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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 ?

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    $\begingroup$ 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
    Commented Jul 4, 2016 at 15:07
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    $\begingroup$ @abx: The pull-back bundles may differ as equivariant bundles. That depends on the characters of the invertible functions on $X$. $\endgroup$
    – Friedrich Knop
    Commented Jul 4, 2016 at 15:36
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    $\begingroup$ @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
    Commented Jul 4, 2016 at 15:40

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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.

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  • $\begingroup$ 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
    Commented Jul 5, 2016 at 8:48
  • $\begingroup$ See edit above. $\endgroup$
    – Friedrich Knop
    Commented Jul 5, 2016 at 10:58

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