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--- blog:2023-03-29 [2023/03/29 21:39] – pzhou
+++ blog:2023-03-29 [2023/06/25 15:53] (current) – external edit 127.0.0.1
@@ Line 27: / Line 27: @@
 In the weight $(1,2)$ case, we need to quotient out $(T^*\C^2)^o$, with variables $(x_1, y_1, x_2, y_2)$. We have complex moment map condition being
 $$ (x_1 y_1 + 1) (x_2 y_2 + 1)^2 = \beta $$
-for some generic $\beta \in \C^*$. And then, we just take the usual GIT quotient, quotient by $\C^*$.
+for some generic $\beta \in \C^*$. And then, we just take the usual GIT quotient, quotient by $\C^*$. Now, this subspace is smooth, and I believe that the action is also fine. For example, we can use coordinate,
+$$ z = x_2 y_2 + 1, \quad u = x_1^2 y_2, \quad v = y_1^2 x_2 $$
+subject to $z \in \C^*$
+$$ u v = (z-1) (\beta / z^2 - 1). $$
+so we have three special fibers, $z = \pm \sqrt{\beta}$ and $z=1$, and the fiber become singular.
+This one dimensional example is OK.
+Try another one. Say the weight is $(1,1,2)$, we can complete that to a basis by $(0,1,0), (0,0,1)$, then take the dual basis, that defines a map
+$$ \Z \xto{(1,1,2)} \Z^3 \xto{(1,-1,0), (2,0,-1)} \Z^2 $$
+OK, so what is the space? We should have
+$$ (x_1 y_1 + 1) (x_2 y_2 + 1) (x_3 y_3+1)^2 = \beta $$
+then quotient by $\C^*$. We have base coordinates like
+$$ z_3 = x_3 y_3+1, \quad z_2 = x_2 y_2 + 1. $$
+We have singularity over $z_2 z_3^2 = \beta$. So, we know the singularities downstairs. It shouldn't matter what is the value of $\beta$ as long as it is generic.
+Now the question is: which circle get contracted?
+Of course, there are many invariant functions. You can do the affine quotient, then use the complex moment map to cut-out a smooth piece, no problem. (interesting question how does the skeleton change).
+Let's see, what is the generic fiber? Well, we are just quotienting $(\C^*)^3_{x}$ by $\C^*$. So far, this is very algebraic. And, as you move to a bad divisors, labelled by $1,2,3$, one of the factors $\C^*$ become $xy=0$. So, what happens here? In terms of the quotient.