Differences

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

--- blog:2023-02-10 [2023/02/11 08:10] – pzhou
+++ blog:2023-02-10 [2023/06/25 15:53] (current) – external edit 127.0.0.1
@@ Line 8: / Line 8: @@
 Same thing, if we ask for the multiplicative case. The fiberwise skeleton of a smooth fiber is $(S^1)^2$, then become subcritical for the special fiber.
+===== Fiber of W =====
+Consider the simple case. $Hilb^2_{hor}(C^* x C)$, where $W = y_1^2 + y_2^2 + (x_1- x_2)/(y_1-y_2)$. We reduce it. Let $A = x_1 + x_2$, $B = (y_1-y_2)^2$, $u = (x_1- x_2)/(y_1-y_2)$, and $u^2 B - A^2 \neq 0$. Superpotential is $W = B + u$.
+Or $Y = {x^2 - y^2 z \neq 0}$ and $W = y + z$.
+We need to consider what is the divisor $x^2 - y^2 z = 0 $.
+The fiber over $w$ is $C^2 \ \{x^2 - y^2 (w-y) = 0 \}$
+Is that deleted guy smooth? Namely
+$$ xdx - (2w y - 3y^2) dy = 0 $$
+means $x=0, y=0$ or $x=0, y= 2w/3$. So, what happens here? when $w=0$, the two critical points in the deleted divisor merge. What do you want to delete to make the complement the vanishing cycle the same?
+Fiber better be R^2