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Is there a faithful representation of a surface group of genus $>2$ into $GL(n,\mathbb{C})$ for some $n$ for which, for each conjugacy class of each embedded loop in the fundamental group, the image in $GL(n,\mathbb{C})$ has a power that is a unipotent element?

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No for $n=2$; for large $n$, I think, it is an open problem. Imagine that you have a faithful representation $Mod(g,1)\to GL(n,C)$ (as we know, they exist) which sends a simple loop in the normal surface subgroup $\pi_1(S_g)\subset Mod(g,1)$ to a nontrivial unipotent element. No reason to think that such representations does not exist, but constructing them is another matter. –  Misha Jul 7 '13 at 7:27
    
Is $s_n$ known (asymptotically), where $s_n$ is the (logarithmic?) proportion of elements representing simple loops in the $n$-ball. If it's large enough it might at least prove this does not exist in small dimension. –  YCor Jul 7 '13 at 13:59
    
@Misha: why no for genus 2? (Also, I presume you mean "as we know, they may exist"?) –  Ian Agol Jul 7 '13 at 15:57
    
@IanAgol: Ian, yes, I meant "may exist", in view of Bigelow's theorem. I did not exclude genus 2, but representations to $GL(2,C)$: This can be seen as an application of my work with Gallo and Marden, but could be proven directly as well. I will write a more detailed "answer" when I have time. –  Misha Jul 8 '13 at 5:05
    
@Misha: of course, I mixed up $n$ and $g$ in your comment. The fact that discrete faithful reps. have at most a finite number of parabolics is not hard to see (for $n=2$). –  Ian Agol Jul 8 '13 at 17:13

1 Answer 1

Misha has said it is an open problem, so I hesitate to say this is an "answer". But, the hypothesis that the rep is faithful means that (after replacing the surface group by a finite index subgroup if necessary) the Zariski closure is connected and not solvable. Hence it has a simple factor, call it $G$. The fact that "all" elements are quasi-unipotent means that the traces (of the adjoint representation of $G$)are all bounded, for all embeddings of the trace field into $\mathbb C$. This is impossible, because that means that the image is finite (this argument is used for example, in Tits' well known paper on the existence of free groups in linear non-solvable by finite groups), and connectedness means that the image is trivial.

Now your hypotheses do not mean all elements have bounded trace, but only that all EMBEDDED loops do; if there is a Zariski dense subset of these embedded loops, we are still OK, I think.

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If you can show that the image of simple loops is Zariski dense, you can deduce that all simple factors of the real Zariski closure are compact, in a compact group virtually unipotent elements have finite order, and because everything lies in a finitely generated field, these orders are bounded, so it's OK. –  YCor Jul 7 '13 at 9:50

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