Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

In Bill Goldman's paper "The Symplectic Nature of the Fundamental Groups of Surfaces" (Advances, 54, 200-225, '84) it is stated that the "Zariski tangent space" to a representation space Hom$(\pi, G)/G$ at a representation $\rho$ is the cohomology group $H^1 (\pi; Ad\ \rho)$. Here G is a Lie group, and Ad $\rho$ is the representation of $\pi$ on the Lie algebra of G induced by the adjoint action of G. In the context of Goldman's paper, maybe this is just meant to refer to the case when $\pi = \pi_1 S$ with $S$ a Riemann surface.

My question is: is there some sense in which this is true for other discrete groups $\Gamma$? In general, $Hom(\pi, G)/G$ is only a semi-algebraic set, and not a variety, so maybe the question is not really meaningful. But I'd like to know whether these first cohomology groups behave like tangent spaces in some useful way.

Here are two specific questions. I'm most interested in the case when G = U(n).

Edit: Let me emphasize that I'm really talking about the topological quotient Hom(G, U(n))/U(n), which is a reasonably nice space. Since U(n) is compact, general nonsense implies that this space is Hausdorff, and even better, it's a semi-algebraic set. So in particular, it's homeomorphic to a simplicial complex.

  1. Say $[\rho]\in$ Hom$\left(\Gamma, U(n)\right)/U(n)$ has an open neighborhood homeomorphic to $\mathbb{R}^m$ for some $m$. Is it then true that dim $H^1 (\Gamma; Ad\ \rho) = m$? In other words, does the cohomology group give the "topological" dimension at "smooth" points in Hom$\left(\Gamma, U(n)\right)/U(n)$? When $H^1 (\Gamma; Ad\ \rho) = 0$, a theorem of Weil (Ann. of Math. (2) 80 1964 149--157) says that $[\rho]$ is an isolated point in Hom$\left(\Gamma, U(n)\right)/U(n)$. This is the converse statement for m=0.

  2. If $[\rho]$ is not a smooth point in the above sense, then in any triangulation of Hom$\left(\Gamma, U(n)\right)/U(n)$, we see that $[\rho]$ must not lie in the interior of a maximal simplex. If $\sigma$ is a maximal simplex of dimension m containing $[\rho]$ (in its boundary), is it true that dim $H^1 (\Gamma; Ad\ \rho) > m$? In other words, does the dimension of the "tangent space" jump up at non-smooth points?

Any ideas, references, examples, or counterexamples would be welcomed!

share|improve this question
    
Surface groups are exceptionally nice. I expect 1. is false for arbitrary $\Gamma$ (the representation "scheme" can be pretty wild). See a paper by Lubotzky and Magid in the Memoirs. If no one else answers, I'll try to think of something more specific tomorrow. –  Donu Arapura Jun 3 '10 at 22:36
    
I actually just checked out Lubotzky and Magid before posting this. They seem to focus on the case of general linear representations. There the space of conjugacy classes of semi-simple representations does have the structure of a variety. Looking some more, I see that they show that these cohomology groups are the Zariski tangent spaces. So this seems to answer my questions positively in this setting, but I'm really interested in unitary representations. –  Dan Ramras Jun 3 '10 at 23:01
    
If $\Gamma$ is the fundamental group of a compact Kaehler manifold, then by Donaldson ..., your space can be identified with tangent space to the moduli scheme of polystable bundles with trivial Chern classes. And I'm convinced this can be bad i.e nonreduced, but I don't have a specific example in mind. Perhaps you can find something in Goldman and Millson's paper in IHES (1988). I guess I'm out of ideas for now. –  Donu Arapura Jun 4 '10 at 1:57
add comment

3 Answers

up vote 5 down vote accepted

So I don't think I am going to answer your question here, and I think you probably know most of what I am going to write at this point. But perhaps this information will be of use to others that happen to pass by this thread.

Let $G$ be algebraic (including compact, and complex reductive) and denote $R(G)=\mathrm{Hom}(\pi,G)$.

The tangent space of $R(G)$ at $\rho$ always make sense since it is algebraic and it is the twisted cocycles $Z^1(\pi,\mathfrak{g}_{\mathrm{Ad}\rho})$.

Likewise, the tangent space to a conjugation orbit also generally makes sense and is the twisted coboundaries $B^1(\pi,\mathfrak{g}_{\mathrm{Ad}\rho})$.

Both of these are results of Weil from Remarks on cohomology of groups, I believe.

