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Let $S$ be a compact connected Riemann surface, and let $\pi = \pi_1(S)$ be its fundamental group. Let $\pi \to G$ be a surjective homomorphism.

Is $G$ linear? (That is, does $G$ admit a faithful representation $G\to \mathrm{GL}_n(\mathbb{C})$ for some $n>0?$ )

What if the kernel of $\pi$ is finite?

I expect the answer to be negative (when $\mathrm{genus}(S) >1$), but I am not sure where to start looking.

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    $\begingroup$ Hyperbolic surface groups are SQ universal, see en.wikipedia.org/wiki/SQ-universal_group, that is every countable group embeds into their quotient. Thus just take your favorite nonlinear group and embed it in a quotient of a surface group. $\endgroup$
    – Igor Belegradek
    Commented Mar 4, 2020 at 14:39
  • $\begingroup$ That's amazing. Thank you. I knew this was to much to hope for. However, if the kernel is finite, I presume the answer is positive. Or is that also false? $\endgroup$
    – Pat
    Commented Mar 4, 2020 at 14:40
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    $\begingroup$ Surface groups are torsion free (except for the projective plane). $\endgroup$
    – Igor Belegradek
    Commented Mar 4, 2020 at 14:42
  • $\begingroup$ Dani Wise has a very sophisticated result for building linear quotients of surface (and other) groups, called the malnormal special quotient theorem. If you want a very refined linear quotient of a surface group, you may want to look it up. $\endgroup$
    – HJRW
    Commented Mar 4, 2020 at 14:58
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    $\begingroup$ Surface group map onto $F_2$ (clear)... $\endgroup$
    – YCor
    Commented Mar 4, 2020 at 16:28

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