<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">%~MoulinÃ© par MaN_auto v.0.33.8 (9c3f6e85) 2025-05-26 16:52:34
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\title{Automorphism group of $\bar{B}$}
\alttitle{Groupe d'automorphismes de $\bar{B}$}

\author{\firstname{Senthamarai Kannan} \lastname{Subramaniam}}
\address{Chennai Mathematical Institute, Plot H1, SIPCOT IT Park, Siruseri, Kelambakkam, 603103, India}
\email{kannan@cmi.ac.in}

\author{\firstname{Aisha} \lastname{Negi}}
\address[1]{Chennai Mathematical Institute, Plot H1, SIPCOT IT Park, Siruseri, Kelambakkam, 603103, India}
\email{aisha@cmi.ac.in}

\subjclass{14L10, 14L30}
\keywords{\kwd{Automorphism groups} \kwd{wonderful compactification} \kwd{unipotent group}}
\altkeywords{\kwd{Groupes d'automorphismes} \kwd{compactification magnifique} \kwd{groupe unipotent}}

\thanks{We thank the Infosys Foundation for the partial financial support}

\begin{abstract}
Let $G$ be a simple algebraic group of adjoint type over the field of complex numbers, different from $\operatorname{PSL}(2,\mathbb{C})$. Let $\bar{G}$ be the wonderful compactification of $G$ constructed by C.~De~Concini and C.~Procesi. Let $\bar{B}$ be the scheme theoretic closure of a Borel subgroup $B$ of $G$ in $\bar{G}$. Then we prove that the connected component, containing the identity automorphism of the group of all algebraic automorphisms of $\bar{B}$ is $B\times B$.
\end{abstract}

\begin{altabstract}
Soit $G$ un groupe algÃ©brique simple de type adjoint sur le corps des nombres complexes, diffÃ©rent de $\operatorname{PSL}(2,\mathbb{C})$. Soit $\bar{G}$ la compactification magnifique de $G$ construite par C.~De~Concini et C.~Procesi. Soit $\bar{B}$ l'adhÃ©rence de Zariski d'un sous-groupe borÃ©lien $B$ de $G$ dans $\bar{G}$. Nous prouvons que la composante connexe, contenant l'automorphisme identitaire du groupe de tous les automorphismes algÃ©briques de $\bar{B}$ est $B\times B$.
\end{altabstract}

\dateposted{2025-06-24}
\begin{document}

\maketitle

\section{Introduction}

Let $G$ be a semisimple algebraic group of adjoint type over the field $\mathbb{C}$ of complex numbers.

Let $\sigma \colon G\to G$ be an automorphism of order two and let $H=G^\sigma$ be the fixed point subgroup. C.~De~Concini and C.~Procesi constructed a very important compactification of $G/H$ (see~\cite[Theorem 3.1, p.~14]{MR0718125}); it is known as the wonderful compactification.

For the special case of the automorphism $\sigma \colon G\times G\to G \times G$ given by $\sigma\parens[\big]{(x,y)}=(y,x)$ for every $(x,y)\in G\times G$, we have $(G\times G)^\sigma=\bigtriangleup(G)$. Hence, we have the wonderful compactification $\bar{G}$, of the semisimple algebraic group $G$ of adjoint type.

Let $T$ be a maximal torus of $G$ and $B$ be a Borel subgroup of $G$ containing $T$. Let $W=N_G(T)/T$ be the Weyl group of $G$ relative to $T$. For $w\in W$, let $n_w\in N_G(T)$ be a representative of $w$. M.~Brion and P.~Polo studied the closures $\bar{Bn_wB}$ in $\bar{G}$ (see~\cite{MR1789463}). These varieties are known as large Schubert varieties. They prove that the large Schubert varieties are normal and Cohen--Macaulay.

M.~Brion proved that the connected component containing the identity automorphism, of the automorphism group of $\bar{G}$ is $G\times G$ if $\PSL(2,\mathbb{C})$ is not a factor of $G$ (see~\cite[Example~2.4.5, p.~75--76]{MR2326138}).

I.~Biswas, S.~S.~Kannan and D.~S.~Nagraj proved that if $G$ is a simple algebraic group of adjoint type different from $\PSL(2,\mathbb{C})$, the connected component containing the identity automorphism, of the automorphism group of $\bar{T}$ is $T$ (see~\cite{MR3377673}).

So, it is interesting to study the automorphism group of the large Schubert varieties.

In this paper, we prove that if $G$ is a simple algebraic group of adjoint type different from $\PSL(2,\mathbb{C})$, then $\aut^{\circ}(\bar{B})=B\times B$.

Note that for $G=\PSL(2,\mathbb{C})$, we have $\bar{B}=\mathbb{P}^2$ and so $\aut^{\circ}(\bar{B})=\PSL(3,\mathbb{C})$ (by~\cite[Theorem~2, p.~75]{MR1334091}).

The organisation of the paper is as follows. Section~\ref{section2} consists of notation and preliminaries. In Section~\ref{section3}, we prove auxiliary lemmas. In Section~\ref{section4}, we discuss the linearization of line bundles for the action of unipotent groups following~\cite{knop1989local}. In Section~\ref{section5}, we prove that the set of all $(\gamma_1,\gamma_2)\in X(T)\oplus X(T)$ such that both $\gamma_1$ and $\gamma_2$ are nonzero for which there is a one dimensional unipotent subgroup $U(\gamma_1,\gamma_2)$ of $\aut^\circ(\bar{B})$ normalized by $T\times T$ with character ($\gamma_1$,$\gamma_2$) is empty. In Section~\ref{section6}, we prove that $\aut^\circ(\bar{B})=B\times B$.

\section{Notation and preliminaries}\label{section2}

In this section, we set up notation and preliminaries. For more details on algebraic groups and Lie algebras, we refer to~\cite{MR0396773} and~\cite{MR0499562}.

Let $G$ be a simple algebraic group of adjoint type over $\mathbb{C}$, different from $\PSL(2,\mathbb{C})$. Let $T$ be a maximal torus of $G$ and $B$ be a Borel subgroup of $G$ containing $T$. Let $X(T)$ denote the group of all characters of $T$.

Let $R(G)\subseteq X(T)$ be the root system of $(G,T)$ and $R^+(G)\subseteq R(G)$ be set of positive roots associated to $(B,T)$. Let $R^-(G)=-R^+(G) $ be the set of negative roots. Let $S=\{\alpha_1,\dotsc,\alpha_n\}$ be the set of simple roots. Let $s_\alpha\in W$ be the reflection corresponding to $\alpha \in R(G)$. Let $\alpha_\circ$ be the highest root. Let $\mathfrak{g}$, $\mathfrak{b}$ and $\mathfrak{t}$ be Lie algebras of $G$, $B$ and $T$ respectively.

Let $\braces[\big]{\mathrm{x}_\alpha, \mathrm{h}_\beta: \alpha \in R(G), \; \beta \in S}$ be the Chevalley basis of $\mathfrak{g}$ with respect to $T$.

Let $E=X(T)\otimes_\mathbb{Z} \mathbb{R}$. $E$ has a positive definite bilinear form denoted by $(\,\cdot\,,\cdot\,)$ which is induced from the Killing form of $\mathfrak{g}$. For $\lambda \in E$ and $\alpha\in R(G)$, define
\[
\langle \lambda,\alpha \rangle\coloneqq\frac{2(\lambda,\alpha)}{(\alpha,\alpha)}.
\]
Let $\{\omega_1,\dotsc,\omega_n\}$ be the set of fundamental weights dual to $\{\alpha_1,\dotsc,\alpha_n\}$ i.e.\ $\langle \omega_i,\alpha_j \rangle=\delta_{ij}$ ($1\leq i,j\leq n$). Then, the weight lattice (respectively, root lattice) $\wedge$ (respectively, $\wedge_r$) is a lattice with basis $\omega_1,\dotsc,\omega_n$ (respectively, $\alpha_1,\dotsc,\alpha_n$).

Since $G$ is of adjoint type, we have $X(T)=\wedge_r$.

Let $\lambda\in\wedge$. Recall that $\lambda$ is a dominant weight if it is a non-negative integral combination of the fundamental weights.

Let $\alpha\in R(G)$. For the conjugation action of $T$ on $G$, let $U_\alpha$ be the one dimensional $T$-stable root subgroup of $G$ on which $T$ acts through $\alpha$. Note that $\mathfrak{g}_\alpha=\mathbb{C}\mathrm{x}_\alpha$ is the Lie algebra of $U_\alpha$. Let $U$ be the unipotent radical of $B$. Then we have, $U=\prod_{\alpha\in R^+(G)}U_\alpha$.

\subsection{Wonderful compactification of \texorpdfstring{$G$}{G}}

Now consider the automorphism
\[
\sigma \colon G \times G \to G\times G
\]
of order two given by
\[
\sigma\parens[\big]{(x,y)}=(y,x), \quad (x,y)\in G\times G.
\]
The fixed point subgroup $(G\times G)^{\sigma}$ is equal to $\bigtriangleup(G)=\braces[\big]{(x,x):x\in G}$. Hence we have $(G\times G)/(G\times G)^{\sigma}=(G\times G)/\bigtriangleup{(G)} \simeq G$.

Let
\[
\Pi \colon \hat{G}\to G
\]
be the simply connected covering of $G$. Let $\hat{T}$ be a maximal torus in $\hat{G}$ such that $\Pi(\hat{T})=T$ (see~\cite[Corollary~C, p.~136]{MR0396773}). Since $\Lie(\hat{G})=\mathfrak{g}$, we have $X(\hat{T})=\wedge$. For a dominant character $\chi$ of $\hat{T}$, let $V(\chi)$ be the irreducible $\hat{G}$-module of highest weight $\chi$.

