Let $X$ be a compact Riemann surface with genus $2$ and $M^2$ the moduli space of stable principal $SL(2)$bundles of rank $r$. We know that $M^2$ is a complex projective variety of dimention $r^2(g1)+1=r^2+1$. I have to prove that if $p \in M^2$ is a smooth point then $T_pM^2 \simeq H^1(X,\mathfrak{sl}(2))$. With $\mathfrak{sl}(2)$ we mean the adjoint bundle. I suppose that $H^1(X,\mathfrak{sl}(2))$ is the sheaf cohomology with coefficients the holomorphic sections of $\mathfrak{sl}(2)$ and they can be regarded as an infinite dimentional Lie algebra. How can I prove the isomorphism $T_pM^2 \simeq H^1(X,\mathfrak{sl}(2))$? If I have $M$ the moduli space of stable principal $G$bundles ($G$ is a simple Lie group) over a compact Riemann surface $X$, how can I prove, in general setting, that $T_pM \simeq H^1(X,\mathfrak{g})$?

$\begingroup$ why do you happen to have to prove that? $\endgroup$ – Fernando Muro Oct 22 '13 at 9:02

3$\begingroup$ @FernandoMuro Because I read it but I don't know how can I do it... $\endgroup$ – Ste Oct 22 '13 at 9:29

$\begingroup$ The cohomology should be on $X$, not on $M^2$. Also the cohomology group is finite dimensional. Finally, the cohomology group does not (obviously) have any nontrivial structure of Lie algebra. It might have a structure of module over the Lie algebra $H^0(X,\mathfrak{sl}_{2,E})$, where $E$ is the original bundle and where $\mathfrak{sl}_{2,E}$ is the associated adjoint bundle. But the stability condition means this Lie algebra will just be $\mathbb{C}$, so the module structure is pretty useless. $\endgroup$ – Jason Starr Oct 22 '13 at 12:44

$\begingroup$ @JasonStarr I'm sorry... So I have some questions: 1) How can I prove that $T_pM \simeq H^1(X,\mathfrak{g})$? Why the stability condition means that $H^0(X,\mathfrak{sl}_{2,E}) \simeq \mathbb{C}$? $\endgroup$ – Ste Oct 22 '13 at 13:06

$\begingroup$ This question is clearly stated as if it were an exercise. I think it's not, but the person who asked chose to offer no motivation after being asked about it. I think this is against the phylosophy of this forum. $\endgroup$ – Fernando Muro Oct 22 '13 at 13:22
This is the general method of deformation theory. The two immediate papers that will give you sufficient detail are:
Holomorphic Vector Bundles on a Compact Riemann Surface (by NarasimhanSeshadri)
Stable Principal Bundles on a Compact Riemann Surface (by Ramanathan)
I am more familiar with the specific scenario of the moduli space of flat or antiselfdual connections on a principal bundle, where everything is translated into an elliptic chain complex, and the Zariski tangent space is the kernel of the linearization of some operator modulo the infinitesimal action.
Use a RiemannHilbert type correspondence: {Flat $G$bundles} ~ { $\pi_1$reps into $G$} (upto some stuff on each side). Now a tangent vector to the moduli space will be an infinitesimal deformation of the representation. If you work out what this means, then you will find an $H^1$ appearing in group cohomology with coefficients in $\frak{g}$. Equivalently, since curves are $K(\pi,1)$'s you can see this using adjoint bundles.
See, e.g., Hitchin's Flat Connections and Geometric Quantization. Also Labourie has some notes on surfaces. Much of this is found in there as well.

1$\begingroup$ I think you are confused. The OP was asking about holomorphic $G$bundles, not flat $G$bundles. $\endgroup$ – abx Apr 5 '17 at 8:20

$\begingroup$ @abx, indeed, I missed the lack of flatness. But this is very interesting (to me): is it known why the results are so similar? $\endgroup$ – Artur Jackson Apr 12 '17 at 23:21

$\begingroup$ The paper below (prop 3) seems to morally explain this: stable principal SL(n) bundles arise from irreducible unitary representations. Note on the stability of Principal Bundles Donghoon Hyeon and David Murphy PDF: ams.org/journals/proc/200413208/S0002993904073861/… $\endgroup$ – Artur Jackson Apr 12 '17 at 23:40