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Indeed,I'm reading the book《representation theory and complex geometry》,there is a proof of the fact that Pic(G)is trivial when G is a simple-connected semisimple algebraic group over C,but the proof is not self-contained,it use some results from representation theory in BGG's article《schubert cells and cohomology of the space G/P. So I'm wondering whether there are other ways to show this fact. And whether the assertion still holds ture when we change the base field C. thanks for all the comments

edited May 29 2012 at 22:22

François G. Dorais♦
23.1k●2●52●116

asked May 15 2012 at 10:43

cheerchan
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Note that the question and header need more editing, but aside from that the tag "algebraic" is not at all useful (and not useful for the few other questions which have that tag). – Jim Humphreys May 15 2012 at 13:45

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I think this fact is "morally" related to the fact that Lie algebra cohomology H^2(g) = 0 for semisimple g, which is rather simple. The relations between the two are standard ideas: topological line bundles are classified by H^2(M,Z) - probably representatives can be chosen invariant so we come to H^2(G) which is the same as H^2(g). These arguments are full of gaps... Probably the main gap that algebraic line bundle may have trivial degree, but be non-trivial, so vanishing of its represenative in H^2(M,Z) does mean that it is trivial. Is there a way to fill them ? – Alexander Chervov May 15 2012 at 14:27

Note that it is easy to see that $\mathrm{Pic}(G)$ is torsion free. Torsion elements in $\mathrm{Pic}(G)$ give rise to étale covers, which must necessarily be trivial as the group is simply connected. In particular see this question: mathoverflow.net/questions/32766/… – Daniel Loughran May 16 2012 at 7:25

@Daniel: I think "torsion free" applies just when you have a projective variety, not an affine algebraic group where Pic can be finite but nontrivial. – Jim Humphreys May 16 2012 at 21:33

@Daniel in the answer you cited Damiano uses exponential sequence - which is in true for analytic category, but NOT in algebraic. If manifold would be compact then by GAGA they coincide, but for non-compact algebraic and analytic are essentially different. If you take elliptic curve and drop out 1 point - you get affine curve - so by exponential sequence Pic^analytic=0, while Pic^algebraic = curve itself+point (as far as I remember). This was quite strange and surprising for me when I learn it. – Alexander Chervov May 17 2012 at 10:47

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Jim Humphreys · Answer 1 · 2012-05-15 17:29:41Z UTC

The question needs a little more detail, including precise references (for instance to the book by Chriss and Ginzburg). Aside from that, there is a fairly long history of related study in a wider context, for example an old article by V.L. Popov (in a journal translated into English): "Picard groups of homogeneous spaces of linear algebraic groups and one-dimensional homogeneous vector fiberings" (Russian), Izv. Akad. Nauk SSSR Ser. Mat. 38 (1974), 294–322.

Popov works over an arbitrary algebraically closed field, where among other things he can show that the Picard group is trivial for a connected simply connected algebraic group.

ADDED: I should add a 1976 reference which is probably more helpful and which also has numerous references back to the original literature. This is a paper by Birger Iversen, "The geometry of algebraic groups", Adv. in Math. 20 (1976), 57-85. An important source for example is the work of Chevalley in the 1950s.

In any case, what proof of triviality you like best will depend a lot on what you already know about algebraic groups and algebraic geometry. Different approaches are possible.

Concerning the use of representation theory by BGG (and others), it should be emphasized that only the most elementary characteristic-free ideas are actually involved (as in Chevalley's seminars). To study line bundles for a connected semisimple group in any characteristic, it's natural to associate them with characters of a maximal torus and related geometrically constructed finite dimensional representations. In turn, being "simply connected" in Chevalley's general sense just involves the position of the root lattice inside the full weight lattice. While the fine details of representation theory in prime characteristic involve questions still not fully answered, none of this is needed for the study of Picard groups.

Bart Van Steirteghem · Answer 2 · 2012-05-15 16:32:05Z UTC

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You can find a proof of the fact that $\mathrm{Pic}(G)$ is trivial when $G$ is a simply-connected semisimple algebraic group over $\mathbb{C}$ in Section 4 (Proposition 4.6) of Local properties of algebraic group actions by F. Knop, H. Kraft, D. Luna and T. Vust [in: "Algebraische Transformationsgruppen und Invariantentheorie" (H. Kraft, P. Slodowy, T. Springer eds.) DMV-Seminar 13, Birkhäuser Verlag (Basel-Boston) (1989), pp. 63-76].

answered May 15 2012 at 16:32

Bart Van Steirteghem
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	@Bart: This is a useful reference, but note that the authors confine their treatment to characteristic 0 (where the notion of "simply connected" is for instance more straightforward). They do however remark that their main results seem to be true for any characteristic, leaving it to the reader to sort that out. – Jim Humphreys May 15 2012 at 17:32
	thanks for your useful answer – cheerchan May 16 2012 at 3:19

Are there other ways to show Pic(G)is trivial when G is a simple-connected semisimple algebraic groups over C?

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