Peng Zhou

well, two months passed.

what did I learn today? with Alexei's talk yesterday and the discussion today.

what I

fk, a month has passed.

what do I want? I want raising and lowering operator, which is adding and removing strand operator, which is a special case of gluing an extra guy and put some extra strands operator.

How does sl2 work? We know that, by KLRW algebra, the planar version, modulo some stop, there is only one way to realize the categorification.

I bet, we can do a purely downstairs theory, even define it.

So, how do we define it? Suppose, we say that hom betewen T-branes follows KLRW algebra, then what?

all the homs are in degree zero?

1. after discussion with M yesterday, I realized I need more terms in the differential, ok, not bad. I should write up some examples, for other people to get it.

2. I don't know how it is related with representations of $gl(1|1)$. I think it is about Alexander polynomial. (is it about oriented knot?)

so you found a secret rule to define the differential, ok, good for you!

how to prove that it works? they are guesses, though very solid ones.

1. you need to define a functor;

2. you need to prove an excision lemma, things are only dependent on the boundary, which reduces this question to the many stop case.

ok, why it is about $gl(1|1)$? Vera told me, we have $E,F, H, C$.

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.

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

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Now, what is the expectation? We should use the KLRW algebra as the benchmark to tell me what grading on the endo of the T-brane I should get. Basically, dot has positive grading 2, crossing with a puncture has grading 1.

First question, you claim that, you have an $S^1$-family of symplectic form, show me. Consider $\C^2 / \mu_n$, with weight $(1,-1)$, $n=2$. We consider the coordinate ring $$ \C[x,y]^{\mu_2} = \C[xy, x^2, y^2] $$ when you blow-up, you put in the ratio coordinate $u = x/y$ and $v = y/x$, you find that this happens to be invariant since $n=2$. So, you have two patches, one with coordinate $(x/y, y^2)$ call it $U_y$ (it does not mean $y \neq 0$, it just means when $x,y$ both goes to 0, $x$ goes to zero faster, the other with coord $(y/x, x^2)$ call it $U_x$. Is this the total space of $O(-2)$? yes (tangent bundle to fano has section, since we have many automorphism)

So, now, what is the hol'c symplectic form? We can use the old one, like $$ \Omega = dx \wedge dy = (1/2) d(x/y) \wedge d(y^2) = (1/2) d x^2 \wedge d(y/x). $$ It is the one that makes sense in local coordinate.

Suppose we do $\C^2 / \mu_3$, weight $(1,-1)$, then the invariants are $x^3, y^3, xy$. How do we blow-up? Well, we can consider more invariant functions, like $x^2/y, x/y^2, y^2/x, y/x^2$. So, we have 3 patche, with coordinate

• $(y/x^2, x^3)$.
• $(y^2/x, x^2/y)$
• $(y^3, x/y^2)$

I don't know how I get it, I guessed it, and it worked.

So, we have local coordinates, and we have holomorphic symplectic form. (why it is non-degenerate? well, just check it locally. )

We define $\omega_\theta = Re(e^{i\theta} \Omega)$.

What does extra grading mean? Well, you can have if you have a Kahler manifold $M$, you can have $H^*(M)$ equipped with an extra grading $p-q$, Hodge grading. Is it possible to get this from the $S^1$-family? In the case of $\P^n$, all the weight grading are zero, because you only have $(i,i)$ class. That's not the case for elliptic curve, where you have $(1,0)$ and $(0,1)$.

Hey, I made some progress today. About Koszul duality, at least, I have a concrete conjecture. Let's state it as 'The symplectic Koszul duality for category O'. It goes as following: take a 3d N=4 gauge theory, some compact group $G$ acting on some representation $T^*N$. You can form two LG A-models, let's call that $(M_{H, \alpha}, W_{H,\beta})$ and $(M_{C, \beta}, W_{C, \alpha})$, where $\alpha$ and $\beta$ are some parameters. Then you prove that the two wrapped Fukaya categories are equivalent (after passing to $\Z/2$ grading? or invent a mixed version of the Fukaya category, where the hom space has a second grading)

What does mixed mean for toric variety constructible sheaf? This is quite useful, since Fukaya category for $T^*X$, $X$ toric is the same as constructible sheaf on $X$. First of all, we only consider unipotent monodromy, second of all, only an affine toric variety. Unipotent is somehow essential, because it create fiiltration on the nearby cycle. Jordan decomposition. Then, you can do fake Frobenius twist. Somehow, the fake Frob twist can be made isomorphic with the old one. Example, on $\C^2$, with basis $e_1, e_2$, we had unipotent monodromy $e_1 \mapsto e_1, e_2 \mapsto e_2+e_1$.There is nothing holy about $e_2$, if we change it to $e_2 + x e_1$, it still works. The Frob pullback monodromy is like, the old monodromy, but $p$ times. We need to find a matrix $A$, such that $$ A \begin{pmatrix} 1 & 1 \cr 0 & 1 \end{pmatrix}^p A^{-1} = \begin{pmatrix} 1 & 1 \cr 0 & 1 \end{pmatrix} $$ The trouble is that, $A$ is not unique, but only well-defined up to a stabilizers of the two matrices. But, once you fix $A$, you can diagonalize the matrix, and find eigenspaces. That eigenspace splitting is the weight decomposition.

So, the data of a Frobenius eigensheaf for the toric variety, is somehow and upgrade of the weight filtration to a weight $\Z$ 'grading'. like find a basis adapt to a flag.

What's the analog here? First of all, it is nothing about toric variety. We will have $T^*(Gr(k,n))$. If anything, it should be about a hyperKahle rotation. Namely, we have an $S^1$ worth of family for symplectic structure (together with Kahler structure).

Let's do an example. Consider $T^*\C$, with base coord $z = x + i y$, fiber coord $w = u + i v$, $x,y,u,v$ are real coordiante. Then, we have $$\lambda_\C = w dz = (u d x - v dy) + i (v dx + u dy)$$ If we take $$ \lambda_\theta = Re(e^{i\theta} \lambda_\C) = \cos(\theta) (u d x - v dy) - \sin(\theta) (v dx + u dy) $$

So, what's going on? It is like, we apply $e^{i\theta}$ rotation of the fiber (only the fiber, not the base, not $(1,1)$ weight rotation, not $(1,-1)$ weight Ham rotation, just one factor). I don't think it is going to be invariant.

The monodromy of $S^1$-family of symplectic structure on a fixed space. What is 'winding'?

In the space of symplectic structure, we have a loop.

It is so annoying that I cannot pin this done. Let me try again.

Let's say, just in the hypertoric case (which I know I am cheating, but whatever). You have two spaces, why there is a matching of Fukaya categories for two seemingly unrelated spaces? Even just for $T^*P^{n-1}$ and $\C^2 / \mu_n$ resolution, why?

I know you will say, Gale duality, but I don't understand why it worked.

Let's try to just guess, what the Lagrangian correspondence will look like.

We have complex moment maps, $\mu: M = T^*\P^2 \to \C^2$ and $\mu^\vee: M^\vee = \C^2/\mu_3 \to \C$.