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Let $S$ be a Riemann surface with genus $g>0$. Let $M$ be the mapping class group of $S$. $Hom(\pi_1(S),Gl(n, \mathbb{C}))$ is the representation space of fundamental group of $S$

Question: Is there an integer $n>1$, such that there exists an irreducible representation $\rho \in Hom(\pi_1(S),Gl(n, \mathbb{C}))$ with $\gamma(\rho)$ conjugate to $\rho$ by an element in $Gl(n, \mathbb{C})$, for all $\gamma\in M$ ?

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  • $\begingroup$ arxiv.org/pdf/1612.02508.pdf describes the representations whose equivalence class is preserved by {\bf finite} subgroups of the mapping class group. $\endgroup$ – ThiKu Aug 19 '17 at 9:38
  • $\begingroup$ Thank you for your reference. However, I am especially interested in the $Gl(n, \mathbb{C})$-representation. $\endgroup$ – Feng Hao Aug 19 '17 at 10:41
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There are counterexamples as soon as $g > 1$.

Let $n$ be the number of surjective homomorphisms $\pi_1(S) \to A_5$, up to $S_5$-conjugacy. (We can see that $n \geq 1$ using the fact that $A_5$ can be generated for two elements.)

Then there is a homomorphism $\pi_1(S) \to A_5^n$, where we take the product of all these homomorphisms. I claim this homomorphism is surjective.

Proof: Let $H$ inside $A_5^n$ be the image. Its projection onto each factor is surjective, and these surjective morphisms are distinct up to automorphisms of $A_5$. All such subgroups are all of $A_5^n$, by induction on $n$: Given that it is true for $A_5^{n-1}$, the projection of a subgroup $H$ onto the first $n-1$ factors must have image $A_5^{n-1}$, and the kernel is a normal subgroup of $A_5$, hence either $A_5$ or $1$. If $A_5$ then $H= A_5^n$, if $1$ then there is an $n$th homomorphism $H= A_5^{n-1} \to A_5^n$ which is not equal to any of the other ones up to an automorphism, which is false.

Now if we take $V$ an irreducible representation of $A_5$ which is stable under the action of $S_5$ (like the standard representation), then $V^{ \otimes n}$ is an irreducible representation of $A_5^n$ and thus by composition an irreducible representation of $\pi_1$. Any element of the mapping class group permutes the homomorphisms $\pi_1(S)$ (and maybe also acts on them by elements of $S_5$) and so preserves $V^{\otimes n}$, so $V^\otimes n$ is invariant under the mapping class group.

We could also run this argument with the Mathieu group $M_{11}$, which has no outer automorphisms at all, so any irreducible representation works.

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  • $\begingroup$ Why should the orbit of an irreducible representation be finite? It is known that the mapping class group action on the Hitchin component (of the character variety) is properly discontinuous, thus has finite stabilizers. But all Hitchin representations are irreducible. $\endgroup$ – ThiKu Aug 19 '17 at 9:52
  • $\begingroup$ @ThiKu I didn't claim it was. I used "if" in the second paragraph, and in the third paragraph I spoke only of representations that factor through a finite group. $\endgroup$ – Will Sawin Aug 19 '17 at 9:54
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    $\begingroup$ This has nothing to do with the question, the question asks about irreducible representation with $n>1$. If you add irreducible representations the result is not irreducible anymore. $\endgroup$ – Anton Mellit Aug 19 '17 at 21:29
  • $\begingroup$ @WillSawin Thank you for your answer. As Anton mentioned. I want to find an irreducible representation. Actually, I hope that the answer is "no". $\endgroup$ – Feng Hao Aug 20 '17 at 3:33
  • $\begingroup$ @FengHao - Sorry, missed that. Let me try to answer the actual question... $\endgroup$ – Will Sawin Aug 20 '17 at 6:46

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