# Is the irreducible locus of the character variety a principal bundle in Zariski topology?

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

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).

• I think you can use Luna slice theorem here: In works in the algebraic category and will yield a local trivialization. – Moishe Kohan Jan 9 at 17:40
• Doesn't Luna just give you étale local triviality? – Sean Lawton Jan 9 at 17:43