Differences

This shows you the differences between two versions of the page.

--- blog:2025-01-02 [2025/01/03 21:25] – pzhou
+++ blog:2025-01-02 [2025/01/04 07:14] (current) – [What is the slice?] pzhou
@@ Line 35: / Line 35: @@
 $$\gdef\ncal{\mathcal{N}}$$
 Now, they consider something really weird (Is that already in MVy paper? https://arxiv.org/pdf/math/0206084) Here we have some basic story for Nilpotent orbit and slices for $GL_m$. It is always healthy to learn some basic rep theory. Here we go:
+==== Mirkovic-Vybornov ====
   - Let $D$ be an $N$ dimensional vector space. Let $\ncal$ be the nilpotent cone in $gl(N)$. If $x \in \ncal$, we can ask for its Jordan block type, which is an un-ordered partition of $N$. Denote this partition by $\mu=(\mu_1 \geq \mu_2 \cdots \geq \mu_m > 0)$. Then, we can do the dual partition $\mu^\vee=(m=\mu^\vee_1 \geq \mu^\vee_2 \geq \mu^\vee_n > 0).$, let $\vec a$ be a permutation of $\mu^\vee$. Now, we are ready to consider a particular of n-flag variety. $$ F_{\vec a} = \{0 =F_0 \In F_1 \cdots F_n \mid \dim(F_i/F_{i-1}) = a_i \}, \quad T^*F_{\vec a} = \{(u,F) \mid u (F_i) \In F_{i-1} \}. $$ Well, I don't understand why such an endomorphism $u$ provides the cotangent direction. I can tell this is true for Grassmannian, how about 2-step flag? Well, we can first say that, the space of flags is transitive under the global $GL(D)$ action, so any infinitesimal action is generated by $End(D)$. There are certain parabolic sub-algebra $\frak p$ preserving the partial flag, so the tangent space is $\frak g / \frak p$, and its dual is $\frak p^\perp$, namely those dual $(\frak g)^\vee$ element that vanishes on $\frak p$. How does that translate to $u(F_i) \In F_{i-1}$? The things in $\frak p$ are those $x \in \frak g$, where $x(F_i) \In F_i$. The way dual $\frak g$ is identified with $\frak g$ is via taking trace. So, if we want to have an element $u \in \frak p^\perp$, the necessary-sufficient condition is that $Tr(u x) = 0$ for all $x \in \frak p$. Then, by an explicit calculation of trace with basis, we can see the cotangent fiber is parametrized by this, $\frak p^\perp \cong \frak n_p$.
   - Now I am very confused. Given a Jordan block type, meaning a partition, we can have many ordered partition corresponding to it. What's the meaning of the ordered partition? OK, just like $sl_3$ Weyl group acting on the weight lattice. There are different 'singular block' I would say, where we can have highest weight be $(a,a,b)$ or $(a,b,b)$, for $a>b$. Do they corresponds to different representation of $sl_3$? I think so. (weight lattice of $sl_3$ is the diagonal quotient of that for $gl_3$).
+  - Let $P$ be a parabolic Lie subalgebra of $gl(N)$, and $a=(a_1,\cdots, a_n)$ be ordered partition where $a_i$ is the size of the $i$-th block (Borel corresponds to all $a_i=1$). Let $\frak n_P$ be the nil-radical of $P$, and we consider $G \cdot \frak n_P$ by adjoint action. Claim that this is the closure of some nilpotent orbit $O_\lambda$.
+  - Example: Consider $G/P$ is $(N,N)$ 2-step flag, ie. $G/P = Gr(N, 2N)$, a generic element in $\frak n_P$ is a block-upper triangular matrix of rank $N$, which can be conjugated to $$\begin{pmatrix} 0 & I_N \cr 0 & 0 \end{pmatrix}$$, this in turn can be put into a Jordan form, which has $N$ many $2 \times 2$ blocks. So, given an ordered partition $a$, we form the un-ordered partition, $\mu_a$, then do the transposition for the dual partition, that gives the nilpotent class for the cotangent fiber.
+  - Another example: consider $G/P$ is the full flag, so the $\vec a=(1,\cdots, 1)$. Then we take a generic element in $n_P$ is a regular nilpotent element.
+OK, so we have learned how to resolve $\overline{O_\mu}$  nilpotent orbit closure, which involves a choice of ordered partition.
+==== What is $G_N$? ====
+This is a very weird subset of the full Grassmannian. There is no finite dimensional analog. So, we have two cuts, one is the determinant cut, the total singularity is positive $N \geq 0$; the second is that, in each direction we have some positivity constraint. Imagine we have $S_m$ acting on $\Z^m$ by permutation, then we have some 'hard wall' given by $\Z^m_+$. It is sort of a truncation, filtration? Given the cone $\Z^m_+$, and given a level $N$, we have a finite many lattice points, hence Weyl orbit, hence T-fixed points. This cone is closed under addition. So we are good.
+Using convolution space to resolve is also OK. The Bott-Samuelson resolution?
+==== What is the slice? ====
+Consider the $gl(N)$ nilpotent cone, cone in the sense of invariant under $\C^*$, but not closed under addition.
+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.
+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 $G(K)$.
+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.
+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?
+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