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Let $F$ be a closed oriented surface of negative Euler characteristic. Let $X^i(F)$ be the subset of the $SL_2\mathbb{C}$-character variety of the fundamental group of $F$ corresponding to irreducible representations. The mapping class group of $F$, $\mathcal{M}(F)$ acts on $X^i(F)$. Let $[\rho]\in X^i(F)$, is the orbit of the $[\rho]$ under the action of the mapping class group, $$\mathcal{M}(F).[\rho]$$ Zariski dense in $X(F)$?

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    $\begingroup$ Yes, but I am not sure it is in the literature. You can prove it by imitating Goldman's proof of ergodicity of the MCG action on the $SU(2)$-character variety in combination with some results from my paper written with Gallo and Marden on monodromy of Schwarzian differential equations. Check with Bill, he might tell you if the result was written down by somebody. On the other hand, while the action on $SU(n)$-character variety is still ergodic, there are non-Zariski dense orbits of Zariski dense representations (when $n$ is large). $\endgroup$ – Misha Jun 1 '19 at 5:11
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    $\begingroup$ A cheeky thing to point out is that $\rho$ should be infinite. $\endgroup$ – Ian Agol Jun 2 '19 at 2:02
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    $\begingroup$ The correct requirement in such question is to assume that the image of $\rho$ is Zariski dense (over the complex numbers). $\endgroup$ – Misha Jun 2 '19 at 18:46
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    $\begingroup$ @Misha What is a counterexample in $SU(n)$ for $n$ large? It seems like an interesting phenomenon. $\endgroup$ – Will Sawin Jun 4 '19 at 19:27
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Here is a partial answer (showing there exists a Zariski dense collection of irreducible representations with the requisite property).

In Topological Dynamics on Moduli Spaces II by Previte, and Xia it is shown in Theorem 1.4 that if the image of a $SU(2)$-representation of a surface group ($g\geq 2$) is dense in $SU(2)$ (which implies it's irreducible) then the mapping class group orbit of that representation in the $SU(2)$-character variety is dense (the result is stated in terms of relative character varieties but the closed case is included for genus at least 2 by fixing boundary values to be 2).

Since the $SU(2)$-character variety is Zariski dense in the $SL(2,\mathbb{C})$-character variety, the property you want follows for the conjugacy classes of unitary representations with dense image.

Remark 1: In Representations of surface groups with finite mapping class group orbits, by Biswas, Koberda, Mj and Santharoubane, it is shown that for large $n$ there are $SL(n,\mathbb{C})$-representations with infinite image that are actually fixed by the entire mapping class group!

Remark 2: In Surface group representations in $SL_2(\mathbb{C})$ with finite mapping class orbits by Biswas, Gupta, Mj, and Whang, it is shown (Theorem C) a representation has bounded mapping class group orbit in the character variety if and only if it is unitary up to conjugacy (if the genus $g\geq 2$).

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As some further evidence for a positive answer to your question, there is a paper of Cantat and Loray that proves a relative version of this for mapping class group the 4-holed sphere (rel boundary). In this case, the variety of representations (with fixed holonomy around each boundary curve) is a cubic surface of the form $$x^2+y^2+z^2+xyz= Ax+By+Cz+D,$$ where $A,B,C,D$ depend on the holonomy of the boundary components, and $x,y,z$ are the traces of simple loops meeting pairwise in two points. In Theorem D of the paper, they prove that there is no invariant curve. Hence any orbit must be either finite or Zariski dense.

I believe one could leverage this theorem into a proof in the closed case. The idea would be to find for an irreducible (Zariski dense) representation $\rho$ a 4-holed sphere subsurface whose orbit under the relative mapping class group of this subsurface is Zariski dense. Then the Zariski closure would be at least two dimensional. Repeating this for other subsurfaces by induction, I believe one could show that the mapping class group orbit is Zariski dense.

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  • $\begingroup$ Thanks. That seems like it will work. $\endgroup$ – Charlie Frohman Jun 5 '19 at 1:27

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