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I am sorry to give a bounty to such a crappy question but an answer would help me a lot.

I am stuck with the following simple (i guess but) technical problem.

Let $G$ be a complex reductive group, $\mathcal{O}:=\mathbb{C}[[t]]$ the formal power series, $\mathcal{K}=\mathbb{C}((t))$ the formal Laurant series, and $P^+$ the set of dominant integral (co)-weights of $G$. Then there exists a Bruhat decomposition $$G(\mathcal{K})=\bigsqcup_{\lambda} G(\mathcal{O})\lambda G(\mathcal{O})$$ Let furher denote $G(\mathcal{K})^\lambda:= G(\mathcal{O})\lambda G(\mathcal{O})$ and $\overline{G(\mathcal{K})^\lambda}:= \bigsqcup_{\mu\in P^+, \mu \leq \lambda}G(\mathcal{K})^\mu$. Denote by $Gr:=G(\mathcal{K})/G(\mathcal{O})$ the affine Grassmannian and define $Gr^\lambda$ and $\overline{Gr^\lambda}$ similiarly.

Now let $\lambda,\mu,\nu\in P^+$ and $w\in W$ such that $\lambda:= w(\nu-\mu)\in P^+$. (Of course $W$ is the Weyl Group of $G$).

Then one has the multiplication map $$m:\overline{G(\mathcal{K})^\lambda}\times_{G(\mathcal{O})} Gr^\mu\to \overline{Gr^{\lambda+\mu}}$$

Now my question:

Is it true that $$m([g,x])\in Gr^\nu\Rightarrow g\in G(\mathcal{K})^\lambda$$ or even more $$m^{-1}(Gr^\nu)=[G(\mathcal{O})\lambda, [w\mu]] $$

I can prove both statements in some cases with simple calculations but i wonder if there is a "nice" proof in the general case. I would appreciate any hint or reference.

I talked with some experts who where prettey sure that both things hold (at least in nice cases whaterever this could be) and that the answer should be somehow contained in the work of Iwahor and Matsumoto (On some Bruhat decomposition and the structure of the Hecke rings of p-adic Chevalley groups), however i am not able to figure it out myself.

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    $\begingroup$ 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 $= \rho$. $\endgroup$
    – Allen Knutson
    Commented Jun 21, 2013 at 11:22
  • $\begingroup$ 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. $\endgroup$
    – Oliver Straser
    Commented Jun 21, 2013 at 12:07
  • $\begingroup$ (Oops, not $H^0$, $H^{top}$.) $\endgroup$
    – Allen Knutson
    Commented Jun 22, 2013 at 12:19

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