This question is closely related to Peter Crooks question. Strata of the Affine Grassmannian

Let $G$ be a complex reductive group, $\mathcal{K}:= \mathbb{C}((t))$, $\mathcal{O}:= \mathbb{C}[[t]]$ and $Gr=G(\mathcal{K})/G(\mathcal{O})$ be the corresponding affine grassmannian.

The $G(\mathcal{O})$-orbits on $Gr$ may be indexed by the integral dominant coweights $P^+$. For $\lambda\in P^+$ it is well known that for every the $G(\mathcal{O})$-orbit $G(\mathcal{O}) [\lambda]= G(\mathcal{O}) \lambda G(\mathcal{O})/G(\mathcal{O})$ there exists an equivariant affine bundle $G(\mathcal{O}) [\lambda]\to G/P_\lambda$ where $P_\lambda\subset G$ is the parabolic subgroup of $G$ with Levi-factor $\lambda$.

Is there also an equivariant vector bundle $G(\mathcal{O}) [\lambda]\to G/P_\lambda$?

I think I read this somewhere but I can't remember where.


1 Answer 1


Yes, though it's more naturally an affine bundle than a vector one. One good way to think about it is the existence of loop rotation on both $G(\mathcal{O})$ and $G(\mathcal{K})$, that is, an action of $\mathbb{C^*}$ induced by the action on $\mathbb{C}((t))$ that gives $t$ weight 1. Under this action, if we consider $g\in G(\mathcal{O})$, its limit $\lim_{s\to 0}s\cdot g$ exists and lies in $G(\mathbb{C})$. In particular, the induced $\mathbb{C}^*$ on $Gr$ preserves $G(\mathcal{O})$ orbits, and these are even closed under taking $\lim_{s\to 0}s\cdot g$, which must lie in the orbit $G[t^\lambda]$, since $[t^\lambda]$ is fixed by this action. The map you're looking for is precisely $g\mapsto \lim_{s\to 0}s\cdot g$, and the fiber is the orbit of $[t^\lambda]$ under the "unipotent radical" of $G(\mathcal{O})$ (which is a finite dimensional affine space).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.