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--- blog:2023-07-25 [2023/07/25 20:30] – pzhou
+++ blog:2023-07-25 [2023/07/25 23:48] (current) – pzhou
@@ Line 27: / Line 27: @@
 http://math.stanford.edu/~conrad/210CPage/handouts/repring.pdf
-What are you saying? Here,  $G$  is a connected compact Lie group,  $T$  maximal torus. We get a 'root system'  $\Phi(G,T)$  in  $X^*(T)$ . There is a co-root system,  $\Phi^\vee$ , which pairs with  $\Phi$  integrally. OK, we have root lattice, character lattice, coroot lattice, cocharacter lattice. But then, we also have the weight lattice and co-weight lattice. What's going on here?
+What are you saying? Here,  $G$  is a connected compact Lie group,  $T$  maximal torus.
+Well, from the notion of a Lie algebra  $\frak g$  alone, we should get the notion of root  $\Phi$ , coroot  $\Phi^\vee$ , then the weight lattice  $P$  (integral dual to the  $\Z \Phi^\vee$ ) and  $Q$  coweight lattice, integral dual to  $\Z\Phi$ . All that come for free, if you give me a Lie algebra. Now, if you give me a group form, then I have a torus, then I can do character  $X^*(T)$  and cocharacters. We should have  $\Z \Phi \In X^*(T) \In P$ , and  $\Z \Phi^\vee \In X_*(T) \In P^\vee$ .
+Now, if  $G$  is simply connected, then  $T$  is big (others are a quotient of it), so  $X^*(T) = P$  (yes, lots of representations), and  $X_*(T)$  (is small) is  $\Z \Phi^\vee$ . If  $G$  has trivial center, then it is the other way around,  $X^*(T) = \Z\Phi$ , the adjoint type.
+{{:blog:pasted:20230725-135403.png}}
+First thing I learned, the derived subgroup (generated by commutator), is not necessary of adjoint-type. Why we want it? To kill as much torus as possible? Does this guy have finite  $\pi_1$ ? Should be. Is it semi-simple? OK, there are some torus part, and there are some semi-simple part,then we marry them together.
+Ex 1:  $PGL_n$ , it is $(pt \times SL_n)/\mu_n$
+Ex 2:  $GL_n$ , it is $ (\C^* \times SL_n)/\mu_n$
+Well, why do we need to take the universal cover? Why cannot we just take the derived subgroup, then times the torus, and quotient by something? Well,  $Z$  can only be a torus? That makes it even more suspicious!
+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.