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For any abelian variety $A$, there is a dual abelian variety $\hat{A}$ which parametrizes degree zero line bundles.

Is it possible to expect similar duality for group varieties (suppose over $\mathbb{C}$ for simplicity)? I noticed that there is a notion of Cartier duality, but I feel this is not what I am looking for.

Besides, what is wrong if one try to look at the moduli space of degree zero line bundles on group variety just as the case in abelian variety?

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The "degree" of a line bundle is not defined in general, it is better to study $\textrm{Pic}^0$, the connected component of the indentity of the Picard scheme. For abelian varieties this is exactly the dual abelian variety.

As noted in the comments, the Picard group of a linear algebraic group $G$ is finite, so $\textrm{Pic}^0(G)=0$ and hence it is not very interesting. By Chevalley's theorem, we may write any algebraic group $H$ as $$0 \to G \to H \to A \to 0,$$ where $G$ is linear algebraic and $A$ is abelian. I claim that $\textrm{Pic}^0(H) = \textrm{Pic}^0(A) = \widehat{A}$, so one can reduce to the case of abelian varieties. Here is a sketch of a proof. By [1, Prop. 6.10], we have an exact sequence $$0 \to \textrm{Pic}(A) \to \textrm{Pic}(H) \to \textrm{Pic}(G).$$ As already noted, $\textrm{Pic}(G)$ is finite. Hence when we restrict this exact sequence, to the identity component, we obtain $\textrm{Pic}^0(H) = \textrm{Pic}^0(A)$, as required.

[1] Sansuc - Groupe de Brauer et arithmétique des groupes algébriques linéaires.

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    $\begingroup$ The group is not isomorphic to $G\times A$, even as a variety. For instance, a nontrivial extension of $A$ by $\mathbb{G}_m$ is not a product. $\endgroup$
    – abx
    Commented Feb 20, 2014 at 21:19
  • $\begingroup$ Hmm. I think it should still be possible to prove the result by using a suitable fibration argument. I'll work it out if the OP shows an interest. $\endgroup$
    – Daniel Loughran
    Commented Feb 20, 2014 at 21:43
  • $\begingroup$ Yes, I am very interest to see how does this reduced to the abelian variety. Thank you in advance for your clear explanation. $\endgroup$
    – Li Yutong
    Commented Feb 21, 2014 at 0:19
  • $\begingroup$ Done. Though you will need to have a look at the reference yourself and fill in some of the steps which I ommited. $\endgroup$
    – Daniel Loughran
    Commented Feb 21, 2014 at 10:56
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For linear algebraic groups $G$, the picard group is the group of characters on $G$ which is finitely generated (and will be trivial for semi-simple groups). So one does not talk of moduli space of degree zero line bundles for them. If you are talking of group varieties that are neither affine nor projective then Chevalley's theorem states that they have unique maximal normal closed linear subgroups making the quotient an abelian variety. So it will reduce to studying on abelian varieties.

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    $\begingroup$ Shafarevich's theorem? I thought everybody called it Chevalley's structure theorem, see here. $\endgroup$
    – abx
    Commented Feb 20, 2014 at 5:30
  • $\begingroup$ You are right. it is Chevalley's theorem. I have seen it in only one place, in Shafarevich's book BASIc Algebraic Geometry. I mixed it up. $\endgroup$
    – P Vanchinathan
    Commented Feb 20, 2014 at 9:38
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    $\begingroup$ The Picard group of a connected semisimple algebraic group need not be trivial in general. It is trivial if and only if the group is simply connected. $\endgroup$
    – Daniel Loughran
    Commented Feb 20, 2014 at 9:59
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    $\begingroup$ I also think your description of the Picard group is wrong, i.e. it is not the group of characters. For example the Picard group of $\mathbb{G}_m$ is trivial, whereas its character group is $\mathbb{Z}$. The Picard group of a linear algebraic group is in fact finite, I believe. $\endgroup$
    – Daniel Loughran
    Commented Feb 20, 2014 at 11:25
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    $\begingroup$ To Daniel: I think the correct statement is about equivariant line bundles: For a closed subgroup $H\subset G$ of a connected linear algebraic group $G$: the group of characters of $H$ is isomorphic to G-line bundles on $G/H$. Is that correct? $\endgroup$
    – P Vanchinathan
    Commented Feb 20, 2014 at 14:07

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