Peng Zhou

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====== 2023-09-12 ======

The new exciting thing today is, the raising and lowering operator. 

===== raising and lowering =====

Let $E$ be the functor for adding strands, and $F$ the functor for removing strand, then we should
  * define the functor $F$
  * check that $F$ and $E$ are adjoint
  * test it on T-branes. 
  * understand Rouquier's condition geometrically.

The setup is the following: $\gdef\lcal{\mathcal L}$
  * we have a smooth closed surface $S$ with boundary, and there are some stops on the boundary. 
  * We can consider $k$ disjoint Lagrangians $L_1, \cdots, L_k$, which defines an object in $Sym^k(S) \RM \Delta$
  * We can consider $k$ disjoint Lagrangians $\lcal_1, \cdots, \lcal_k$ in $S \times \C^*$, which defines an object in $Sym^k(S \times \C^*) \RM \Delta$. We assume these $\lcal_i = L_i \times F_i \In S \times \C^*$ are of product type. 

I immediately run into a trouble, which is, how to deal with the fact that, when I add another strand, the superpotential would be very different (there are interaction among many many terms). 

I could say the following: either don't deal with the fiberwise non-compact space (no, I need T-brane).

==== Challenge 1 ====


define $E_L$. My Lagrangian is sitting in $Sym^k(B \times \C^*) \RM \Delta_B$. Then, I want to concatenate a strand, an object, near a specific stop. We can certainly do it for object, (hom from T to each strand you want to delete, you get a dg vector space as coef) and we can do it for morphism as well. it doesn't do much.

define $F_R$. Suppose you want to subtract a strand. ok, this is really interesting. On the level of objects, you can wrap the $T$ brane, and count the intersections with the many strands. On the level of morphism, it works as well, by counting disks. 

Try see if we have $E_L F_R \to id$, and we have $id \to F_R E_L$. In the first case, you subtract, then you add. oh, it is then like tautology, because it is like $T \otimes hom(T, L_i) \to L_i$, sure. when you add and then substract, you always have that $id \to Hom(T,T)$. 

==== Challenge 2 ====
See if you can recover $NH_k$? But this is already done. Let's double check. 

how to see $T^{k-1} \to (k) Hom(T, T) T^{k-1}$, this is like $(k) NH_{k-1} \C[x]$, OK, roughly speaking $\C[S_k]$ as $\C[S_{k-1}]$ module is $k$ dimensional. OK. fine.


==== Challenge 3 ====

Reproduce Rouquier.