Differences

This shows you the differences between two versions of the page.

--- blog:2023-02-24 [2023/02/25 09:01] – pzhou
+++ blog:2023-02-24 [2023/06/25 15:53] (current) – external edit 127.0.0.1
@@ Line 58: / Line 58: @@
 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 $Tot(O(-1))$.