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.
  • $\begingroup$ Minor comment/typo: You say in your comments that $SL_2(\mathbb{C})$ is rank 2, you mean it is rank 1. $\endgroup$ Jan 9, 2021 at 16:52

1 Answer 1


First, let's assume that the genus $g$ of $\Sigma$ is greater than or equal to 2 (otherwise the irreducible locus might be empty if $G$ is non-abelian).

Then for most choices of $G$, the answer is no, since there are irreducible representations that have centralizers larger than the center of $G$ (these are called "bad representations").

To account for this, you want to restrict to the so-called "good locus", that is, the set of representations whose centralizers are equal to the center of the $G$.

In that case, I believe an argument similar to Lemma 2.2, given Theorem 3, should show that the map $$Hom^{good}(\pi_1(\Sigma),G)\to Hom^{good}(\pi_1(\Sigma),G)//G$$ is a $PG$-bundle in the analytic topology ($PG=G/Z(G)$).

I expect that the action of $PG$ is in fact scheme-theoretically free on the good locus and thus, by a similar argument as Corollary 2.2.8, the map should also be a $PG$-bundle in the sense of Definition 4.8 (étale locally trivial).

I don't know off the top of my head if there are cases where these bundles are locally trivial in the Zariski topology, but I highly doubt it. Some anecdotal evidence for this is the development of tools to address fibrations on orbit-type strata of representation/character varieties that are locally trivial in the analytic topology but not the Zariski topology for computing E-polynomials (see here for example).

Lastly, please accept my apology for this terse and choppy "answer". I got an email notification about this question and thought I would give a quick response off the top of my head (I am not really participating in MO these days). I hope it helps (at least to give you some direction). Feel free to email me if you have questions (I may not be checking MO).

  • $\begingroup$ I think you can use Luna slice theorem here: In works in the algebraic category and will yield a local trivialization. $\endgroup$ Jan 9, 2021 at 17:40
  • 1
    $\begingroup$ Doesn't Luna just give you étale local triviality? $\endgroup$ Jan 9, 2021 at 17:43

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