Note that any $\sigma$-stable maximal torus of $G\times G$ is of the form $T'\times T'$ for some maximal torus~$T'$ in $G$. As in~\cite[p.~3]{MR0718125} define
\[
(T\times T)_1=\braces[\bigg]{(t_1,t_2)\in T\times T:\sigma\parens[\big]{(t_1,t_2)}=(t_1^{-1},t_2^{-1})}=\braces[\big]{(t,t^{-1}):t\in T}
\]
and
\[
(T\times T)_0=\braces[\bigg]{(t_1,t_2)\in T\times T:\sigma\parens[\big]{(t_1,t_2)}=(t_1,t_2)}=\braces[\big]{(t,t):t\in T}=\bigtriangleup (T).
\]
Observe that any character of $T\times T$ is of the form $(\chi,\psi)$ where $\chi,\psi \in X(T)$ as $X(T\times T)\simeq X(T)\oplus X(T)$. Moreover we have, $\sigma^*\parens[\big]{(\chi,\psi)}=(\psi,\chi)$. The root system $R(G\times G)$ of $G\times G$ relative to $T\times T$ is $\braces[\big]{(\alpha,0):\alpha\in R(G) }\cup\braces[\big]{(0,\alpha):\alpha\in R(G) }$. $B\times B^-$ is a Borel subgroup of $G \times G$ with the property as in~\cite[Lemma 1.2, p.~4]{MR0718125}. Further, for a dominant character $(\chi,\psi)$ of $T\times T$ relative to $B\times B^-$, the irreducible representation $V\parens[\big]{(\chi,\psi)}$ is equal to $V(\chi)\otimes V(-\psi)^*$, where $V(\chi)$, (respectively, $V(-\psi)$) is the irreducible $G$-module with highest weight $\chi$ (respectively, $-\psi$). Further $V(\chi,\psi)^{\bigtriangleup(G)}\neq \{0\}$ iff $(\chi,\psi)$ is a special weight. That is, $V(\chi,\psi)^{\bigtriangleup(G)}\neq \{0\}$ iff $\chi=-\psi$. Further $(\chi,\psi)$ is regular special iff $\chi$ is regular dominant and $\chi=-\psi$. Thus for any regular special $(\chi,\psi)$, we have $V(\chi,\psi)=V(\chi)\otimes V(\chi)^*=\End\parens[\big]{V(\chi)}$, where $\chi$ is a regular dominant weight.

Now, let $\chi$ be a regular dominant weight. Let
\[
\pi \colon \End\parens[\big]{V(\chi)}\setminus\{0\} \to\mathbb{P}\parens[\big]{\End\parens[\big]{V(\chi)}}
\]
be the natural map and let $x_\circ=\pi(\id)$, where $\id$ is the multiplicative identity element of $\End\parens[\big]{V(\chi)}$. Then we have
\[
\bar{G}=\bar{(G\times G).x_\circ}\subseteq \mathbb{P}\parens[\big]{\End\parens[\big]{V(\chi)}}.
\]
By~\cite[p.~10]{MR0718125}, $\bar{G}$ is independent of the regular dominant character $\chi$. In this paper we take $\chi=2\rho$, where
\[
\rho= \frac{1}{2}\sum_{\alpha\in R^+(G)}\alpha=
\sum_{i=1}^{n}\omega_i.
\]

For $w\in W$ we have the large Schubert variety $X(w)=\bar{Bn_wB}$ (see~\cite{MR1789463}). For $w=1$, we have $X(1)=\bar{B}$.

Let $\aut(\bar{B})$ be the group of all automorphisms of the variety $\bar{B}$. Let $\aut^{\circ}(\bar{B})$ be the connected component of $\aut(\bar{B})$ containing the identity automorphism. By~\cite[Theorem~3.7, p.~17]{MR0217090}, $\aut^{\circ}(\bar{B})$ is a linear algebraic group.

By~\cite[Theorem~2, p.~75]{MR1334091}, we have $\aut^{\circ}\parens[\big]{\mathbb{P}\parens[\big]{\End\parens[\big]{V(2\rho)}}}=\PSL\parens[\big]{\End\parens[\big]{V(2\rho)}}$. Consider the action of $G\times G$ on $\mathbb{P}\parens[\big]{\End\parens[\big]{V(2\rho)}}$. Note that $\bar{G}$ is $G\times G$ stable. Hence, we have a homomorphism
\[
\phi \colon G\times G \to \textstyle{\aut^\circ}(\bar{G}).
\]
Consider the action of $B\times B$ on $\bar{B}$ given by $(b_1,b_2).x=b_1xb_2^{-1}$ for $b_1,b_2\in B$ and $x\in \bar{B}$. Thus, we have a morphism of algebraic groups
\[
\phi_B \colon B\times B \to \textstyle{\aut^\circ}(\bar{B})
\]
given by the action of $B\times B$ on $\bar{B}$. Let $(b_1,b_2)\in \Ker(\phi_B)$. Then we have
\[
(b_1,b_2).b=b_1bb_2^{-1}=b
\]
for all $b\in B$. For $b=e$, we have $b_1=b_2$. Moreover, we have $b_1\in Z(B)=Z(G)$ (see~\cite[Corollary~B, p.~140]{MR0396773}). Since $G$ is of adjoint type, we have $b_1=e$. Thus, $\phi_B$ is injective. Therefore,
\[
\phi_B \colon B\times B \to \phi_B(B\times B)
\]
is bijective. Since the base field is $\mathbb{C}$, by~\cite[p.~44]{MR0396773}
\[
(\dd\phi_B)_{(e,e)} \colon \mathfrak{b}\oplus \mathfrak{b} \to \Lie\parens[\big]{\phi_B(B\times B)}
\]
is an isomorphism. Therefore, $\phi_B$ is birational. Hence by Zariski Main Theorem~\cite[Theorem~5.2.8, p.~85]{MR1642713}, $\phi_B$ is an isomorphism.

\section{Auxiliary lemmas}\label{section3}

In this section, we prove some auxiliary lemmas that will be useful to prove the main theorem.

Let $A=\aut^{\circ}(\bar{B})$.

From~\cite[p.~126--127]{MR0396773}, any one dimensional unipotent group is isomorphic to the additive group~$\mathbb{G}_a$.
\begin{defi}\label{un group}
Let $(\gamma_1,\gamma_2)\in X(T\times T)$. A $T\times T$-stable subgroup $U(\gamma_1,\gamma_2)$ of $A$ (for the conjugation action) for which there is an isomorphism
\[
u_{(\gamma_1,\gamma_2)} \colon \mathbb{G}_a\to U(\gamma_1,\gamma_2)
\]
such that
\[
(s,t)u_{(\gamma_1,\gamma_2)}(a)(s,t)^{-1}=u_{(\gamma_1,\gamma_2)}(\gamma_1(s)\gamma_2(t).a)
\]
for all $(s,t)\in T\times T$ and $ a\in \mathbb{C}$ is called a one dimensional unipotent subgroup with character $(\gamma_1,\gamma_2)$.
\end{defi}

Let $\Gamma$ be the set of nonzero weights of the action of $T\times T$ on $\Lie \parens[\big]{\aut^\circ(\bar{B})}$. Then for any ($\gamma_1$,$\gamma_2$)$\in$ $\Gamma$ there exists a unipotent one-parameter subgroup $U(\gamma_1,\gamma_2)$ normalized by $T\times T$ such that $\Lie\parens[\big]{U(\gamma_1,\gamma_2)}\subseteq \parens[\big]{\Lie\parens[\big]{\aut^\circ{(\bar{B})}}}_{(\gamma_1,\gamma_2)}$ (see~\cite[Lemma~C, p.~96]{MR0396773}). Any such $U(\gamma_1,\gamma_2)$ is one dimensional unipotent subgroup with character ($\gamma_1$,$\gamma_2$)(see Definition~\ref{un group}).

Let $\Gamma_0\coloneqq\Gamma\setminus \parens[\big]{X(T)\times\{0\}\cup \{0\}\times X(T)}$.

The key result of the paper is to prove that $\Gamma_0=\emptyset$ (see Corollary~\ref{cor non existence} and Lemma~\ref{non existence for G2}). The details of proof are developed in Sections~\ref{section3}, \ref{section4} and~\ref{section5}.

Assume on the contrary that $\Gamma_0\neq\emptyset$. From now onwards, we fix $(\gamma_1,\gamma_2)\in \Gamma_0$ to be a maximal element relative to the dominant ordering on $X(T\times T)$ given by $R^+(G\times G)$ (for details about dominant ordering, see~\cite[p.~47]{MR0499562}). (Note: from~\cite{MR0217090}, $\aut^{\circ}(\bar{B})$ is a linear algebraic group. Hence, $\aut^{\circ}(\bar{B})$ is isomorphic to a closed subgroup of $\GL(q,\mathbb{C})$ for some $q\in \mathbb{N}$).

