. Since then, substantial progress was made \cite{Lu1}\cite{L1}\cite{L2}\cite{HSTZ}\cite{S}. In \cite{L1}, it is shown that the two problems are somehow related, namely it is proved that if $g_\muhat$ is non-zero then $U_\muhat(q)$ is also non-zero (the converse is false). A 10 years old conjecture due to Saxl states that if $\mu^1=\mu^2$ is a staircase partition then $g_\muhat$ is non-zero for any partition $\mu^3$. The main theorem of this paper is the following theorem (as predicted by the Saxl conjecture and the results of \cite{L1}), which is proved in \S\ref{s:mainthm}. \begin{theo}[Analogue of the Saxl conjecture for unipotent characters] If $\mu^1=\mu^2$ is a staircase partition, then $U_\muhat(q)\neq 0$ for any partition $\mu^3$. \label{theointro}\end{theo} In a second part we are interested in the unipotent characters whose tensor-square contains all the unipotent characters. By the above theorem, the unipotent characters attached to the staircase partitions do satisfy this property. By \cite{HSTZ}\cite{L1}, we know that the Steinberg characters ${\rm St}$ also satisfy this property. We make the following conjecture (see Conjecture \ref{conj}). \begin{conj} Let $\mu=(\mu_1,\mu_2,\dots)$ be a partition of $n$. (1) If $\mu_1\leq \lceil n/2\rceil$, then for any partition $\tau$ of $n$ we have $$ U_{(\mu,\mu,\tau)}(q)\neq 0. $$ (2) If $\mu_1>\lceil n/2\rceil$, then $$ U_{(\mu,\mu,(1^n))}(q)= 0. $$ \label{conj0}\end{conj} We verify this conjecture for $n\leq 8$ using Mattig's experimental data \cite{Lu}. Theorem \ref{theointro} is also an evidence for this conjecture as the staircase partitions do satisfy the condition on $\mu_1$. The results of this paper and \cite{L1} suggests that Conjecture \ref{conj0} could be reduced to a statement on Kronecker coefficients (see \S\ref{tensor}). However we could not find a precise reasonable conjectural statement on Kronecker coefficients that would imply Conjecture \ref{conj0}. \section{Preliminaries} We denote by $\calP$ the set of all partitions and for a non-negative integer $n$, we let $\calP_n$ be the subset of partitions of size $n$. For a partition $\lambda=(\lambda_1,\lambda_2,\dots,\lambda_r)$ with $\lambda_1\geq\lambda_2\geq\cdots\geq\lambda_r$ we denote by $\ell(\lambda)=r$ its length and by $|\lambda|=\lambda_1+\cdots+\lambda_r$ its size. \subsection{Roots of star-shaped graphs}\label{ss:roots} Let us start by explaining briefly why root systems appear in our context. Crawley-Boevey \cite{CB} associated to any $k$-tuple $\mathcal{O}=(\mathcal{O}_1,\dots,\mathcal{O}_k)$ of adjoint orbits of $\gl_n(K)$ (with $K$ an algebraically closed field) a star-shaped graph $\Gamma_{\mathcal{O}}$ with $k$ branches (whose lengths are determined by the Jordan type of the orbits). Using the theory of quiver representations, he found a very nice solution in terms of the root system of $\Gamma_{\mathcal{O}}$ of the Deligne-Simpson problem which is to determine for which $k$-tuples of adjoint orbits $\mathcal{O}$ the following equation has a solution $$ X_1+\cdots+X_k=0,\hspace{1cm}X_1\in\mathcal{O}_1,\dots, X_k\in\mathcal{O}_k. $$ In \cite{hausel-letellier-villegas2}, we use Fourier transforms to link the counting (over finite fields) of the number of solutions of the above equation with the multiplicity of the trivial character of $\GL_n(\F_q)$ in a tensor product of irreducible characters of $\GL_n(\F_q)$. We use this in \cite{L1} to study tensor products of unipotent characters of $\GL_n(\F_q)$. We now recall some definitions for an arbitrary graph (see \cite[Chapter 5]{kac} for more details). Assume given a finite graph $\Gamma=(I,\Omega)$ where $I$ is the set of vertices and $\Omega$ the set of edges. We assume that $\Gamma$ has no loops. For $i\in I$ we let $\be_i$ be the element of $\Z^I$ defined as $$ (\be_i)_j=\delta_{i,j} $$ Denote by ${\mathfrak C}=(c_{ij})_{i,j\in I}$ the Cartan matrix of $\Gamma$, i.e. $$ c_{ij}=\begin{cases}2 &\text{ if } i=j \\ -n_{ij}&\text{ otherwise}\end{cases} $$ where $n_{ij}$ is the number of edges between the vertices $i$ and $j$. The Cartan matrix defines a symmetric bilinear form $(\,,\,)$ on $\Z^I$ by $$ (\be_i,\be_j)=c_{ij}. $$ For $i\in I$, we have the fundamental reflection $s_i:\Z^I\rightarrow\Z^I$ $$ s_i(\bv)=\bv-(\bv,\be_i)\be_i,\hspace{1cm}\bv\in\Z^I. $$ If $n_{ij}$ is at most $1$ (which will be our case), then the $i$-th coordinate of $s_i(\bv)$ equals $$ (\sum_j v_j)-v_i $$ where $j$ runs over the vertices which are connected to $i$ by an edge, and the $j$-th coordinate of $s_i(\bv)$, for $j\neq i$, remains unchanged (i.e. $(s_i(\bv))_j=v_j$). We say that $\bv\in\Z^I$ is a \emph{dimension vector} if the coordinates of $\bv$ are all non-negative. \begin{exam}Consider the graph {\footnotesize{ \begin{center} \begin{tikzpicture}[scale=.4, mycirc/.style={circle,fill=black!,inner sep=2pt, minimum size=.1 cm} ] \node[mycirc,label=45:{},label=45:{$4$} ] (n1) at (0,0) {}; \node[mycirc,label=45:{$2$} ] (n2) at (5,0) {}; \node[mycirc,label=45:{$2$} ] (n3) at (-5,0) {}; \node[mycirc,label=45:{$1$} ] (n4) at (0,-5) {}; \draw (n1) -- (n2) -- (n3); \draw(n1)--(n4); \end{tikzpicture} \end{center}}} \noindent with vector dimension $\bv$ whose coordinates are as indicated on the graph. Then if $s_0$ denotes the reflection at the central vertex, then the coordinate of $s_0(\bv)$ at the central vertex is $2+2+1-4=1$ and the coordinates at the other vertices remain unchanged. \label{exD4}\end{exam} The Weyl group $W$ of $\Gamma$ is the subgroup of automorphisms $\Z^I\rightarrow\Z^I$ generated by the fundamental reflections $\{s_i\,|\, i\in I\}$. A vector $\bv\in\Z^I$ is called a \emph{real root} if $\bv=w(\be_i)$ for some $i\in I$ and $w\in W$. \begin{exam}The dimension vector {\footnotesize{ \begin{center} \begin{tikzpicture}[scale=.4, mycirc/.style={circle,fill=black!,inner sep=2pt, minimum size=.1 cm} ] \node[mycirc,label=45:{},label=45:{$2$} ] (n1) at (0,0) {}; \node[mycirc,label=45:{$1$} ] (n2) at (5,0) {}; % \node[mycirc,label=45:{$0$} ] (m1) at (10,0) {}; \node[mycirc,label=45:{$1$} ] (n3) at (-5,0) {}; % \node[mycirc,label=45:{$0$} ] (m2) at (-10,0) {}; \node[mycirc,label=45:{$1$} ] (n4) at (0,-5) {}; %\node[mycirc,label=45:{$0$} ] (m3) at (0,-10) {}; \draw (n1) -- (n2) -- (n3); \draw(n1)--(n4); \end{tikzpicture} \end{center}}} \noindent is a real root as it can be obtained from $\be_0$ (where $0$ is the labeling of the central vertex) by applying first to $\be_0$ the reflections at the other vertices to get the dimension vector with coordinate $1$ everywhere and then by applying the reflection $s_0$. \end{exam} The set of \emph{fundamental imaginary} roots $M$ is defined as the subset of $\bv\in\Z^I\backslash\{0\}$ with connected support such that for all $i\in I$ we have $$ (\be_i,\bv)\leq 0. $$ In the next proposition we will give a more explicit necessary and sufficient condition for a dimension vector to be in $M$ in the case of star-shaped graph. Recall that an imaginary root is a vector which is of the form $w(\bv)$ or $w(-\bv)$ for some $\bv\in M$ and $w\in W$. A root is said to be \emph{positive} if its