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We define the affine Grassmannian to be the quotient $Gr = GL_n(\mathbb{C}((t)))/GL_n(\mathbb{C}[[t]])$ where $\mathbb{C}((t))$ is the field of formal Laurent series and $\mathbb{C}[[t]]$ is the ring of formal power series. (The affine Grassmannian can be defined more generally, but here we restrict to a special case.) If we let $B$ be the Borel subgroup of upper triangular matrices in $GL_n(\mathbb{C})$, $T$ a maximal torus, then the Weyl group $W=N(T)/T$ is just $S_n$. Let $\widetilde{W} = \mathbb{Z}^{n-1} \rtimes W$ denote the affine Weyl group. Then for $i = 1, 2,...,n-1$ the affine permutations in $\widetilde{W}$ correspond to the usual permutation matrices in $GL_n(\mathbb{C})$, namely the identity matrix with columns $i$ and $i+1$ interchanged. The matrix for the affine permutation $s_0$ has ones along the diagonal in rows $2, 3, ..., n-1$, has $t$ in the right hand corner, and $t^{-1}$ in the bottom left corner. Let $I$ denote the Iwahori subgroup, that is, the inverse image of $B$ under the reduction map $GL_n(\mathbb{C}((t))) \rightarrow GL_n(\mathbb{C})$. Then $I$ is the set of upper triangular matrices mod $t$. I read somewhere that $GL_n(\mathbb{C}((t)))$ has a decomposition

$GL_n(\mathbb{C}((t))) = \cup IwGL_n(\mathbb{C}[[t]])$

where $w$ varies across the affine permutation matrices. This decomposition is supposed to induce the Bruhat decomposition of the affine Grassmannian into Schubert cells. Now something here is wrong. Since $I \subset GL_n(\mathbb{C}[[t]])$, and the determinant of any affine permutation matrix is 1 or -1, we have that for any $w \in \widetilde{W}$ the determinant of any matrix in $IwGL_n(\mathbb{C}[[t]])$ has power series determinant, but the matrix

t^{-1} 0

0 t^{-1}

is in $GL_n(\mathbb{C}((t)))$ with determinant $t^{-2}$ and inverse

t 0

0 t

So, my question is, what is wrong here? What is the correct decomposition and indexing set?

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1 Answer 1

up vote 3 down vote accepted

There are two discrepancies: first, the "strict" affine Bruhat decomposition applies to SL(n), not GL(n); second, the decomposition you're wanting, in the strict case, would be $G=\bigcup_w IwI$, that is, with Iwahori on both sides. (This would make it a disjoint union.)

To adapt to GL(n), the affine Weyl group needs to be enlarged to include the obvious extra diagonal elements.

This doesn't depend so much on $\mathbb C((t))$ and $\mathbb C[[t]]$, as it applies to the field of fractions of a (maybe complete) discrete valuation ring.

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OK, yes, I was thinking that the double cosets would be IwI, rather than IwK, but I was confused because I thought we were working over GL(n). So, if I want the indexing set to be affine permutation matrices, I need to work with SL(n); that gives Gr = U IwK, where K = SL(n, C[[t]]) and the indexing set is diagonal matrices in C((t)) with entries whose exponents are integers that sum to 0. I guess now the question is: what is the indexing set for the decomposition of GL(n)? –  Mehta Jan 5 '12 at 0:05
    
I didn't see where you defined affine permutation matrix, but anyway, it should be a monomial matrix with powers of t. If you want the $IwK$ decomposition (which I would call $IwP$, but never mind) to be disjoint, let $w$ only vary over $diag(t^{d_1}, t^{d_2}, \ldots)$ with $(d_1 \geq d_2 \geq \ldots d_n) \in \integers^n$. –  Allen Knutson Jan 16 '12 at 23:25

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