Peng Zhou

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====== 2023-09-16 ======

The removing strand operator is not that simple: taking intersections, and putting in the object. There must be some interesting differentials correcting it.

Let $L^k$ be a k-tuple of Lagrangians in $\Sym^k(\Sigma)$, avoiding a stop. Let $E$ be the raising operator, i.e., $F$ the lowering operator. Let $F$ be adding a brane, by adding a T-brane, and $E$ be $Hom(T, -)$. (note the change of notation).

Then, $F$ is easy to achieve, thanks to the stop, we can just add a brane there. But, $E$, its right-adjoint, is a bit difficult. The adjoint condition basically involves solving an equation. What we want, is a concrete, purely Fukaya category like functor. 

===== Removing a Strand =====
If we are just dealing with $gl(1|1|)$ quiver Hecke algebra, then it is a purely algebraic game. 

In the case of T-brane, on a disk with two stops (linear KLR), the problem is completely solved. We just have $k$ output with degree $0, -1, \cdots, 1-k$, with the $i$-th term maps to the $i-1$-th term, with the obvious map. $q$-degree can be deduced.

But, what is the holomorphic disk counting proof? And, what if we have something other than a T-brane? You cannot just say how it acts on object. 

Indeed, suppose we have a test object $L'$ with $(k-1)$-tuple of disjoint Lagrangian, and we have $L' + T$ hom to some $L^k$. There is a clear count of all possible intersections. Here already, we have differentials, it is not possible to eliminate differentials. The differential can involve a $T$-intersection point, or not. If it does, then we need to explicitly remember it. 

However, this depends on a choice of a test object, and the disk counting uses that test object. Is there a universal test object? Let me draw some examples with different test objects.