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Suppose $S$ is a scheme, and $G$ a smooth $S$-group scheme.

Then there exists an algebraic stack BG called the classifying stack of $G$, defined as the quotient stack $[S/G]$ where $G$ acts trivially on $S$. I was wondering what is $Pic(BG)$.

Is it true that $Pic(BG)= H^1(k,G)$ when $S=Speck$ the spectrum of a field?

What can we say in general?

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  • $\begingroup$ As a complement to abx's answer, $H^1(k,G)$ appears as set of connected components of $[\ast/G]$, i.e., it is homotopical information. (both the quotient construction $[\ast/G]$ and the cohomology $H^1$ depend on a topology, and you have to use the same topology on both sides to get the statement) On the other hand, $\operatorname{Pic}([S/G])$ should be the $G$-equivariant first Chow group of $S$, so this is cohomological information. They are of fairly different nature. $\endgroup$ Mar 25, 2015 at 16:54

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$\mathrm{Pic}([S/G])$ is the group of line bundles on $S$ together with a $G$-linearization. When $S=\mathrm{Spec}(k)$, any line bundle on $S$ is trivial, and a $G$-linearization is given by a character $G\rightarrow \mathbb{G}_m$, so $\mathrm{Pic}[S/G]=\mathrm{Hom}(G,\mathbb{G}_m)$. No relation with $H^1(k,G)$, which is not a group when $G$ is not abelian.

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