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I must be missing something. How can the Picard rank of a compact Kahler manifold drop below its Hodge number $H^{1,1}(X)$? For simplicity, let $X$ be a K3 surface.

After all, we have the standard sequence of maps: $$H^1(X,\mathcal{O}_X) \rightarrow H^1(X, \mathcal{O}^\times_X) \rightarrow H^2(X,\mathbb{Z}) \xrightarrow{i_*} H^2(X,\mathbb{C}) \xrightarrow{j_*} H^2(X,\mathcal{O}_X)$$ Here $H^1(X,\mathcal{O}_X) = 0$ (by regularity of $X$) and we have an injection $Pic(X) = H^1(X,\mathcal{O}^\times_X)\hookrightarrow H^2(X,\mathbb{Z})$ (the latter is torsion free in this case by the universal coefficient theorem).

Furthermore, by the exponential exact sequence, $Pic(X)$ should surject onto the subgroup of $H^2(X,\mathbb{Z})$ annihilated by the composite $j_* \circ i_*$. It's easy to see that this is precisely the subgroup consisting of those classes $\alpha \in H^2(X,\mathbb{Z})$ which land on a Hodge class in $H^{1,1}(X) \subset H^2(X,\mathbb{C})$.

Finally, we know that $dim_\mathbb{C}(H^{1,1}(X)) = 20$, and I believe (?) that the map $i_* \colon H^2(X,\mathbb{Z}) \rightarrow H^2(X,\mathbb{C})$ induced by the inclusion $\mathbb{Z} \subset \mathbb{C}$ is just the tensor product with $\mathbb{C}$. Thus in conclusion I don't see how the Picard rank $\rho(X)$ could be low -- it seems like it should be 20 in every case.

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  • $\begingroup$ Could this have something to do with it: the Hodge decomposition on $H^2(X, \mathbb{Z}) \otimes \mathbb{C}$ might fail to "lift" nicely to a decomposition on $H^2(X, \mathbb{Z})$. In particular, some high-rank subgroup of $H^2(X, \mathbb{C})$ could restrict to yield a much-lower-rank subgroup of $H^2(X, \mathbb{Z})$. $\endgroup$
    – BD107
    Commented May 31, 2016 at 1:10
  • $\begingroup$ I added the algebraic-geometry tag to make the question more visible. $\endgroup$
    – Nefertiti
    Commented Jun 1, 2016 at 8:03

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This is a confusing point that is very instructive to wrestle with. Let me also stick to $K3$s, although some of what I write is general.

Everything you wrote above is true (up to the sentence "Thus in conclusion").

The answer is basically what you wrote in the comment. This is hard to explain without a picture, but let me try to give my point of view anyway.

Think of $H^2(X,\mathbf C)$ as a fixed vector space with the fixed lattice $H^2(X,\mathbf Z)$ inside it. (Even though it is not quite accurate, I like to actually draw the lattice $\mathbf Z^2$ inside $\mathbf R^2$ to make this concrete.)

Then $H^{1,1}(X)$ is a complex subspace of $H^2(X,\mathbf C)$ that is "moving around in all possible ways" inside $H^2(X,\mathbf C)$ as we vary the complex structure of our surface. In particular, it is easy to arrange that $H^{1,1}(X)$ misses all the nonzero points of our lattice --- just take a "random" complex subspace. More generally, one can arrange for $H^{1,1}(X) \cap H^2(X,\mathbf Z)$ to have any chosen rank between 0 and 20.

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  • $\begingroup$ Yes, I see this now. We should let the complex subspace be "irrational" or "imaginary", etc. Of course, if $X$ is projective, then $\rho(X)$ is at least 1 -- we have the cohomology class of the hyperplane bundle. $\endgroup$
    – BD107
    Commented Jun 1, 2016 at 16:30

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