Peng Zhou

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 $\mu^\vee: M^\vee = \C^2/\mu_3 \to \C$.

It is useful to recap what I did today, or these days.

I have been thinking about Koszul duality a lot these days.

• where does the mysterious grading come from. What is mixed category? The terminology is weird that, mixed constructible sheaf does not form a mixed category. you need to take a (not full) subcategory.
• why the two spaces, Coulomb branch and Higgs branch, are related? I don't see the relation at all. You can say, categorically, one is about $Map(S^1, V/G)_{dR}$ the other is $Map(S^1_{dR}, V/G)$. But concretely, on a low brow way, what are we talking about?

You want to say, for the same theory, the Coulomb branch and Higgs branch, are just two shadows of the same object, the theory. Therefore, the category of line operators in the two branches are the same. Just some equivalence of categories (but where are the generators?) The BDGH paper. let me shut-up and read.

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I read Tom Braden's paper on mixed category for toric varieties. At least one idea is useful, namely where does the extra grading come from.

The idea is that, given a vector space with a unipotent operator, then you can split the vector space into Jordan blocks. Each Jordan block has a Z-indexed-filtration, with the index well-defined upto a global shift.

Suppose you have T^*P^n, with the stratification given by P^0, P^1, ... , like in category O. Then we want to equip the wrapped Fukaya category's cocore's endomorphism with extra grading.

Combinatorially, we can try to match the $End(T)$ algebra with the combinatorially defined $A(V)$ algebra. $A(V)$ algebra has extra grading. You can say that this is like $T$-equivariant cohomology, where the generator has cohomologically grading $2$. Then, we forget the cohomological grading and only keep the extra grading. Then the question is, why the endomorphism of $T$ brane is the same as $A(V)$. You can say that Gammage-Mcbreen-Webster solved the problem, but not quite. Local system is naturally identified with coherent sheaves module over torus; Only unipotent local system is identified with module near the identity of the torus; and only unipotent system with an explicit Frobenius action allows you to have a $\Z$-index filtration. So, the Fukaya category needs a graded lift.

There should be a version of the A-side, whose endomorphism of co-core is isomorphic to $A(V)$ as an ungraded algebra, then we can choose some graded lift. Well, the crossing over (between strands of different color) should have grading $1$, dot having grading $2$. That's the grading convention of $A(V)$.

Now, what's the relation between

• Frobenius on a variety $X$ defined over $\F_q$. we get, mixed constructible sheaf.
• algebraic Lagrangian skeleton in a algebraic symplectic variety, over $\F_q$. we should get mixed microlocal sheaf.
• holomorphic Lagrangian skeleton in a holomorphic symplectic variety, over $\C$. we should get mixed microlocal sheaf.
• Ask Saito, what's the analog in $\C$ for the mixed constructible sheaf?

In principle, all these question should be answered by these Koszul duality people. Koszul duality is about, two holomorphic skeletons, and a duality correspondence that sends cocore to core. It has been proven in the realization of mixed DQ module; but not yet in the relatization of Fukaya category.

First of all, Reeb chord, is not like Morse critical point. So, we don't necessarily have a canonical grading on it. It depends on a choice of holomorphic symplectic structure. There is some Conley-Zehnder index that I don't quite know. Maybe only $\Z/2$-grading is canonical. But that is Morse grading.

Let's dream a bit. Consider the 3d mirror between $T^*\P^2$ and $A_2$ surface, the smoothing of $\Z^2/\mu_3$ (we say $A_1$ surface is the smooth $T^*\P^1$).

What's the classical story of Koszul duality? Quadratic duality.

• symmetric algebra vs. anti-symmetric algebra.
• symmetric algebra but with a bit more quadratic relations; less anti-symmetric algebra. I don't have a geometric intuition for what does that mean. I don't understand why quadratic plays a big role here.

There is another way to think about this. Say, you have some semi-simple ring $R_0 = \oplus k e_\alpha$, and then some graded ring $R = R_0 \oplus R_1 \cdots $. So we have some 'augmentation', $R \to R_0$, most likely sending all the $R_{>0} \to 0$. Then, we should have some $R^! = End_{R-mod}(R_0)$. That might serve as the dual algebra. maybe graded? diagonally graded?

OK fine.

How about the algebra $A(V)$, it is positively graded, with degree $0$ part semi-simple. When the conormal intersects with the 'diagonal Lagrangian' (ok, I don't really know what does that mean? what is this diagonal Lagrangian in $T^*\R^n$? There is no good looking one. We can say

Suppose we have a region $\Delta_\alpha \In \eta + V$ and dual region $\Delta^\vee_\alpha \In V^\vee_\xi$, They don't necessarily intersect. Projectively, they might intersect.

Consider the following construction. Starting from $T^*\C^N$, with Hamiltonian $(\C^*)^N$-action, with alg moment map to $\C^N$ with singularity hyperplanes. OK, take a slice $\eta + V_\C \In \C^N$ (taking complex moment map), then take a quotient.

Consider SES $$ V_\Z \to \Z^N \to (V^\perp)^*_\Z $$ and $$ V^\perp \to (\Z^N)^* \to V^*_\Z $$ We want to say that $(V^\perp_\Z)^* \otimes \C$ is like the dual Lie algebra $Lie(L)^*$ of $L \In T=(\C^*)^N$. Then, $Lie(T)^* \cong \Z^N$.

• We are alg symp reduction by $L = (V^\perp_\Z)\otimes \C^*$ to get $M_H(V)$.
• We are alg symp reduction by $(V_\Z)\otimes \C^*$ to get $M_H(V)$.

It is sort of wrong to say $(V_\Z)\otimes \C^*$ and $(V^\perp_\Z)\otimes \C^*$ are two different subgroups of the same torus. They are different. Although the dual torus of $\C^*$ is $\C^*$, it is not the same one. So, we should not say that they are reduction from the same torus.

However, maybe we can take the product $M_H(V) \times M_H(V^\vee)$. We can say that, we have $T^*(\C^{2N})$, and do Hamiltonian reduction by a Lagrangian torus $V_{\C^*} \times V^\perp_{\C^*}$, and is left with still a product space.

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