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Hi everyone,

Let $X$ be a smooth projective variety over a field $k$ and let $L$ be a line bundle on $X$. I'm reading the article Heights for line bundles on arithmetic varieties and there one speaks of $\textrm{Pic}^L(X)$. What is that? And above that, if $L$ and $K$ are algebraically equivalent line bundles, why is $\textrm{Pic}^L(X) = \textrm{Pic}^K(X)$? (Here algebraic equivalence boils down to $L = K \mod \textrm{Pic}^0(X)$.) Maybe it is even better to just ask why $\textrm{Pic}^L(X) = \textrm{Pic}^K(X)$ when $L$ and $K$ are isomorphic.

I'm guessing that taking $L=\mathcal{O}_X$ will give the connected component of the Picard scheme. So I ask, is $\textrm{Pic}^L(X)$ also an abelian variety?

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I removed the LaTeX from the title, since it does not display properly on the front page. – Harry Gindi Apr 9 '10 at 2:19
up vote 4 down vote accepted

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 $Pic^L(X)$ -- 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 $Pic^{[L]}(X) = Pic^0(X)$, 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 $Pic^0(X)$ just by mapping a line bundle $M$ to $M - L$. 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 $Pic^0(X)$ 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 $L/\overline{k}$ to be any degree $n$ line bundle, we get $Pic^{[L]}(X) = Pic^n(X) = Alb^n(X)$, a torsor whose $k$-rational points parameterize $k$-rational divisor classes of degree $n$. (Note that here when I write $Pic^0(X)$ I am talking about the Picard variety rather than the degree $0$ part of the Picard group. More careful notation would be $\underline{\operatorname{Pic}}^0(X)$.)

In particular, if $X$ is a genus one curve, then there is a canonical isomorphism $X \cong Pic^1(X)$, so $Pic^1(X)$ 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

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Minor nitpick: the abelian variety structure on $Pic^1$ requires the rational point to be chosen as part of the datum, not just something that exists. – S. Carnahan Apr 9 '10 at 5:31
True; text modified accordingly. – Pete L. Clark Apr 9 '10 at 6:05
$Pic^0(X)$ need not be an abelian variety if $k$ has positive characteristic (since it need not be reduced). – ulrich Apr 9 '10 at 10:06
@unknown: fair enough. Let's define Pic^0(X) to be the reduced subscheme of the Picard scheme in that case. – Pete L. Clark Apr 9 '10 at 15:36

I'd guess it means the set of line bundles algebraically equivalent to L, modulo linear equivalence. In that case, for L = O_X you get Pic^0 and, in general, Pic^L is a principal homogeneous space for Pic^0.

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+1: That would be my guess as well. On a curve the Neron-Severi group is just Z, so Pic^n(X) is common notation to denote line bundles of degree n, or algebraically equivalent to any given line bundle of degree n. On a general variety X, the Neron-Severi group is finitely generated, and Pic^L(X) would then be the fiber over the class of [L] in NS(X). – Pete L. Clark Apr 9 '10 at 0:16

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