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 ${\rm SL}_2$.

Since I erred in my initial comment, to compensate here is a characteristic-free argument over general fields $k$ (that also adapts to work over general commutative rings). As in Paul Levy's answer, we will show that beyond the trivial rank-1 case, PGL$_2$ is obtained precisely for the short root of adjoint type B$_n$ with $n \ge 2$ (and one could make the statement more uniform including rank 1 via type B$_n$ for $n \ge 1$ if we declare that the roots for type ${\rm{B}}_1 = {\rm{A}}_1$ are short, or perhaps both long and short).

The simply connected case always gives ${\rm{SL}}_2$ for the general reason that the derived group of any torus centralizer in a simply connected semisimple group is always simply connected. A proof of this general fact, based on the characterization of "simply connected" in terms of simple positive coroots being a basis of the cocharacter lattice, is sketched in Corollary 9.5.11 of https://www.ams.org/open-math-notes/omn-view-listing?listingId=110663 (where the argument is given over fields, and a reference is provided to make the argument work over rings).

Now consider a general split connected semisimple $k$-group $G$ that is (absolutely) simple with split maximal $k$-torus $S$, and let $f:\widetilde{G} \rightarrow G$ be the split simply connected central cover. The preimage $\widetilde{S} := f^{-1}(S)$ is a split maximal $k$-torus of $\widetilde{G}$, and $\Phi(G,S)= \Phi(\widetilde{G},\widetilde{S})$ via the finite-index inclusion ${\rm{X}}(S) \subset {\rm{X}}(\widetilde{S})$ due to the centrality of the subgroup scheme $\ker f$ (even though ${\rm{Lie}}(f)$ may not be surjective).

Here is the main trick to avoid too much case-work: by the settled simply connected case we know $[\widetilde{G}_a, \widetilde{G}_a] = {\rm{SL}}_2$ for all $a \in \Phi$, and this maps onto $[G_a,G_a]$ via a central isogeny. In particular, if the group scheme $\ker f$ has odd degree then we're done (as SL$_2$ has scheme-theoretic center $\mu_2$ of order 2). Thus, the case of type A$_2$ (i.e., PGL$_3$ and SL$_3$) always gives SL$_2$ since 3 is odd. This will dispose of nearly all remaining cases below because most vertices in Dynkin diagrams are adjacent to one with the same length and such a pair of vertices along with the edge joining them is the A$_2$ diagram.

Let $n>1$ be the rank of the reduced and irreducible root system $\Phi$. We may assume that our root $a$ belongs to a chosen basis $\Delta$ of $\Phi$. We may also assume $G$ is not simply connected or else there's nothing to do, so we're not in types G$_2$ and F$_4$. We shall now show that we get SL$_2$ away from the short roots of B$_n$ and the long roots of type C$_n$ (and then we will analyze these remaining cases). Being away from those two classes of cases, note that we're not in type ${\rm{B}}_2 = {\rm{C}}_2$. Thus, $n \ge 3$ and in the Dynkin diagram the root $a \in \Delta$ is adjacent to another root $b \in \Delta$ with the same length. Thus, the codimension-2 subtorus $S_{a,b} = (\ker a \cap \ker b)^0_{\rm{red}} \subset S$ killed by $a$ and $b$ has centralizer whose derived group $G_{a,b}$ is of type A$_2$ (with split maximal $k$-torus $S' = S \cap G_{a,b} = a^{\vee}(\mathbf{G}_m)b^{\vee}(\mathbf{G}_m)$ and Dynkin diagram having nodes $a|_{S'}, b|_{S'}$). The original commutator subgroup of interest for $(G, S, a)$ agrees with the one for $(G_{a,b}, S', a|_{S'})$. This reduces our task to the settled case of type A$_2$!

It remains to consider long roots of type C$_n$ with $n \ge 2$ and short roots of type B$_n$ with $n \ge 2$. We will reduce these to consideration of adjoint type ${\rm{B}}_2 = {\rm{C}}_2$ (concretely, SO$_5$). Using the action of the Weyl group, we may arrange that our root $a$ corresponds to the unique node that is long for type C and short for type B. Let $b \in \Delta$ be the unique node adjacent to $a$ in the diagram. The center of simply connected type B$_n$ is $a^{\vee}(\mu_2)$ (this is the key fact, as also noted in Paul Levy's comment and answer), so by open cell considerations we see that the group $G_{a,b}$ made as above but now for the nodes $a$ and $b$ as just defined is also of adjoint type when we are in type B.

Keeping in mind what we are aiming to show (that we get SL$_2$ for the long roots of adjoint type C$_n$ with $n \ge 2$ and PGL$_2$ for the short roots of adjoint type B$_n$ with $n \ge 2$), since the simply connected case gives only SL$_2$ in general and in types B and C the only options are simply connected and adjoint type (as the fundamental group of the root system has order 2) we can pass to $G_{a,b}$ to reduce to the case of adjoint type ${\rm{B}}_2={\rm{C}}_2$. (It doesn't matter that for type C this passage to rank 2 might leak back to the simply connected case, since that case always gives SL$_2$ anyway.)

Letting $\{a, b\}$ be a basis for type ${\rm{B}}_2 = {\rm{C}}_2$ with $a$ short and $b$ long, we want to show that $[G_a, G_a] = {\rm{PGL}}_2$ and $[G_b, G_b] = {\rm{SL}}_2$. The center of $\widetilde{G}$ is $a^{\vee}(\mu_2)$ as noted already (contained in the split maximal torus $a^{\vee}(\mathbf{G}_m)$ of $[\widetilde{G}_a, \widetilde{G}_a] = {\rm{SL}}_2$), so for $G = \widetilde{G}/Z_{\widetilde{G}} = \widetilde{G}/a^{\vee}(\mu_2)$ we have $[G_a,G_a] = [\widetilde{G}_a,\widetilde{G}_a]/a^{\vee}(\mu_2) = {\rm{SL}}_2/\mu_2 = {\rm{PGL}}_2$. Likewise, $[\widetilde{G}_b,\widetilde{G}_b]$ meets the rank-2 split maximal torus $\widetilde{S} = \mathbf{G}_m^{\Delta^{\vee}}$ in precisely its own split maximal torus $b^{\vee}(\mathbf{G}_m)$, so its intersection with $a^{\vee}(\mathbf{G}_m)$ is trivial. In particular, its intersection with $a^{\vee}(\mu_2)$ is trivial, so the central isogeny ${\rm{SL}}_2 = [\widetilde{G}_b, \widetilde{G}_b] \twoheadrightarrow [G_b,G_b]$ has trivial (scheme-theoretic) kernel and thus is an isomorphism.