Today, I did

1. more study on the slices of affine Grassmannian

I run into a note by Vasily Krylov's note on slices.

$\gdef\Gr{\mathcal{Gr}}$ There are two key new ideas,

• one is that $\Gr$ is $Bun_G(\P^1)$ with a trivialization on $\P^1 \RM 0$
• the other is that, there is a $\C^*_t$-action on $\Gr$, either viewed as rotation $z$, as $G((z) )/G[[z] ]$, or viewed as acting on the domain $\P^1$. The limit that $t \to 0$ is like zooming in at $z=0$, and $t \to \infty$ is zooming in at $z=\infty$. We have $$(\Gr)^{\C^*_t} = \sqcup_{\mu \in \Lambda_+} G z^\mu$$ where $\Lambda_+$ is the dominant cocharacters in $T \In G$.

Given $\lambda \in \Lambda_+$, we have $$\Gr^\lambda = \{ x \in \Gr \mid \lim_{t \to 0} t \cdot x \in G z^\lambda\}. $$ $$\Gr_\lambda = \{ x \in \Gr \mid \lim_{t \to \infty} t \cdot x \in G z^\lambda\}. $$ $$ W_\lambda = \{ x \in \Gr \mid \lim_{t \to \infty} t \cdot x = z^\lambda\} $$

It probably is a good idea to twist $\C^*_t$ with some $\C^*$ subgroup in $T\In G$, compatible with the choice of $B \In G$. (is there some $\C^*$-action on $GL_n$ by conjugation, where the attracting manifold is $B$?) This hopefully will break the symmetry of all the $T$-fixed points on $\Gr$. The goal for the additional twist is to cut down the fixed loci, so that they become points, maybe the intersection of these stable and unstable manifold will be the open loci $W_\mu^\lambda$ on the nose.

You see, the $G$-action and $\C^*_t$ action on $\Gr$ does not generate $G(O)$-action at all, the $\C^*_t$-action is really powerful. Its effect is hard to see on $Bun_G(\P^1)$, but is visible on the 'trivialization' data.

In the example of $T^*\P^1$, let $G=GL(2)$. We have $\mu = (1,1)$ and $\lambda = (2,0)$.

Claim 1: We have $G \cdot z^{\lambda} \cong \P^1$ (view $z^\lambda$ as $z^\lambda G(O) / G(O)$). So that $$ W_\mu^\lambda = (T^*\P^1)_{aff}. $$

Confusion, what's the dimension of the nilpotent cone for $gl(2)$? Well, it is 2 complex dimension, because $\dim_\C gl(2) - 2 = 2$.

I think this is $Gr(1,n)$, $T^*Gr(1,n)$ is a resolution of some strata of $\mathcal{N}_{gl(n)}$. Which strata?

First, we need to say what is the transverse slice in the nilcone of $gl(n)$. The top strata is given by partition $\lambda$ of $n$ as $n=2+1+1+\cdots+1$, the bottom strata is given by partition $n=1+\cdots+1$. They only point in the bottom strata is $0$. So, the slice is just the top strata's closure itself.

Second, I would like to embed this to affine Grassmannian. We turn partition to dominant weights $\lambda = (2,1,\cdots,1,0)$ and $\mu=(1,\cdots,1)$. So $W_\mu \cap \overline{Gr^\lambda}$ (which contains the point $z^\mu$) is isomorphic to $(T^*Gr(1,n) )_{aff}$.

In this case, the comparison to the nilpotent slice is easy, the top strata is given by partition $n=n$, and the bottom is given by $n=1+\cdots+1$.

So, in the affine Gr slice presentation, we have $\lambda = (n,0,0,\cdots, 0)$ and $\mu=(1,\cdots,1)$ for $G=GL(n)$.