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Let $A$ be an abelian variety defined over $\mathbf{C}$ (of dimension $>1$) and let $\Theta_A$ be the holomorphic tangent sheaf of $A$.

Q: 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 $h^1(A,\Theta_A)=h^0(A,\mathcal{O}_A)=1$.

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$\Theta_A$ is trivial of rank $g=\dim A$. So it's $H^1$ has dimensional $g^2$. Note that when $g>1$, this is bigger than the dimension of the moduli space of abelian varieties, in case you were wondering that. –  Donu Arapura Apr 15 '12 at 21:11
    
Thanks Donu for the quick answer. So could you give me more details on how you get the $g^2$? –  Hugo Chapdelaine Apr 15 '12 at 21:18
    
So with what you said you need to compute $H^1(A,\mathcal{O}_A)$ –  Hugo Chapdelaine Apr 15 '12 at 21:21
    
Hugo, sorry I have to run. The computation of the last thing should be in Mumford's abelian varieties for example. –  Donu Arapura Apr 15 '12 at 21:27
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The tangent bundle $\Theta_A$ is trivial of rank $g$, as Donu notes, and so $\dim H^1(A,\Theta_A) = g H^1(A,\mathcal O_A)$. The fact that $H^1(A,\mathcal A)$ has dimension $g$ is a standard fact. One way to prove it is by Hodge symmetry: it has the same dimension as $H^0(A,\Omega^1_A)$, and the latter has dimension $g$ because an every holomorphic one-form on an abelian variety is necessarily invariant, and any $g$-dimensional Lie group has a $g$-dimensional space of invariant one-forms. Regards, –  Emerton Apr 16 '12 at 4:26
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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!

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Thanks Donu, so you are using the Dolbeault's isomorphism $H^1(A,\mathcal{O}_A)≃H_{\overline{\partial}}^{0,1}(A)$. –  Hugo Chapdelaine Apr 16 '12 at 13:19
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Donu, Is there a way using (only) deformation theory to see that an abelian variety with complex multiplication can be defined over number field? When $X$ is a smooth projective variety over $\amthbf{C}$ such that $H^1(X,\Theta_X)=0$ then since the Kodaira-Spencer map is trivial we see readily that $X$ can be defined over a number field. I know about the classical proof for elliptic curves which uses the $j$-invariant but I'm wondering if there exists some refinement of the deformation theory argument that I have just explained. –  Hugo Chapdelaine Apr 16 '12 at 13:25
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Hugo: (1) yes, Dolbeault. (2) That's an interesting thought. It seems entirely plausible to me that there should be a deformation theory for varieties with endomorphisms, and that CM abelian vars may be rigid. But I haven't seen such theory/computation worked out. –  Donu Arapura Apr 16 '12 at 14:00
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