1
$\begingroup$

Given a variety abelian $ A $ defined over an algebraically closed field of characteristic $ 0 $, Mumford define $ Pic^0(A)$= $L \in Pic (A) | T^*_x{L}L = L \ for \ all \ x \ in A$ , where $T_x$ is translation by x.

I wonder if this coincides with the usual definition: $ Pic ^ 0 ( A )$ is the connected component of identity in $ Pic (A) $?

$\endgroup$

2 Answers 2

8
$\begingroup$

If I'm not mistaken, you've copied down Mumford's definition incorrectly: it should be the set of all line bundles $L$ such that $T_x^* L \cong L$ for all $x \in A$.

Once you make this correction: yes, this turns out to be the connected component of the identity in $\operatorname{Pic}(A)$. If you read further on in the book, you'll probably find this out. If not, try for instance Milne's notes on abelian varieties.

$\endgroup$
3
  • $\begingroup$ Yes, you're right, it was just way to write $\endgroup$
    – Flávio
    Commented Apr 13, 2011 at 18:38
  • $\begingroup$ You mean the connected component of identity in $ Pic (A) $, no? $\endgroup$
    – Flávio
    Commented Apr 13, 2011 at 18:47
  • $\begingroup$ @Flavio: yes.... $\endgroup$ Commented Apr 13, 2011 at 21:09
2
$\begingroup$

Although probably a bit late, I'd like to point out that you can find a beautiful exposition of the theory of the Picard scheme in the survey article by S. L. Kleiman with the same title, which is part 5 of the volume "Fundamental Algebraic Geometry" edited by Fantechi et al. and pubished by the AMS.

In particular, your question is answered in detail in $\S 9.5$ ("The connected component of the identity").

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .