Picard group of $(SL(n)\times SL(m))$-orbits Let $\mathbb{P}^N$ be the projective space of $n\times m$ matrices with complex entries modulo scalar. Consider the $(SL(n)\times SL(m))$-action on $\mathbb{P}^N$ given by $((A,B),Z)\mapsto AZB^{T}$. Now, consider the matrix
$$
J_k = \left(
\begin{array}{cc}
I_{k} & 0 \\ 
0 & 0
\end{array} 
\right)
$$
and let $X_k\subset\mathbb{P}^N$ be the orbit of $J_k$. I would like to ask if anyone knows how to compute the Picard group of $X_k$.
Thank you very much.
 A: Let $G={\rm SL}(n)\times {\rm SL}(n')$ and let $H\subset G$ denote the stabilizer of $J_k$ in $G$. We write ${\frak X}(G)$ for the character group of $G$. Then ${\frak X}(G)=0$. We have a canonical isomorphism ${\rm Pic}\,X_k\cong {\frak X}(H)$;
see this answer.
See also  Proposition 6.10 in Sansuc's paper Groupe de Brauer et arithmétique des groupes algébriques linéaires sur un corps de nombres.
For $g\in G={\rm SL}(n)$, $g'\in{\rm SL}(n')$,
we write $g=\begin{pmatrix} A&B\\C&D \end{pmatrix}$,
$g'=\begin{pmatrix} A'&B'\\C'&D' \end{pmatrix}$,
where $A,A'$ are $k\times k$ square matrices,
$D$ is an $(n-k)\times(n-k) $-square matrix,
and $D'$ is an $(n'-k)\times(n'-k)$-square matrix.
To compute $H$, we solve the equation
$$ g\, J_k\, g^{\prime\, T}=\lambda J_k, \quad\lambda\in K^\times,$$
where $K$ is the base field.
We obtain
$$C=0,\quad  C'=0,\quad  AA^{\prime\, T}=\lambda I_k.$$
We assume that $k\ge 1$. If $n>k$ and $n'>k$, then the character group ${\frak X}(H)$
is generated by the the characters:
$$d_A=\det A,\quad d'_A=\det A',\quad \lambda,\quad d_D=\det D,\quad d'_D=\det D',$$
satisfying the relations (written additively)
$$ d_A+d_D=0,\quad d'_A+d'_D=0,\quad  d_A+d'_A=k\lambda. $$
We see that ${\frak X}(H)$ is a free abelian group with two generators $d_A$ and $\lambda$, and hence ${\rm Pic}\,X_k\simeq {\Bbb Z}\oplus {\Bbb Z}$.
If $n=k$ but $n'>k$, then we have $d_D=0$, and hence $d_A=0$. We see that ${\frak X}(H)$ is a free abelian group with one generator $\lambda$,
and hence ${\rm Pic}\,X_k\cong {\Bbb Z}$.
Similarly, if $n'=k$ but $n>k$, then we have $d'_D=0$, and again
${\rm Pic}\,X_k\cong {\Bbb Z}$.
Finally, if  $n=k$ and  $n'=k$, then we have $d_D=0$ and $d'_D=0$, whence
$d_A=0$, $d'_A=0$, and $k\lambda=0$.
We see that ${\frak X}(H)\cong {\Bbb Z}/k {\Bbb Z}$ with generator $\lambda$,
and hence  ${\rm Pic}\,X_k\cong {\Bbb Z}/k {\Bbb Z}$ in this case.
