Let $\Sigma$ be a compact orientable surface and let $G$ be a reductive algebraic group (say, $G=\mathrm{SL}_n(\mathbb{C})$ for simplicity). The representation variety is $$ X_G(\Sigma) = \mathrm{Hom}(\pi_1(\Sigma), G). $$

The group $G$ also acts on $X_G(\Sigma)$ by conjugation. This action is not free or closed but if we restrict to the open set $X_G(\Sigma)^{irr} \subseteq X_G(\Sigma)$ of irreducible representations, then it is known that the action of the inner automorphism group $\mathrm{Inn}(G) = G/Z(G)$ is closed and free. Moreover, by Luna stratification theorem, it gives rise to an $\mathrm{Inn}(G)$-principal bundle **in the étale topology**
$$
\mathrm{Inn}(G) \to X_G(\Sigma)^{irr} \to X_G(\Sigma)^{irr}/G.
$$

I was wondering whether this map is actually locally trivial in the **Zariski topology** or not. Or, at least, do we have an equality in the Grothendieck ring of algebraic varieties
$$
[X_G(\Sigma)^{irr}]=[X_G(\Sigma)^{irr}/G][\mathrm{Inn}(G)]?
$$

**Some remarks**

- If $\mathrm{Inn}(G)$ is connected, the previous equation holds for $E$-polynomials (alternating sum of Hodge numbers) since the monodromy of the quotient map is trivial in cohomology.
- Even though $\mathrm{SL}_n(\mathbb{C})$ is an special group (every étale principal bundle is Zariski locally trivial), $\mathrm{Inn}(\mathrm{SL}_n(\mathbb{C}))=\mathrm{PGL}_n(\mathbb{C})$ is not special.
- In the rank 2 case, $G=\mathrm{SL}_2(\mathbb{C})$, I think I can manage to give a very pedestrian proof of this by stratifying $X_G(\Sigma)^{irr}$ according to the Jordan forms of its elements and using the eigenspaces to `trivialize' the action. Since obtaining the eigenspaces can be done algebraically in this rank on certain Zariski open sets, you are done.