TITLE: Tangent Space to Moduli Space of Vector Bundles on Curve
QUESTION [9 upvotes]: Let $X$ be a curve of genus $g \geq 2$.  Using Geometric Invariant Theory, we can construct a moduli space $\mathcal{M}(r,d)$ of vector bundles on $X$ of rank $r$ and degree $d$.  The details of this construction are a bit over my head for now, however I would like to at least be able to prove that the dimension of this moduli space is $r^{2}(g-1)+1$.  
I'm lacking understanding of one key fact.  In Michael Thaddeus' paper (http://www.math.columbia.edu/~thaddeus/papers/odense.pdf) he mentions that the tangent space to $\mathcal{M}(r,d)$ at any stable bundle $E$ satisfies the following
$T_{E} \mathcal{M}(r,d) \simeq H^{1}(\rm{End}E)$
Can anyone help me understand this?  I don't understand Thaddeus' argument.  Given the above fact, it's trivial to apply Hirzebruch-Riemann-Roch and complete the derivation of the dimension of the moduli space.

REPLY [2 votes]: Warning: I have seen the construction of the moduli of (stable) vector bundles as a manifold via a symplectic quotient (in a course with notes available here), rather than as a scheme via GIT quotient, which I don't understand well. However, my understanding is that the two are analagous, and this is supported based on the fact that the argument Thaddeus appears to be making is the same as in the symplectic case. In particular, both for both types of quotient $X//G$, we first find some subset $Y \subset X$ where $G$ acts nicely enough to take a nice quotient, and $X//G:= Y/G$.
I think the argument in mind is that whenever you have a (sufficiently nice) group $G$ acting (sufficiently nicely) on a (sufficiently nice) scheme $X$ (so that the quotient makes sense as an orbit space), the projection map $T_x Y \rightarrow T_{[x]}(X//G)$ has kernel equal to the image of the infinitesimal action of $G$ on $Y$, $\mathfrak{g}\rightarrow T_x Y$ induced by the action of $G$ on $Y$. In Thaddeus' argument, he demonstrates an exact sequence:
$$\mathfrak{gl}(\chi,\mathbb{C})\rightarrow T_EY \rightarrow H^1(\operatorname{End}(E)) \rightarrow 0$$
After this, he observes that the first map is the infinitesimal action, so we have that $H^1(\operatorname{End}(E))$ is isomorphic to the cokernel of the infinitesimal action, which gives the result we wanted by the general theory mentioned above.