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--- blog:2023-03-29 [2023/03/29 22:19] – pzhou
+++ blog:2023-03-29 [2023/06/25 15:53] (current) – external edit 127.0.0.1
@@ Line 38: / Line 38: @@
 $$ \Z \xto{(1,1,2)} \Z^3 \xto{(1,-1,0), (2,0,-1)} \Z^2 $$
 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 $$
+$$ (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 $z_2 z_3^3 = \beta$. Say $\beta = e^{i \theta}$.
+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.