\begin{lemm}\label{alpha=nu1+nu2}
Let $(\nu_1,\nu_2)\in X(T)\oplus X(T)$ be such that there exists a one dimensional $T \times T$-stable unipotent subgroup $U(\nu_1,\nu_2)$ of $A$ with character $(\nu_1,\nu_2)$ satisfying:
\begin{enumerate}\romanenumi
\item either every element of $U(\nu_1,\nu_2)$ commutes with every element of $U_{(\beta,0)}$ for all $\beta\in R^+(G)$;
\item or every element of $U(\nu_1,\nu_2)$ commutes with every element of $U_{(0,\beta)}$ for all $\beta\in R^+(G)$.
\end{enumerate}
Let $\alpha=\nu_1+\nu_2$. Then, we have:
\begin{enumerate}\romanenumi
\item $\alpha\in R^+(G)$;
\item $U(\nu_1,\nu_2).x_\circ=U_{(\alpha,0)}.x_\circ=U_{(0,\alpha)}.x_\circ.$
\end{enumerate}
\end{lemm}
\begin{proof}
\mbox{}\\*[-2em]
\begin{step}[ {\boldmath{$U(\nu_1,\nu_2).x_\circ\nsubseteq\bar{T}$}}]\label{alpha=nu1+nu2_step1}
Assume on the contrary that $U(\nu_1,\nu_2).x_\circ\subseteq\bar{T}$. Since $T\times T$ normalizes $U(\nu_1,\nu_2)$, for every $u\in U(\nu_1,\nu_2),$ $(s,t)\in T\times T$ there exists $u' \in U(\nu_1,\nu_2)$ such that $u.\parens[\big]{(s,t).x_\circ}=(s,t).(u'.x_\circ)$. Then by assumption $\bar{T}=\bar{(T\times T).x_\circ}$ is $U(\nu_1,\nu_2)$-stable. From~\cite[Theorem~3.1, p.~786]{MR3377673}, we have $\aut^{\circ}(\bar{T})=T$. Hence the action of $U(\nu_1,\nu_2)$ on $\bar{T}$ is trivial. Also we have either every element of $U(\nu_1,\nu_2)$ commutes with every element of $U_{(\beta,0)}$ for every $\beta \in R^+(G)$ or every element of $U(\nu_1,\nu_2)$ commutes with every element of $U_{(0,\beta)}$ for every $\beta \in R^+(G)$. Thus the action of $U(\nu_1,\nu_2)$ on $\bar{B}=\bar{UT}=\bar{TU}$ is trivial (by identifying $B$ as $(B\times1).x_\circ$ or $(1\times B).x_\circ$). This gives a contradiction to $U(\nu_1,\nu_2)$ being a one dimensional $T\times T$-stable unipotent subgroup of $A$.
\end{step}
\begin{step}[Proof of the lemma]
For any $g\in G$, observe that $(g,1).x_\circ=(1,g^{-1}).x_\circ$. Therefore, for every $\beta\in R^+(G)$,
\[
U_{(\beta,0)}.x_\circ=U_{(0,\beta)}.x_\circ.
\]
Note that $U(\nu_1,\nu_2).x_\circ\cap \parens[\big]{(B\times 1).x_\circ}$ is a nonempty $\bigtriangleup(T)$-stable open subset of $U(\nu_1,\nu_2).x_\circ$. From Step~\ref{alpha=nu1+nu2_step1}, we have
\[
U(\nu_1,\nu_2).x_\circ\cap (B\times 1).x_\circ \nsubseteq {(T\times 1).x_\circ}.
\]
Note that, $\restr{(\nu_1,\nu_2)}{\bigtriangleup(T)}=\nu_1+\nu_2=\alpha$.
\begin{enonce*}[remark]{Claim}
$\nu_1+\nu_2$ is non trivial.
\end{enonce*}
Take $u_{(\nu_1,\nu_2)}(a).x_\circ\in (U(\nu_1,\nu_2).x_\circ\cap (B\times 1).x_\circ)\setminus (T\times 1).x_\circ$ i.e.\ $u_{(\nu_1,\nu_2)}(a).x_\circ=(ut_0,1).x_\circ$ for some $u\in U\setminus{\{1\}}$, $a\in \mathbb{C}^\times$ and $t_0\in T$. Then,
\begin{equation}\label{3.1}
\begin{split}
(tut_0t^{-1},1).x_\circ 
&amp; = \parens[\big]{(t,t)(ut_0,1)}.x_\circ				\\
&amp; = (t,t).\parens[\big]{(ut_0,1).x_\circ}				\\
&amp; = (t,t).\parens[\big]{u_{(\nu_1,\nu_2)}(a).x_\circ}	\\
&amp; = u_{(\nu_1,\nu_2)}\parens[\big]{(\nu_1+\nu_2)(t).a}.x_\circ.
\end{split}
\end{equation}
If $\nu_1$+$\nu_2$ is the trivial character, then for every $t\in T$, we have
\begin{equation}\label{3.2}
(t,t).\parens[\big]{u_{(\nu_1,\nu_2)}(a).x_\circ}= u_{(\nu_1,\nu_2)}(a).x_\circ.
\end{equation}
Therefore, by~\eqref{3.1} and~\eqref{3.2}, we have
\[
(tut_0t^{-1},1).x_\circ= (ut_0,1).x_\circ.
\]
Hence, we have
\[
tut_0t^{-1}=ut_0
\]
for all $t\in T$, which is not possible because $u\neq 1$ and $C_G(T)=T $ (see~\cite[Corollary, p.~155]{MR0396773}). Therefore, $\nu_1+\nu_2$ is nontrivial.

Since $U(\nu_1,\nu_2).x_\circ\cap(B\times 1).x_\circ$ is $\bigtriangleup (T)$ stable and $\nu_1+\nu_2$ is nontrivial,
\[
U(\nu_1,\nu_2).x_\circ\cap (B\times 1).x_\circ=U(\nu_1,\nu_2).x_\circ.
\]
Let $R^+(G)=\{\beta_1,\dotsc, \beta_N\}$. Write $u=u_{\beta_1}(a_1)\dotsm u_{\beta _N}(a_N) $ where $a_i \in \mathbb{C}$ for all $1\leq i\leq N$. On the other hand $(t,t).\parens[\big]{U(\nu_1,\nu_2).x_\circ}=U(\nu_1,\nu_2).x_\circ \subseteq (B\times 1).x_\circ$ for all $t\in T$. Hence, we have $tut_0t^{-1}.x_\circ\in U(\nu_1,\nu_2).x_\circ$. If $\abs[\big]{\braces[\big]{i\in \mathbb{N}:a_i\neq 0}}\geq2$, then dimension of $\bigtriangleup(T).\parens[\big]{(ut_0,1).x_\circ}\geq 2$. This is a contradiction to $\bigtriangleup(T).\parens[\big]{(ut_0,1).x_\circ}\subset U(\nu_1,\nu_2).x_\circ$ because $U(\nu_1,\nu_2).x_\circ$ is one dimensional. Therefore $u=u_{\beta}(c)$ for some $\beta \in R^+(G)$ and $c\in \mathbb{C}^\times$. Hence, we have
\begin{equation}\label{eq x}
U(\nu_1,\nu_2).x_\circ =U_{(\beta,0)}t_0.x_\circ.
\end{equation}
Since $x_\circ \in U(\nu_1,\nu_2).x_\circ =U_{(\beta,0)}t_0.x_\circ$, we have $t_0=1$. Further from weight consideration in~\eqref{eq x}, $\beta=\nu_1+\nu_2=\alpha$.\qedhere
\end{step}
\let\qed\relax
\end{proof}
\begin{coro}\label{commuting theorem}
We have:
\begin{enumerate}\romanenumi
\item \label{commuting theorem_1} either every element of $U(\gamma_1,\gamma_2)$ commutes with every element of $U_{(\beta,0)}$ for all $\beta \in R^+(G)$;
\item or every element of $U(\gamma_1,\gamma_2)$ commutes with every element of $U_{(0,\beta)}$ for all $\beta \in R^+(G)$.
\end{enumerate}
\end{coro}
\begin{proof}
Since ($\gamma_1$,$\gamma_2$) is a maximal element of $\Gamma_0$, we have either~\eqref{commuting theorem_1} holds or there exists $\beta\in R^+(G)$ such that there is a one dimensional unipotent subgroup $U'$ in $[U(\gamma_1,\gamma_2),U_{(\beta,0)}]$ normalized by $T\times T$ with character $(\gamma_1,\gamma_2) +(\beta,0)$. By maximality of $(\gamma_1,\gamma_2) $ in $\Gamma_0$, $(\gamma_1,\gamma_2) +(\beta,0)=(0,\gamma_2)$, so $(\gamma_1,\gamma_2) =(-\beta,\gamma_2)$. Note again by using maximality, we have every element of $U'$ commutes with every element of $U_{(\alpha,0)}$ for all $\alpha\in R^+(G)$. Now by Lemma~\ref{alpha=nu1+nu2}, we have $\gamma_2\in R^+(G)$. So to prove the corollary it suffices to prove that if $(\gamma_1,\gamma_2) +(\beta,0)=(0,\alpha)$ for some $\alpha,\beta\in R^+(G)$, then every element of $U(\gamma_1,\gamma_2)$ commutes with every element $U_{(0,\gamma)}$ for all $ \gamma \in R^+(G)$. This follows from maximality of $(\gamma_1$,$\gamma_2)$ in $\Gamma_0$.
\end{proof}
\begin{coro}\label{alpha is a positive root}
Let $\alpha=\gamma_1+\gamma_2$. Then we have
\begin{enumerate}\romanenumi
\item $\alpha\in R^+(G)$;
\item $U(\gamma_1,\gamma_2).x_\circ=U_{(\alpha,0)}.x_\circ=U_{(0,\alpha)}.x_\circ $.
\end{enumerate}
\end{coro}
\begin{proof}
By Corollary~\ref{commuting theorem}, either every element of $U(\gamma_1,\gamma_2)$ commutes with every element of $U_{(\beta,0)}$ for all $\beta \in R^+(G)$, or every element of $U(\gamma_1,\gamma_2)$ commutes with every element of $U_{(0,\beta)}$ for all $\beta \in R^+(G)$. Therefore from Lemma~\ref{alpha=nu1+nu2} we have $\alpha\in R^+(G)$, and $U(\gamma_1,\gamma_2).x_\circ=U_{(\alpha,0)}.x_\circ=U_{(0,\alpha)}.x_\circ $.
\end{proof}

\section{Linearization of \texorpdfstring{$i^*\bigl(\mathcal{O}(1)\bigr)$}{i*(O(1))} for \texorpdfstring{$U(\gamma_1,\gamma_2)$}{U(gamma1, gamma2)}}\label{section4}

In this section, we prove linearization of $i^*\parens[\big]{\mathcal{O}(1)}$ for the action of $U(\gamma_1,\gamma_2)$, following the arguments and results from~\cite{knop1989local}. Further, we recall a Lemma on weight vectors from~\cite{MR0396773}.

\begin{lemm}\label{line bundle is linearisable}
Let $H=(T\times T)U(\gamma_1,\gamma_2)$. Let $X=\bar{B}\subseteq \mathbb{P}\parens[\big]{\End\parens[\big]{V(2\rho)}}$. Let $i \colon X \hookrightarrow \mathbb{P}\parens[\big]{\End\parens[\big]{V(2\rho)}}$ be the inclusion, and let $\mathcal{L}=i^*\parens[\big]{\mathcal{O}(1)}$. Then $\mathcal{L}$ is $H$-linearizable.
\end{lemm}
\begin{proof}
By~\cite[Proposition~4.5, p.~74, and Proposition~2.4, p.~67]{knop1989local} and the proof of~\cite[Proposition~2.4, p.~67]{knop1989local}, $\mathcal{L}^{\otimes\abs*{\Pic(H)}}$ is $H$-linearizable. Further, since $\mathbb{C}[H]$ is a unique factorization domain, $\Cl(H)=\{0\}$ (see~\cite[Proposition~6.2, p.~131]{MR0463157}). By~\cite[Corollary~6.16, p.~145]{MR0463157}, since $H$ is smooth, we have $\Pic(H)=\Cl(H)$. Therefore $\abs[\big]{\Pic(H)}=1$. Hence $\mathcal{L}$ is $H$-linearizable.
\end{proof}

\begin{coro}\label{lifting action}
Let $X=\bar{B}\subseteq \mathbb{P}\parens[\big]{\End\parens[\big]{V(2\rho)}}$. Let $\hat{X}\subseteq \End\parens[\big]{V(2\rho)}$ be the cone over $X$. Let $\id\in \End\parens[\big]{V(2\rho)}$ be the identity endomorphism. Let $V$ be the $ \mathbb{C}$-linear span of $(B\times B).(\id)$. Then we have:
\begin{enumerate}\romanenumi
\item \label{lifting action_1} $V=\mathbb{C}$-linear span of $\hat{X}$;
\item \label{lifting action_2} there is a rational representation $\Psi \colon H\to \GL(V)$ such that $\pi\parens[\big]{\Psi(h).x}=h.\pi(x)$ for every $h\in H$ and $x\in \hat{X}\setminus\{0\}$.
\end{enumerate}
\end{coro}
\begin{proof}
Proof of~\eqref{lifting action_1} follows from the fact that $\bar{B}=\bar{(B\times B).x_\circ}$.