coordinates are all non-negative. An imaginary positive root is of the form $w(\delta)$ with $\delta\in M$. The Weyl group $W$ preserves thus the set of positive imaginary roots. From now we assume that $\Gamma$ is a star-shaped graph with $3$ legs as follows \begin{center} \begin{tikzpicture}[scale=.4, mycirc/.style={circle,fill=black!,inner sep=2pt, minimum size=.1 cm} ] \node[mycirc, label=220:{$0$}] (n) at (-20,0) {}; \node[mycirc, label=315:{$[1,1]$}] (n1) at (-16,5) {}; \node[mycirc ,label=315:{$[1,2]$}] (n2) at (-10,5) {}; \node[circle] (n3) at (-5,5) {$\cdots$}; \node[mycirc ,label=315:{$[1,r_1-1]$}] (n4) at (0,5) {}; \node[mycirc ,label=315:{$[1,r_1]$}] (n5) at (6,5) {}; %%%%%%%%%%%%%%%5 \node[mycirc, label=315:{$[2,1]$}] (m1) at (-16,0) {}; \node[mycirc ,label=315:{$[2,2]$}] (m2) at (-10,0) {}; \node[circle] (m3) at (-5,0) {$\cdots$}; \node[mycirc ,label=315:{$[2,r_2-1]$}] (m4) at (0,0) {}; \node[mycirc ,label=315:{$[2,r_2]$}] (m5) at (6,0) {}; %%%%%%%%%%%%%%%5 \node[mycirc, label=315:{$[3,1]$}] (k1) at (-16,-5) {}; \node[mycirc ,label=315:{$[3,2]$}] (k2) at (-10,-5) {}; \node[circle] (k3) at (-5,-5) {$\cdots$}; \node[mycirc ,label=315:{$[3,r_3-1]$}] (k4) at (0,-5) {}; \node[mycirc ,label=315:{$[3,r_3]$}] (k5) at (6,-5) {}; \draw (n)--(n1) -- (n2) -- (n3) --(n4) --(n5); \draw (n)--(m1) -- (m2) -- (m3) --(m4) --(m5); \draw (n)--(k1) -- (k2) -- (k3) --(k4) --(k5); \end{tikzpicture} \end{center} \noindent where $I=\{0\}\cup\{[i,j]\,|\, 1\leq i\leq 3,\, 1\leq j\leq r_i\}$. The reflection $s_0$ at the central vertex $0$ acts on $\bv=(v_i)_{i\in I}\in\Z^I$ as $$ s_0(\bv)_i=\begin{cases}v_{[1,1]}+v_{[2,1]}+v_{[3,1]}-v_0&\text{ if } i=0,\\ v_i&\text{ otherwise}\end{cases} $$ and the other reflections $s_{[i,j]}$ act on $\bv$ as $$ s_{[i,j]}(\bv)_r=\begin{cases}v_{[i,j-1]}+v_{[i,j+1]}-v_{[i,j]}&\text{ if } r=[i,j]\\ v_r&\text{otherwise}.\end{cases} $$ with $v_{[i,r_i+1]}=0$. For $\bv=(v_i)_i\in\Z^I$ define the integer \begin{equation}\label{eq:delta} \delta(\bv):=v_0-\sum_{i=1}^3(v_0-v_{[i,1]}). \end{equation} For a triple $\muhat=(\mu^1,\mu^2,\mu^3)$ of partitions of $n$ of length respectively $r_1+1, r_2+1,r_3+1$, we let $\bv_\muhat$ be the dimension vector of $\Gamma$ with coordinate $n$ at the central vertex $0$ and $n-\sum_{s=1}^j\mu^i_j$ at the vertex $[i,j]$. For example if $\muhat=((1^3),(1^3),(1^3))$, then the coordinates of $\bv_\muhat$ are as indicated on the following graph. {\footnotesize{ \begin{center} \begin{tikzpicture}[scale=.4, mycirc/.style={circle,fill=black!,inner sep=2pt, minimum size=.1 cm} ] \node[mycirc, label=120:{$3$}] (n) at (-20,0) {}; \node[mycirc, label=45:{$2$}] (n1) at (-15,5) {}; \node[mycirc ,label=45:{$1$}] (n2) at (-10,5) {}; % \node[mycirc ,label=45:{$0$}] (n3) at (-5,5) {}; % \node[circle] (n3) at (-5,5) {$\cdots$}; %%%%%%%%%%%%%%%5 \node[mycirc, label=45:{$2$}] (m1) at (-15,0) {}; \node[mycirc ,label=45:{$1$}] (m2) at (-10,0) {}; % \node[mycirc ,label=45:{$0$}] (m3) at (-5,0) {}; % \node[circle] (m3) at (-5,0) {$\cdots$}; %%%%%%%%%%%%%%%5 \node[mycirc, label=45:{$2$}] (k1) at (-15,-5) {}; \node[mycirc ,label=45:{$1$}] (k2) at (-10,-5) {}; % \node[mycirc ,label=45:{$0$}] (k3) at (-5,-5) {}; % \node[circle] (k3) at (-5,-5) {$\cdots$}; \draw (n)--(n1) -- (n2); %--(n3) ; \draw (n)--(m1) -- (m2);%--(m3) ; \draw (n)--(k1) -- (k2) ;%--(k3) ; \end{tikzpicture} \end{center}}} We have the following proposition. \begin{prop}\label{prop:imaginaryroot}\cite[Proposition 20]{L1} A vector is in $M$ if and only if it is the form $\bv_\muhat$ for some triple of partitions $\muhat$ and \begin{equation} \delta(\bv_\muhat)=n-\mu^1_1-\mu^2_1-\mu^3_1\geq 0. \label{eq:d}\end{equation} \end{prop} Let $\N^I$ be the set of dimension vectors of $\Gamma$ and let $(\N^I)^*$ be the subset of dimension vector with non-increasing coordinates along each leg, and let $(\N^I)^{**}\subset (\N^I)^*$ be the subset of dimension vectors of the form $\bv_\muhat$. We have the following result. \begin{prop}\ \begin{itemize} \item[(i)] The subgroup $H$ of $W$ generated by the set of reflections $\{s_i\,|\, i\in I\backslash\{0\}\}$ preserves $(\N^I)^*$. \item[(ii)] For any $\bv\in(\N^I)^*$, there exists a dimension vector $\bu\in(\N^I)^{**}$ such that $\bv=h(\bu)$ for some $h\in H$. More precisely for each $i=1,2,3$, the sequence $$ \sigma^i(\bu):=(u_0-u_{[i,1]}, u_{[i,1]}-u_{[i,2]},\dots, u_{[i,r_i-1]}-u_{[i,r_i]}, u_{[i,r_i]}) $$ is the non-increasing sequence obtained by re-ordering the coordinates of $\sigma^i(\bv)$. \end{itemize} \label{ordering}\end{prop} \begin{proof} \noindent(i) It is sufficient to show that $s_{[i,j]}((\N^I)^*)=(\N^I)^*$ for any $[i,j]$. Note that $\bv\in (\N^I)^*$ if and only if the coordinates of $ \sigma^i(\bv)$ are non-negative for all $i$. Furthermore, from the definition of $s_{[i,j]}$, we see that the coordinates of $\sigma^i(s_{[i,j]}(\bv))$ are obtained from those of $\sigma^i(\bv)$ by permuting the $j$-th coordinate with the $(j+1)$-th coordinate, and that $\sigma^k(s_{[i,j]}(\bv))=\sigma^k(\bv)$ for $k\neq i$, hence the result. \noindent(ii) As the effect of the reflection $s_{[i,j]}$ on the sequence $\sigma^i(\bv)$ is to permute the $j$-th coordinate with the $(j+1)$-th coordinate, we can obtain a vector dimension $\bu$ such that the sequences $\sigma^i(\bu)$, with $i=1,2,3$, are partitions. \end{proof} For a triple $\muhat=(\mu^1,\mu^2,\mu^3)\in\calP_n^3$, we denote by $\Gamma_\muhat$ the graph with $3$ legs as above and with $r_i:=\ell(\mu^i)-1$. Notice that $\bv_\muhat$ is then a dimension vector of $\Gamma_\muhat$. \subsection{Representations of the symmetric group and symmetric functions}\label{symmetric-section} For a partition $\mu$ of size $n$ we denote by $\chi^\mu$ the corresponding irreducible character of the symmetric group $S_n$ and by $\chi^\mu_\lambda$ its value at an element of cycle-type $\lambda$. The trivial character is $\chi^{(n)}$ and the sign character $\chi^{(1^n)}$. We denote by $R_n$ the $\Z$-module generated by all irreducible characters of $S_n$ and we consider the ring $$ R=\bigoplus_{n=0}^{+\infty}R_n $$ with $R_0=\Z$ and with product defined by $$ f_n\cdot f_m={\rm Ind}_{S_n\times S_m}^{S_{n+m}}(f_n\times f_m). $$ Then $R$ is a commutative, associative, graded ring with an identity element. It is equipped with the usual scalar product $\left\langle\,,\,\right\rangle$ making the basis of irreducible characters an orthonormal basis. Let $\x=\{x_1,x_2,\dots\}$ be an infinite set of variables and let $\Lambda=\Lambda(\x)$ be the ring of symmetric functions. For a partition $\mu=(\mu_1,\mu_2,\dots)$ we let $$ p_\mu=p_\mu(\x)=(x_1^{\mu_1}+x_2^{\mu_1}+\cdots)(x_1^{\mu_2}+x_2^{\mu_2}+\cdots)\cdots $$ be the corresponding power sum symmetric function. It is equipped with a scalar product (Hall pairing) such that $$ \left\langle p_\lambda,p_\mu\right\rangle=\delta_{\lambda\mu}z_\lambda $$ where $z_\lambda$ denotes the size of the centralizer in $S_{|\lambda|}$ of an element of $S_{|\lambda|}$ of cycle-type $\lambda$. The \emph{Frobenius characteristic map} \cite[Chap. I, \S 7]{macdonald} is a ring isomorphism ${\rm ch}:R\rightarrow\Lambda$ defined by $$ {\rm ch}(f)=\sum_{|\mu|=n}z_\mu^{-1}f_\mu p_\mu, $$ for any $f\in R_n$ where $f_\mu$ denotes the value of $f$ at the conjugacy class of cycle-type $\mu$. Moreover, ${\rm ch}$ is an isometry with respect to the scalar products. Recall also that the Frobenius characteristic map of the irreducible character $\chi^\mu$ is the Schur symmetric function $s_\mu=s_\mu(\x)$. Recall that the base change matrix between the two base $\{s_\mu\}_\mu$ and $\{p_\mu\}_\mu$ is given by the character table $\{\chi^\lambda_\mu\}$ of symmetric groups, more precisely \begin{equation} s_\lambda=\sum_{|\mu|=|\lambda|}\chi^\lambda_\mu\frac{p_\mu}{z_\mu}. \label{Sym}\end{equation} Given two partitions $\lambda$ and $\mu$ respectively of size $n$ and $m$, and a partition $\nu$ of $n+m$, the Littlewood-Richardson coefficient $c_{\lambda\mu}^\nu$ is defined by \begin{align} c_{\lambda\mu}^\nu:&=\left\langle \chi^\nu,{\rm Ind}_{S_n\times S_m}^{S_{n+m}}(\chi^\lambda\boxtimes\chi^\mu)\right\rangle\label{LRdef}\\ &=\left\langle s_\nu,s_\lambda s_\mu\right\rangle.