Let $H = L^{2}(S^{1},\mathbb{C}^{n})$, $H_{0}\subseteq H$ the subset of maps that extend holomorphically to the unit disc, and $H_{m} = z^{m}H_{0}$. Consider the affine Grassmannian for $GL_{n}$ in the lattice model:

$\mathcal{Gr}_{n} = \left\{ V\subseteq H:\, zV\subseteq V\right\}$

Let me describe two possible filtrations of this object. The first, is by the subspaces whose lattice diagrams have all their sticks and dots concentrated in a band of width $2i$ about zero:

$\mathcal{Gr}_{n}^{i} = \left\{V \in \mathcal{Gr}_{n}:\, H_{i}\subseteq V\subseteq H_{-i}\right\}$

I think this can also be identified as the union of orbits of elements $f \in \mathrm{GL}_{n}(\mathcal{K})$ such that $f(z) = \sum_{\vert k\vert \leq m} A_{k}z^{k}$.

The second filtration I can think of comes from the decomposition of the affine Grassmannian into $\mathcal{Gr}_{\lambda}$'s. For each coroot $\lambda = (\lambda_{1},\dots,\lambda_{n})$ satisfying $\lambda_{1}\geq \lambda_{2}\geq\dots\geq \lambda_{n}$, we get a finite dimensional variety $\mathcal{Gr}_{\lambda}$ from the orbit of $GL_{n}(\mathcal{O})$ through $\text{diag}(z^{\lambda_{1}},\dots,z^{\lambda_{n}})\cdot H_{0}$.

If a coroot $\lambda = (\lambda_{1},\dots,\lambda_{n})$, define its height by $\text{ht}(\lambda) = \sum_{i} \lambda_{i}$. Let $\check{T}/W$ denote the set of Weyl orbits of the coroot lattice, for which I always pick representatives in the positive Weyl chamber, and define:

$X_{i} = \coprod_{\lambda \in \check{T}/W,\, \text{ht}(\lambda) \leq i} \mathcal{Gr}_{\lambda}$

My question: Is $X_{i}$ the same space as $\mathcal{Gr}_{m}^{i}$? I think the main difficulty I'm having with answering this for myself is that I'm not sure how to compute to which $\mathcal{Gr}_{\lambda}$ a given lattice $V$ belongs. Any reference or explanation for this would be much appreciated!



This statement isn't true; as you've currently defined everything, your $X_i$ are infinite dimensional (in fact they are unions of connected components.) You want instead to take the $X_i$ unions of $\lambda$ with $i\geq\lambda_1$ and $\lambda_n\geq -i$.

Let $B$ be the Borel for $GL_n(\mathcal{K})$. Both $\mathcal{Gr}_n^i$ and $X_i$ are $B$ invariant (as they are $GL_n(\mathcal{O})$ invariant), so by Bruhat decomposition we just need to check that they contain the same set of points of the form $\operatorname{diag}(z^{\lambda_1},\cdots,z^{\lambda_n})\cdot H_0.$ $\mathcal{Gr}_n^i$ contains such a point if and only if all the $\lambda_i$ are from $-i$ to $i$, and $\mathcal{Gr}_{\lambda'}$ contains such a point if and only if $\lambda$ and $\lambda'$ are Weyl conjugate. This immediately gives what you want.

(As for an explicit description of which $\mathcal{Gr}_{\lambda}$ a given lattice is contained in: This is determined by the "dimensions" of the intersections of the lattice with $H_i$ (where by dimension I really mean valuation of $z$ in the determinant.) This is because this is a $G(\mathcal{O})$ invariant property and then we use Bruhat decomposition as above.)

If you would prefer to avoid using Bruhat decomposition, for $GL_n$ all of this can be done explicitly by explicit diagonalization starting from dimensions of intersections with $H_i$. Sorry if any of this is sketchy; I don't have much time right now, if necessary I can give more details later.

  • $\begingroup$ I probably should have noticed that there were infinitely many $\mathcal{Gr}_{\lambda}$'s in each piece. Even for $GL_{2}$ there are the coroots $\text{diag}(z^{n},z^{-n})$. Thanks for your answer! $\endgroup$ – James Mracek Jan 24 '15 at 15:41
  • $\begingroup$ Can you elaborate on what you mean by the valuation of $z$ in the determinant? $\endgroup$ – Tyler Holden Feb 4 '15 at 17:05

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