Proof of~\eqref{lifting action_2}. Let $R=\oplus_{d\in \mathbb{Z}_{\geq 0}} H^0(\bar{B},\mathcal{L}^{\otimes d})$. Since $\mathcal{L}=i^*\parens[\big]{\mathcal{O}(1)}$ is very ample, $H^0(X,\mathcal{L})$ is base point free. By~\cite[p.~259]{MR2810322}, $i_X \colon X\hookrightarrow\mathbb{P}\parens[\big]{H^0(X,\mathcal{L})^*}$ is a closed immersion. Further, $i_X$ is $H$-equivariant. From~\cite[Corollary~3.5, p.~679]{kannan2002projective}, $X$ is projectively normal relative to $\mathcal{L}$.

Hence,
\[
\Sym\parens[\big]{H^0(X,\mathcal{L})} \to R
\]
is surjective and $R=\mathbb{C}[\hat{X}]$. Hence $\hat{X}$ is a closed subvariety of $H^0(X,\mathcal{L})^*$. From Lemma~\ref{line bundle is linearisable}, $\mathcal{L}$ is $H$-linearizable. Therefore, $H^0(X,\mathcal{L})$ is a rational $H$-module (see~\cite[Lemma, p.~67]{knop1989local}). Hence $\Sym\parens[\big]{H^0(X,\mathcal{L})}$ is a $H$-algebra.

Further $\hat{X}$ is a $H$ stable subvariety of $H^0(X,\mathcal{L})^*$ follows from the following commutative diagram:
\[
\begin{tikzcd}
\hat{X}\setminus\{0\} \arrow[r,hookrightarrow,"i_{\hat{X}}"] \arrow[d,"\pi\restriction{\hat{X}\setminus\{0\}}"] &amp; H^0(X,\mathcal{L})^* \setminus\{0\}\arrow[d, "\pi"]\\
X \arrow[r,hookrightarrow,"i_X"] &amp; \mathbb{P}\parens[\big]{H^0(X,\mathcal{L})^*}.
\end{tikzcd}
\]
Recall that $i_X$ is $H$-equivariant and clearly $\pi$ is $H$-equivariant. Then from the above diagram, $\hat{X}\setminus\{0\}$ is $H$-stable.
Hence from part~\eqref{lifting action_1}, $V$ is a $H$-submodule of $H^0(X,\mathcal{L})^*$. Hence~\eqref{lifting action_2} follows from the above commutative diagram.
\end{proof}

Proof of the following lemma is along the lines of proof~\cite[Lemma, p.~102]{MR0396773}.

\begin{lemm}\label{weight vectors}
Let $T'$ be a torus. Let $V_1$ be a finite dimensional rational $T'$-module. Let $V_2\subset V_1$ be a $T'$ submodule. Let $v\in V_2\setminus\{0\}$. Write $v=\sum_{i=1}^{r}v_i$, where each $v_i\in V_1$ is a weight vector of weight $\mu_i$ $(1\leq i\leq r)$ with $\mu_1,\dotsc,\mu_r$ are distinct weights. Then $v_i\in V_2$ for every $1\leq i\leq r$.
\end{lemm}

\section{Proof of \texorpdfstring{$\Gamma_0=\emptyset$}{Gamma0= emptyset}}\label{section5}

In this section, we prove that $\Gamma_0$ is empty.

\begin{lemm}\label{gamma1 gamma2 are positive}
We have:
\begin{enumerate}\romanenumi
\item $\gamma_1\gneq0$ and $ \gamma_2\gneq 0$.
\item Let $\alpha=\gamma_1+\gamma_2$, $k=\langle 2\rho,\alpha\rangle$. Then both $2\rho-k\gamma_1$ and $2\rho-k\gamma_2$ are weights of $V(2\rho)$.
\end{enumerate}
\end{lemm}
\begin{proof}
By Corollary~\ref{alpha is a positive root}, we have $\alpha\in R^+(G)$ and
\[
U(\gamma_1,\gamma_2).x_\circ=U_{(\alpha,0)}.x_\circ=U_{(0,\alpha)}.x_\circ.
\]
Let $M$ be the set of all weights of $V(2\rho)$. Then, we have $\dim\parens[\big]{V(2\rho)_\mu} =  \dim\parens[\big]{V(2\rho)_{\sigma(\mu)}}$ for $\sigma\in W$, $\mu\in M$ (see~\cite[Theorem, p.~113]{MR0499562}). Further $-2\rho=w_0(2\rho)$ is the lowest weight in $M $, where $w_0$ is the longest element of $W$. Hence we have $\dim\parens[\big]{V(2\rho)_{-2\rho}}=1$. Note that for every $\mu\in M$, there is a basis $\braces[\big]{v_{\mu,i}:1\leq i\leq m_\mu}$ of $V(2\rho)_\mu$ and a basis $\braces[\big]{v_{-\mu,i}:1\leq i\leq m_\mu}$ (dual to $\braces[\big]{v_{\mu,i}:1\leq i\leq m_\mu}$) of $V(2\rho)^*_{-\mu}$ such that
\[
\id=v_{-2\rho}\otimes v_{2\rho}+\sum_{\mu\in M\setminus\{2\rho\}} \, \sum_{i=1}^{m_\mu}v_{-\mu,i}\otimes v_{\mu,i}.
\]
Note that $v_{2\rho}$ (respectively, $v_{-2\rho}$) is a highest weight vector (respectively, lowest weight vector) of $V(2\rho)$.

Define, $\tau_\alpha\coloneqq\exp(\mathrm{x}_\alpha)\exp(-\mathrm{x}_{-\alpha})\exp(\mathrm{x}_\alpha)$. Therefore, we have
\[
(\tau_\alpha,1).\id= (\tau_\alpha,1).\parens*{v_{-2\rho}\otimes v_{2\rho}+\sum_{\mu\in M\setminus\{2\rho\}} \, \sum_{i=1}^{m_\mu}v_{-\mu,i}\otimes v_{\mu,i}}.
\]

Also $\langle -2\rho, \alpha \rangle\in \mathbb{Z}_{&lt;0} $ and $\tau_\alpha(v_{-2\rho})\in V(2\rho)_{s_\alpha(-2\rho)}$ (see~\cite[p.~114]{MR0499562}). So, we have
\[
\tau_\alpha(v_{-2\rho})=\frac{\mathrm{x}_\alpha^{\langle2\rho,\alpha\rangle}}{\langle 2\rho, \alpha \rangle\mathrel{!}}. v_{-2\rho}.
\]
Hence, we have $v_{s_\alpha(-2\rho)}\otimes v_{2\rho}$ appears in $(\tau_\alpha,1).\id$ with non zero coefficient. Also by~\cite[Lemma~C, p.~96]{MR0396773}, we have $u_{\alpha}(a)= \exp(a\mathrm{x}_\alpha)$, for all $a\in \mathbb{C}$. We have
\begin{equation}\label{5.1}
\begin{split}
\parens[\big]{u_\alpha(a),1}.\id
&amp; = \parens[\big]{u_\alpha(a).v_{-2\rho}} \otimes v_{2\rho}+\sum_{\mu\in M\setminus\{2\rho\}} \, \sum_{i=1}^{m_\mu}\parens[\big]{u_\alpha (a).v_{-\mu,i}} \otimes v_{\mu,i} \\
&amp; = \parens*{\parens*{1+a\mathrm{x}_\alpha+\frac{a^2\mathrm{x}_\alpha^2}{2\mathrel{!}}+\dotsb+\frac{a^{\langle 2\rho, \alpha\rangle}\mathrm{x}_\alpha^{\langle 2\rho, \alpha\rangle}}{\langle 2\rho, \alpha\rangle\mathrel{!}}}.v_{-2\rho}} \otimes v_{2\rho} \\
&amp; \quad + \sum_{\mu\in M\setminus\{2\rho\}} \, \sum_{i=1}^{m_\mu}\parens[\big]{u_\alpha(a).v_{-\mu,i}}\otimes v_{\mu,i}
\end{split}
\end{equation}
for all $a\in \mathbb{C}$. For $a=1$, $v_{s_\alpha(-2\rho)} \otimes v_{2\rho}$ appears in the expression of $u_{(\alpha,0)}(1).\id=\parens[\big]{u_\alpha(1),1}.\id$ with nonzero coefficient.

Let $X$, $\hat{X}$, $V$ and $H$ be as in Corollary~\ref{lifting action}. Then by Corollary~\ref{lifting action}, there is a homomorphism $\Psi \colon H\to \GL(V)$ such that $\pi\parens[\big]{\Psi(h).x}=h.\pi(x)$ for every $h\in H$ and $x\in \hat{X}\setminus\{0\}$. Therefore, using Corollary~\ref{alpha is a positive root}, we have $\alpha\in R^+(G)$ and
\[
H.x_\circ=(T\times T)U_{(\alpha,0)}.x_\circ=(T\times T)U_{(0,\alpha)}.x_\circ.
\]
Hence, by Corollary~\ref{lifting action} we have
\[
H.\id=(T\times T)U_{(\alpha,0)}.\id=(T\times T)U_{(0,\alpha)}.\id.
\]
Hence, by~\eqref{5.1}, there is an element $u\in U(\gamma_1,\gamma_2)\setminus\{1\}$ such that the coefficient of $v_{s_\alpha(-2\rho)} \otimes v_{2\rho}$ in expression of $u.\id$ is non zero. Further, by Lemma~\ref{weight vectors}, $u.\parens[\big]{\sum_{i=1}^{m_\mu}v_{-\mu,i}\otimes v_{\mu,i}}\in V$ for all $\mu\in M$. Let $\mu$ be a weight of $V(2\rho)$ such that $v_{s_{\alpha}(-2\rho)} \otimes v_{2\rho}$ appears in
\begin{equation}\label{5.2}
u.\parens*{\sum_{i=1}^{m_\mu}v_{-\mu,i}\otimes v_{\mu,i}}
\end{equation}
with nonzero coefficient. Further
\begin{equation}\label{5.3}
u.\parens*{\sum_{i=1}^{m_\mu}v_{-\mu,i}\otimes v_{\mu,i}} = \sum_{l\in \mathbb{Z}_{\geq 0}}\frac{\mathrm{x}^l}{l\mathrel{!}}.\parens*{\sum_{i=1}^{m_\mu}(v_{-\mu,i}\otimes v_{\mu,i})} \in \sum_{l\in \mathbb{Z}_{\ge 0}} V(2\rho)^*_{-\mu+l\gamma_1}\otimes V(2\rho)_{\mu+l\gamma_2}.
\end{equation}
Hence, by~\eqref{5.2} and~\eqref{5.3}, we have
\begin{equation}\label{5.4}
s_\alpha(-2\rho)=-2\rho+\langle 2\rho, \alpha\rangle \alpha=-\mu+l_0\gamma_1
\end{equation}
and
\begin{equation}\label{5.5}
2\rho=\mu+l_0\gamma_2
\end{equation}
for some $l_0 \in \mathbb{Z}_{\geq 0}$. From~\eqref{5.4} and~\eqref{5.5} we have following:
\begin{itemize}
\item $\langle 2\rho,\alpha \rangle=l_0$;
\item $ 2\rho-\langle 2\rho,\alpha \rangle\gamma_2$ is a weight of $V(2\rho)$;
\item $\gamma_2\gneq0$.
\end{itemize}
Similarly for $\gamma_1$, consider $\parens[\big]{1,u_\alpha(1)}.\id$. Because $-2\rho$ is a weight of $V(2\rho)$, $v_{2\rho} \otimes v_{-s_\alpha(2\rho)}$ appears with nonzero coefficient in the expression of $u^\prime.\id$ for some $u^\prime\in U_{(0,\alpha)}$. Hence $ 2\rho-\langle 2\rho,\alpha \rangle\gamma_1$ is a weight of $V(2\rho)$ and $\gamma_1\gneq 0$.
\end{proof}

