I still want to understand what Cautis-Kamnitzer-Licata did.

One motivation is the geometric of geometric Satake, which says $Rep(G)^{fd}$ and $Perv_{G^\vee(O)}(Gr_{G^\vee})$ are related, simple representation $V(\lambda)$ goes to IC sheaf $IC_\lambda$. The guess is, if $\lambda$ is miniscule, then $Gr_\lambda$ is smooth projective, and the graded dg category $Coh_{\C^*}(Gr_\lambda)$ can be used to do categorification of $V(\lambda)$. Recall $\C^*$ acts on $G^\vee(K)$ and $G^\vee(O)$ by rotating the domain disk.

There is something called convolution Grassmannian, just sequences of nested lattices. It also have $\C^*$-action, and it is a (Bott-Samuelson) resolution of singular space $Gr_{\sum_i \lambda_i}$. Conjecturally, this convolution Grassmannian corresponds to categorified tensor product.

How to discuss the 'relative position' of two elements in Grassmannian? I mean $G\RM (G/P \times G/P) = P \RM G / P$. If $G = GL(n), P=P_{n_1,n_2}$, then I know $G = \sqcup BwB$, but many $wB$ can be absorbed into $P$, so we have classification by double coset $W_P \RM W / W_P$. This is like 'shuffle'. Now similarly, $W_{aff}$ contains a copy of $W$ and a copy of $\Lambda$, but $W_{G(O)}=W$, so after quotienting, we are left with just $\Lambda/W = \Lambda_+$.

A state is a sequence of miniscule $\lambda_i$, which corresponds to a convolution Grassmannian's Coh category. A braid corresponds to an invertible functor from one to another convolution Grassmannian.

• Is the convolution Grassmannian a moduli stack of a type of objects inside some category? I tried to use Legendrian sheaves to get this, but it is not natural at all. Pause this thought.

Here is a puzzle:

• 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?

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:

1. 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, $\frak p^\perp \cong \frak n_p$.
2. 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 $gl_3$).
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 $O_\lambda$.
4. 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.
5. 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.

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?

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