Let $A$ be an abelian variety defined over $\mathbf{C}$ (of dimension $>1$) and let $\Theta_A$ be the holomorphic tangent sheaf of $A$.
Question. How does one compute $H^1(A,\Theta_A)$ ?
If $A$ has dimension $1$ then using Serre's duality one finds that $H^1(A,\Theta_A)\simeq H^0(A,\omega_A^2)$ where $\omega_A$ is the canonical line bundle of $A$. Since $\omega_A\simeq\mathcal{O}_A$ one finds that $h^1(A,\Theta_A)=h^0(A,\mathcal{O}_A)=1$.
Hugo,
Although this was already discussed in the comments, perhaps I can write few more details here. The material can be found in many books such as Mumford's Abelian Varieties or the book on the same by Birkenhake and Lange.
Claim $\dim H^1(A,\Theta)= g^2$.
The first thing to observe is that $A$ is a group, so a basis for the tangent space at $0$ can be translated to give a global basis. Thus the tangent bundle $\Theta=\mathcal{O}_A^g$ where $g=\dim A$. Thus $H^1(A,\Theta)= H^1(A,\mathcal{O}_A)^g$. So this reduces the claim to checking $\dim H^1(A,\mathcal{O}_A)=g$. For this, let me use the Hodge theorem (alternatives can be found in the above refs.). Write $A$ as the quotient of $\mathbb{C}^g$ by a lattice. The Euclidean metric induces a Kähler metric on $A$, with respect to which $H^1(A,\mathcal{O}_A)$ can be realized as the space of harmonic forms of type $(0,1)$. These are necessarily invariant under the group, because the metric is. $\lbrace d\bar z_1,\ldots, d\bar z_g\rbrace$ give a basis for the invariant $(0,1)$-forms, and they are clearly harmonic. So this proves the claim.
Finally, by Kodaira-Spencer, $H^1(A,\Theta)$ is the space of first order deformations of $A$. As noted above, the moduli space of principally polarized abelian varieties has dimension only $g(g+1)/2$. Which means that roughly half these deformations are nonalgebraic!
