• problem with torsion in $\pi_1(G)$.

So, what's the difference between $GL_2, PGL_2, SL_2$? $GL_2$ is the father of all, $PGL_2$ and $SL_2$ each take out some abelian part out of it. $PGL_2$ totally killed the central subgroup, by quotienting out it, whereas $SL_2$ tries to take a slice, that intersects minimally with it.

Then, $T_{SL} \to T_{GL} \to T_{PGL}$ the composition is a finite cover. $$ \Gamma \to T_{SL} \to T_{PGL} $$ This has many immediate consequences

• The cocharacter lattice $Hom(\C^*,T_{SL}) \to Hom(\C^*, T_{PGL})$ is injective. This is because $Hom(S^1,-)$ is only left exact, not right exact, hence loses surjectivity, we are led to $Hom^1(S^1, - )$. If the target is a discrete group $\Gamma$, then $Hom^1(S^1, \Gamma) = \Gamma$.

$$ X_*(T_{SL}) \to X_*(T_{PGL}) \to \Gamma $$

• The character lattice, $Hom(T_{PGL}, \C^*) \to \Hom(T_{SL}, \C^*)$, seems to be injective. Is it because $Hom(-, \C^*)$ is exact? Here, if $\Gamma$ is a finite abelian group, then $Hom(\Gamma, \C^*)$ is just the 'Pontryagin dual'.

$$ X^*(T_{PGL}) \to X^*(T_{SL}) \to \Gamma^\vee $$

So far so good. Now because of $SL_n \to PGL_n$ is surjective, hence, we have any representation of $PGL_n$ goes to rep of $SL_n$, no loss of information.

How about the functions on $T_{PGL}$ and $T_{SL}$? If we ignore $W$-invariance, we have more functions on $T_{SL}$. Basically, the deck transformation is transparent to $W$ action. This also matches with $X^*(T_{PGL}) \to X^*(T_{SL})$. I would also say, $\Gamma = \pi_1(PGL_n)$, which is also the center of $SL_n$. They are canonically the same. (no funny dual involved). Indeed, this $\Gamma$ is also the deck transformation of $SL_n \to PGL_n$. the thing you quotient by become the fundamental group of the base.

So, we should have $K^{PGL_n}(pt) \into K^{SL_n}(pt)$, just because $SL_n$ has more representations. How does the folding affect the quotient? Well, say for $T_{SL_2}$ coordinate $z$ and $T_{PGL_2}$ coordinate $w=z^2$, $W$ action changes $z$ to $1/z$, hence $w$ to $1/w$. The thing that I don't quite understand is: when we fold $z$ to $1/z$, there are two fixed points $1, -1$, but these should get quotient out by $\Z_2$. But then, new fixed points pop up.

What's the class function ring for $G=PGL_2$? They are function on $GL_2$, that is invariant under rescaling and conjugation. So, we need to divide out by determinant, which is a homogenous degree $2$ guy. We basically only know about eigenvalues, say $a = (x_1^2+x_2^2)/(x_1x_2)$.

OK, fine. Now, how about $T_{PGL_2}/W$? What does $T$ know? T knows $x_1/x_2$ and $x_2/x_1$. So you imagine, we should get $x_1/x_2 + x_2/x_1$, but indeed we get that. So, what is the problem?

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.

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.