MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
$\Theta_A$ 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
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
up vote 5 down vote accepted


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!

share|cite|improve this answer
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
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
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.