VGIT take 2, this time with superpotential on the B-side. On one hand, this story is well understood by other people, not me; on the other hand, the story of window on A-side is not so well understand, and we don't know what the mirror of taking quotient is.

Well, the mirror of LG model is another LG model. The most foundational one is the one proven by Nadler, mirror of pair-of-pants. (indeed, this is the building block of everything)

The next step is to turn on some group action on the B-side, and turn on some fibration on the A-side. (or vice versa)

But why the recipe works? Big question, ask Vivek.

Now, Vivek says: one should take categorical fixed point, namely consider the equivariant category. Say $Coh(X) \cong Fuk(Y)$, and $T$ acts on $X$, with $Y \to T^\vee$ (I don't know why that is the case, and how to generalize this to non-abelian group. Ask Teleman), then we have $$ Coh(X)^T \cong Fuk(Y)^T \cong Fuk(\wt Y) $$ where $\wt Y$ is pullback of $Y$ along the universal cover $\wt T \to T$.

That's a good first step, and necessary. Now, we need to pass to categorical GIT quotient, namely, throw away some 'unstable loci' on the B-side, and correspondingly, do some stop removal on the A-side.

This is well understood in the toric setting, when the A-side is given by skeleton.

But, if the A-side is given by fibration with superpotential (in Mina's word, some log term of the superpotential), then it is necessary to unfold.

Is there a way to remove unstable loci?

I don't know how to do it. For example, mirror symmetry says $\C^2$ is mirror to $(\C^*)^2$ with superpotential $W = x+y$, and we have $W^{-1}(R) = \{x+y=R\}$ a pair-of-pants, with skeleton as that, two circle plus an arc connecting them. If we want to describe the mirror to $\C^2 \RM \{0\}$, I don't know how to describe it geometrically. Sure, you can express mirror of $O_0$ as certain Lagrangian, and then quotient out that Lagrangian, but this is not geometrical.

(We can do skeleton, which is more symplectic geometry, but let's not go there yet.)

Suppose we have, on the B-side, a quotient $\C^*$ acting on $\C^2$ by weight $(1,-1)$. Then, on the A-side, we have a map $\pi: (\C^*)^2 \to \C^*_t$, of the form $t = x/y$. Then, we have $W = y(1+t)$, where $y$ is the fiber coordinate.

The B-side quotients (two of them, though isomorphic by chance) are both $\C$. The A-side fibers are of two form, one is $t \to \infty$ and one is $t \to 0$.

If you have to ask, what is the window on the B-side, we say it is generated by a single line bundle $O(k)$ for some $k$. On the A-side, what it means to take a fiber and then do the Fukaya category? It is exactly like, taking a sectororial neighborhood of a downstairs window Lagrangian. Mina's T-brane. It is perverse schober in an interesting way.

Now, let's consider something cooler. How about $\C^*$ acting on $\C^3$, with weight $(1,1,-2)$? Well, we have $t = xy/z^2$, and $W = x+y+z = t z^2 / y + y + z = y(1 + z/y + t(z/y)^2)$. A-side Fiber is $(\C^*)^2$. But, do we know how many singular $t$ are there? Let's compute. It is like, for how many $t$ do we have singular fiber for $x^2 + x + t=0$? Sadly, just one $t$, $t=1/4$. Then, how do we know that the window size have to be $2$? What I thought was wrong! It is useful to understand examples!

Let $t_0$ denote the singular value of $t$. As $t \to t_0$, the fiber A-model degenerate in the sense that the superpotential (hence the stop) changes topology. Previously, the fiberwise superpotential has regular fiber being $\C \RM \{2 pts\}$, now at $t=t_0$, the two points coalesed. There is a vanishing arc in the fiberwise stop. And, that vanishing arc bring out a Lagrangian in the fiber. Now, apply $t$-monodromy to that fiber Lagrangian, we get a twist.

But, how does the A-model know about the window size? It is easy for the B-model to know about window sizes, by looking at the Koszul resolution. For me, I need to use the window skeleton.