\begin{coro}\label{cor commuting}
We have:
\begin{enumerate}\romanenumi
\item \label{cor commuting_1} there are no roots $\alpha,\beta\in R^+(G)$ such that $(\gamma_1,\gamma_2)+(\beta,0)=(0,\alpha)$;
\item \label{cor commuting_2} there are no roots $\alpha,\beta\in R^+(G)$ such that $(\gamma_1,\gamma_2)+(0,\beta)=(\alpha,0)$.
\end{enumerate}
In particular, every element of $U(\gamma_1,\gamma_2)$ commutes with every element of $U\times 1$ and every element of\/ $1\times U$.
\end{coro}
\begin{proof}
Proof of~\eqref{cor commuting_1}: if $(\gamma_1,\gamma_2)+(\beta,0)=(0,\alpha)$ for some $\alpha,\beta\in R^+(G)$, then $\gamma_1=-\beta$ and $\gamma_2=\alpha$ which contradicts $\gamma_1\gneq0$ by Lemma~\ref{gamma1 gamma2 are positive}. Similar proof can be given for~\eqref{cor commuting_2}.
\end{proof}

\begin{lemm}
We have $\alpha_\circ=\gamma_1+\gamma_2$.\label{alpha equals to alpha0 }
\end{lemm}
\begin{proof}
Let $\alpha=\gamma_1+\gamma_2$. Let $u\in U$. By Corollaries~\ref{alpha is a positive root} and~\ref{cor commuting} we have
\begin{equation}\label{5.6}
\begin{split}
\parens[\big]{u_{(\gamma_1,\gamma_2)}(1)(u,1)}.x_\circ
&amp; = \parens[\big]{(u,1)u_{(\gamma_1,\gamma_2)}(1)}.x_\circ 	\\
&amp; = (u,1).\parens[\big]{u_{(\gamma_1,\gamma_2)}(1).x_\circ}	\\
&amp; = (u,1).\parens[\big]{u_{(\alpha,0)}(a).x_\circ}
\end{split}
\end{equation}
and
\begin{equation}\label{5.7}
\begin{split}
\parens[\big]{u_{(\gamma_1,\gamma_2)}(1)(1,u^{-1})}.x_\circ
&amp; = \parens[\big]{(1,u^{-1})u_{(\gamma_1,\gamma_2)}(1)}.x_\circ	\\
&amp; = (1,u^{-1}).\parens[\big]{u_{(\gamma_1,\gamma_2)}(1).x_\circ}	\\
&amp; = (1,u^{-1}).\parens[\big]{u_{(0,\alpha)}(b).x_\circ}
\end{split}
\end{equation}
for some $a,b\in\mathbb{C}^\times$. Since
\[
(u,1).x_\circ=(1,u^{-1}).x_\circ,
\]
we have
\[
u_{(\gamma_1,\gamma_2)}(1).\parens[\big]{(u,1).x_\circ} = u_{(\gamma_1,\gamma_2)}(1).\parens[\big]{(1,u^{-1}).x_\circ}
\]
for every $u\in U$.

From~\eqref{5.6} and~\eqref{5.7}
\begin{equation}\label{5.8}
(u,1).\parens[\big]{u_{(\alpha,0)}(a).x_\circ} = (1,u^{-1}).\parens[\big]{u_{(0,\alpha)}(b).x_\circ}
\end{equation}
for every $u\in U$.

Take $u=1$,
\[
\parens[\big]{u_{\alpha}(a),1}.x_\circ = \parens[\big]{1,u_\alpha(b)}.x_\circ.
\]
Therefore, we have
\[
\parens[\big]{u_{\alpha}(a), u_\alpha(b)^{-1}}.x_\circ=x_\circ.
\]
Hence, we have $b=-a$. Using~\eqref{5.8}, we have
\[
(1,u^{-1}).\parens[\big]{u_{(\alpha,0)}(a).x_\circ} = (1,u^{-1}).\parens[\big]{u_{(0,\alpha)}(-a).x_\circ} = (u,1).\parens[\big]{u_{(\alpha,0)}(a).x_\circ}.
\]
Hence, $(u,u)$ fixes $u_{(\alpha,0)}(a).x_\circ$ for every $u\in U$. Thus, we have $u_{(\alpha,0)}(a)\in Z(U)=U_{\alpha_\circ}$. Since $a\neq 0$, we have $\alpha=\alpha_\circ$.
\end{proof}

\begin{lemm}\label{langle with alpha0 is 1}
Let $(\nu_1,\nu_2)\in \mathrm{X}(T)\oplus \mathrm{X} (T)$ be a character of $T\times T$ satisfying
\begin{enumerate}\romanenumi
\item \label{langle with alpha0 is 1_1} $\nu_1\gneq 0$, $\nu_2\gneq 0$ and $\alpha_\circ=\nu_1+\nu_2$;
\item $2\rho-\langle 2\rho,\alpha_\circ\rangle\nu_1$ and\/ $2\rho-\langle 2\rho,\alpha_\circ\rangle\nu_2$ are weights of $V(2\rho)$.
\end{enumerate}
Then we have, $\langle \nu_1,\alpha_\circ \rangle=1$ and $\langle \nu_2,\alpha_\circ \rangle=1$.
\end{lemm}
\begin{proof}
By hypothesis~\eqref{langle with alpha0 is 1_1}, we have $\langle\nu_i,\alpha_\circ\rangle\in \mathbb{Z}_{\geq0}$ for every $i=1,2$ and $ \langle \nu_1,\alpha_\circ\rangle+\langle \nu_2,\alpha_\circ\rangle=\langle \nu_1+\nu_2,\alpha_\circ\rangle=\langle\alpha_\circ,\alpha_\circ\rangle=2$. Therefore, $\langle\nu_i,\alpha_\circ\rangle\in \{0,1,2\}$ for $i=1,2$.

Assume on the contrary that $\langle\nu_i,\alpha_\circ\rangle=2 $ for some $i=1,2$. Without loss of generality, we may assume that $\langle\nu_1,\alpha_\circ\rangle=2 $. Then
\[
\begin{split}
s_{\alpha_\circ}(2\rho -\langle 2\rho,\alpha_\circ\rangle \nu_1)
&amp; = 2\rho-\langle 2\rho,\alpha_\circ\rangle\alpha_\circ-\langle 2\rho,\alpha_\circ\rangle\nu_1+\langle 2\rho,\alpha_\circ\rangle\langle\nu_1,\alpha_\circ\rangle \alpha_\circ 	\\
&amp; = 2\rho-\langle 2\rho,\alpha_\circ\rangle\alpha_\circ-\langle 2\rho,\alpha_\circ\rangle\nu_1+\langle 2\rho,\alpha_\circ\rangle2 \alpha_\circ 									\\
&amp; = 2\rho +\langle 2\rho,\alpha_\circ\rangle(\alpha_\circ-\nu_1)			\\
&amp; = 2\rho+\langle 2\rho,\alpha_\circ\rangle\nu_2
\end{split}
\]
is a weight of $V(2\rho) $ and $ 2\rho+\langle 2\rho,\alpha_\circ\rangle\nu_2&gt;2\rho$ ($\because \ \nu_2\gneq 0$ and $ \langle 2\rho,\alpha_\circ\rangle\in \mathbb{Z}_{&gt;0}$). This gives a con\-tra\-dic\-tion to $2\rho$ being highest weight of $V(2\rho)$. Therefore, we have
\[
\langle \nu_1, \alpha_\circ \rangle = 1 \qquad \text{and} \qquad \langle \nu_2, \alpha_\circ \rangle = 1. \qedhere
\]
\let\qed\relax
\end{proof}
\begin{coro}\label{ cor langle with alpha0 is 1}
We have $\langle \gamma_1,\alpha_\circ \rangle=1$ and $\langle \gamma_2,\alpha_\circ \rangle=1$.
\end{coro}
\begin{proof}
By Lemmas~\ref{gamma1 gamma2 are positive} and~\ref{alpha equals to alpha0 }, $(\gamma_1$,$\gamma_2)$ satisfies hypothesis of Lemma~\ref{langle with alpha0 is 1}. Hence, we have $\langle \gamma_1,\alpha_\circ \rangle=1$ and $\langle \gamma_2,\alpha_\circ \rangle=1$.
\end{proof}

