Peng Zhou

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blog:2025-01-02 [2025/01/04 01:24] – [Skew Howe paper] pzhoublog:2025-01-02 [2025/01/04 07:14] (current) – [What is the slice?] pzhou
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 Using convolution space to resolve is also OK. The Bott-Samuelson resolution? Using convolution space to resolve is also OK. The Bott-Samuelson resolution?
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 +==== What is the slice? ====
 +Consider the $gl(N)$ nilpotent cone, cone in the sense of invariant under $\C^*$, but not closed under addition. 
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 +MVy gives construction for transverse slice $T_x$ to $x \in O_\lambda$, and we have $T_{x,\mu} = T_x \cap \overline O_\mu$, the transverse slice $S_\lambda^\mu$. However, I have no intuition what is the shape of the slice, or its resolution.
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 +Next, we consider the congruence subgroup $L^-G = \in G[z^{-1}]$, which is are section of group $G$ that passes through $e \in G$ at $z=\infty$. So, I guess we can view $L^- G$ as a subgroup in $G(K)$. 
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 +So, we have torus fixed point $L_\lambda$, which is the diagonal lattice. Consider the almost trivial $G=GL(1)$ case. The things in $L^- G$ is only, trivial $e$. Too boring. $G=GL(2)$, let $u=z^{-1}$. We consider $G[u]$, but a non-vanishing algebra section over $\C$ is basically just constant, and we require the matrix at $u=0$ is identity, so the determinant is basically $1$. That is $SL_2[u]$. It is not too hard to get a lot of elements here. 
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 +We define $T_\lambda = L^-G \cdot L_\lambda$, $L_\lambda$ is certain non-negative lattice in $\C^M( (z) )$. 
 +What kind of subsets is $T_\lambda$? Will the group $L^-(G)$ be transverse to $G[z]$ inside $G( (z) )$. Is this so called MV-cycle? 
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 +No, I am barking at the wrong tree. This $L^- G$ is a finite co-dim subgroup of $G[z^{-1}]$. Suppose $g(z) \in L^- G$, and we wonder what is $g(z) L_\lambda$, then it probably can have a lot 
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blog/2025-01-02.1735953886.txt.gz · Last modified: 2025/01/04 01:24 by pzhou