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Let $S$ be a closed surface of genus larger than 1, $G$ be a compact, simply connected simple Lie group with finite center.

Consider the representation variety $M(S,G)=Rep(\pi_1(S), G)$. Witten´s formula gives a way to calculate the volume of this space.

Now Let $X$ be the subset of this variety consisting of irreducible representations of the surface group $\pi_1(S)$ to $G$, and let´s call it the ¨irreducible¨ representation variety of surface group.

My questions are:

  1. Can we say anything about the distribution of $X$? For example, is it an open submanifold of Zariski dense part of $M(S,G)$? And is it connected?

  2. Do we have some formulas, similar to Witten´s, to calculate the volume of $X$ w.r.t symplectic form that restricted from $M(S,G)$?

Answers and References are very welcome. Thanks a lot.

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    $\begingroup$ Actually, fixing a complex structure on $S$ gives a natural complex structure on $M(S,G)$: it is the moduli space of $G$-principal semi-stable bundles on $S$ (Narasimhan-Seshadri-Ramanathan). Then $X$ is the locus of stable bundles; it is Zariski open in $M(S,G)$, the complement being a closed subvariety. In particular the volume of $X$ is the same as the volume of $M(S,G)$. $\endgroup$
    – abx
    Commented Aug 15, 2018 at 15:13
  • $\begingroup$ @abx So if we choose another complex structure on S, is the conclusion the same? $\endgroup$
    – BiM
    Commented Aug 15, 2018 at 15:19
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    $\begingroup$ Yes. The symplectic form is independent of the choice of the complex structure. $\endgroup$
    – abx
    Commented Aug 15, 2018 at 15:37

1 Answer 1

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Let $\Sigma$ be a closed orientable surface of genus $g\geq 2$, and $\pi$ its fundamental group. Let $X(\pi, G)=Hom(\pi,G)/G$ be the conjugation quotient for $G$ a compact Lie group whose derived subgroup $DG$ is simply connected.

From:

Ho, Nan-Kuo; Liu, Chiu-Chu Melissa. Connected components of spaces of surface group representations.II. Int. Math. Res. Not. (2005), no. 16, 959–979,

we know $X(\pi,G)$ is connected (irreducible). Moreover, the reducible locus is a proper Zariski closed subset.

The fact that a free group of rank $g\geq 2$ surjects onto $\pi$, implies the irreducible locus is non-empty.

Thus, the locus of irreducible representations is connected, open, and dense.

Since the reducible locus is proper and Zariski closed, any volume calculation for $X(\pi, G)$ will be the same for the irreducible locus.

However, even if $G$ is simply-connected and simple (as you postulate), the irreducible locus will generally have orbifold singularities (unless $G=SU(n)$). So it is not a submanifold of the locus of Zariski-dense representations (those whose image is Zariski dense).

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