blog

2026-03-11 should I be a BS artist?

The first time I heard about this phrase, it is used by Prof Z about some work by Prof B. It is half joking, half disapproval. One either needs to painstakingly verify lots of details, or one can wave ones' way out. Through my academic career, I have seen many handwaver, but also more solid prover.

What should I say? The handwaver usually are better at convince an idea, a desire, a plan, which is most of the time what the reader want. The next step is for the reader to believe or not. If the reader know it must be true, no doubt about it, so fine, we can skip the verification step. The prover, taking much more pain, to show all things works, but people usually just glance over without even bother to read or follow.

What should I be? I cannot be 100% prover. I will knowingly leave a gap, a jumpable gap, not requiring to use your wing-of-belief to fly over, just for the sake of speed. Since I don't have much time. Then, should the proof be considered as finished? For the sake of writing the paper and submit it, yes. For the sake of understanding how it works, maybe not.

So, the answer is no. You should write a clear statement with a clear proof. Put doubt and confusion in the world is negative contribution.

2026-02-28 (2)

Previously, we have said that $M_H$ and $M_C$ are $n=0$-shifted symplectic stack. And there is no natural way to shift the $n$ around, so naturally the home for 3d MS is about 0-shifted symp stack.

Now, suppose we are given two hol'c symp Lagrangians, then their intersection is $-1$-shifted symp mfd. $$ X \times_{T^*X} X = T^*[-1] X = Spec_X(Sym(T[1]X)) $$

We somehow need to assign a category to the derived intersection. I don't think that is the way to go, at least not on the 2A-side. One need seriously consider wrapping.

If we consider path space between two Lagrangians, then it carries a natural closed action one-form $\alpha$. Crit cohomology of the path space, $H^*_{crit}(Path(L_1, L_2), \alpha)$, suppose to be Floer cohomology. How to categorify this?


What is the input to a 3d category?

2026-02-28

Here is some vague thought about 3d mirror symmetry (following Ben G and Justin H).

Let $G$ and $G^L$ be Langlands dual varieties. Assume $(G, M)$ is S-dual to $(G^L, M^L)$, where $M$ is some nice G-Ham space, same for $M^L$. Example $$ M = pt, \quad M^L= Whit_{G^L}(T^*G^L) $$ where $Whit_{G^L}$ is one-side symp reduction by $U$ with generic character.

We can consider endomorphism type spaces $$ M/G \times_{T^*[1]BG} M/G \leftrightarrow M^L/G^L \times_{T^*[1]BG^L} M^L/G^L$$ In this example, we should get what? The space $T^*(BG)$ 3d mirror to $Whit_{G^L}^2(T^*G^L)$.

OK, if $M$ is a usual 0-shifted symp stack, then $[M/G]$ should be viewed as a 1-shifted Lagrangian in the 1-shifted cotangent bundle $T^*[1]BG = [g^*/G]$, that's what G-Ham space gives you. So that the intersection is like 0-shifted symp space.

And the terminology about Coulomb branch, and bi-Whittaker reduction is that $$ M_C(G, M) = Whit_{G^L}(M^L) = Whit_{G^L}(T^*G^L) \times_{T^*[1]BG^L} M^L/G^L. $$ $$ M_H(G, M) = M / / G = 0/G \times_{T^*[1]BG} M/G $$ $$ M_H(G, M) \leftrightarrow M_C(G, M) $$

So Coulomb branch 3d mirror to Higgs branch is a special case of relative Langlands, where one slot is about the basic dual setup.

2026-02-24

I am recording this past discussion with Ben G. Here (n+1)-dim TFT will be given by some n-category, we use $nA$ or $nB$ as the corresponding category.

Let $G$ and $G^\vee$ be Langlands dual groups. Assume relative Langlangds says there are some equivalence of 3-cats $$ 3A_G \cong 3B_{G^{\vee}}, \quad 3IndPerv(BG) \cong 3IndCoh(BG^\vee). $$ where objects in $3A_G$ comes from Ham G-space $M = T^*X$, where $X$ are $G$-spherical varieties, which means Borel acts with finitely many objects, same for $3B_{G^L}$. In the following, we will denote objects by $X$ instead of $T^*X$.

