5
$\begingroup$

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?

$\endgroup$
6
  • $\begingroup$ Just to check, you mean finite dimensional, not finite codimensional Schuberts, right? (Not that I know the answer either way.) $\endgroup$ Apr 28, 2011 at 19:42
  • $\begingroup$ The proof of Proposition 13.2.19 in Kumar's "Kac-Moody groups, their flag varieties and representation theory" might provide what you want. $\endgroup$ Apr 28, 2011 at 20:06
  • $\begingroup$ Yes, I mean finite-dimensional ones. $\endgroup$
    – Ben Webster
    Apr 28, 2011 at 20:32
  • $\begingroup$ I would try Olivier Matthieu's monograph (in French) - Asterisque 159-160. $\endgroup$ Apr 28, 2011 at 20:48
  • $\begingroup$ I suspect you want to define your Schubert variety as a single, rather than as a double quotient. $\endgroup$ Apr 28, 2011 at 22:06

1 Answer 1

6
$\begingroup$

[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.

$\endgroup$
1
  • $\begingroup$ 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. $\endgroup$ Apr 28, 2011 at 22:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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