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--- blog:2023-07-25 [2023/07/25 22:35] – [More on this] pzhou
+++ blog:2023-07-25 [2023/07/25 23:48] (current) – pzhou
@@ Line 47: / Line 47: @@
 OK, so what? You have some finite group acting on the coordinate ring of the torus. You take invariant.
+So, what is this? Well, I think the best way is not to quotient, but remember the action.
+So, what is the answer, for $G = PGL_n$? The answer for K-Coulomb? What is the naive answer? The base should be still $K^{G}(pt)$. Well, we have $T_{sc} / W$. That is $\prod_{k=1}^{n-1} \C_k$, with $\mu_n$ acting with different powers (indicated in the lower indices).
+BFM says it is $ T_{sc} \times T_{ad} / W$, fibered to the base $T_{ad}/W$.
+What is $B_G^{G^\vee}$. So, if $G=PGL_2$ and $G^\vee = SL_2$, you have the quotient torus $T_{ad}$ and subtorus $T_{sc}$. Then, the root is like $u=x_1/x_2$ on $T_{ad}$, and on $T_{sc}$, it is like $x_1/x_2=x_1^2 = z^2$. So, we have
+$$ \C[u^{\pm 1}, z^{\pm 1}, \frac{z^2-1}{u-1}]^{S_2} $$
+Look, what happens when $u=-1$, and $z=\pm 1$? These are also fixed points of $W$, but there is no blow-up to resolve it.
+but why is $u=-1$ also a fixed point? That is a fake one.