Let $G$ be a split, almost-simple connected reductive group over a field $F$ with split maximal torus $T$. I am trying to understand precisely the groups $[G\_{\alpha}, G\_{\alpha}]$, where $\alpha$ is a root, $G\_{\alpha} = \mathcal{Z}\_G(T\_{\alpha})$, and $T\_{\alpha} = (\mathrm{ker} \ \alpha)^{\circ}$. Sometimes this group is isomorphic to $SL\_2$, and sometimes to $PGL\_2$. 


Given a root $\alpha$, let's say ``in coordinates'' (for example, $G = PSO(2n)$ and $\alpha = e\_{n-1} + e\_n$, or $G = SL(10) / \mu\_2$, and $\alpha = e\_2 - e\_3$), how can I tell which group $[G\_{\alpha}, G\_{\alpha}]$ is? 


I am especially interested in any isogeny of type $A\_n$, although really I will be interested in all types $A$ through $G$, and all isogenies. Perhaps you can give me a reference for how to do this computation in general, or an explanation for how to do this?



Centralizers of subtori in reductive groups, derived subgroups
We can ask this question for an arbitrary nilpotent element of the Lie algebra ${\mathfrak g}$ of $G$ (not just root elements as in the question). I'm sure this must already be written down somewhere, but it's also (mostly) quite easy to work out.


In characteristic zero or large enough, any nilpotent element $e$ of ${\mathfrak g}$ lies in an $\mathfrak{sl}\_2$-subalgebra which can be described as ${\rm Lie}(H)$ where $H\leq G$ is either ${\rm SL}\_2$ or ${\rm PGL}\_2$. If $G$ is of adjoint type then it is easy to say when $H\cong{\rm PGL}\_2$: when $e$ is *even*, i.e. the weights in the weighted Dynkin diagram are all $0$ or $2$. For classical types this holds iff the orders of all Jordan blocks have the same parity i.e. are all even or all odd. (Here I am thinking of ${\mathfrak g}$ as a classical Lie algebra even though I am only assuming that $G$ is isogenous to a classical group.)


Non-adjoint type cases can then be dealt with as follows. Clearly, if $H={\rm PGL}\_2$ then $e$ has to be even (by projecting to the adjoint quotient of $G$). For $G={\rm SL}\_n/\mu\_m$ where $m$ is odd, we can see directly that $H={\rm PGL}\_2$ if all Jordan blocks of $e$ are odd and $H={\rm SL}\_2$ otherwise; if $G={\rm SL}\_n/\mu\_m$ where $m$ is even then $H={\rm PGL}\_2$ for any even $e$. If $G={\rm Sp}\_{2n}$ then $H={\rm PGL}\_2$ if and only if all Jordan blocks of $e$ are odd. For $G={\rm SO}\_{2n}$ we have $H={\rm PGL}\_2$ iff all Jordan blocks are of odd order. For $G={\rm Spin}\_n$ or either of the remaining intermediate types $D\_{2n}$ there is an awkward but probably feasible calculation involving the kernel of the adjoint quotient. For example, I think it is true that when $G={\rm Spin}\_{2n+1}$ we have $H={\rm PGL}\_2$ iff all Jordan blocks are of odd order and there is an even number of Jordan blocks of order $8n\pm 3$.


For the exceptional types, there are (modulo possible miscounting) 2 (non-zero) even orbits in type $G\_2$; 7 in type $F\_4$; 9 in type $E\_6$; 23 in type $E\_7$; 26 in type $E\_8$. The adjoint case already completes the picture except for the simply-connected group in type $E\_7$ - I'll leave that as an exercise!


So we can answer your question in this more general setting: if $G={\rm SO}\_{2n+1}$ then short root elements are even and so the group you describe will be ${\rm PGL}\_2$; for all other $G$, or for long root elements, the group you describe is ${\rm SL}\_2$.