\label{LR1} \end{align} Choose a total order on the set of all partitions and denote by ${\bf T}^o$ the set of non-increasing sequences of partitions $\omega^o=\omega^1\omega^2\cdots\omega^r$. The size of $\omega^o$ is defined as $$ |\omega^o|=\sum_{i=1}^r|\omega^i| $$ and we denote by ${\bf T}^o_n$ the elements of size $n$. It will be convenient to write the elements of ${\bf T}^o$ in the form $(\omega^1)^{n_1}(\omega^2)^{n_2}\cdots(\omega^s)^{n_s}$ with $\omega^1>\omega^2>\cdots>\omega^s$. We will need the following generalization of the Littlewood-Richardson coefficients: For $\omega^o=(\omega^1)^{n_1}(\omega^2)^{n_2}\cdots(\omega^s)^{n_s}\in{\bf T}^o_n$ and for a partition $\nu$ of $n$, we generalize (\ref{LRdef}) as $$ c_{\omega^o}^\nu:=\left\langle \chi^\nu,{\rm Ind}_{S_{\omega^o}}^{S_n}\left((\chi^{\omega^1})^{\boxtimes n_1}\boxtimes\cdots\boxtimes(\chi^{\omega^s})^{\boxtimes n_s}\right)\right\rangle $$ where $$ S_{\omega^o}=\prod_{i=1}^s(S_{|\omega^i|})^{n_i}\subset S_n. $$ For a partition $\mu$ we denote by $V_\mu$ an irreducible representation of $S_{|\mu|}$ affording the character $\chi^\mu$ and we put $$ V_{\omega^o}=\bigotimes _{i=1}^s (V_{\omega^i})^{\otimes n_i}. $$ This is an $S_{\omega^o}$-module. For a partition $\mu$ of $|\omega^o|$ we have \begin{align*} c_{\omega^o}^\mu&={\rm dim}\,{\rm Hom}_{S_n}\left(V_\mu,{\rm Ind}_{S_{\omega^o}}^{S_n}(V_{\omega_o})\right)\\ &={\rm dim}\,{\rm Hom}_{S_{\omega^o}}\left(V_\mu,V_{\omega_o}\right). \end{align*} The last equality is by Frobenius reciprocity. Now the group $$ W_{\omega^o}:=\prod_{i=1}^s S_{n_i} $$ acts on the space $V_{\omega^o}$ by the permutation action of $S_{n_i}$ on $(V_{\alpha^i})^{\otimes n_i}$ and can be regarded as a subgroup of the normalizer of $S_{\omega^o}$ in $S_n$ (which is the semi-direct product $S_{\omega^o}\rtimes W_{\omega^o}$). Therefore it acts on the space $$ \mathbb{C}_{\omega^o}^\mu:={\rm Hom}_{S_{\omega^o}}\left(V_\mu,V_{\omega_o}\right) $$ as $$ w\cdot f(v):=w\cdot f(w^{-1}\cdot v) $$ for $f\in\mathbb{C}_{\omega^o}^\mu$, $v\in V_{\omega^o}$ and $w\in W_{\omega^o}$. We now explain how to express the term $$ \Tr\left(w\,,\, \mathbb{C}_{\omega^o}^\mu\right) $$ with $w\in W_{\omega^o}$ in terms of Schur functions. We consider a total ordering on the set of pairs $(d,\lambda)$ where $d$ is a strictly positive integer and where $\lambda$ is a non-zero partition. Define ${\bf T}$ to be the set of \emph{types}, namely the set of sequences $$ (d_1,\omega^1)^{m_1}(d_2,\omega^2)^{m_2}\cdots(d_s,\omega^s)^{m_s} $$ with $(d_1,\omega^1)>(d_2,\omega^2)>\cdots>(d_s,\omega^s)$. The size of a type $\omega=\{(d_i,\omega^i)^{n_i}\}_i$ is $$ |\omega|:=\sum_{i=1}^sn_id_i|\omega^i| $$ and we denote by ${\bf T}_n$ the set of types of size $n$. Given a family $\{f_\lambda\}_\lambda$ of symmetric functions indexed by partitions, we define for any type $\omega=\{(d_i,\omega^i)^{n_i}\}_i$, a symmetric function $$ f_\omega:=\prod_if_{\omega^i}(\x^{d_i})^{n_i} $$ where $\x^d$ stands for the set $\{x_1^d,x_2^d,\dots\}$. Notice that giving $\omega^o=(\omega^1)^{n_1}\cdots(\omega^s)^{n_s}\in{\bf T}^o_n$ and a conjugacy class of $W_{\omega^o}$ (or equivalently a partition of $n_i$ for all $i$) defines a type $\omega\in{\bf T}_n$. \begin{exam} If $\omega^o=(1^2)^4(3^1)^6\in{\bf T}^o_{26}$, then $W_{\omega^o}=S_4\times S_6$ and if we choose the partitions $(2,2)$ and $(3,2,1)$, then the associated type is $$ \omega=(2,(1^2))^2(3,(3^1))(2,(3^1))(1,(3^1))\in{\bf T}_{26}. $$ \end{exam} We have the following proposition which generalizes Formula (\ref{LR1}). \begin{prop}\cite[Proposition 6.2.5]{L2} Let $\omega^o=(\omega^1)^{n_1}\cdots(\omega^s)^{n_s}\in{\bf T}^o_n$ and $w\in W_{\omega^o}$, then $$ \Tr\left(w\,,\, \mathbb{C}_{\omega^o}^\mu\right)=\left\langle s_\mu, s_\omega\right\rangle $$ where $\omega$ is the type defined from the pair $(\omega^o,w)$. \label{Prop13}\end{prop} \begin{exam}\label{ex} Consider $\omega^o=(1^2)(1)^2$ and $\mu=(3,1)$. Then $W_{\omega^o}=S_2$ since the unique partition $(1)$ of $1$ has multiplicity $2$. Using the above proposition, we wish to compute $$ \Tr\left(\sigma, \mathbb{C}^{(3,1)}_{(1^2)(1)^2}\right) $$ where $\sigma$ is the non-trivial element of $S_2$. Notice that the type $\omega$ associated to $(\omega^o,\sigma)$ is $\omega=(1,(1^2))(2,1)$ and so \begin{align*} s_\omega&=s_{(1^2)}(\x)s_{(1)}(\x^2)\\ &=s_{(1^2)}p_2(\x). \end{align*} Using Formula (\ref{Sym}) to express $p_2$ in terms of Schur functions together with Littlewood-Richardson coefficients, we compute \begin{align*} \left\langle s_{(3,1)}(\x),s_\omega\right\rangle&=\left\langle s_{(3,1)}(\x),s_{(1^2)}(\x)s_{(2)}(\x)-s_{(1^2)}(\x)s_{(1^2)}(\x) \right\rangle\\ \\ &= \left\langle s_{(3,1)}(\x),s_{(3,1)}(\x)-s_{(2^2)}(\x)-s_{(1^4)}(\x)\right\rangle=1. \end{align*} This shows that $\Tr\left(\sigma, \mathbb{C}^{(3,1)}_{(1^2)(1)^2}\right)=1$, equivalently, $\mathbb{C}^{(3,1)}_{(1^2)(1)^2}$ is the trivial representation. \end{exam} \subsection{Unipotent characters of \texorpdfstring{$\GL_n(\F_q)$}{GL\_n(F\_q)}}\label{unipotent} Denote by $B$ the subgroup of $\GL_n$ of upper triangular matrices and denote by $\mathbb{C}[\GL_n(\F_q)/B(\F_q)]$ the $\mathbb{C}$-vector space with basis the set $\GL_n(\F_q)/B(\F_q)$. The group $\GL_n(\F_q)$ acts by left multiplication on the latter set and so acts on $\mathbb{C}[\GL_n(\F_q)/B(\F_q)]$. The \emph{unipotent representations} of $\GL_n(\F_q)$ are defined as the irreducible constituents of $\mathbb{C}[\GL_n(\F_q)/B(\F_q)]$ and they are naturally parametrized by the irreducible characters of $S_n$ and so by the partitions of $n$. We will denote by $\calU^\mu$ the unipotent character corresponding to the partition $\mu$. Then $$ \calU^{(n^1)}={\rm Id}_{\GL_n(\F_q)},\hspace{1cm}\calU^{(1^n)}={\rm St} $$ where ${\rm St}$ denotes the Steinberg character of $\GL_n(\F_q)$. Given a triple $\muhat=(\mu^1,\mu^2,\mu^3)$ partitions of size $n$, we consider the multiplicities $$ g_\muhat=\left\langle \chi^{\mu^1}\otimes\chi^{\mu^2}\otimes\chi^{\mu^3},1\right\rangle_{S_n},\hspace{1cm} U_\muhat(q):=\left\langle \calU^{\mu^1}\otimes\calU^{\mu^2}\otimes\calU^{\mu^3},1\right\rangle_{\GL_n(\F_q)}. $$ Recall that the first one is a non-negative integer known as a \emph{Kronecker coefficient} while the second one is a polynomial in $q$ with non-negative integer coefficients (see \cite[Theorem 4]{L1}). \begin{theo}\cite[Proposition 6]{L1} If $g_\muhat\neq 0$ then $U_\muhat(q)\neq 0$. In fact the term $g_\muhat$ contributes to the constant term of $U_\muhat(q)$ (i.e. $g_\muhat\leq U_\muhat(0)$). \label{L1-1}\end{theo} The converse of the above theorem is false. For instance if $\muhat=((1^n),(1^n),(1^n))$, then $g_\muhat=0$ but $U_\muhat(q)\neq 0$ (see \cite[\S 3.6]{L1} for $n=3$). \subsection{Generalities on tensor products of unipotent characters} We consider the set $\overline{\calP}$ of triples of partitions of the same size and we repeat what we did in \S \ref{symmetric-section} with $\overline{\calP}$ instead of $\calP$. Namely we choose a total ordering on $\overline{\calP}$ and we denote by $\overline{\mathbf{T}}_n^o$ the set of sequences $\omegahat^o=(\omegahat^1)^{n_1}(\omegahat^2)^{n_2}\cdots(\omegahat^s)^{n_s}$ of elements of $\overline{\calP}$ with $\omegahat^1>\cdots>\omegahat^s$ such that $\sum_in_i|\omega^i|=n$. Define $$ {\mathbf{S}}_{\omegahat^o}:=\prod_{i=1}^s({\mathbf{S}}_{|\omegahat^i|})^{n_i}\subset {\mathbf{S}}_n,\hspace{1cm}{\mathbf{V}}_{\omega^o}:=\bigotimes_{i=1}^s({\mathbf{V}}_{\omegahat^i})^{\otimes n_i},\hspace{1cm}W_{\omegahat^o}:=\prod_{i=1}^s S_{n_i} $$ where for $n\in\N^*$ and $\muhat=(\mu^1,\mu^2,\mu^3)\in\overline{\calP}$ we put $$ {\mathbf{S}}_n=S_n\times S_n\times S_n,\hspace{1cm}{\mathbf{V}}_\muhat=V_{\mu^1}\boxtimes V_{\mu^2}\boxtimes V_{\mu^3}. $$ Define the $W_{\omegahat^o}$-module $$ \C^\muhat_{\omegahat^o}:={\rm Hom}_{{\mathbf{S}}_{\omegahat^o}}\left({\mathbf{V}}_{\omegahat^o},{\mathbf{V}}_\muhat\right). $$ \begin{rema}Notice that we have a natural map $\overline{{\mathbf{T}}}_n^o\rightarrow ({\mathbf{T}}_n^o)^3, \omegahat^o\mapsto (\omega^o_1,\omega^o_2,\omega^o_3)$ where $\omega^o_i$ is obtained by mapping each $\omegahat^j\in\overline{\calP}$ in $\omegahat^o=(\omegahat^1)^{n_1}\cdots(\omegahat^s)^{n_s}$ to its $i$-th coordinate. Notice that the multiplicities of partitions in $\omega_i^o$ may be larger than $n_1,n_2,\dots,n_s$ : For instance if we take $\omegahat^o=((1^2),(2^1),(2^1))((2^1),(1^2),(2^1))$, then $\omega_1^o=\omega_2^o=(1^2)(2^1)$ and $\omega^o_3=(2^1)^2$. As vector spaces, we have $$ \C^\muhat_{\omegahat^o}=\bigotimes_{i=1}^3\mathbb{C}^{\mu^i}_{\omega_i ^o} $$ where $\muhat=(\mu^1,\mu^2,\mu^3)$. Via the diagonal embedding of $W_{\omegahat^o}$ in $W_{\omega^o_1}\times W_{\omega^o_2}\times W_{\omega^o_3}$, this is an isomorphism of $W_{\omegahat^o}$-modules. \end{rema} We have the following result. \begin{theo} \cite[Corollary 5]{L1} Let $\muhat=(\mu^1,\mu^2,\mu^3)\in\overline{\calP}$ be of size $n$ (i.e. $|\mu^1|=|\mu^2|=|\mu^3|=n$). Assume that there exists $\omegahat^o=(\omegahat^1)^{n_1}\cdots(\omegahat^s)^{n_s}\in\overline{\mathbf{T}}_n^o$ such that the two following conditions are satisfied: \begin{itemize} \item[(1)]The dimension vector $\bv_{\omegahat^i}$ is a root of $\Gamma_{\omegahat^i}$ for all $i=1,\dots,s$, \item[(2)] $\left\langle \C^\muhat_{\omegahat^o},1\right\rangle_{W_{\omegahat^o}}\neq 0$. \end{itemize} Then $U_\muhat(q)\neq 0$. \label{L1-3}\end{theo} \begin{lemm}Theorem \ref{L1-3} implies Theorem \ref{L1-1}. \end{lemm} \begin{proof} If we take $\omegahat^o=((1),(1),(1))^n$, then the condition (1) of the above theorem is satisfied as the graph is the Dynkin diagram $A_1$ with dimension vector $1$ at the unique vertex. Moreover ${\mathbf{S}}_{\omegahat^o}=({\mathbf{S}}_1)^n$ is the trivial subgroup of ${\mathbf{S}}_n$, $W_{\omegahat^o}=S_n$ and ${\mathbf{V}}_{\omegahat^o}=\mathbb{C}$. Therefore for $\muhat=(\mu^1,\mu^2,\mu^3)$ of size $n$, $$ \C^\muhat_{\omegahat^o}={\rm Hom}(\mathbb{C},V_{\mu^1}\boxtimes V_{\mu^2}\boxtimes V_{\mu^3})=V_{\mu^1}\boxtimes V_{\mu^2}\boxtimes V_{\mu^3} $$ as $S_n$-modules (for the diagonal action). Therefore $\left\langle \C^\muhat_{\omegahat^o},1\right\rangle_{W_{\omegahat^o}}=g_\muhat$ and so if it is non-zero then so is $U_\muhat(q)$ by Theorem \ref{L1-3}. \end{proof} \begin{rema}Theorem \ref{L1-3} also implies that if $\bv_\muhat$ is a root then $U_\muhat(q)\neq 0$ as in this case we can take $\muhat$ for $\omegahat^o$. \label{L1-2}\end{rema} \begin{exam}Let us give an example where the condition of Theorem \ref{L1-3} is satisfied but $\bv_\muhat$ is not a root and $g_\muhat= 0$. Moreover this example will be used later. Consider $$ \muhat=((2^2),(2^2),(3,1)). $$ Then $(\Gamma_\muhat,\bv_\muhat)$ is as follows {\footnotesize{ \begin{center} \begin{tikzpicture}[scale=.4, mycirc/.style={circle,fill=black!,inner sep=2pt, minimum size=.1 cm} ] \node[mycirc,label=45:{},label=45:{$4$} ] (n1) at (0,0) {}; \node[mycirc,label=45:{$2$} ] (n2) at (5,0) {}; % \node[mycirc,label=45:{$0$} ] (m1) at (10,0) {}; \node[mycirc,label=45:{$2$} ] (n3) at (-5,0) {}; % \node[mycirc,label=45:{$0$} ] (m2) at (-10,0) {}; \node[mycirc,label=45:{$1$} ] (n4) at (0,-5) {}; %\node[mycirc,label=45:{$0$} ] (m3) at (0,-10) {}; \draw (n1) -- (n2) -- (n3); \draw(n1)--(n4); \end{tikzpicture} \end{center}}} We can check that the corresponding Kronecker coefficient $g_\muhat$ vanishes (and so the condition of Theorem \ref{L1-1} is not satisfied) and that $\bv_\muhat$ is not a root (by applying first the central reflection to $\bv_\muhat$ and then reflections at other vertices). Let us see that the condition of Theorem \ref{L1-3} is satisfied. Take $\omegahat^o=\omegahat^1(\omegahat^2)^2$ with $\omegahat^1=((1^2),(1^2),(1^2))$ and $\omegahat^2=((1),(1),(1))$. Then $\bv_{\alpha^1}$ is the longest root of the Dynkin diagram $D_4$ and $\omegahat^2$ is the unique positive root of $A_1$. We have $$ \mathbb{C}_{(1^2)(1)^2}^{(2^2)}=\mathbb{C},\hspace{1cm}\mathbb{C}_{(1^2)(1)^2}^{(3,1)}=\mathbb{C}. $$ We need to compute the action of $W_{\omegahat^o}=S_2$ on these two spaces. Denote by $\sigma$ the non-trivial element of $S_2$, by Example \ref{ex} we have $$ {\rm Tr}\left(\sigma\,,\, \mathbb{C}^{(3,1)}_{(1^2)(1)^2}\right)=1. $$ Therefore the action of $S_2$ on $\mathbb{C}_{(1^2)(1)^2}^{(3,1)}$ is trivial and so $S_2$ acts trivially on $$ \C_{\omegahat^o}^\muhat=\mathbb{C}_{(1^2)(1)^2}^{(2^2)}\otimes\mathbb{C}_{(1^2)(1)^2}^{(2^2)}\otimes\mathbb{C}_{(1^2)(1)^2}^{(3,1)} $$ as the tensor square of $\mathbb{C}_{(1^2)(1)^2}^{(2^2)}$ must be the trivial $S_2$-module (in fact a calculation shows that $\mathbb{C}_{(1^2)(1)^2}^{(2^2)}$ is the sign representation of $S_2$). We thus deduce that $$ \left\langle \C^\muhat_{\omegahat^o},1\right\rangle_{W_{\omegahat^o}}=1. $$ The condition of Theorem \ref{L1-3} is thus satisfied and so we have $U_\muhat(q)\neq 0$. The result of \cite{L1} is more precise and tells us that $$ U_\muhat(q)=\left\langle \C^\muhat_{\omegahat^o},1\right\rangle_{W_{\omegahat^o}}=1 $$ which is consistent with the experimental data. \label{example}\end{exam} \section{An analogue of the Saxl conjecture for unipotent