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 ? 



Uniqueness of descent
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.


