• 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. 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?