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--- blog:2025-01-02 [2025/01/03 18:16] – pzhou
+++ blog:2025-01-02 [2025/01/04 07:14] (current) – [What is the slice?] pzhou
@@ Line 19: / Line 19: @@
   * later they uses Nakajima quiver variety, and took the B-model there, what's the relation?
   * Does it matter whether we do compactification or not? Which one is more natural? So far in the paper sl2 and slm, they do the compactified version.
+They posted three papers in a row, 0902.179x. In these papers, their main examples is $T^*Gr(k,N) $. This is the Higgs branch of the$ [N]-(k)$ quiver. What's the relation to previous work?
+The spherical object and the  $\P^n$  object. https://arxiv.org/pdf/math/0507040, I have no idea what is the Atiyah-class and the Kodaira-Spencer class. Is there a categorical notion for these classes?
+===== Skew Howe paper=====
+In this paper, https://arxiv.org/pdf/0902.1795, they consider the (derived) equivalence
+ $Coh( Gr_{\lambda} \wt \times Gr_{\mu}) \to Coh( Gr_{\mu} \wt \times Gr_{\lambda}).$
+How does this go about? It is not geometrical, doing a fiber product or stuff. One can express it as FM kernel, but it is not useful unless the kernel is geometrical.
+Let  $G = GL_m$ . The affine Grassmannian for  $Gr_G$  is $G(K)/G(O) $. Given an element$ M $in$ G(O) $, it is an$ m \times m $matrix with entries in$ O $, such that its determinant is invertible element in$ O $, in particular one can plug in$ z=0 $to get$ G $. For an element in$ G(K)$, if we do determinant, we would get  $z^n$ ,  $n \in \Z$ . If we do  $SL_m$ 's affine Grassmannian, we just get the 'boring' piece where  $n=0$ ; if we do  $PGL_m$ , then we get the quotient up version,  $\Z/m\Z$  many components. Hmm, it seems to be related to  $\pi_1(G)$ .
+ $\gdef\ncal{\mathcal{N}}$
+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