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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!

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  • $\begingroup$ 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 $char=0$. $\endgroup$
    – Bernie
    May 12, 2015 at 10:21

2 Answers 2

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Let me add a little to rfauffar's correct answer, just to point out that you can see the answers to your questions without knowing anything about the bounds he mentions (or in fact much of anything besides the Hodge decomposition).

  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 $H^2(X,\mathbf Z) \subset H^{1,1}(X)$.
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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 $\binom{2g}{2}$.

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