characters}\label{s:mainthm} Fix a positive integer $d$, put $n=\sum_{i=1}^d i$ and consider the partition $\xi_d=(d,d-1, \ldots , 1)$ of $n$. Recall that the Saxl conjecture says that for any partition $\tau$ of $n$ we have $$ g_{(\xi_d,\xi_d,\tau)}\neq 0. $$ The Saxl conjecture implies the following theorem which is the main theorem of this paper. \begin{theo}\label{thm:analoguesaxl} For any partition $\tau$ of $n$ we have $$ U_{(\xi_d,\xi_d,\tau)}(q)\neq 0. $$ \label{maintheo}\end{theo} Let us choose a partition $\tau=(\tau_1,\tau_2, \ldots, \tau_s)$ of $n$. The proof of this theorem is divided into three cases : (1) $\tau_1\leq n-d$ and $d\geq 7$, (2) $\tau_1 \leq n-d$ and $d<6$, (3) $\tau_1>n-d$. To ease the notation we will denote the multipartition $(\xi_d,\xi_d,\tau)$ by $\tauhat$. \subsection{The case \texorpdfstring{$\tau_1\leq n-d$ and $d \geq 7$}{tau\_1 ≤ n-d and d ≥ 7}} The aim of this section is to prove the following result. \begin{prop} If $\tau_1\leq n-d$ and $d\geq 7$ then the dimension vector $\bv_\tauhat$ is an imaginary root of $\Gamma_\tauhat$. \label{theo1}\end{prop} Together with Remark \ref{L1-2}, this implies that Theorem \ref{maintheo} is true when $\tau_1\leq n-d$ and $d\geq 7$. \begin{lemm} If $\tau_1\leq n-2d$, then $\bv_\tauhat$ is a fundamental imaginary root of $\Gamma_\tauhat$. \end{lemm} \begin{proof}From Equation \eqref{eq:d}, we have $\delta(\bv_\tauhat)=n-(2d+\tau_1)$. If $\tau_1\leq n-2d$, then $$ \delta(\bv_\tauhat)=n-(2d+\tau_1)\geq 0 $$ from which we deduce that $\bv_\tauhat$ is a fundamental imaginary root by Proposition \ref{prop:imaginaryroot}. \end{proof} To prove Proposition \ref{theo1} we are thus reduced to prove the following result: \begin{prop}If $n-2d<\tau_1\leq n-d$ and $d\geq 7$ then the dimension vector $\bv_\tauhat$ is an imaginary root of $\Gamma_\tauhat$. \label{theo1'}\end{prop} \begin{proof}Let us first explain the strategy of the proof. We assume that $$ n-2d< \tau_1\leq n-d $$ and we write $\tau_1=n-2d+k$ for some integer $1\leq k\leq d$. Notice that $$ \delta(\bv_\tauhat)=n-2d-\tau_1=-k $$ and so $\bv_\tauhat$ is not a fundamental imaginary root. To see that $\bv_\tauhat$ is an imaginary root, we will construct an element $w\in W$ such that $$ \delta(w(\bv_\tauhat))\geq 0, $$ and $w(\bv_\tauhat)$ has the form $\bv_\muhat$ i.e. such that $w(\bv_\tauhat)$ is a fundamental imaginary root. As $W$ preserves positive imaginary roots, this will prove that $\bv_\tauhat$ is a positive imaginary root. To construct such a $w$ we will use the following process. We apply the central reflection $s_0$ to $\bv_\muhat$. This will have the good effect of increasing the $\delta$ value but will have the wrong effect of creating a dimension vector which is not in the form $\bv_\muhat$, for some $\muhat$, anymore. Therefore we will apply reflections at other vertices to get a dimension vector of the form $\bv_\muhat$ using Proposition \ref{ordering}(ii). At the end of the first step of the process we will have a dimension vector $\bv_\tauhat^{(1)}$ of the form $\bv_\muhat$ such that $\delta(\bv_\tauhat^{(1)})>\delta(\bv_\tauhat)$. If $\delta(\bv_\tauhat^{(1)})<0$, we repeat the process with $\bv_\tauhat^{(1)}$ and we keep going until we get a dimension vector $\bv_\tauhat^{(m)}$ with a non-negative $\delta$. We now explain this in detail. We first assume that $d\geq 8$. Notice that $(\Gamma_\tauhat,\bv_\tauhat)$ is as follows \begin{center} \begin{tikzpicture}[scale=.4, mycirc/.style={circle,fill=black!,inner sep=2pt, minimum size=.1 cm} ] \node[mycirc, label=120:{$\substack{n}$}] (n) at (-25,0) {}; \node[mycirc, label=45:{$\substack{n-d}$}] (n1) at (-19,5) {}; \node[mycirc, label=45:{$\substack{n-2d+1}$}](n2) at (-15,5) {}; \node[mycirc, label=45:{$\substack{n-3d+3}$}] (n6) at (-10,5) {}; \node[circle] (n3) at (-5,5) {$\cdots$}; \node[mycirc ,label=45:{$\substack{1}$ }] (n4) at (0,5) {}; % \node[mycirc ,label=45:{$\substack{0}$}] (n5) at (6,5) {}; %%%%%%%%%%%%%%%5 \node[mycirc, label=45:{$\substack{n-d}$}] (m1) at (-19,0) {}; \node[mycirc, label=45:{$\substack{n-2d+1}$}] (m2) at (-15,0) {}; \node[mycirc, label=45:{$\substack{n-3d+3}$}] (m6) at (-10,0) {}; \node[circle] (m3) at (-5,0) {$\cdots$}; \node[mycirc ,label=45:{$\substack{1}$ }] (m4) at (0,0) {}; % \node[mycirc ,label=45:{$\substack{0}$}] (m5) at (6,0) {}; %%%%%%%%%%%%%%%5 \node[mycirc, label=45:{$\substack{2d-k}$}] (k1) at (-19,-5) {}; \node[mycirc ,label=45:{$\substack{2d-k-\tau_2} $}] (k2) at (-15,-5) {}; \node[mycirc, label=45:{$\substack{2d-k-\tau_2-\tau_3}$}] (k6) at (-10,-5) {}; \node[circle] (k3) at (-5,-5) {$\cdots$}; \node[mycirc ,label=45:{$\substack{\tau_r}$}] (k4) at (0,-5) {}; % \node[mycirc ,label=45:{$\substack{0}$}] (k5) at (6,-5) {}; \draw (n1) -- node[below]{$\substack{\textcolor{red}{d-1}}$} (n2) -- node[below]{$\substack{\textcolor{red}{d-2}}$} (n6)--node[below]{$\substack{\textcolor{red}{d-3}}$} (n3) --node[below]{$\substack{\textcolor{red}{2}}$} (n4);% --(n5); \draw (m1) --node[below]{$\substack{\textcolor{red}{d-1}}$} (m2) -- node[below]{$\substack{\textcolor{red}{d-2}}$} (m6)--node[below]{$\substack{\textcolor{red}{d-3}}$} (m3) --node[below]{$\substack{\textcolor{red}{2}}$} (m4);% --(m5); \draw (k1) -- node[below]{$\substack{\textcolor{red}{\tau_2}}$} (k2) --node[below]{$\substack{\textcolor{red}{\tau_3}}$}(k6)-- node[below]{$\substack{\textcolor{red}{\tau_4}}$}(k3) --node[below]{$\substack{\textcolor{red}{\tau_{r-1}}}$}(k4);% --(k5); \draw (n) -- node[below]{$\substack{\textcolor{red}{d}}$} (n1); \draw (n) -- node[below] {$\substack{\textcolor{red}{d}}$} (m1); \draw (n) -- node[below] {$\substack{\textcolor{red}{n-2d+k}}$} (k1); \end{tikzpicture} \end{center} where the black color integers indicate the coordinates of $\bv_\tauhat$ and the red color integers on the edges denote the successive differences between the coordinates of $\bv_\tauhat$, i.e. $\sigma^i(\bv_\tauhat)$. We now apply the central reflection $s_0$ and we get $(\Gamma_\tauhat,s_0(\bv_\tauhat))$ \begin{center} \begin{tikzpicture}[scale=.4, mycirc/.style={circle,fill=black!,inner sep=2pt, minimum size=.1 cm} ] \node[mycirc, label=120:{$\substack{n-k}$}] (n) at (-25,0) {}; \node[mycirc, label=45:{$\substack{n-d}$}] (n1) at (-19,5) {}; \node[mycirc, label=45:{$\substack{n-2d+1}$}](n2) at (-15,5) {}; \node[mycirc, label=45:{$\substack{n-3d+3}$}] (n6) at (-10,5) {}; \node[circle] (n3) at (-5,5) {$\cdots$}; \node[mycirc ,label=45:{$\substack{1}$ }] (n4) at (0,5) {}; % \node[mycirc ,label=45:{$\substack{0}$}] (n5) at (6,5) {}; %%%%%%%%%%%%%%%5 \node[mycirc, label=45:{$\substack{n-d}$}] (m1) at (-19,0) {}; \node[mycirc, label=45:{$\substack{n-2d+1}$}] (m2) at (-15,0) {}; \node[mycirc, label=45:{$\substack{n-3d+3}$}] (m6) at (-10,0) {}; \node[circle] (m3) at (-5,0) {$\cdots$}; \node[mycirc ,label=45:{$\substack{1}$ }] (m4) at (0,0) {}; % \node[mycirc ,label=45:{$\substack{0}$}] (m5) at (6,0) {}; %%%%%%%%%%%%%%%5 \node[mycirc, label=45:{$\substack{2d-k}$}] (k1) at (-19,-5) {}; \node[mycirc ,label=45:{$\substack{2d-k-\tau_2} $}] (k2) at (-15,-5) {}; \node[mycirc, label=45:{$\substack{2d-k-\tau_2-\tau_3}$}] (k6) at (-10,-5) {}; \node[circle] (k3) at (-5,-5) {$\cdots$}; \node[mycirc ,label=45:{$\substack{\tau_r}$}] (k4) at (0,-5) {}; % \node[mycirc ,label=45:{$\substack{0}$}] (k5) at (6,-5) {}; \draw (n1) -- node[below]{$\substack{\textcolor{red}{d-1}}$} (n2) -- node[below]{$\substack{\textcolor{red}{d-2}}$} (n6)--node[below]{$\substack{\textcolor{red}{d-3}}$} (n3) --node[below]{$\substack{\textcolor{red}{2}}$} (n4);% --(n5); \draw (m1) --node[below]{$\substack{\textcolor{red}{d-1}}$} (m2) -- node[below]{$\substack{\textcolor{red}{d-2}}$} (m6)--node[below]{$\substack{\textcolor{red}{d-3}}$} (m3) --node[below]{$\substack{\textcolor{red}{2}}$} (m4);% --(m5); \draw (k1) -- node[below]{$\substack{\textcolor{red}{\tau_2}}$} (k2) --node[below]{$\substack{\textcolor{red}{\tau_3}}$}(k6)-- node[below]{$\substack{\textcolor{red}{\tau_4}}$}(k3) --node[below]{$\substack{\textcolor{red}{\tau_{r-1}}}$}(k4);% --(k5); \draw (n) -- node[below]{$\substack{\textcolor{red}{d-k}}$} (n1); \draw (n) -- node[below] {$\substack{\textcolor{red}{d-k}}$} (m1); \draw (n) -- node[below] {$\substack{\textcolor{red}{n-2d+k}}$} (k1); \end{tikzpicture} \end{center} We have $s_0(\bv_\tauhat)\in (\N^I)^*$ and $\delta(s_0(\bv_\tauhat))=0$ but $s_0(\bv_\tauhat)$ is not anymore in $(\N^I)^{**}$ as we now see in detail. Since $\tau$ is a partition of $n$ we have $\tau_1+\tau_2\leq n$ and so $2d-k\geq\tau_2$. Moreover as $d\geq 8$, we have \begin{equation} n=\frac{d(d+1)}{2}\geq 4d \label{eq1}\end{equation} from which we get $n-2d\geq 2d-k$. Therefore the successive differences between the coordinates of the dimension vector on the last leg define a partition as $n-2d\geq 2d-k\geq\tau_2$, i.e. $$\sigma^3(s_0(\bv_\tauhat))=(n-2d+k,\tau_2, \tau_3, \ldots, \tau_r)$$ is a partition. This is not the case for the first two legs if $k>1$, i.e. $$\sigma^1(s_0(\bv_\tauhat))=\sigma^2(s_0(\bv_\tauhat))=(d-k,d-1,d-2,\ldots , 2,1)$$ is not a partition if $k>1$. But we may apply some element in $H$ to obtain a dimension vector $\bv^{(1)}_\tauhat$ in $(\N^I)^{**}$, i.e. move the first $d-k$ to come after $d-k+1$, so that the sequence becomes $(d-1,d-2, \ldots , d-k+1,d-k,d-k,d-k-1, \ldots,1)$ (see proof of Proposition \ref{ordering}(ii)). More explicitly, we apply $s_{[2,k-1]}\cdots s_{[2,1]}s_{[1,k-1]}\cdots s_{[1,1]}$ to $s_0(\bv_{\tauhat})$ to get $(\Gamma_\tau,\bv^{(1)}_\tauhat)$ with $\bv^{(1)}_\tauhat\in (\N^I)^{**}$: \begin{center} \begin{tikzpicture}[scale=.4, mycirc/.style={circle,fill=black!,inner sep=2pt, minimum size=.1 cm} ] \node[mycirc, label=90:{$\substack{n-k}$}] (n) at (-30,0) {}; \node[mycirc, label=45:{$\substack{n-k-d+1}$}] (n1) at (-24,5) {}; \node[mycirc, label=45:{$\substack{n-k-2d+3}$}] (n2) at (-20,5) {}; \node[mycirc, label=45:{$\substack{n-k-3d+6}$}] (n6) at (-15,5) {}; \node[circle] (11) at (-10,5) {$\cdots$}; \node[circle] (n3) at (-5,5) {$\cdots$}; \node[mycirc ,label=45:{$\substack{1}$ }] (n4) at (0,5) {}; % \node[mycirc ,label=45:{$\substack{0}$}] (n5) at (6,5) {}; %%%%%%%%%%%%%%%5 \node[mycirc, label=45:{$\substack{n-k-d+1}$}] (m1) at (-24,0) {}; \node[mycirc, label=45:{$\substack{n-k-2d+3}$}] (m2) at (-20,0) {}; \node[mycirc, label=45:{$\substack{n-k-3d+6}$}] (m6) at (-15,0) {}; \node[circle] (22) at (-10,0) {$\cdots$}; \node[circle] (m3) at (-5,0) {$\cdots$}; \node[mycirc ,label=45:{$\substack{1}$}] (m4) at (0,0) {}; % \node[mycirc ,label=45:{$\substack{0}$}] (m5) at (6,0) {}; %%%%%%%%%%%%%%%5 \node[mycirc, label=45:{$\substack{2d-k}$}] (k1) at (-24,-5) {}; \node[mycirc ,label=45:{$\substack{2d-k-\tau_2 }$}] (k2) at (-20,-5) {}; \node[mycirc ,label=45:{$\substack{2d-k-\tau_2-\tau_3 }$}] (k6) at (-15,-5) {}; \node[circle] (33) at (-10,-5) {$\cdots$}; \node[circle] (k3) at (-5,-5) {$\cdots$}; \node[mycirc ,label=45:{$\substack{\tau_r}$ }] (k4) at (0,-5) {}; %\node[mycirc ,label=45:{$\substack{0}$}] (k5) at (6,-5) {}; \draw (n3)--node[below]{$\substack{\textcolor{red}{d-k}}$} (11); \draw (n1) -- node[below]{$\substack{\textcolor{red}{d-2}}$}(n2) --node[below]{$\substack{\textcolor{red}{d-3}}$}(n6)--node[below]{$\substack{\textcolor{red}{d-4}}$}(11); \draw (n3) --node[below]{$\substack{\textcolor{red}{2}}$}(n4);% --node[below]{$\substack{\textcolor{red}{1}}$}(n5); \draw (m1) -- node[below]{$\substack{\textcolor{red}{d-2}}$}(m2) -- node[below]{$\substack{\textcolor{red}{d-3}}$}(m6)--node[below]{$\substack{\textcolor{red}{d-4}}$}(22); \draw(m3) --node[below]{$\substack{\textcolor{red}{2}}$}(m4);% --node[below]{$\substack{\textcolor{red}{1}}$}(m5); \draw (22)--node[below]{$\substack{\textcolor{red}{d-k}}$} (m3); \draw (k2)--node[below]{$\substack{\textcolor{red}{\tau_3}}$}(k6)-- node[below]{$\substack{\textcolor{red}{\tau_4}}$}(33); \draw(k3) --node[below]{$\substack{\textcolor{red}{\tau_{r-1}}}$}(k4);% --(k5); \draw(33)--node[below]{} (k3); \draw (n) -- node[below] {$\substack{\textcolor{red}{d-1}}$} (n1); \draw (n) -- node[below] {$\substack{\textcolor{red}{d-1}}$} (m1); \draw (n) -- node[below] {$\substack{\textcolor{red}{n-2d}}$} (k1); \draw (k1) -- node[below]{$\substack{\textcolor{red}{\tau_2}}$} (k2); \end{tikzpicture} \end{center} As shown just before the graph, in the first two legs there are two consecutive edges labelled by $d-k$ and we have $$ \delta(\bv^{(1)}_\tauhat)=-k+2. $$ Therefore if $k=1$ or $2$ then $\bv^{(1)}_\tauhat$ is a fundamental imaginary root and we are done. If $k\geq 3$ we keep going with the process, i.e. we apply $s_0$ first and then use the reflections of $H$ to re-order the labels of the edges in a non-decreasing way from right to left since $s_0(\bv^{(1)}_\tauhat)\in (\N^I)^*$. Then we get a dimension vector $\bv^{(2)}_\tau\in(\N^I)^{**}$ and $(\Gamma_\tauhat,\bv^{(2)}_\tau )$ as follows \begin{center} \begin{tikzpicture}[scale=.35, mycirc/.style={circle,fill=black!,inner sep=2pt, minimum size=.1 cm} ] \node[mycirc, label=120:{$\substack{n-2k+2}$}] (n) at (-30,0) {}; \node[mycirc, label=45:{$\substack{n-d-2k+4}$}] (n1) at (-24,5) {}; \node[mycirc ,label=45:{}][mycirc, label=45:{$\substack{n-2d-2k+7}$}] (n2) at (-20,5) {}; \node[mycirc, label=45:{$\substack{n-3d-2k+11}$}] (n6) at (-15,5) {}; \node[circle] (11) at (-10,5) {$\cdots$}; \node[circle] (n3) at (-5,5) {$\cdots$}; \node[mycirc ,label=45:{$\substack{1}$ }] (n4) at (0,5) {}; % \node[mycirc ,label=45:{$\substack{0}$}] (n5) at (6,5) {}; %%%%%%%%%%%%%%%5 \node[mycirc, label=45:{$\substack{n-d-2k+4}$}] (m1) at (-24,0) {}; \node[mycirc, label=45:{$\substack{n-2d-2k+7}$}] (m2) at (-20,0) {}; \node[mycirc, label=45:{$\substack{n-3d-2k+11}$}] (m6) at (-15,0) {}; \node[circle] (22) at (-10,0) {$\cdots$}; \node[circle] (m3) at (-5,0) {$\cdots$}; \node[mycirc ,label=45:{$\substack{1}$}] (m4) at (0,0) {}; % \node[mycirc ,label=45:{$\substack{0}$}] (m5) at (6,0) {}; %%%%%%%%%%%%%%%5 \node[mycirc, label=45:{$\substack{2d-k}$}] (k1) at (-24,-5) {}; \node[mycirc ,label=45:{$\substack{2d-k-\tau_2 }$}] (k2) at (-20,-5) {}; \node[mycirc ,label=45:{$\substack{2d-k-\tau_2 -\tau_3}$}] (k6) at (-15,-5) {}; \node[circle] (33) at (-10,-5) {$\cdots$}; \node[circle] (k3) at (-5,-5) {$\cdots$}; \node[mycirc ,label=45:{$\substack{\tau_r}$ }] (k4) at (0,-5) {}; % \node[mycirc ,label=45:{$\substack{0}$}] (k5) at (6,-5) {}; \draw (n3)--node[below]{$\substack{\textcolor{red}{d-k+1}}$} (11); \draw (n1) --node[below] {$\substack{\textcolor{red}{d-3}}$} (n2) --node[below] {$\substack{\textcolor{red}{d-4}}$} (n6)--node[below] {$\substack{\textcolor{red}{d-5}}$} (11); \draw (n3) --node[below]{$\substack{\textcolor{red}{2}}$}(n4);% --node[below]{$\substack{\textcolor{red}{1}}$}(n5); \draw (m1) --node[below] {$\substack{\textcolor{red}{d-3}}$} (m2) --node[below] {$\substack{\textcolor{red}{d-4}}$} (m6)--node[below] {$\substack{\textcolor{red}{d-5}}$} (22); \draw(m3) --node[below]{$\substack{\textcolor{red}{2}}$}(m4);% --node[below]{$\substack{\textcolor{red}{1}}$}(m5); \draw (22)--node[below]{$\substack{\textcolor{red}{d-k+1}}$} (m3); \draw (k2)--node[below] {$\substack{\textcolor{red}{\tau_3}}$} (k6)-- node[below] {$\substack{\textcolor{red}{\tau_4}}$} (33); \draw(k3) --node[below] {$\substack{\textcolor{red}{\tau_{r-1}}}$} (k4);% --(k5); \draw(33)--node[below]{ } (k3); \draw (n) -- node[below] {$\substack{\textcolor{red}{d-2}}$} (n1); \draw (n) -- node[below] {$\substack{\textcolor{red}{d-2}}$} (m1); \draw (n) -- node[below] {$\substack{\textcolor{red}{n-2d-k+2}}$} (k1); \draw (k1) -- node[below]{$\substack{\textcolor{red}{\tau_2}}$} (k2); \end{tikzpicture} \end{center} Notice that from (\ref{eq1}) we have $$ n-2d-k+2\geq 2d-k $$ and so $n-2d-k+2\geq \tau_2$ and so the successive differences of the coordinates on the last leg form a partition, i.e. $\sigma^3(\bv^{(1)}_\tauhat)=(n-2d-k+2,\tau_2,\tau_3,\ldots , \tau_r)$ is a partition. We have $$ \delta(\bv^{(2)}_\tauhat)=-k+4. $$ So if $k\leq 4$ then $\bv^{(2)}_\tauhat$ is a fundamental imaginary root and the process stops. If $k>4$ we start again the process and for $0\leq m\leq \lceil k/2\rceil$ we end up, after $m$ iterations of our labelling algorithm, with a dimension vector $\bv_\tauhat^{(m)}\in(\N^I)^{**}$ which has the form \begin{center} \begin{tikzpicture}[scale=.35, mycirc/.style={circle,fill=black!,inner sep=2pt, minimum size=.1 cm} ] \node[mycirc, label=120:{$\substack{n-mk\\+m(m-1)}$}] (n) at (-30,0) {}; \node[mycirc, label=45:{$\substack{n-d \\ -mk+m^2}$}] (n1) at (-24,5) {}; \node[mycirc ,label=45:{}] (n2) at (-20,5) {}; \node[mycirc ,label=45:{}] (n6) at (-15,5) {}; \node[circle] (11) at (-10,5) {$\cdots$}; \node[circle] (n3) at (-5,5) {$\cdots$}; \node[mycirc ,label=45:{$\substack{1}$ }] (n4) at (0,5) {}; % \node[mycirc ,label=45:{$\substack{0}$}] (n5) at (6,5) {}; %%%%%%%%%%%%%%%5 \node[mycirc, label=45:{$\substack{n-d\\ -mk+m^2}$}] (m1) at (-24,0) {}; \node[mycirc ,label=45:{ }] (m2) at (-20,0) {}; \node[mycirc ,label=45:{ }] (m6) at (-15,0) {}; \node[circle] (22) at (-10,0) {$\cdots$}; \node[circle] (m3) at (-5,0) {$\cdots$}; \node[mycirc ,label=45:{$\substack{1}$}] (m4) at (0,0) {}; % \node[mycirc ,label=45:{$\substack{0}$}] (m5) at (6,0) {}; %%%%%%%%%%%%%%%5 \node[mycirc, label=45:{$\substack{2d-k}$}] (k1) at (-24,-5) {}; \node[mycirc ,label=45:{$\substack{2d-k-\tau_2 }$}] (k2) at (-20,-5) {}; \node[mycirc ,label=45:{ }] (k6) at (-15,-5) {}; \node[circle] (33) at (-10,-5) {$\cdots$}; \node[circle] (k3) at (-5,-5) {$\cdots$}; \node[mycirc ,label=45:{$\substack{\tau_r}$ }] (k4) at (0,-5) {}; % \node[mycirc ,label=45:{$\substack{0}$}] (k5) at (6,-5) {}; \draw (n3)--node[below]{$\substack{\textcolor{red}{d-k+m-1}}$ } (11); \draw (n1) --node[below] { } (n2) --(n6)--(11); \draw (n3) --node[below]{ }(n4);% --node[below]{ }(n5); \draw (m1) --node[below] { } (m2) -- (m6)-- (22); \draw(m3) --node[below]{$\substack{\textcolor{red}{2}}$ }(m4) ;%--node[below]{ }(m5); \draw (22)--node[below]{$\substack{\textcolor{red}{d-k+m-1}}$ } (m3); \draw (k2)--node[below]{$\substack{\textcolor{red}{\tau_3}}$} (k6)--node[below]{$\substack{\textcolor{red}{\tau_4}}$} (33); \draw(k3) --node[below]{$\substack{\textcolor{red}{\tau_{r-1}}}$} (k4);% --(k5); \draw(33)--(k3); \draw (n) -- node[below] {$\substack{\textcolor{red}{d-m}}$} (n1); \draw (n) -- node[below] {$\substack{\textcolor{red}{d-m}}$} (m1); \draw (n) -- node[below] {$\substack{\textcolor{red}{n-2d}\\ \textcolor{red}{-(m-1)k+m(m-1)}}$} (k1); \draw (k1) -- node[below]{$\substack{\textcolor{red}{\tau_2}}$} (k2); \end{tikzpicture} \end{center} Note that we can check that $s_0(\bv^{(j)}_\tauhat)\in (\N^I)^*$ for all $j =0,1,2,\ldots , \lceil k/2\rceil$ inductively. The key point in the induction process is that we never need to use reflections in $H$ on the last leg to re-organise the labels on the edges in a non-increasing way, i.e. when $m\leq\lceil k/2\rceil$ we always have $$ n-2d-(m-1)k+m(m-1)\geq 2d-k\geq \tau_2, $$ equivalently, $\sigma^3(\bv^{(m)}_\tauhat)=(n-2d-(m-1)k+m(m-1),\tau_2, \ldots , \tau_r)$ is a partition. Indeed, as the sequence $(n-mk+m(m-1))_{m\leq\lceil k/2\rceil}$ is decreasing, we only need to check the above inequality for $m=\lceil k/2\rceil$, i.e. we need to check that \begin{equation} 2d^2-14d\geq\begin{cases}k^2-6k &\text{if } k\text{ is even,}\\k^2-6k+1&\text{ if } k\text{ is odd.}\end{cases} \label{ineq2}\end{equation} But these inequalities are true under our assumption $0\leq k\leq d$ and $d\geq 8$. We have $$ \delta(\bv_\tauhat^{(m)})=-k+2m $$ and so for $m=\lceil k/2\rceil$ we get that $\bv^{(m)}_\tauhat$ is a fundamental imaginary root which proves the theorem for $d\geq 8$. The first part of the proof works as the inequality (\ref{eq1}) still holds for $d=7$. However the end of the proof needs to be modified as the inequalities (\ref{ineq2}) are not verified anymore for $d=7$ and $k=7$. We can check this case by hand as follows: $\delta(\bv_\tauhat^{(5)})=1$ for $\tau=(21,7)$, $\delta(\bv_\tauhat^{(4)})=0$ for $\tau=(21,6,1)$ and $\delta(\bv_\tauhat^{(4)})=1$ for all $\tau=(21,\tau_2,\ldots , \tau_l)$ such that $\tau_2 \leq 5$. \end{proof} \subsection{Remaining case \texorpdfstring{$\tau_1\leq n-d$ and $d \leq 6$}{tau\_1 ≤ n-d and d ≤ 6}} It remains to verify the theorem for the cases where $d\leq 6$ and $\tau_1 \leq n-d$. In this range, there exist elements $\bv_\tauhat$ which are not imaginary roots; rather, they are real roots or correspond to a non-zero Kronecker coefficient. Recall that if $\bv_\tauhat$ is a root (respectively, if $g_\tauhat\neq 0$), then by Remark \ref{L1-2} (respectively, by Theorem \ref{L1-1}), $U_\tauhat(q)\neq 0$. The findings are summarized in the following table. \begin{figure}[H] \centering \begin{tabular}{|c|c|c|c|c|} \hline $d$& $n$ & $\tau=(\tau_1,\tau_2,\dots)$ & $\bv_\tauhat$ &$U_{\tauhat}(q)\neq 0$ because \\ \hline\hline $6$&$21$ & \makecell{$\substack{ \tau_1 \leq 15\ \text{except}\ (15,6)}$} & Imaginary root & $\bv_\tauhat$ is a root \\ \hline $6$&$21$ & \makecell{$\substack{ (15,6)}$} & $\substack{\text{Under reflections,}\\ \text{equivalent to the real root} \\ ((1^3),(1^3),(2,1)) }$ & $\bv_\tauhat$ is a root \\ \hline\hline $5$&$15$ & \makecell{$\substack{ \tau_1 \leq 10 \\ \text{except}\ (9,6),(10,4,1),(10,5)}$} & Imaginary root &$\bv_\tauhat$ is a root \\ \hline $5$&$15$ & \makecell{$\substack{ (10,4,1)}$} & $\substack{\text{Under reflections,}\\ \text{equivalent to the real root} \\ ((1^2),(1^2),(1^2)) }$ &$\bv_\tauhat$ is a root \\ \hline $5$&$15$ & \makecell{$\substack{ (10,5)}$} & Not a root& $g_\tauhat=141$ \\ \hline $5$&$15$ & \makecell{$\substack{ (9,6)}$} & $\substack{\text{Under reflections,}\\ \text{equivalent to the real root} \\ ((1^3),(1^3),(2,1)) }$&$\bv_\tauhat$ is a root \\ \hline\hline $4$&$10$ & \makecell{$\substack{\tau_1\leq 5\ \text{except}\ (5,4,1),(5^2) }$} & Imaginary root &$\bv_\tauhat$ is a root \\ \hline $4$&$10$ & \makecell{$\substack{(5,4,1)}$} & $\substack{\text{Under reflections,}\\ \text{equivalent to the real root} \\ ((1^2),(1^2),(1^2)) }$ &$\bv_\tauhat$ is a root \\ \hline $4$&$10$ & \makecell{$\substack{(5^2)}$} & Not a root &$g_\tauhat=6$\\ \hline $4$&$10$ & \makecell{$\substack{(6,1^4)}$} & Imaginary root &$\bv_\tauhat$ is a root \\ \hline $4$&$10$ & \makecell{$\substack{(6,2,1^2)}$} & $\substack{\text{Under reflections,}\\ \text{equivalent to the real root} \\ ((1^2),(1^2),(1^2)) }$ &$\bv_\tauhat$ is a root \\ \hline $4$&$10$ & \makecell{$\substack{ (6,2^2) }$} & Not a root & $g_\tauhat=39$\\ \hline $4$&$10$ & \makecell{$\substack{ (6,3,1) }$} & Not a root & $g_\tauhat=54$\\ \hline $4$&$10$ & \makecell{$\substack{(6,4)}$} & Not a root & $g_\tauhat=15$\\ \hline\hline $3$&$6$ & \makecell{$\substack{\tau_1=1,2\ \text{except}\ (2^2,1^2), (2^3)}$} & Imaginary root &$\bv_\tauhat$ is a root \\ \hline $3$&$6$ & \makecell{$\substack{(2^2,1^2)}$} & $\substack{\text{Under reflections,}\\ \text{equivalent to the real root} \\ ((1^2),(1^2),(1^2)) }$ &$\bv_\tauhat$ is a root \\ \hline $3$&$6$ & \makecell{$\substack{(2^3)}$} & Not a root &$g_\tauhat=2$ \\ \hline $3$&$6$ & \makecell{$\substack{(3,1^3)}$} & Not a root &$g_\tauhat=4$\\ \hline $3$&$6$ & \makecell{$\substack{(3,2,1)}$} & Not a root &$g_\tauhat=5$\\ \hline $3$&$6$ & \makecell{$\substack{(3^2)}$} & Not a root &$g_\tauhat=2$\\ \hline\hline $2$&$3$ & \makecell{$\substack{(1^3)}$} & Not a root &$g_\tauhat=1$\\ \hline\hline $1$&$1$ & \makecell{$\substack{(1)}$} & Real root &$\bv_\tauhat$ is a root \\ \hline \end{tabular} \end{figure} \subsection{The case \texorpdfstring{$\tau_1>n-d$}{tau\_1 > n-d}} Recall that the dominance order $\trianglelefteq$ is defined as follows: for two partitions $\lambda$ and $\mu$ of size $n$, $\lambda \trianglelefteq \mu$ if $\lambda_1+\cdots +\lambda_i \leq \mu_1+\cdots +\mu_i$ for all $i$. In \cite{Ik}, it is proved that if $\xi_d$ is comparable with $\tau$, namely $\tau \trianglelefteq \xi_d$ or $\xi_d \trianglelefteq \tau$, then the Kronecker coefficient $\delta_\tauhat \neq 0$ and so by Theorem \ref{L1-1}, $U_\tauhat(q)\neq 0$. Hence in the case $\tau_1>n-d$, to prove Theorem \ref{maintheo} it is sufficient to prove the following result. \begin{lemm}If $\tau_1>n-d$, then $\xi_d\trianglelefteq \tau$. \end{lemm} \begin{proof}We need to prove that $$ \tau_1+\cdots+\tau_i\geq d+(d-1)+\cdots+(d-i+1) $$ for all $i$. By assumption on $\tau_1$ note that $$ \tau_1+\cdots+\tau_i\geq (n-d+1)+(i-1)=\frac{d(d-1)}{2}+i. $$ Therefore it is enough to prove that $$ \frac{d(d-1)}{2}+i\geq \frac{i(2d-i+1)}{2} $$ which is equivalent to $$ (d-i)^2\geq d-i. $$ Note that because of the assumption $\tau_1>n-d$, the length of $\tau$ is at most $d$ and so $d-i\geq 0$ from which we deduce the above inequality. \end{proof} \section{The tensor square of unipotent characters}\label{tensor} In this section we are interested in the unipotent characters of $\GL_n(\F_q)$ whose tensor square contains as a direct summand all the unipotent characters of $\GL_n(\F_q)$. We saw in the previous section that this is the case of the unipotent characters attached to the staircase partitions. We also know that the Steinberg characters satisfy this property \cite[Proposition 33]{L1} (see also \cite{HSTZ} for other groups). In this section we give a conjectural necessary and sufficient condition for a unipotent character to verify this property. \subsection{Main result and conjecture} The aim of this section is to bring some evidences for the following conjecture. \begin{conj} Let $\mu=(\mu_1,\mu_2,\dots)$ be a partition of $n$, then (i) $U_{(\mu,\mu,\tau)}(q)\neq 0$ for all partitions $\tau$ of $n$ if and only if $\mu_1\leq \lceil n/2\rceil$. (ii) If $\mu_1>\lceil n/2\rceil$, then $$ U_{(\mu,\mu,(1^n))}(q)= 0. $$ \label{conj}\end{conj} We can check that the above conjecture is correct for $n\leq 8$ from Mattig's experimental data \cite{Lu}. Notice that if $\mu'\trianglelefteq\mu$ and if $\mu_1\leq \lceil n/2\rceil$ then $\mu'_1\leq \lceil n/2\rceil$ and so the first assertion of Conjecture \ref{conj} implies the conjecture below. \begin{conj}If for some partition $\mu$ of $n$ we have $$ U_{(\mu,\mu,\tau)}(q)\neq 0,\hspace{.5cm}\text{ for all partitions }\tau, $$ then for all partitions $\mu'\trianglelefteq \mu$, we have $$ U_{(\mu',\mu',\tau)}(q)\neq 0,\hspace{.5cm}\text{ for all partitions }\tau. $$ \label{weak}\end{conj} Notice that the set $\{\mu\,|\, \mu_1\leq \lceil n/2\rceil\}$ has a maximal element $\mu^{\rm max}$ with respect to the dominance order. If $n=2k$ is even, $\mu^{\rm max}$ is the partition with two parts $(k,k)$ and if $n=2k+1$ is odd, it is the partition $(k+1,k)$. \begin{theo}For any partition $\mu\trianglelefteq\mu^{\rm max}$ we have $$ U_{(\mu,\mu,(1^n))}(q)\neq 0. $$ \label{theo2}\end{theo} \begin{proof} Put $$ \omegahat^o:=\begin{cases}((1^2),(1^2),(1^2))^k&\text{ if } n=2k,\\ ((1^2),(1^2),(1^2))^k((1),(1),(1))&\text{ if }n=2k+1.\end{cases} $$ By Theorem \ref{L1-3} it is enough to prove that \begin{equation} \left\langle \C_{\omegahat^o}^{(\mu,\mu,(1^n))},1\right\rangle\neq 0. \label{ine}\end{equation} Put $$ \omega^o:=\begin{cases}(1^2)^k&\text{ if } n=2k,\\ (1^2)^k(1)&\text{ if }n=2k+1\end{cases} $$ so we have $$ \C_{\omegahat^o}^{(\mu,\mu,(1^n))}=\mathbb{C}_{\omega^o}^\mu\otimes\mathbb{C}_{\omega^o}^\mu\otimes\mathbb{C}_{\omega^o}^{(1^n)}. $$ To prove (\ref{ine}) it is enough to prove the following two statements : (1) For all partitions $\mu$ of $n$ such that $\mu\trianglelefteq \mu^{\rm max}$, we have $\mathbb{C}_{\omega^o}^\mu\neq 0$, (2) $\mathbb{C}_{\omega^o}^{(1^n)}$ is the trivial $S_k$-module. By Proposition \ref{Prop13} we have $$ c_{\omega^o}^\mu:={\rm dim}\, \mathbb{C}_{\omega^o}^\mu=\left\langle s_\mu,s_{\omega^o}\right\rangle. $$ Notice that $$ s_{\omega^o}:=\begin{cases}(s_{(1^2)})^k&\text{ if }n=2k,\\ (s_{(1^2)})^k s_{(1)}&\text{if }n=2k+1.\end{cases} $$ We focus on proving (1) and (2) for even \(n = 2k\); the case of odd \(n\) will be addressed briefly, as the method is essentially the same. By Pieri's formula, for two partitions $\lambda$ and $\mu$ of size $r$ we have $$ \mathbb{C}^\mu_{\lambda (1^r)}\neq 0 $$ if $\mu$ can be obtained from $\lambda$ by adding $r$ boxes with no two in the same row. In this case it is of dimension $1$. We now prove (1) by induction on $k$. The case $k=1, 2$ are easy to verify. We thus assume that (1) is true for $k 0$ when $\tau\trianglelefteq (m-1,m-1)$. Let $\mu$ be a partition of $2m$ such that $\mu\trianglelefteq (m,m)$. Then $\mu$ is one of the following kind : (i) $\mu=(m,m)$, (ii) $\mu=(m,\mu_2,\dots,\mu_\ell)$ with $\mu_2