Wednesday, 8 June 2011

ag.algebraic geometry - For a line bundle L on a smooth projective variety X, what is meant by Pic^L(X)

Let me address the last part of your question.



Let X be a smooth, projective variety over an arbitrary ground field k.



I want to write Pic[L](X) instead of PicL(X) -- i.e., to make explicit that the variety depends only on the Neron-Severi class of L -- for reasons which will become clear shortly.



Suppose first that L is algebraically equivalent to 0. Then Pic[L](X)=Pic0(X), so certainly it is an abelian variety.



Next suppose that L is a k-rational line bundle on X. Then Pic[L](X) is not literally an abelian variety, because it is a nonidentity coset of a group rather than a group itself. However, it is canonically isomorphic to the abelian variety Pic0(X) just by mapping a line bundle M to ML. So it might as well be an abelian variety, really.



Finally, supose that L is not itself k-rational but that its Neron-Severi class L is rational -- i.e., L is given by a line bundle over the algebraic closure which is algebraically equivalent to each of its Galois conjugates. Then Pic[L](X) is a well-defined principal homogenous space of the Picard variety Pic0(X) but need not have any k-rational points. For instance, suppose that X is a curve. Then the Galois action on the Neron-Severi group is trivial, so taking L/overlinek to be any degree n line bundle, we get Pic[L](X)=Picn(X)=Albn(X), a torsor whose k-rational points parameterize k-rational divisor classes of degree n. (Note that here when I write Pic0(X) I am talking about the Picard variety rather than the degree 0 part of the Picard group. More careful notation would be underlineoperatornamePic0(X).)



In particular, if X is a genus one curve, then there is a canonical isomorphism XcongPic1(X), so Pic1(X) can be endowed with the structure of an abelian variety iff X has a k-rational point.



Some further material along these lines can be found in Section 4 of



http://math.uga.edu/~pete/wc2.pdf

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