Example: $G = G_m, G^L = G_m^\vee$. As dual objects $$ G_m \sim pt $$ $$ A^1 \sim (A^1)^\vee $$ Then we have $$ 3A(G_m, G_m) = 2Loc(G_m) = 2Coh(B G_m^\vee) = Rep(G^\vee)-mod $$

In general, $G$ acts on $pt$ is 4d dual to $G^\vee$ acts $T^*G / /_\chi N$ twisted Whittaker reduction. Quick sanity check $$ (T^*pt/G, T^*pt/G)_G = T^*(pt/G), \quad (T^*G / /_\chi N, T^*G / /_\chi N)_{G^\vee} = T^*(N \RM G / N) $$ If we do $2B$ on the left, we $Rep(G)-Mod$, if we do $2A$ on the right, we get $2A$ on the multiplicative pure Coulomb branches.

What the heck is the $Rep(G)-mod$? For $G=SL_2$, we know $Rep(G)$ is monoidally generated by the fundamental rep $F$, with and $F$ is self-dual. so we have $id \to F^R F$ and $F F^R \to id$. We also have $F^R F \to FF^R$ (some braiding functor?). We have a 2-category (of B-type) $$ Rep(SL_2)-Mod = \{F: C \leftrightarrow C: F^R \mid [id_C \to F^R F \to F F^R \to id_C] = 2 \} $$ Where does such cat $C$ some from? Say $C = Coh(X/SL_2)$.

2026-02-27

Talked with Kifung and Conan about how 3d MS will transport brane to brane. There are some partial success, but still more to be understood.

One interesting thing is that they somehow don't need to specify which side is 3dA-side and which is 3d B-side. Their prescription is somewhat simple, just take fiber product of Lagrangians in (possibly shifted) (real or complex) symplectic manifold, then apply Fuk or Coh to the fiber product.

One thing we need to be worried about is that: mirror symmetry depends on a choice of torus, otherwise it won't be functorial. So, it is wrong to say, I give you a complex manifold, and then I find 'the' mirror of it. What might be better said is that, I give you a Kahler manifold with roughly, a Lagrangian torus fibration, then the mirror is another Kahler manifold with torus manifold. Then we can check if the two sides are mirror to each other.

→ Read more...

2026-01-28

Talking with Hansol about conic fibration over some hypersurface. He mentioned Seidel's paper on Lefschetz fibration's suspension, which I am reading now.

→ Read more...

2025-11-04

I need to write more stuff down, otherwise it is only in my head, mysterious even to my collaborator.

reading the paper that is based on my phd work, makes me a bit happy and sad. happy that my work is useful, sad that I am still in the jobless state. But at least, I am doing math and discussing math with interesting people.

2025-10-23

to do:

  • what's holomorphic Floer theory? In relation to NAHT.
  • what's the idea / statement of geometric langlands?
  • what zastava?

no, I am not the kind of mathematician that writes proofs, I am rather the one that discovers structures and building bridges.

2025-09-05

how to describe perverse schober on $(D, 0)$? it is a machine that, input a disk with stop / singular Lagrangian skeleton / holomorphic function on an open subset, and output a category; input a morphism of object, output a functor; finally, input a 2-morphisms between 1-morphism, and output a natural transformation.

but the problem is, what is a 2-mor in the category of disk with stops? I can view disk with stops as Lagrangian in $T^*X$, or constructible sheaves in X. But these are 1-categories, hom between these objects are just set, at best chain complexes.

How can we move these sectors? closed embedding, open restriction, and then non-char deformation. cobordism as higher morphism doesn't give me much at all.

