11
$\begingroup$

Background

Let $S_{g,n}$ be an oriented surface of genus $g$, with $n$ punctures. We explicitly prohibit the non-hyperbolic cases:

  • $g=0$, $n=0,1,2$.
  • $g=1$, $n=0$.

Let $\Gamma$ be the fundamental group of $S_{g,n}$; a group arising this way is called a Fuchsian group (as opposed to some authors, we don't require that $\Gamma$ comes with an embedding into $PSL_2(\mathbb{R})$).

Let $G$ be a complete reductive algebraic group over a field $k$. The representation algebra $Rep(\Gamma,G)$ is defined such that $k$-algebra maps $$ Rep(\Gamma,G)\rightarrow R$$ are naturally equivalent to group maps $$ \Gamma\rightarrow G(R)$$ The representation variety is then $X_{\Gamma,G}:=Spec(Rep(\Gamma,G))$, despite the fact that this scheme can be non-reduced and hence not really a 'variety'.

Question

For arbitrary groups $\Gamma$, the scheme $X_{\Gamma,G}$ can be quite bad. It is non-reduced $G=PSL_2(\mathbb{C})$ and for $\Gamma$ some Artinian groups (Kapovich-Millson, 1999) or for $\Gamma$ the fundamental group of some 3-manifolds (Kapovich, 2001).

For $\Gamma=\mathbb{Z}^2=\pi_{1,0}$ (one of the excluded cases), the representation variety $X_{\mathbb{Z}^2,G}$ is the commuting scheme of $G$. The reducedness of the commuting scheme is still an open question.

Despite this, it seems like it might be possible there are general theorems about nice properties of $X_{\Gamma,G}$, when $\Gamma$ is Fuchsian. For example, if $g=0$, then $\pi_{0,n}=F_{n-1}$, the free group on $n-1$ generators. Then $X_{F_{n-1},G}=G^{n-1}$, which is a smooth variety.

For $\Gamma$ Fuchsian, is it known whether $X_{\Gamma,G}$ is

  • reduced?
  • normal?
  • smooth?

I have a vested interest in their normality, but that seems like the question least likely to be addressed directly.

$\endgroup$
2
  • 2
    $\begingroup$ As soon as n (the # of punctures) is at lest one, then your "Fuchsian group" is a free group. $\endgroup$ May 28, 2011 at 21:40
  • 1
    $\begingroup$ I hadn't realized. I guess that means the interesting case is $n=0$. $\endgroup$ May 30, 2011 at 4:14

1 Answer 1

11
$\begingroup$

Answers to all questions are negative, however, $Hom(\Gamma, G)$ is smooth away from representations $\rho$ such that the centralizer of $\rho(\Gamma)$ in $G$ is finite. (This is due to Andre Weil, "Remarks on cohomology of groups", Annals of Math., 1964, but since then it was reproven by many others in a variety of ways. For instance, Goldman has a very clear proof using Poincare duality and vanishing of $H^2$. However, if you allow Fuchsian groups with torsion instead of just surface groups, Weil's result is the best. Incidentally, there is a way to have an interesting theory of representations of fundamental groups of surfaces with boundary: You just have to consider relative representation varieties where you fix the conjugacy classes of images of boundary loops. Weil deals with these as well.)

i. There are (real) non-reduced examples: $G=PU(2,1)$ and $\rho$ is a discrete and faithful representation that lands in $PU(1,1)$, see e.g. paper of Goldman and Millson http://www2.math.umd.edu/~wmg//LocalRigidity.pdf

The key reason is that the dimension of Zariski tangent space to $Hom(\Gamma, G)$ at all these representations is larger than the actual dimension of the representation variety. On the other hand, over algebraically closed fields it will be reduced.

ii. Even if you look at the reduced scheme, both smoothness and normality will fail at the trivial representation to a nonabelian group, like $SL(2)$. The analytical germ of $Hom(\Gamma, G)$ at the trivial representation is given by: Take the vector space $Z^1(\Gamma, {\mathfrak g})$ (with trivial action of $\Gamma$ on the Lie algebra) and impose the quadratic equations given by vanishing of the cup-products $[\omega \cup \omega], \omega\in Z^1$. See
Goldman's paper "Representations of fundamental groups of surfaces", reference [5] in the paper by Goldman and Millson linked above. There is actually a much stronger result by Goldman-Millson and Simpson which describes local singularities of representation varieties of Kahler groups in terms of cup product on $H^1$.

However, it is an open problem if $Hom(\Gamma, G)$ is smooth at representations with finite centralizers, where $\Gamma$ is a Kahler group.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.