most recent 30 from http://mathoverflow.net 2013-06-19T18:48:11Z http://mathoverflow.net/feeds/question/27454 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/27454/equivariance-of-vector-bundles-over-g-b Faisal 2010-06-08T12:18:33Z 2010-06-10T05:58:20Z

<p>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 <a href="http://www.math.harvard.edu/~lurie/papers/bwb.pdf" rel="nofollow">here</a>). 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.</p> <p>More generally, can an arbitrary vector bundle $V$ on $X$ be made $G$-equivariant? If so, in how many ways?</p>

http://mathoverflow.net/questions/27454/equivariance-of-vector-bundles-over-g-b/27462#27462 t3suji 2010-06-08T13:25:40Z 2010-06-08T13:25:40Z

<p>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 $G$.</p>

http://mathoverflow.net/questions/27454/equivariance-of-vector-bundles-over-g-b/27472#27472 Bugs Bunny 2010-06-08T14:25:29Z 2010-06-10T05:58:20Z

<p>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.</p>