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!



Cohomological interpretation of G-equivariant line bundles
See Theorem 4.2.2 in <https://www-fourier.ujf-grenoble.fr/~mbrion/lin.pdf> 


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