Unfortunately the tangent space to a quotient is not always the quotient of the tangent spaces, so it is not true that the tangent space to $R(G)/G$ is always $H^1(\pi,\mathfrak{g}_{\mathrm{Ad}\rho})$. Note that if $G$ is compact this quotient is the orbit space, and if $G$ is complex reductive it is the GIT quotient.

An easy example is the following: Let $\pi=\mathbb{Z}$ and let $G=\mathrm{SL}(n,\mathbb{C})$. Then the GIT quotient $R(G)/G$ is $\mathbb{C}^{n-1}$, parameterized by coefficients of the characteristic polynomial, and so is smooth. Thus the tangent space to the identity character is $\mathbb{C}^{n-1}$. On the other hand, the orbit of the identity is trivial and it is a smooth point in $R(G)=G$. So the cocycles are $\mathfrak{g}=\mathbb{C}^{n^2-1}$ and the coboundaries are trivial. Thus the first cohomology is also $\mathbb{C}^{n^2-1}$ which is much bigger than the correct result of $\mathbb{C}^{n-1}$.

However the following is true. If $\rho \in R(G)$ is smooth (always true for irreducible reps in surface groups, and all reps in twisted surface groups and all reps in free groups), and also have a closed conjugation orbit, then using an appropriate Slice Theorem (Luna, or Mostow) one can prove that the tangent space at the equivalence class $[\rho]$ is the tangent space at 0 in a quotient of cohomology: $T_0(H^1(\pi,\mathfrak{g}_{\mathrm{Ad}\rho})/C)$, where $C$ is the centralizer of $\rho$ in $G$. But in general there are known counter examples. See here.

Returning to the above example, we then have $\mathfrak{sl}(n,\mathbb{C})/C=\mathbb{C}^{n-1}$ since $C=\mathrm{SL}(n,\mathbb{C})$ (still parameterized by the coefficients of the characteristic polynomial, but now instead of the determinant being fixed, the trace is fixed).

If the centralizer is acts trivially one recovers the usual statement that the tangent space is $H^1(\pi,\mathfrak{g}_{\mathrm{Ad}\rho})$. For example, this holds for irreducible representations and $G=\mathrm{GL}(n,\mathbb{C})$ or $\mathrm{SL}(n,\mathbb{C})$ (or their maximal compact subgroups), and $\pi$ is a free group or a (twisted) surface group. But even when $\pi$ is a free group and $G=\mathrm{PSL}(2,\mathbb{C})$ and $\rho$ is an irreducible representation there are counter examples (namely there are irreducible singularities in that case).

There is always an open dense subset of smooth irreducibles and a further subset of those whose centralizer is equal to the center of $G$. These are called "good" representations (a la Millson). The set of good representations modulo G is a manifold, and the set of all irreducibles forms an orbifold (for general $G$ irreducible means that $\rho(\pi)$ is not contained in a parabolic subgroup).

Let me try to briefly address (1). Whenever the dimension of the tangent space is the dimension of the space, in the strong topology there is a neighborhood that looks like a ball. The converse is not true in general.

Think of $x^2=y^3$ which has a singularity at $(0,0)$ algebraically but the neighborhood of $(0,0)$ in the variety is homeomorphic to a Euclicean neighborhood. On the other hand $xy=0$ also has a dimension jump in tangent space at $(0,0)$ but no neighborhood is homeomorphic to a ball. Anyway, singularities in an algebraic setting can be mild or wild or something in between (like orbifold type).

With respect to (1), if $[\rho]$ is in the interior of a maximal simplex, I think it might still depend on $[\rho]$. Anyway, in general (as the example $x^2=y^3$ shows) knowing that a point in a semi-algebraic set has a neighborhood homeomorphic to a ball does not tell you it is not singular.

Note: A recent paper of Millson and Kapovich shows that singularities in character varieties can get as bad as you can imagine. See here.

For $\pi$ a free group however, I believe generally (for all but a finite number of counter examples) the situation is this: reducibles are singular and have no neighborhood homeomorphic to a ball. For $\pi$ a free group and $G$ equal to $\mathrm{GL}(n,\mathbb{C})$ or $\mathrm{SL}(n,\mathbb{C})$ this more or less has been established. See my paper here. Also the irreducibles are either smooth or admit orbifold singularieties. But there are orbifolds that are homeomorphic to manifolds (but not always), so it seems that one could have an orbifold singularity (with an irreducible) that happens to have a neighborhood homeomorphic to a ball (this would give a negative answer to (1)). I don't however have an example to show this for certain.

