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As far as I know, the main representability result for the relative Picard functor $Pic_{X/k}$, for a noeth. sep. scheme of finite type over a field $k$ is:

If $X$ is proper then $Pic_{X/k}$ is representable by a $k$-scheme loc. of finite type. (This is attributed to Murre and Oort in Bosch-Lüttkebohmert-Raynaud)

I am interested in what can be said once the requirement of properness is dropped, e.g. what can be said for quasi-projective varieties?

Representability is probably to much to ask for (even as an algebraic space), but do you have references or know of examples where the Picard functor of a non-projective quasi-projective variety is representable?

Is there a weaker sense of representability in which sense the "open" Picard functor is representable?

Is the group somehow controlled by (the group of $k$-points of) representable objects. (I have the naive impression that if $X$ is my quasi-projective variety, then a proper hypercovering of $X$ should be able to compute $H^1(X,\mathcal{O}_X^*)$, and that then one might be able to use representability theorems for proper/projective maps, but I know nearly nothing about the involved technical requirements.)

Edit: I should have added that I do not want to assume resolution of singularities.

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  • $\begingroup$ Where does your "naive impression" related to proper hypercovers come from? It sounds dubious. Proper hypercovers are good for descent with abelian etale sheaves due to the proper base change theorem, but when you bring in line bundles over the structure sheaf you enter the world of quasi-coherent sheaves, for which it doesn't seem like proper hypercoverings (without some kind of flatness conditions which seem impractical) will be useful -- but I'd be happy to hear to the contrary. My recollection is that nothing much can be said without properness conditions on $X$ over $k$. $\endgroup$
    – BCnrd
    Commented Jul 30, 2010 at 3:10
  • $\begingroup$ Hehe, they come precisely from browsing through your notes on cohomological descent and from the "general concept" (which apparently is none) that cohomology can be computed via surjective proper hypercoverings. By "naive" I guess I ment "uninformed". I've also seen "simplicial Picard" groups mentioned in a few places and I guess I was hoping for comparison theorems for the simplicial Picard group of a hypercovering of X with the Picard group of X $\endgroup$
    – Lars
    Commented Jul 30, 2010 at 8:10
  • $\begingroup$ Dear Lars: For what you want, perhaps some non-proper simplicial methods can be used? (Not quite sure what kind of examples or applications you have in mind with your question, so this suggestion is a bit vague.) One place to look is at the computation of the Picard groups of some Artin stacks. $\endgroup$
    – BCnrd
    Commented Jul 30, 2010 at 10:34
  • $\begingroup$ Thanks Brian! One thing I would like to be able to do is, for a line bunde L on U (smooth quasi-projective), to determine whether L is trivial by pulling back to some suitable alteration, or proper hypercovering of U. Also, do you have a reference for non-proper simplicial methods? $\endgroup$
    – Lars
    Commented Jul 30, 2010 at 14:56
  • $\begingroup$ If $f:U'\to U$ is an alteration with $U$, $U'$ both smooth then $f$ is flat outside some $Z\subset U$ of codimension $\ge2$, so you can verify triviality of $L$ on the complement, where there is faithfully flat descent. So if you do the usual simplicial thing with successive alterations you should get $Pic(U)$ injecting into $Pic(U_\bullet)$. $\endgroup$
    – Tony Scholl
    Commented Jul 30, 2010 at 19:26

1 Answer 1

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The first thing to consider is the case of affine curves : let $k$ be an algebraically closed field, $C/k$ a smooth affine curve, $\bar{C}/k$ its smooth projective compactification, $\bar{C}=C\cup{p_0,p_1,...,p_n}$, $J=J(\bar{C})$ the jacobian, $\theta:C\rightarrow J$ the map induced by the choice of the base point $p_0$. Then $Pic^0(C)$ is identified with the quotient $J/\langle\theta(p_1),\ldots,\theta(p_n))\rangle$. This is always divisible but depends somehow on what this subgroup of the groups of rational points of an abelian variety look like (does it land in the torsion, etc.).

Let's think about it over $\mathbb{C}$ : there you have the quotient of a complex torus by a finitely generated subgroup : when this subgroup is not discrete the quotient does look like it is not representable as the $\mathbb{C}-$points of a scheme.

*Edit : * As Emerton pointed out in the comments, in this case the correct "geometric" object is the 1-motive associated to C. But there is a general construction of Picard 1-motives associated to varieties over a field of characteristic 0 due to Barbieri-Viale and Srinivas, which encode the $Pic^0$ geometrically :

Albanese and Picard 1-motives Luca Barbieri-Viale - Vasudevan Srinivas Mémoires de la SMF 87 (2001), vi+104 pages

http://arxiv.org/abs/math/9906165

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  • $\begingroup$ In this context, perhaps one is better of looking at the 1-motive given by $\mathbb Z\langle p_1,\ldots,p_n\rangle \to J,$ rather than actually forming the quotient. Of course, it depends on what you want to do! $\endgroup$
    – Emerton
    Commented Aug 5, 2010 at 16:35
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    $\begingroup$ Yes indeed ! In fact, this points the way to the general answer : there is a very nice general construction of Picard and Albanese 1-motives due to Barbieri-Viale and Srinivas. I will add the reference in my answer. $\endgroup$
    – Simon Pepin Lehalleur
    Commented Aug 5, 2010 at 16:57
  • $\begingroup$ I assume you mean "does NOT look like it is represented as the $\mathbb{C}$-points of a scheme". $\endgroup$
    – David E Speyer
    Commented Sep 20, 2010 at 20:45

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