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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asked Apr 28, 2011 at 19:31
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$\begingroup$ Just to check, you mean finite dimensional, not finite codimensional Schuberts, right? (Not that I know the answer either way.) $\endgroup$– David E SpeyerCommented Apr 28, 2011 at 19:42
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$\begingroup$ The proof of Proposition 13.2.19 in Kumar's "Kac-Moody groups, their flag varieties and representation theory" might provide what you want. $\endgroup$– Peter McNamaraCommented Apr 28, 2011 at 20:06
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$\begingroup$ Yes, I mean finite-dimensional ones. $\endgroup$– Ben Webster ♦Commented Apr 28, 2011 at 20:32
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$\begingroup$ I would try Olivier Matthieu's monograph (in French) - Asterisque 159-160. $\endgroup$– Alexander WooCommented Apr 28, 2011 at 20:48
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$\begingroup$ I suspect you want to define your Schubert variety as a single, rather than as a double quotient. $\endgroup$– Peter McNamaraCommented Apr 28, 2011 at 22:06
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[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.
answered Apr 28, 2011 at 22:03
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$\begingroup$ 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. $\endgroup$ Commented Apr 28, 2011 at 22:05