Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $\mathcal{K} = \mathbb{C}((t)), \mathcal{O}=\mathbb{C}[[t]]$, $G=SL_2$ (or any semisimple group), and $\text{Gr}_G=G(\mathcal{K})/G(\mathcal{O})$; there is a left action of $G(\mathcal{O})$ on $\text{Gr}_G$. Let $X_*(T)=\text{Hom}(\mathbb{C}^{\times},T)$ (note that there is a natural embedding of $X_*(T)$ inside $G(\mathcal{K})$). Let $B$ be a Borel subgroup. Let $X^*(T) = \text{Hom}(T, \mathbb{C}^{\times})$.

Choose $\lambda \in X_*(T)$ to be dominant, and (abusing notation), let $\lambda$ also denote the image of $\lambda \in X_*(T) \subset G(\mathcal{K})$ in the quotient $\text{Gr}_G$. Define $\text{Gr}^{\lambda} = G(\mathcal{O}) \cdot L_{\lambda}$.

(1) How can we prove that $\text{Gr}_G$ is the disjoint union of $\text{Gr}^{\lambda}$ (as $\lambda$ ranges across the dominant weights)?

This is stated on the fourth paragraph of page $4$, http://arxiv.org/pdf/math/0401222v4.pdf .

(2) Let $\mu \in X_*(T)$ be another dominant weight. Why is the following statement true: $\text{Gr}^{\mu} \subset \overline{\text{Gr}^{\lambda}}$ if and only if $\lambda - \mu$ is a sum of positive co-roots?

I'm guessing we need to construct a set of limit points (to show that $L_{\mu} \in \overline{\text{Gr}^{\lambda}} \Leftrightarrow \lambda - \mu$ is a sum of positive co-roots) - but I'm having trouble.

This is stated in Remark $2.2$ on pg $4$ of http://arxiv.org/pdf/math/0401222v4.pdf .

share|improve this question
    
Vinoth: It's worth asking the authors in this case, though of course they might not add much to what's in their paper. –  Jim Humphreys Aug 28 '13 at 20:55

2 Answers 2

up vote 1 down vote accepted
+100

My knowledge about this is cumming from p-adic groups and not from ind/prog groups, so the following (especially part III) might be inaccurate or incomplete.

I. Proof that $\bigcup Gr_\lambda=Gr$ for the $GL_n$ (or similarly $SL_n$) case.

This is basically Gauss elimination posses. We have to proof that any matrix can bring to a diagonal form with ascending powers of $t$ on the diagonal by multiplying it from both side by a matrix in G(\mathcal{O}). in other words we are allowed to make the following elementary operations:

a) switching 2 rows or colons

b) multiply a row or a colon be an element in $\mathcal O^\times$.

c) adding one row (or colon) to another after multiplying in by element in $\mathcal O$

Using (a) we can put the entry with the lowest valuation (i.e. with the lowest power of $t$ ) to be $a_{11}$. then using (c) we can kill all the rest of the first row and colon. Using (b) we can normalize $a_{11}$ to be a power of $t$.

We continue by induction.

II. Proof that $ Gr_\lambda \cap Gr_\mu=\emptyset$ for the case of $SL_2$.

consider the action of $G(\mathcal K)$ on the set of $\mathcal O$ submodulus in $\mathcal K^2$. let $f(g)$ be the maximal valuation of $g(\mathcal O^2)$. it is easy to see that $f$ separate the $Gr_\lambda$.

A similar proof is poseble for $GL_n$. I believe you can find in http://www.math.uchicago.edu/~mitya/langlands.html (Course on representations of p-adic groups)

III. description of $\overline {Gr_\lambda}$ for the case of $SL_2$.

as we sow above $G_{\lambda}=f^{-1}(\lambda)$. We need to show that the closure of $f^{-1}(\lambda)$ is $f^{-1}(\{\mu| \mu\geq \lambda\})$. If instead of $f$ we consider the valuation on $\mathcal K$ this statement is obvious. So I believe it easily follows from the ind-variety structure on the affine grassmannian

I think that the general case of (III) can be deduce from the $SL_2$ case. I think also that (I) and (II) are easily generalized to general groups if you know some structure theory, but I'll have to think about it.

share|improve this answer

This may be a little bit lazy but the statement you want is given on page 227 in [Lu]. The proof is given on page 228.

[Lu]=Lusztig, George Singularities, character formulas, and a q-analog of weight multiplicities. Analysis and topology on singular spaces, II, III (Luminy, 1981), 208–229, Astérisque, 101-102,

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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