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S Jul 10, 2013 at 13:09 history bounty ended CommunityBot
S Jul 10, 2013 at 13:09 history notice removed CommunityBot
Jul 2, 2013 at 12:27 history edited Oliver Straser CC BY-SA 3.0
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Jul 2, 2013 at 12:04 history edited Oliver Straser CC BY-SA 3.0
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S Jul 2, 2013 at 11:59 history bounty started Oliver Straser
S Jul 2, 2013 at 11:59 history notice added Oliver Straser Authoritative reference needed
Jun 22, 2013 at 12:19 comment added Allen Knutson (Oops, not $H^0$, $H^{top}$.)
Jun 21, 2013 at 12:07 comment added Oliver Straser My question somehow breaks down to the following: Given $h\in G(\mathcal{O})$, $\lambda,\mu,\nu$ and $w$ as above. I would like to show: $\lambda h \mu\in G(\mathcal{K})^\nu\Leftrightarrow$ there exists $h_1,h_2\in G(\mathcal{O})$, s.t. $h_2 \cdot w\cdot h_2=h$ and $\lambda h_1\lambda^{-1}\in G(\mathcal{O})$ as well as $\mu^{-1} h_2\mu\in G(\mathcal{O})$. I was hoping that some kind "Bruhat decomposition" would yield such a result.
Jun 21, 2013 at 11:22 comment added Allen Knutson Geometric Satake says that the $H^0$ of the $\nu$-fiber of $m$ should be the intertwining space $Hom(V_\lambda \otimes V_\mu,V_\nu)$. If I'm reading it right, your second question is about the PRV conjecture, which only said (in this language) that that fiber is nonempty, not that it's irreducible. I think it'll be reducible for e.g. $G=SL(3)$, $\lambda = \mu = \nu =$ the highest root $= \rho$.
Jun 21, 2013 at 8:55 history asked Oliver Straser CC BY-SA 3.0