If we drag out a thimble from $t=t_0$ to a regular fiber, and ask for the boundary Lagrangian in the fiber, we will get the basic generator, and doing more $t$ rotation give me more.

Is there something about 'maximal number of twists' such that we still have 'linearly independent' Lagrangians?

I believe, one should take a 'tropical fiber' of the universal LG A-model.

It is OK, we have many phases in the GIT quotient, also in the A-model side.

Different fibers have equivalent A-models, and we just have equivalences of categories. (what is the story for non CY case?)

The question is, why GIT quotient corresponds to taking far away fiber?

Let's recall first: we have total space mirror symmetry, not necessarily about toric stuff. We have partially torus action on B-side, and we have fibration to dual torus on A-side.

What does the basic mirror symmetry teach me? Let $T = \C^*$ be the B-side torus, $T^\vee$ is the A-side torus. We have $Coh(T) = Fuk(T^\vee)$. Well, then I don't understand, $Coh_T(T)$ should be a point, but $Fuk(\wt T^\vee)$ is nothing. I think one needs to manually unwrap the skeleton, so put some stop at infinity.

The point is, $T$-eqvariant(ible) objects on B-side (line bundle here) matches with $T$-equivariantible object on the $A$-side, things that is not changed by tensoring by local system, since the line Lagrangian is contractible.

Taking $\C^*$ quotient on the B-side, means we only look at the equivariant degree 0 part of the morphism. That effect say, the structure sheaf behaves like a point, so done. On the A-side, we do the unwrapping, that avoids too many unwanted self-intersections (which corresponds to non-zero equivariant degree morphism).

Now, consider $(\C^*)^2$ on the B-side, mirror to the dual $(\C^*)^2$ on the A-side. Then, structure sheaf goes to some positive Lagrangian on the A-side. Suppose we have $(\C^*)$ acts by weight $(a,b)$ on $(\C^*)^2$ (with a,b coprime), then we do certain unwrapping. On the A-side, if we use sheaf model, then the Lagrangian is a cotangent fiber over a point.

OK, here we have a free quotient on the B-side. I should take unwrapping and put stop at infinity, coz you never wrap past that stop. It is clear, from these basic example, that you should never take the 'fiber' It just happen to be right.

Can I not do ANY unwrapping, just take the equivariant degree $0$ part of the hom space, where the equivariant degree is determined by the map to dual $\C^*$? OK, I guess, you first take those objects that is fixed (up to isomorphism) under the $\C^*$ action on the A-side (by tensor local system), then you restrict the morphism to be degree 0.

That was only OK, if you have free quotient. The stack quotient and GIT quotient, and the naive quotient are the same

Now, consider non-free action. Consider $\C^*$ acting on $\C$. We get either a point or $\emptyset$. I think previously I am very much, deeply confused about direction of unrolling. That unrolled direction, on A-side, is dual Lie algebra space for the GIT quotient space on the B-side.

Consider next, $\C^*$ acting on $\C^2$ with weight $(1,-2)$. We know how to do stacky quotient on both sides. Yes. we should also take localization (stop removal), and then take slices.

But, how do I know if there is a part of the skeleton that should be removed? maybe we have a skeleton model that has no obvious part needs to be removed.

In the skeleton approach, there is nothing explicit wrong, it is only my viewpoint is wrong, which is internal to me.

However, on the A-side, when we make the size of the coefficient really big or small, then take the fiber, it indeed have the correct tropical image. The tropical limit is what I want.

Consider the example of $\C^3 / \C^*$ by weight (1,1,-1). Then, $t = xy/z$, and $W = x + y + xy/t$. The fiberwise $W$ always are well-defined, for $t \neq 0$. There is no special moment in $t$ (weird!) OK, maybe when $t = 0$, then $W$ become $x + y + xy/t$, hence mirror is $Tot(O(-1))$. And for $t=\infty$, $W =x + y$.

The A-side fiber always matches the 'most positive' GIT chamber, in the above it is $Tot(O(-1))$.