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 $\gamma\in M$ ?



Mapping class group and representation of fundamental group of Riemann surfaces
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.