\begin{lemm}\label{non existence}
Let G be of type different from $A_1$ and $G_2$. Then there are no weights $\nu_1,\nu_2$ such that
\begin{enumerate}\romanenumi
\item \label{non existence_1} $\nu_1\gneq 0$, $\nu_2\gneq 0$ and $ \nu_1+\nu_2=\alpha_\circ$;
\item \label{non existence_2} $2\rho-\langle 2\rho, \alpha_\circ\rangle\nu_1$ and\/ $2\rho-\langle 2\rho, \alpha_\circ\rangle\nu_2$ are weights of $V(2\rho)$;
\item \label{non existence_3} $\langle \nu_1,\alpha_\circ\rangle=1$ and $\langle \nu_2,\alpha_\circ\rangle=1$.
\end{enumerate}
\end{lemm}
\begin{proof}
Assume on the contrary such $\nu_1$ and $\nu_2$ exist. Using~\eqref{non existence_1} and~\eqref{non existence_3} one can show there exist $\alpha_i,\alpha_j\in S$ $(1\leq i,j\leq n)$ ($i,j$ need not be distinct) such that
\begin{equation}\label{5.9}
\alpha_i\leq \nu_1, \quad \alpha_j\leq \nu_2,
\end{equation}
both $\langle \alpha_\circ,\alpha_i\rangle$ and $\langle \alpha_\circ,\alpha_j\rangle$ are positive. Further, since
\[
\langle \nu_1-\alpha_i,\alpha_\circ\rangle=0 \qquad \text{and} \qquad \langle \nu_2-\alpha_j,\alpha_\circ\rangle=0,
\]
we have
\begin{equation}\label{5.10}
\alpha_i\nleq \nu_1-\alpha_i \qquad \text{and} \qquad \alpha_j\nleq \nu_2-\alpha_j.
\end{equation}
Let $\beta_1,\beta_2\in R^+(G)$ of maximal height such that
\begin{equation}\label{5.11}
\alpha_i\leq \beta_1\leq \nu_1 \qquad \text{and} \qquad \alpha_j\leq \beta_2\leq \nu_2.
\end{equation}
Let
\[
\nu_1=\beta_1+\nu_1' \qquad \text{and} \qquad \nu_2=\beta_2+\nu_2'
\]
for some $\nu_1'\geq 0$, $\nu_2'\geq 0$.

By maximality of $\beta_1$, and $\beta_2$, we have
\[
\langle\nu_1', \beta_1\rangle\geq 0 \qquad \text{and} \qquad \langle\nu_2', \beta_2\rangle\geq 0.
\]
Hence, we have
\begin{equation}\label{5.12}
\langle\nu_1, \beta_1\rangle\geq 2 \qquad \text{and} \qquad \langle\nu_2, \beta_2\rangle\geq 2.
\end{equation}
\newstepset
\begin{step}
We prove that if one of the following inequalities
\begin{equation}\label{5.13}
\langle 2\rho,\beta_1\rangle\lneq \langle 2\rho, \alpha_\circ\rangle
\end{equation}
or
\begin{equation}\label{5.14}
\langle 2\rho,\beta_2\rangle\lneq \langle 2\rho, \alpha_\circ\rangle
\end{equation}
holds, then we will arrive at contradiction.

Without loss of generality assume that $\langle 2\rho,\beta_1\rangle\lneq \langle 2\rho, \alpha_\circ\rangle$. Since $2\rho-\langle 2\rho,\alpha_\circ \rangle \nu_1$ is a weight of $V(2\rho)$, $s_{\beta_1}(2\rho-\langle 2\rho,\alpha_\circ \rangle \nu_1)$ is also a weight of $V(2\rho)$. Consider
\[
s_{\beta_1}\parens[\big]{2\rho-\langle 2\rho,\alpha_\circ \rangle \nu_1} = 2\rho -\langle 2\rho,\alpha_\circ \rangle \nu_1-\langle 2\rho,\beta_1 \rangle\beta_1+\langle 2\rho,\alpha_\circ \rangle \langle \nu_1,\beta_1 \rangle \beta_1.
\]
By~\eqref{5.9} and~\eqref{5.10}, coefficient of $\alpha_i$ in $s_{\beta_1}(2\rho-\langle2\rho,\alpha_\circ \rangle\nu_1)-2\rho$ is
\[
-\langle 2\rho,\alpha_\circ \rangle-\langle 2\rho,\beta_1 \rangle+\langle 2\rho,\alpha_\circ \rangle \langle \nu_1,\beta_1 \rangle.
\]
By assumption on $\beta_1$, from~\eqref{5.12} and~\eqref{5.13}
\[
-\langle 2\rho,\alpha_\circ \rangle-\langle 2\rho,\beta_1 \rangle+\langle 2\rho,\alpha_\circ \rangle \langle \nu_1,\beta_1 \rangle\gneq 0.
\]

Thus, we have $s_{\beta_1}(2\rho-\langle2\rho,\alpha_\circ \rangle\nu_1)-2\rho\nleq 0$. This is a contradiction to $2\rho$ being highest weight of $V(2\rho) $. Hence such $\nu_1,\nu_2$ does not exist if either~\eqref{5.13} or~\eqref{5.14} holds.
\end{step}

\begin{step}
Note that if either $\beta_1$ or $\beta_2$ is a long root then clearly one of the inequalities~\eqref{5.13}, \eqref{5.14} will hold. If $G$ is simply laced case, all roots are of same length and hence both $\beta_1$ and $\beta_2$ are long roots. Therefore, we may assume $G$ is not simply laced and of type different from $G_2$.

If both $\beta_1$ and $\beta_2$ are short roots it is not evident that one of the inequalities~\eqref{5.13}, \eqref{5.14} will always hold. We prove case by case that one of the inequalities~\eqref{5.13}, \eqref{5.14} will always hold.

For the ordering of simple roots, we follow the convention in~\cite[p.~57--58]{MR0499562}.

\begin{enonce*}[remark]{Type \boldmath{$B_2$}}
In this case, $\alpha_\circ=\alpha_1+2\alpha_2=2\omega_2$ where $\alpha_1$ is a long root and $\alpha_2$ is a short root. Here $i=j=2$ as in~\eqref{5.9}. Let $\beta_1, \beta_2$ be as in~\eqref{5.11}. Since $\beta_\circ=\alpha_1+\alpha_2$ is the highest short root, we have $\beta_1\leq \beta_\circ$ and $\beta_2\leq \beta_\circ$. Hence, we have either $\beta_1=\alpha_2$ or $ \beta_2=\alpha_2$. Moreover $\langle 2\rho,\alpha_2\rangle=2$ and $\langle 2\rho,\alpha_\circ\rangle=4$.
\end{enonce*}

\begin{enonce*}[remark]{Type \boldmath{$B_n$; \ $n \geq 3$}}
In this case, $\alpha_\circ=\alpha_1+2\alpha_2+\dotsb+2\alpha_n=\omega_2$ where $\alpha_1,\dotsc,\alpha_{n-1}$ are long roots and $\alpha_n$ is a short root. Here, we have $i=j=2$ as in~\eqref{5.9}. Let $\beta_1, \beta_2$ be as in \eqref{5.11}. As $\beta_\circ=\alpha_1+\alpha_2+\dotsb+\alpha_n$ is the highest short root, $\beta_1\leq \beta_\circ$ and $\beta_2\leq \beta_\circ$. Then either $\beta_1\leq \alpha_2+\dotsb+\alpha_n$ or $\beta_2\leq \alpha_2+\dotsb+\alpha_n$. Hence, we have either $\langle2\rho,\beta_1\rangle\leq\langle2\rho,\alpha_2+\dotsb+\alpha_n\rangle $ or $\langle2\rho,\beta_2\rangle\leq\langle2\rho,\alpha_2+\dotsb+\alpha_n\rangle$. Moreover $\langle2\rho,\alpha_2+\dotsb+\alpha_n\rangle=4n-6$ and $\langle2\rho,\alpha_\circ\rangle=4n-4$.
\end{enonce*}

\begin{enonce*}[remark]{Type \boldmath{$C_n$; $n\geq 3$}}
In this case, $\alpha_\circ=2\alpha_1+2\alpha_2+\dotsb+2\alpha_{n-1}+\alpha_n=2\omega_1$ where $\alpha_1,\dotsc,\alpha_{n-1}$ are short roots and $\alpha_n$ is a long root. Here, we have $i=j=1$ as in~\eqref{5.9}. Let $\beta_1, \beta_2$ be as in \eqref{5.11}. As $\beta_\circ=\alpha_1+2\alpha_2+\dotsb+2\alpha_{n-1}+\alpha_n$ is the highest short root, $\beta_1\leq \beta_\circ$ and $\beta_2\leq \beta_\circ$. Then either $\beta_1\leq \alpha_1+\dotsb+\alpha_{n-1}$ or $\beta_2\leq \alpha_1+\dotsb+\alpha_{n-1}$. Hence, we have either $\langle2\rho,\beta_1\rangle\leq\langle2\rho,\alpha_1+\dotsb+\alpha_{n-1}\rangle $ or $\langle2\rho,\beta_2\rangle\leq\langle2\rho,\alpha_1+\dotsb+\alpha_{n-1}\rangle$. Moreover $\langle2\rho,\alpha_1+\dotsb+\alpha_{n-1}\rangle=2n-2$ and $\langle2\rho,\alpha_\circ\rangle=2n$.
\end{enonce*}

\begin{enonce*}[remark]{Type \boldmath{$F_4$}}
$\alpha_\circ=2\alpha_1+3\alpha_2+4\alpha_3+2\alpha_4=\omega_1$ where $\alpha_1,\alpha_2$ are long roots and $\alpha_3$ and $\alpha_4$ are short roots. Here, we have $i=j=1$ as in~\eqref{5.9}. Let $\beta_1, \beta_2$ be as in \eqref{5.11}. As $\beta_\circ=\alpha_1+2\alpha_2+3\alpha_3+2\alpha_4$ is the highest short root, $\beta_1\leq \beta_\circ$ and $\beta_2\leq \beta_\circ$. Then one of the $\beta_1,\beta_2$ is of the form $\alpha_1+\alpha_2+ a\alpha_3+b\alpha_4$ for some $a,b\in \mathbb{Z}_{\geq 0}$. Then possible choices for $a$ are $0$, $1$, $2$, $3$ and for $b$ are $0$, $1$, $2$.

