# Deforming line bundles on abelian varieties

Let $X$ ba an abelian variety over $\mathbb C$. I would like to understand how line bundles on $X$ deform. The obstructions to deform line bundles lie in $$\textrm{Ext}^2(L,L)=H^2(X,\mathscr O_X).$$ Is this the trivial vector space? If so, this would in particular imply the smoothness of the Picard scheme. I know this is very classical but I cannot find a reference.

Note that what I ask is equivalent to asking about the surjectivity of the map $$\textrm{Pic }X\overset{c_1}{\longrightarrow}H^2(X,\mathbb Z).$$

Also, I would be curious to know if the answer changes in positive characteristic...

Thank you!

• I think one has $dim(H^2(\mathcal{O}_X))=\frac{g(g-1)}{2}$ if $dim(X)=g$. The smoothness of the Picard scheme can be proved in many different ways. For example with the help of the tangent space ($=H^1(\mathcal{O}_X)$) or by using a result of Cartier, which states that "good" group schemes are always smooth in $char=0$. May 12, 2015 at 10:21

1. The cotangent bundle of an abelian variety is trivial of rank $g=\operatorname{dim} X$, since $X$ is an algebraic group. Hence $\Omega^2_X = \bigwedge^2 \mathcal O_X$ is the trivial bundle of rank ${g \choose 2}$. Therefore $$\operatorname{dim} H^2(\mathcal O_X) = \operatorname{dim} H^0(\Omega^2_X) = {g \choose 2}$$
where the first equality is "Hodge symmetry" $h^{p,q}=h^{q,p}$.
1. For any variety, the image of $c_1$ always lands in $H^2(X,\mathbf Z) \cap H^{1,1}(X)$. So the $c_1$ map could only be surjective if $H^2(X,\mathbf Z) \subset H^{1,1}(X)$. On the other hand, the Hodge decomposition says that $$H^2(X,\mathbf C) = H^2(X,\mathbf Z) \otimes \mathbf C = H^{0,2} \oplus H^{1,1} \oplus H^{2,0}$$ and the middle summand is a complex subspace. Since we know from the first question that the two outer summands are nonzero, it cannot be the case that $H^2(X,\mathbf Z) \subset H^{1,1}(X)$.
As the comment says, $\dim H^2(X,\mathcal{O}_X)=\binom{g}{2}$. In general (see Birkenhake and Lange's book Abelian Varieties, Theorem 3.5.5), $$\dim H^q(X,\mathcal{O}_X)=\binom{g}{q}.$$ Moreover, the first Chern class map $\mbox{Pic}(X)\to H^2(X,\mathbb{Z})$ is never surjective, since $$\mbox{rank} (c_1(\mbox{Pic}(X)))\leq g^2$$ (see Exercise 2.6 (5) of Birkenhake-Lange) and $$H^2(X,\mathbb{Z})\simeq\wedge^2\mathbb{Z}^{2g}$$ which is of rank $\binom{2g}{2}$.