Timeline for Picard group of $(SL(n)\times SL(m))$-orbits

7 events

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Jan 18, 2021 at 10:55	comment	added	Mikhail Borovoi		You are right. Now ${\frak X}(G)$ is a free abelian group of rank 2 with generators $d=\det g$ and $d'=\det g'$. The restriction map sends $d$ to $d_A+d_D$ and it sends $d'$ to $d'_A+d'_D$. Since ${\rm Pic}\, X_k\cong{\rm coker}[{\frak X}(G)\to{\frak X}(H)]$, in the case $n>k$, $n'>k$ we obtain ${\rm Pic}\, X_k\simeq {\Bbb Z}^2$.
Jan 18, 2021 at 10:03	comment	added	user114666		I think that looking at the cokernel $d_D,d'_{D}$ should not appear anymore.
Jan 18, 2021 at 10:01	comment	added	Mikhail Borovoi		No. Then ${\frak X}(H)$ is generated by $d_D,\ d'_D,\ d_A,\ d'_A,\ \lambda$ with one relation $d_A+d'_A=k\lambda$, and hence ${\rm Pic}\, X_k\simeq {\Bbb Z}^4$.
Jan 18, 2021 at 9:42	comment	added	user114666		Thank you very much. If we take $G = GL(n)\times GL(n')$ instead of $G = SL(n)\times SL(n')$ as in your previous answer then in the case $k < n$ and $k < n'$ we have that the Picard group is generated by $d_{A},d_{A'},\lambda$ with the relation $d_{A}+d_{A'} = k\lambda$ right?
Jan 18, 2021 at 8:22	history	edited	Mikhail Borovoi	CC BY-SA 4.0	added 21 characters in body
Jan 18, 2021 at 8:14	history	edited	Mikhail Borovoi	CC BY-SA 4.0	Mistakes corrected.
Jan 18, 2021 at 7:59	history	answered	Mikhail Borovoi	CC BY-SA 4.0