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.
Here is a puzzle:
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:
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