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--- blog:2023-02-24 [2023/02/25 07:18] – pzhou
+++ blog:2023-02-24 [2023/06/25 15:53] (current) – external edit 127.0.0.1
@@ Line 32: / Line 32: @@
 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 $W = x+y+z = t z^2 / y + y + z$.
+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?
+===== Forget the window for now =====
+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.
+===== But why the wrong answer works?  =====
+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 $Tot(O(-1))$.