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Let $G$ be a complex semisimple group, $B$ a Borel subgroup of $G$ and $X=G/B$ the flag variety of $G$. If $G$ is simply connected, then every line bundle $L$ on $X$ can be made $G$-equivariant (see the proof of Theorem 1 here). Is this also true if $G$ isn't simply connected? By identifying $X$ with the flag variety of the universal covering group $\widetilde{G}$ of $G$, we can at least get a $\widetilde{G}$-action on $L$, but I haven't been able to figure out if this action factors through to yield a $G$-action.

More generally, can an arbitrary vector bundle $V$ on $X$ be made $G$-equivariant? If so, in how many ways?

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up vote 5 down vote accepted

A vector bundle does not have to be $G$-equivariant. For instance, Horrock-Mumford bundle on $P^4$ is equivariant only with respect to Heisenberg group and not full $SL_5$. Pull it back to the flag variety of $SL_5$ to get a bundle of rank 2, not $SL_5$-equivariant. I am sure someone will tell you easier examples than that.

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I am answering the second question, btw, while t3suji answered the first... – Bugs Bunny Jun 8 '10 at 14:26
Thank you both. – Faisal Jun 9 '10 at 21:59

No, the action of $\tilde G$ on a line bundle $L$ does not have to factor through $G$ (why would it?) Simplest example: $X={\mathbb P}^1$, $L=O(1)$, $G=PGL(2)$, $\tilde G=SL(2)$. Look at the action of $\tilde G$ on $H^0(X,L)={\mathbb C}^2$: it does not factor through $G$.

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I brought up this example a similar context:… – Allen Knutson Jun 10 '10 at 15:30

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