\begin{case}[$a=0$]
Then for any choice of $b$, $\alpha_1+\alpha_2+ a\alpha_3+b\alpha_4$ is not a short root because $\alpha_1$+$\alpha_2$ is a long root and $\alpha_1+\alpha_2+\alpha_4$, $\alpha_1+\alpha_2+2\alpha_4$ are not roots.
\end{case}

\begin{case}[$a=1$]
Then $b=0$ or $1$ because $\alpha_1+\alpha_2+\alpha_3+2\alpha_4$ is not a root.
\end{case}
\begin{case}[$a=2$]
Then $b=1$ because $\alpha_1+\alpha_2+2\alpha_3$ and $\alpha_1+\alpha_2+2\alpha_3+2\alpha_4$ are long roots.
\end{case}
\begin{case}
Further $a=3$ is not possible since, $\alpha_1+\alpha_2+3\alpha_3$, $\alpha_1+\alpha_2+3\alpha_3$+$\alpha_4$ and $\alpha_1+\alpha_2+3\alpha_3+2\alpha_4$ are not roots.
\end{case}
Thus, either $\beta_1$ or $\beta_2$ is in $\braces[\big]{\alpha_1+\alpha_2+\alpha_3,\alpha_1+\alpha_2+\alpha_3+\alpha_4,\alpha_1+\alpha_2+2\alpha_3+\alpha_4}$. Further, we have $\langle 2\rho,\alpha_1+\alpha_2+\alpha_3\rangle=10$, $\langle 2\rho,\alpha_1+\alpha_2+\alpha_3+\alpha_4\rangle=12$ and $\langle 2\rho,\alpha_1+\alpha_2+2\alpha_3+\alpha_4\rangle=14$. As $\langle 2\rho,\alpha_\circ\rangle=16$, for all possible cases, one of the inequalities~\eqref{5.13}, \eqref{5.14} always hold.\qedhere
\end{enonce*}
\end{step}
\let\qed\relax
\end{proof}

\begin{coro}
If $G$ is of type different from $A_1$ and $G_2$, then $\Gamma_0$ is empty.\label{cor non existence}
\end{coro}
\begin{proof}
Assume that $\Gamma_0\neq \emptyset$. Let $(\gamma_1,\gamma_2)$ be maximal element of $\Gamma_0$ as before. Then by Lemma~\ref{gamma1 gamma2 are positive}, Lemma~\ref{alpha equals to alpha0 } and Corollary~\ref{ cor langle with alpha0 is 1}, $\gamma_1$,$\gamma_2$ satisfies following properties:
\begin{enumerate}\romanenumi
\item $\gamma_1\gneq 0$, $\gamma_2\gneq 0$ and $ \gamma_1+\gamma_2=\alpha_\circ$;
\item $2\rho-\langle 2\rho, \alpha_\circ\rangle\gamma_1$ and $2\rho-\langle 2\rho, \alpha_\circ\rangle\gamma_2$ are weights of $V(2\rho)$;
\item $\langle \gamma_1,\alpha_\circ\rangle=1$ and $\langle \gamma_2,\alpha_\circ\rangle=1$.
\end{enumerate}
But this is a contradiction to Lemma~\ref{non existence}. Hence $\Gamma_0$ is empty.
\end{proof}

\begin{lemm}
Let $G$ be of type $G_2$. Then $\Gamma_0$ is empty.\label{non existence for G2}
\end{lemm}
\begin{proof}
Let $(\gamma_1,\gamma_2)$ be a maximal element of $\Gamma_0$ as before. Here $\alpha_\circ=3\alpha_1+2\alpha_2=\omega_2$ (see~\cite[p.~57--58]{MR0499562} for convention). Then from Corollary~\ref{ cor langle with alpha0 is 1} we have
\[
\alpha_2\leq \gamma_1 \qquad \text{and} \qquad \alpha_2\leq \gamma_2.
\]
Let $\beta_1,\beta_2\in R^+(G)$ of maximal height such that
\[
\alpha_2\leq \beta_1\leq \gamma_1 \qquad \text{and} \qquad \alpha_2\leq \beta_2\leq \gamma_2.
\]
If either $\beta_1$ or $\beta_2$ is a long root then~\eqref{5.13} or~\eqref{5.14} will hold. Hence, using the same argument as in the proof of Lemma~\ref{non existence}, we will arrive at a contradiction. So assume both $\beta_1$ and $\beta_2$ are short roots. As $\beta_\circ=2\alpha_1+\alpha_2$ is the highest short root, $\beta_1\leq \beta_\circ$ and $\beta_2\leq \beta_\circ$. Then one of the $\beta_1, \beta_2$ is $2\alpha_1+\alpha_2$ and the other is $\alpha_1+\alpha_2$.

Without loss of generality, we may assume $\gamma_1=\beta_1=2\alpha_1+\alpha_2$ and $\gamma_2=\beta_2=\alpha_1+\alpha_2$. Then $\langle2\rho, \alpha_\circ \rangle =6$, $\langle2\rho,\gamma_1\rangle=10$ and $\langle2\rho,\gamma_2\rangle=8$. Note
\[
\mu_\circ=s_{\gamma_2}(2\rho)=2\rho-8\gamma_2
\]
is a weight of $V(2\rho)$. So, $\mathrm{x}_{(0,\alpha_\circ)}(v_{-\mu_\circ}\otimes v_{\mu_\circ})$ appears in $u_{(0,\alpha_\circ)}(1).\id$ with nonzero coefficient. Note that $U(\gamma_1,\gamma_2).\id=U_{(0,\alpha_\circ)}.\id$ by Corollary~\ref{alpha is a positive root} and Lemma~\ref{alpha equals to alpha0 }. There exists a weight $\mu$ of $V(2\rho)$ such that
\[
\mathrm{x}_{(0,\alpha_\circ)}(v_{-\mu_\circ}\otimes v_{\mu_\circ})=c.(v_{-\mu_\circ}\otimes v_{\mu_\circ+\alpha_\circ})
\]
appears in $u_{(\gamma_1,\gamma_2)}(a)(v_{-\mu}\otimes v_{\mu})$ with nonzero coefficient for some $a,c\in \mathbb{C}^\times$.

Hence, there exists $l\in\mathbb{Z}_{\geq 0}$ such that
\[
\mu+l\gamma_2=\mu_\circ+\alpha_\circ
\]
and
\[
-\mu+l\gamma_1=-\mu_\circ.
\]
Hence, we have $l=1$. Thus,
\[
\mu=2\rho-6\alpha_1-7\alpha_2
\]
is a weight of $V(2\rho)$. Now consider
\[
s_{\alpha_1+\alpha_2}(\mu)=\mu-\langle \mu,\alpha_1+\alpha_2\rangle(\alpha_1+\alpha_2)=2\rho+\alpha_1
\]
as $\langle \mu,\alpha_1+\alpha_2\rangle=-7$.
Since $\mu$ is a weight of $V(2\rho)$, $s_{\alpha_1+\alpha_2}(\mu)$ is also a weight of $V(2\rho)$. But $2\rho+\alpha_1&gt;2\rho$ which is a contradiction to $2\rho$ being highest weight of $V(2\rho)$. Hence, we have $\Gamma_0=\emptyset$.
\end{proof}

Thus, we have the following.
\begin{coro}\label{Unipotent subgroup}
Any $T\times T$-stable one dimensional unipotent subgroup of $A$ is of the form $U(\alpha,0)$ or $U(0,\alpha)$ for some $\alpha\in R^+(G)$.
\end{coro}

\section{Proof of Main Theorem}\label{section6}

In this section, we prove that $\aut^\circ(\bar{B})=B\times B$.
\begin{lemm}\label{multiplicity free}
Let $\alpha\in R^+(G)$. Then $U_{(\alpha,0)}$ (respectively, $U_{(0,\alpha)}$) are only one dimensional $T\times T$ stable unipotent subgroup of $A$ with character $(\alpha,0)$ (respectively, $(0,\alpha)$).
\end{lemm}
\begin{proof}
Define
\[
\Gamma' =
{\left\{
\alpha\in R^+(G) \;\middle\vert\;
\begin{aligned}
&amp; \text{there exist at least two 1-dimensional $T\times T$ stable}	\\[-.5ex]
&amp; \text{unipotent subgroups of $A$ with character $(\alpha,0)$}
\end{aligned}
\right\}}.
\]

Let $\beta\in \Gamma'$ be a maximal element with respect to the dominant ordering. Let $U'(\beta,0)$ be a closed, connected one dimensional unipotent subgroup of $A$ that is normalized $T\times T $ with character $(\beta,0)$ different from $U_{(\beta,0)}$.

By Corollary~\ref{Unipotent subgroup} and the fact $2\beta\notin R^+(G)$, every element of $U_{(\beta,0)}$ commutes with every element of $U'(\beta,0)$. Hence, $H=U'(\beta,0)U_{(\beta,0)}$ is closed connected 2-dimensional $T\times T$-stable subgroup of $A$. Also, notice that $U'(\beta,0).x_\circ \cap (B\times 1).x_\circ$ is a nonempty open subset of $U'(\beta,0).x_\circ $ which is $\bigtriangleup T$-stable and $\restr{(\beta,0)}{\bigtriangleup T}$ is a non trivial character. Hence, we have $U'(\beta,0).x_\circ=U_{(\beta,0)}.x_\circ$. Therefore, $\dim(H.x_\circ)=\dim(U_{(\beta,0)}.x_\circ)=1$. Thus, we have $\dim(H_{x_\circ})=1$.

Also, $H_{x_\circ}$ is a $\bigtriangleup T$-stable subgroup under conjugation. Since $T\times T$ acts on $H$ through $(\beta,0)$, $H_{x_\circ}$ is also $T\times T$-stable.

Since, for any $T\times T$-stable one dimensional unipotent subgroup of $A$, $T\times T$ acts through $(\alpha,0)$ or $(0,\alpha) $ for some $\alpha\in R^+(G)$, every element of $H$ commutes with every element of $U_{(0,\alpha)}$ for all $\alpha\in R^+(G)$. Since the action of $T\times T$ on $H$ is through $(\beta,0)$, every element of $1\times T$ commutes with every element of $H$. Hence, $H_{x_\circ}$ acts trivially on $\bar{B}$ giving a contradiction to $H_{x_\circ}$ being one dimensional. Therefore $\Gamma'$ is empty. Thus, we have for all $\alpha\in R^+(G)$, $U_{(\alpha,0)}$ is the only one dimensional $T\times T$ stable unipotent subgroup of $A$ with character $(\alpha,0)$. Similarly, for all $\alpha\in R^+(G)$, $U_{(0,\alpha)}$ is the only one dimensional $T\times T$ stable unipotent subgroup of $A$ with character $(0,\alpha)$.
\end{proof}

\begin{prop}
$T\times T$ is a maximal torus of $A$.\label{maximal torus}
\end{prop}
\begin{proof}
Let $T'$ be a maximal torus of $A$ containing $T\times T$. Consider the inclusion map
\[
i \colon T\times T \longhookrightarrow T'.
\]
$i$ induces a homomorphism
\[
f=\restr{i^*}{X(T')} \colon X(T')\to X(T\times T).
\]
Further, since $ T'$ is a torus, from~\cite[Exercise~3.2.10(2), p.~46]{MR1642713} $f$ is surjective.

