# What is the Picard group of a Schubert variety in the affine Grassmannian?

I'm not sure I have a lot more to say than the title. Let $G$ be your favorite simple algebraic group over $\mathbb{C}$, and let $$\overline {\mathrm{Gr}}_\lambda= \overline{G(\mathbb{C}[[t]])\cdot t^\lambda \cdot G(\mathbb{C}[[t]])}/ G(\mathbb{C}[[t]]).$$ It's a commonly cited theorem that $\overline {\mathrm{Gr}}_\lambda$ is a projective variety for every $\lambda$, but the usual tricks for finding the Picard group of a Schubert variety in the finite dimensional case don't work (the group $G(\mathbb{C}[[t]])$ is perfect if $G$ is semi-simple). Is this Picard group computed anywhere in the literature?

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Just to check, you mean finite dimensional, not finite codimensional Schuberts, right? (Not that I know the answer either way.) – David Speyer Apr 28 '11 at 19:42
The proof of Proposition 13.2.19 in Kumar's "Kac-Moody groups, their flag varieties and representation theory" might provide what you want. – Peter McNamara Apr 28 '11 at 20:06
Yes, I mean finite-dimensional ones. – Ben Webster Apr 28 '11 at 20:32
I would try Olivier Matthieu's monograph (in French) - Asterisque 159-160. – Alexander Woo Apr 28 '11 at 20:48
Peter- Make that an answer, and I'll accept it. – Ben Webster Apr 28 '11 at 20:53

## 1 Answer

[From the comments] The proof of Proposition 13.2.19 in Kumar's "Kac-Moody groups, their flag varieties and representation theory" appears to provide the requested information.

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Alexander Woo's suggestion of Mathieu's Asterisque 159-160 monograph is also potentially useful, it contains a chapter "Groupe de Picard des varietes de Schubert" but I cannot speak for its contents since I haven't read it. – Peter McNamara Apr 28 '11 at 22:05