How to use SOD? Does SOD give me SES of id -> UV -> T? Maybe we just say, 2-perv is data plus condition. this is like saying, endomorphism algebra has some generators and relations, that is to say, endomorphism algebra is the quotient of some free algebra by some 2-sided ideal. So to define action of the quotient algebra on a vector space, we just need to ask the free algebra act, and such that the relation is satisfied. Instead of saying some linear combination of the free algebra is zero, we say some chain complex of object is acyclic. The question is, how do you build that chain complex.

2025-07-02

i want to translate, wedrich-dyckerhoff, into our own language.

First, what is Beck-Chevalley stuff? I am reading page 10 of DW. There are two ways to go from $(2,1)$ to $(1,2)$ partition on the line, one is break-then-merge, the other is merge-then-break. The two ways are different, obviously. And even better, there is a relation between the two ways, how? We have adjunctions, in our setting, splitting is the right adjoint (splitting is restriction) of merge. (we can have various versions of adjoints, don't worry, or limit oneself).

We have these two adjunctions

  • [ (2) -> (1,1) -> (2) ] ==> [(2) -id-> (2) ]
  • [(1, 1) -id-> (1,1)] ==> [ (1, 1) -> (2) -> (1,1) ]

What's the condition of an $A_1$-schober? cofiber of the 2nd guy is an auto-equivalence. Fiber of the first guy is also an equivalence.

If we want to match with the diagram DW consider, we need to identify the left-adjoint with merge, and right-adjoint with split

We are going to use these adjunctions to get the sweeping move

  • (2,1) -> (1,1,1) -> (1,2)
  • (2,1) -> (1,1,1) -> (1,2) -> (3) -> (1,2)
  • (2,1) -> (1,1,1) -> (2,1) -> (3) -> (1,2)
  • (2,1) -> (3) -> (1,2)

what are we trying to say now? we can say

  • (a+b,c) -> (a,b,c) -> (a,b+c)
  • (a+b,c) -> (a,b,c) -> (a,b+c) -> (a+b+c) -> (a, b+c)
  • (a+b,c) -> (a,b,c) -> (a+b,c) -> (a+b+c) -> (a, b+c)
  • (a+b,c) -> (a+b+c) -> (a,b+c)

There is a natural transformation from the top row to the bottom row, but

Given this commutative square (1,1,1) -> (2,1) or (1,2) -> (3), it is neither a pullback square or a pushforward square.

let me read the construction 3.5. OK, you have something X, living over the $k$-cube. You don't want a section, no. you want a 'fibered map'

you first map a thickened cube, to the cube, where the initial time slice maps to the origin, and the final slice map to identity. Then you want to map the thickend cube to X, so that fiber goes to fiber. what's going on here? You want these arrows, along the time direction, to be 'cocartesian', meaning having the initial-factor-through property, as GPT teaches me.

What is the so called inflation-deflation, on pg 13? it is push-forward from the origin fiber, or pullback to the origin fiber

let me try to understand (3.1.4). We start life with a cube worth of categories, and functors lining the edges. We assume there is a biCartesian fibration, which means there exists (up to contractible choices) a unique lift of an edge if we fix the starting point, or the ending point.

We look at a square in the cube, and project the cube to the square. Say a 3-dim cube, so we are left with 1-dim interval in the fiber. We start with the fiber in the upper right corner. we start with the initial node in that fiber, run co-cartesian extension to the full fiber, then for each fiber position, we have a natural transformation. what's wrong with that? why we need to do deflation? ok, whatever. we still get [1]^{n-1}, in which one direction is in the diagonal base direction, and (n-2) is the fiber direction. And this cube is in the category of functors from this tip to that tip.

the total fiber of (3.1.5) as the BC defect. Question, if we permute the ordering of the index, is it still the same thing? why the first two indices?

If we want to use 'right-adjoint' for the merging, then we are not using $NH_2 \otimes_{1,1} M$ to do extension, rather we are doing $Hom_{1,1}(NH_2, M)$

OK, maybe I want to say, there are two ways of doing restriction, yes. there is a 'left-adjoint restriction', which is, you Reeb flow to the stop, cut, then un-flow a bit. ok, i will do that.

Older entries >>