4
$\begingroup$

Let $G$ be a reductive group, $X$ be a projective variety and $\mathcal L$ an ample $G$ equivariant line bundle on $X$. Then by a descent lemma of Kempf (see Narasimhan, M.S., and Drezet, J.-M.. "Groupe de Picard des variétés de modules de fibrés semi-stables sur les courbes algébriques." Inventiones mathmaticae-1989) the line bundle $\mathcal L$ descends to the quotient $X//G$ iff for any $x \in X^{ss}$ the stabilizer $G_x$ acts trivially on the fiber $\mathcal L_x$.

Now my question is suppose we have a $G$-equivariant line bundle $\mathcal L$ which descends to the quotient $X//G$, is the descent unique ? If not is it unique if $G$ is finite say ?

$\endgroup$
3
  • 2
    $\begingroup$ It cannot be unique in all situations, because there may exist line bundles on $X//G$ which become trivial on $X$. This is the case if the finite group $G$ acts freely on $X$ and admits nontrivial characters. $\endgroup$
    – abx
    Jul 4, 2016 at 15:07
  • 2
    $\begingroup$ @abx: The pull-back bundles may differ as equivariant bundles. That depends on the characters of the invertible functions on $X$. $\endgroup$ Jul 4, 2016 at 15:36
  • 1
    $\begingroup$ @Friedrich Knop: Of course. What I wanted to point out is that the question is ambiguous -- the OP should make precise what he means by unicity of the descent. $\endgroup$
    – abx
    Jul 4, 2016 at 15:40

1 Answer 1

3
$\begingroup$

The answer is yes: if $\pi:X^{ss}\to X/\!/G$ is the quotient morphism then the descended line bundle is $\mathcal L/\!/G:=\pi_*(\mathcal L|_{X^{ss}})^G$. That's a very general construction. The tricky part (due to Kempf) is to show that $\mathcal L/\!/G$ actually is a line bundle.

Edit: Proof: Let wolg. $X=X^{ss}$. A descent of $\mathcal L$ is a line bundle $\mathcal L_0$ on $Y:=X/\!/G$ together with a $G$-isomorphism $\pi^*\mathcal L_0\overset\sim\to\mathcal L$. From this one gets $$ \mathcal L_0\overset\sim\to(\pi_*\pi^*\mathcal L_0)^G\overset\sim\to(\pi_*\mathcal L)^G $$ The left isomorphism follows from $\mathcal L_0\cong\mathcal O_Y$ (locally) and $\mathcal O_Y=(\pi_*\mathcal O_X)^G$ by definition of a categorical quotient.

$\endgroup$
2
  • $\begingroup$ Could you please give me a proof or a reference of the above fact in your answer ? In my case $G$ is finite and hence $X^{ss}=X$. $\endgroup$
    – Mathew
    Jul 5, 2016 at 8:48
  • $\begingroup$ See edit above. $\endgroup$ Jul 5, 2016 at 10:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.