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I believe that the following is well-known, but I cannot find a reference in the literature...

Let $X$ be a smooth variety (in our case $X = M^s(r)$ coarse moduli space of stable rank $r$ vector bundles with trivial determinant over a curve $C$ , so $X$ is not compact !) and let $L$ be a line bundle over $X$. Then we have the standard exact sequence

$$0 \to O_X \to Diff^1(L) \to T_X \to 0$$

now take the long exact sequence of cohomology

$$ ... \to H^1(O_X) \to H^1(Diff^1(L)) \to H^1(T_X) \to H^2(O_X) \to ...$$

Since $Pic(X) = \mathbb{Z}$ , we obtain $H^1(O_X) = 0$. Do we also have $H^2(O_X) = 0$, so that the middle arrow is an isomorphism ? It is natural to expect that

$$dim (H^1(T_X)) = dim (H^1(C, T_C)) = 3g-3,$$

so that any infinitesimal deformation of the moduli comes from a deformation of the curve, but I can find the result only for the coprime case (Narasimhan, Ramanan) and not trivial determinant.

Is this true for trivial determinant as well?

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    $\begingroup$ Certainly there exist smooth varieties such that $H^2(X,\mathcal{O}_X)$ is nonozero. However, the coarse moduli space of stable vector bundles is smooth and unirational, so it might well be that $h^2(X,\mathcal{O}_X)$ is zero. When the degree of the (fixed) determinant is prime to the rank, then $X$ is projective, smooth and unirational, hence $h^2(X,\mathcal{O}_X)$ is zero if the characteristic is $0$ (and probably there is an argument valid in all characteristics). $\endgroup$
    – Jason Starr
    Commented Jan 13, 2016 at 14:49
  • $\begingroup$ Yes in fact that's what I knew $\endgroup$
    – IMeasy
    Commented Jan 13, 2016 at 15:55
  • $\begingroup$ sorry, I meant: Thank you, Jason. Yes in fact that's what I knew. In the case of stable bundles with trivial determinant the variety is still smooth and unirational, but just quasi-projective... $\endgroup$
    – IMeasy
    Commented Jan 13, 2016 at 16:02
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    $\begingroup$ @IMeasy Yes: I agree. I am not claiming that your statement that $H^1(X,\mathcal{O}_X)=0$ is false (I do not know that), but that the argument you give does not suffice to prove it (as far as I understand). $\endgroup$
    – Olivier Benoist
    Commented Jan 13, 2016 at 17:02
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    $\begingroup$ If $g=r=2$ the moduli space is affine. For general $g$ and $r=2$ there is a relatively explicit description of the moduli space (if $C$ is hyperelliptic) due to Desale and Ramanan from which you can probably check whether the vanishing holds. $\endgroup$
    – naf
    Commented Jan 14, 2016 at 5:35

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