Peng Zhou

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blog:2023-07-25 [2023/07/25 20:17] pzhoublog:2023-07-25 [2023/07/25 23:48] (current) pzhou
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 +===== More on this =====
 +http://math.stanford.edu/~conrad/210CPage/handouts/repring.pdf
  
 +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. 
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 +{{:blog:pasted:20230725-135403.png}}
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 +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. 
  
  
  
blog/2023-07-25.1690316275.txt.gz · Last modified: 2023/07/25 20:17 by pzhou