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.