Can one construct the GIT quotient of a projective bundle?

Let $G=PGL(n)$ act on a smooth projective scheme $X$ over $\mathbb{C}$ with nontrivial finite stabilizers ($\cong \mathbb{Z}/2\mathbb{Z}$) only along a divisor $D\subset X$. Furthermore there a is a good quotient $Y=X//G$, with $Y$ also smooth and projective. This is constructed via GIT with a $G$-linearized ample line bundle $\mathcal{L}$ on $X$ and we have the morphism $\pi: X \rightarrow Y$.

There is a $G$-linearized locally free sheaf $\mathcal{E}$ of rank two on $X$, then we have the projective bundle $p: \mathbb{P}(\mathcal{E}^{*})\rightarrow X$ associated to the dual sheaf.

$\bf{Question\,1:}$ is there a $G$-linearized ample line bundle $\mathcal{L}'$ on $\mathbb{P}(\mathcal{E})$ (probably some combination of the pullback of $\mathcal{L}$ and $\mathcal{O}(1)$?) so that we can construct a GIT quotient $f: \mathbb{P}(\mathcal{E}^{*})\rightarrow \mathbb{P}(\mathcal{E}^{*})//G$ such that $p: \mathbb{P}(\mathcal{E}^{*})\rightarrow X$ induces a morphism of quotients $\hat{p}: \mathbb{P}(\mathcal{E}^{*})//G \rightarrow X//G$ with an evident commutative diagram?

If Question 1 has a positive answer:

$\bf{Question\,2:}$ Can we describe $\hat{p}:\mathbb{P}(\mathcal{E}^{*})//G\rightarrow X//G$? I doubt that it is again a projective bundle. My guess is that it is maybe the conic bundle (Brauer Severi variety) associated to the sheaf of algebras associated to the invariant direct image of $\mathcal{E}nd(\mathcal{E})$ on $X//G$. This is because $\mathbb{P}(\mathcal{E}^{*})$ is isomorphic to the Brauer Severi variety associated to $\mathcal{E}nd(\mathcal{E})$ on $X$ and the morphism $\pi: X \rightarrow Y$ is a principal $PGL(n)$-bundle away from $D$. That is the conic bundle $\mathbb{P}(\mathcal{E})//G\rightarrow X//G$ degenerates along the divisor $\pi(D)$. Or is there another description of this space?

• Why do you say that End(E) descends? Are you assuming that the stabilizers of points act trivially (or at least by scalars) on E's fibers? – t3suji Sep 10 '16 at 16:47
• @t3suji : No. I am saying $End(E)^G$, the sheaf of invariants, descends. G acts nontrivially via conjugation on End(E) with the finite stabilizers, so that this bundle will not descend. – Bernie Sep 10 '16 at 22:35
• End(E)^G makes no sense as a sheaf on X: G-invariance is not defined for local sections when G acts non-trivially on X. – t3suji Sep 11 '16 at 5:38
• @t3suji : Ah, thanks. I think misunderstood that the whole time. So it makes sense to talk about the the invariant direct image of $End(E)$ on $X//G$ although this sheaf itself does not descend to $X//G$? The last point just means that there is no sheaf $F$ on the quotient such that $\pi^{*}F\cong End(E)$ as $G$-linearized sheaves? – Bernie Sep 11 '16 at 9:39
• Yes, for any $G$-equivariant sheaf $F$ on $X$, you can consider $F':=(\pi_*(F))^G$ (invariants in the direct image), it is a sheaf on $X//G$. If $F$ descends, then $F'$ is its descent: $F\simeq\pi^*F'$. In general, there is a morphism $\pi^*F'\to F$; it is injective, but not surjective. – t3suji Sep 11 '16 at 14:03