About (2), if the cohomology corresponds to the tangent space (like for free groups, or twisted surface groups), then yes, at a singularity, the dimension must jump, and hence the cohomology does too. But if the point does not correspond to cohomology to begin with, then although the tangent space must still have a dimension jump, the cohomology conceivably could not. Again, I don't have an example to show this for certain.

share|improve this answer
1  
This is an awesome answer, thank you! (Although I am not the OP...) –  Victor Protsak May 30 '13 at 17:42
1  
Thanks Sean! These examples are helpful. –  Dan Ramras May 30 '13 at 22:38
add comment

It is false in general. A counter-example was constructed by J.Huebschmann in section 6 of his paper "Singularities and Poisson geometry of certain representation spaces". However, if, say, $\rho(\Gamma)$ has trivial centralizer, then the Zariski tangent space to $Hom(\Gamma, G)//G$ (you need to use Mumford quotient for the question to make sense) at $[\rho]$ is indeed isomorphic to $H^1(\Gamma, Ad \rho)$. The reason is that the $G$-action at such $\rho$'s admits a local cross-section and then everything works, as it was explained by A.Weil in his 1964 paper "Remarks on cohomology of groups".

share|improve this answer
    
Thanks for the reference, Misha. Section 6 looks very interesting, and appears to be making explicit some issues I've been confused about for a long time... I'll have to look at it carefully. The question was really motivated by some work I did a couple of years ago, giving upper bounds on the dimension (as a CW complex) of the moduli space of $U(n)$ representations of crystallographic groups. –  Dan Ramras May 30 '13 at 22:35
add comment

There are some people around who know much more about this subject than I do, but since none of them has responded so far, let me make a few remarks. By an old theorem of Narasimhan and Seshadri (1965), moduli of flat unitary bundles with irreducible holonomy and of holomorphic stable bundles on a Riemann surface $\Sigma$ are isomorphic. The former space is basically the representation variety you've described and has a canonical symplectic structure; the latter space is a Mumford GIT quotient, so it's an algebraic variety. There is a big machine which implies such a relation between the Kahler/symplectic reduction and a GIT quotient for finite-dimensional group actions, except that the group in this situation is the gauge group, so some care needs to be taken.

Irreducibility of the representation $\rho$ assures that the holonomy of the corresponding connection has trivial centralizer (scalar matrices) and that the corresponding point $[\rho]$ in the moduli space is regular (non-singular and non-orbifold). Without it, you have to be careful about defining the quotient. The first cohomology group

$$\text{H}^1(\Sigma, \operatorname{End}\rho)=\text{H}^1(\pi, \operatorname{End}\rho)$$ classifies $G$-conjugacy classes of infinitesimal deformations of $\rho:\pi\to G$ by abstract Kodaira-Spencer theory, so the answer to Q1 is affirmative: if the point $[\rho]$ is smooth, the dimension of the moduli space at $[\rho]$ is the dimension of the first cohomology. This statement generalizes to arbitrary finitely generated discrete groups $\Gamma$. You can let $\Sigma=\text{K}(\pi,1)$ and consider topological cohomology of $\Sigma$ instead of group cohomology of $\pi$. The case studied by Weil is when $\Sigma$ is a hyperbolic $n$-manifold, so that the moduli space no longer has complex or algebraic structure for $n\geq 3.$ One difficulty in addressing Q2 is that if $\rho$ is reducible, you get an orbifold point, so it doesn't quite make sense to speak about triangulations. Another difficulty is that if the cohomological dimension of $\pi$ is greater than 2 (i.e. $\dim \Sigma\geq 3$), it becomes harder to keep track of the dimensions of higher cohomology groups by means of the Euler characteristic. Representation varieties of 3-manifold groups have been extensively studied (e.g. look at papers of Shalen).

share|improve this answer
    
I actually insist on not being careful in forming the quotient! (I promise I have my reasons...) What I mean is, I'm really interested in the topological quotient space Hom`$(\Gamma, U(n))/U(n)$`, which is a reasonably nice space. General nonsense tells you it's Hausdorff (because U(n) is compact), and it's even semi-algebraic. In particular, it's homeomorphic to an ordinary simplicial complex. Maybe Kodairi-Spencer theory is telling me I need to figure out whether my topological notion of dimension - triangulate and look at the maximal simplices - agrees with some sort of stacky notion? –  Dan Ramras Jun 4 '10 at 4:21
    
I don't have a good feel about what happens at singular points. Do take a look at 3-manifolds people's papers, they've likely answered nearly any question like that you may have. –  Victor Protsak Jun 4 '10 at 4:52
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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