What is the dimension of the moduli space $\mathcal M(r,c_1,c_2)$ of stable rank $r$ vector bundles of fixed chern classes $c_1$ and $c_2$ over an elliptic surface $S$?
In the case $r=2$, I read that $\operatorname{dim} \mathcal M(r,c_1,c_2) = 4c_2 -c_1^2 -3\chi(O_S)$. Is it a consequence of the Donaldson-Uhlenbeck-Yau theorem? I can't find any reference.
Thank you!
As I mentioned in the comments, there are ways to do this using the analytic theorems that you hinted at. However, there are also algebraic techniques, and techniques which combine aspects of both worlds. In each case, it is always easier to compute the dimension of the linearised problem, i.e. the dimension of the tangent space $T_\mathcal{E}\mathcal{M}$ of a given point in the moduli space. These are the infinitesimal deformations. The way in which one views these deformations determines (in my eyes) the approach that one takes. For example, I view deformations of $\mathcal{E}$ as deformations of the holomorphic structure on the underlying smooth vector bundle $E\to S$. They are in bijective correspondence with $\bar{\partial}$-operators which square to zero (i.e. are integrable). The space of general $\bar{\partial}$-operators (not necessarily integrable) is an infinite dimensional vector space modeled on $\Omega^{0,1}(S,\text{End}(E))$, as the difference of two $\bar{\partial}$-operators can be identified with such a $1$-form.
To identify the tangent space $T_\mathcal{E}\mathcal{M}$, then, we must look at those forms in $\Omega^{0,1}(S,\text{End}(E))$ which preserve the integrability condition. These are precisely the ones for which $\bar{\partial}\eta=0$, where this $\bar{\partial}$-operator is the one canonically associated with $\mathcal{E}nd(\mathcal{E})$. Furthermore, we must identify those holomorphic structures which are isomorphic. Translated into this language, this means we need to identify gauge equivalent $\bar{\partial}$-operators. As you can verify, this amounts to the statement that $$T_\mathcal{E}\mathcal{M}=\frac{\ker(\bar{\partial}:\Omega^{0,1}(S,\text{End}(E))\to\Omega^{0,2}(S,\text{End}(E)))}{\text{im }(\bar{\partial}:\Omega^{0,0}(S,\text{End}(E))\to\Omega^{0,1}(S,\text{End}(E)))}=H^1(S,\mathcal{E}nd(\mathcal{E}))$$ Here, we used the Dolbeault theorem to identify Dolbeault cohomology with sheaf cohomology. No matter the approach you choose to take, you find that this is the tangent space of $\mathcal{M}$ at $\mathcal{E}$. Other sources may instead write $\text{Ext}^1(\mathcal{E},\mathcal{E})$, but this is the same thing.
Now, as I mentioned, you can use the Atiyah-Singer index theorem to compute the dimension of this cohomology group. When working over a complex surface, one can use the corollary known as the Hirzebruch-Riemann-Roch theorem, which states $$\chi(S,\mathcal{F})=\int_S\text{ch}(\mathcal{F})\text{td}(S)$$ For $\mathcal{F}=\mathcal{E}nd(\mathcal{E})\cong\mathcal{E}\otimes\mathcal{E}^*$, this becomes $$\chi(S,\mathcal{E}nd(\mathcal{E}))=\int_S\text{ch}(\mathcal{E})\text{ch}(\mathcal{E}^*)\text{td}(S)\implies\\ \dim H^1(S,\mathcal{E}nd(\mathcal{E}))=1+\dim H^2(S,\mathcal{E}nd(\mathcal{E}))-\int_S\text{ch}(\mathcal{E})\text{ch}(\mathcal{E}^*)\text{td}(S)$$ where we used $H^0(S,\mathcal{E}nd(\mathcal{E}))=\mathbb{C}$ since $\mathcal{E}$ is stable. From there you can specify to various situations, to calculate this number concretely. The integral can be expressed in terms of the Chern classes of $\mathcal{E}$ and $S$, and the second cohomology group may be an annoyance.
In the book "The Geometry of Moduli Spaces of Sheaves" by Huybrechts and Lehn, there are various estimates given for the dimensions of moduli spaces of sheaves on algebraic surfaces.