• 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.

Is it just about torsion? Let me run through the thing replacing $SL_n$ by $GL_n$.

$$ \C^* \to GL_n \to PGL_n $$ then, we get $\C^* \to (\C^*)^n \to (\C^*)^{n-1}$, then we get $Hom(\C^*, -)$, to get $$ \Z \to \Z^n \to \Z^{n-1} $$