By Lemma~\ref{multiplicity free}, for every $\alpha\in R^+(G)$, there exist unique characters $\mu_{1,\alpha}$ and $\mu_{2,\alpha}\in X(T')$ such that $f(\mu_{1,\alpha})=(\alpha,0)$ and $f(\mu_{2,\alpha})=(0,\alpha)$ respectively.

Consider the homomorphism of tori
\[
\phi \colon T'\to (\mathbb{C}^{\times})^n
\]
given by $\phi(s) = \parens[\big]{\mu_{1,\alpha_1}(s),\dotsc,\mu_{1,\alpha_n}(s)}$.

\begin{claim}
$\restr{\phi}{T'_{x_\circ}}$ is injective.
\end{claim}
Let $s\in T'_{x_\circ}\cap(\cap_{i=1}^n \Ker (\mu_{1,\alpha_i})$.

For any $i\in\{1,\dotsc, n\}$, $a\in \mathbb{C}^\times$, we have
\begin{equation}\label{1.41}
\parens[\big]{u_{\alpha_i}(a),u_{\alpha_i}(a)}.x_\circ=x_\circ.
\end{equation}
Action of $s$ on both sides gives
\begin{equation}\label{1.42}
\begin{split}
s.x_\circ
&amp; = u_{(\alpha_i,0)}\parens[\big]{\mu_{1,\alpha_i}(s)a}u_{(0,\alpha_i)}\parens[\big]{\mu_{2,\alpha_i}(s)a}.x_\circ \\
&amp; = u_{(\alpha_i,0)}(a)u_{(0,\alpha_i)}\parens[\big]{\mu_{2,\alpha_i}(s)a}.x_\circ,
\end{split}
\end{equation}
because $s\in \Ker(\mu_{1,\alpha_i})$ for all $1\leq i\leq n$. As $s.x_\circ=x_\circ$, from \eqref{1.41} and \eqref{1.42},
\[
\parens[\big]{u_{\alpha_i}(a),u_{\alpha_i}(a)}.x_\circ=u_{(\alpha_i,0)}(a)u_{(0,\alpha_i)}\parens[\big]{\mu_{2,\alpha_i}(s)a}.x_\circ.
\]
Hence, we have $s\in \Ker(\mu_{2,\alpha_i})$ for all $1\leq i\leq n $.

Let $a, b \in \mathbb{C}$. Then we have
\[
s.u_{(\alpha_i,0)}(a).x_\circ=u_{(\alpha_i,0)}\parens[\big]{\mu_{1,\alpha_i}(s)a}.x_\circ=u_{(\alpha_i,0)}(a).x_\circ
\]
and
\[
s.u_{(0,\alpha_i)}(b).x_\circ=u_{(0,\alpha_i)}\parens[\big]{\mu_{2,\alpha_i}(s)b}.x_\circ=u_{(0,\alpha_i)}(b).x_\circ
\]
for all $1\leq i\leq n$. Therefore $s$ fixes every element of $1\times U$ and $U\times 1$. Also $s$ is in $T'$, so $s$ commutes with every element of $T\times T$. Hence the action of $s$ on $\bar{B}$ is trivial. Therefore, $\restr{\phi}{T'_{x_\circ}}$ is injective. Hence, $\dim(T'_{x_\circ})\leq n$.
\begin{claim}
$\dim(T'.x_\circ)=\dim \parens[\big]{(T\times 1).x_\circ}$.
\end{claim}
Note that $T'.x_\circ\cap (B\times 1).x_\circ$ is $\bigtriangleup(T)$-stable open subset of $T'.x_\circ$. Let $(tu,1).x_\circ\in T'.x_\circ\cap (B\times 1).x_\circ$ where $t\in T$ and $u\in U$. Since $\bigtriangleup (T) \subseteq T'$, every element of $\bigtriangleup (T)$ commutes with every element of~$T'$. Hence, for all $s\in T$, we have
\[
\begin{split}
(tu,1).x_\circ
&amp; = (s,s).\parens[\big]{(tu,1).x_\circ} 				\\
&amp; = \parens[\big]{(s,s)(tu,1)(s,s)^{-1}(s,s)}.x_\circ	\\
&amp; = (stus^{-1},1).x_\circ.
\end{split}
\]
Hence, we have
\[
\begin{aligned}
stus^{-1}		&amp;=tu		\\
t^{-1}stus^{-1}	&amp;=u		\\
st^{-1}tus^{-1}	&amp;=u \quad (\because \ s \ \text{and} \ t^{-1} \text{commute})	\\
sus^{-1}			&amp;=u
\end{aligned}
\]
for all $s \in T$.

Hence, we have $u\in C_G(T)=T $ (see~\cite[Corollary, p.~155]{MR0396773}). Thus, we have $ u=1$. Hence $T'.x_\circ\cap (B\times 1).x_\circ$ is contained in $(T\times T).x_\circ$. Therefore, we have $\dim\parens[\big]{(T\times1).x_\circ} = \dim(T'.x_\circ$).

Hence, we have
\[
\begin{split}
\dim(T')
&amp; = \dim(T'.x_\circ) + \dim(T'_{x_\circ}) 	\\
&amp; \leq n+n 									\\
&amp; = \dim(T\times T).
\end{split}
\]
Therefore, we have $T'=T\times T$.
\end{proof}

\begin{theo}\label{main theorem}
We have\/ $\aut^{\circ} (\bar{B})=B \times B$.
\end{theo}
\begin{proof}
As in the beginning we denote $\aut^{\circ}(\bar{B})$ by $A$.
\newstepset
\begin{step}[ {\boldmath{$C_A(T\times T)=T\times T$}}]\label{main theorem_step1}
By Proposition~\ref{maximal torus}, $T\times T$ is a maximal torus of $A$. Hence, if $T\times T\subsetneq C_A(T\times T)$, then by Jordan--Chevalley decomposition, there exists a non trivial unipotent element in $C_A(T\times T)$. Since the base field is $\mathbb{C}$, there is a closed connected positive dimensional unipotent subgroup $U_1$ of $C_A(T\times T)$ (see~\cite[Lemma~C, p.~96]{MR0396773}). Note that $U_1.x_\circ\cap (B\times1).x_\circ$ is a non empty open subset of $U_1.x_\circ$. Since $U_1\subseteq C_A(T\times T)$, we have $g.x=x$ for every $ g\in \bigtriangleup(T)$ and $x\in U_1.x_\circ$. Let $(tu,1).x_\circ\in U_1.x_\circ\cap (B\times 1).x_\circ$ where $t\in T$ and $u\in U$. So for every $s\in T$, we have
\[
\begin{split}
(tu,1).x_\circ
&amp; = (s,s).\parens[\big]{(tu,1).x_\circ}					\\
&amp; = \parens[\big]{(s,s)(tu,1)(s,s)^{-1}(s,s)}.x_\circ		\\
&amp; = (stus^{-1},1).x_\circ.
\end{split}
\]
Hence, we have
\[
\begin{aligned}
stus^{-1}		&amp; = tu		\\
t^{-1}stus^{-1}	&amp; = u		\\
st^{-1}tus^{-1}	&amp; = u \quad (\because \ s \ \text{and} \ t^{-1} \text{commute})	\\
sus^{-1}			&amp; = u
\end{aligned}
\]
for all $s \in T$.

So, we have $u\in C_G(T)=T $ (see~\cite[Corollary, p.~155]{MR0396773}). Hence, we have $ u=1$. Thus, $U_1.x_\circ\cap (B\times 1).x_\circ$ is contained in $(T\times 1).x_\circ$.

Therefore
\[
U_1.x_\circ\subseteq \bar{T}=\bar{(T\times T).x_\circ}.
\]
Thus, $\bar{T}$ is $U_1$-stable. Since $\aut^{\circ}(\bar{T})=T$, the action of $U_1$ on $\bar{T}$ is trivial. Thus $U_1$ fixes $x_0$.

Now let $v\in \Lie(U_1) \setminus\{0\}$. Since $\Ad\parens[\big]{(t,s)}(v)=v$ for every $ (t,s)\in T\times T $, we have $[v,\mathrm{x}]=0$ for all $\mathrm{x}\in \mathfrak{t}\oplus \mathfrak{t}$. For the same reason we have
\[
\ad(v)\parens[\big]{(\mathfrak{g}\oplus\mathfrak{g})_\beta}\subseteq (\mathfrak{g}\oplus\mathfrak{g})_\beta
\]
for all $\beta\in R^+(G\times G)$. Hence $\restr{\ad(v)}{(\mathfrak{b}\oplus \mathfrak{b})}$ is semisimple. Further, since $U_1$ is unipotent, $\restr{\ad(v)}{(\mathfrak{b}\oplus \mathfrak{b})}$ is nilpotent. Thus, we have $\restr{\ad(v)}{(\mathfrak{b}\oplus \mathfrak{b})}=0$. Hence, every element of $U_1$ commutes with every element of $B\times B$. Since $U_1$ fixes $x_\circ$, the action of $U_1$ on $\bar{B}$ is trivial. This is a contradiction to $U_1$ being a positive dimensional subgroup of $A$. Thus, we have $C_A(T\times T)=T\times T$.
\end{step}
\begin{step}
Let $\mathfrak{a}=\Lie(A)$. By Corollary~\ref{cor non existence} and Lemma~\ref{non existence for G2}, we have $\Gamma_0=\emptyset$. Hence, by Lemma~\ref{multiplicity free} and Proposition~\ref{maximal torus}, we have
\[
\mathfrak{a}=\mathfrak{a}^{T\times T}\oplus_{\beta\in R^+(G\times G)} {(\mathfrak{g}\oplus\mathfrak{g})}_{\beta}
\]
where $\mathfrak{a}^{T\times T}$=$\braces[\big]{\mathrm{x}\in \mathfrak{a} : \Ad(t,s)\mathrm{x}=\mathrm{x}, \ \forall (t,s)\in T\times T}$. From Step~\ref{main theorem_step1}, we have $\mathfrak{a}^{T\times T}=\mathfrak{t}\oplus \mathfrak{t}$. Thus, we have $\mathfrak{a}=\mathfrak{b}\oplus\mathfrak{b}$. Further, we have $B\times B\subseteq A$ and $A$ is connected. Hence, we have $A=B\times B$.\qedhere
\end{step}
\let\qed\relax
\end{proof}

\section*{Acknowledgments}

We thank the referee for comments and suggestions which help to improve this paper.

\section*{Declaration of interests}

The authors do not work for, advise, own shares in, or receive funds from any organization that could benefit from this article, and have declared no affiliations other than their research organizations.

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