# Cohomological interpretation of G-equivariant line bundles

In the theory of reductive algebraic groups, there is the following map (notation: $G$ reductive over an algebraically closed field, $T$ a maximal torus, $B$ a Borel, $X(T)$ the characters of $T$, $Pic^G$ denotes the group of isomorphism classes of $G$-equivariant line bundles):

$$X(T) \to Pic^G(G/B), \lambda \mapsto \mathcal O(-\lambda) := G \times^B {k_\lambda}$$

My questions:

1. Is (analogously to $Pic(S) = H^1(S, \mathcal O^\times)$) $Pic^G$ expressible as some cohomology group?
2. Is the above map some kind of connection homomorphism or does it have another cohomological interpretation?

Thank you!

• Could you include a reference to where in Jantzen's book this appears? – Tobias Kildetoft Nov 2 '16 at 12:38
• $Pic^G(X)$ is the same cohomology group of the quotient stack $[X/G]$ in the smooth or fppf topology: $H^1_{fppf}([X/G],\mathcal O^\times)$. – Marc Hoyois Nov 2 '16 at 13:29
• @TobiasKildetoft: the functor $\lambda \mapsto \mathcal O(-\lambda)$ is ubiquitous in Jantzen, I.2.7, for example. I currently don't find the fact that it yields a morphism as stated in my question; I seem to have read this somewhere else. – Jakob Nov 2 '16 at 14:52
• You might mean II.2.7, though I don't see how it appears there either. I asked because he for example does not as far as I can remember use the notation $\operatorname{Pic}^G(G/B)$ anywhere. – Tobias Kildetoft Nov 2 '16 at 18:48
• I have removed the reference to Jantzen to avoid confusion. – Jakob Nov 3 '16 at 19:48

In particular, in your example properness of $X=G/B$ simplifies the left part of the sequence, turning it into $$0\to \hat{G}\xrightarrow{\gamma} Pic_G(X)\to Pic(X)\to Pic(G\times X)$$ where $\hat{G}$ is the group of characters of $G$ and $\gamma$ maps $\chi$ to $X\times_G \chi$. The map $Pic_G(X)\to Pic(X)$ is just the forgetting map, for the definition of the last one, see the text.
Also, if $G$ is affine, as in your case, $Pic(G\times X)=Pic(X)\times Pic(G)$(I write $=$ instead of $\cong$ to emphasise that the isomorphism is given by projection maps) so, applying Proposition 4.2.3, we get $$0\to \hat{G}\to Pic_G(X)\to Pic(X)\to Pic(G)$$
Edit: As Marc Hoyois mentions, if $X$ is a $G$-bundle over $Y$, then $Pic_G(X)=Pic(Y)$, see lemma 3.3.1