# For a line bundle L on a smooth projective variety X, what is meant by Pic^L(X)

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?

• I removed the LaTeX from the title, since it does not display properly on the front page. Apr 9, 2010 at 2:19

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

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

• 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. Apr 9, 2010 at 5:31
• True; text modified accordingly. Apr 9, 2010 at 6:05
• $Pic^0(X)$ need not be an abelian variety if $k$ has positive characteristic (since it need not be reduced).
– naf
Apr 9, 2010 at 10:06
• @unknown: fair enough. Let's define Pic^0(X) to be the reduced subscheme of the Picard scheme in that case. Apr 9, 2010 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.

• +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). Apr 9, 2010 at 0:16