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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?

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    $\begingroup$ One loses nothing by assuming $G$ is semisimple. Then that commutator subgroup is always ${\rm{SL}}_2$ except for precisely when $G = {\rm{PGL}}_2$. Indeed, first one checks this when $G$ is simply connected using that the simple positive coroots relative to a basis of the root system are a basis of the cocharacter lattice precisely in the simply connected case (and that any root is a simple positive root relative to some basis of the root system). In general examine the kernel of the simply connected central cover $\widetilde{G}\rightarrow G$ (which identifies the root systems). $\endgroup$
    – nfdc23
    May 19, 2017 at 14:17
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    $\begingroup$ nfdc23 - Thank your comment, which I need to think about more. I have to say, I am surprised by your response. Based on the amount of time spent on reductive groups of semisimple rank one in the standard texts, I would have assume that $PGL_2$ would play more of a role in the theory. Perhaps I didn't ask the correct question, though. Based on your response, maybe my question should change to: as $[G_{\alpha}, G_{\alpha}] \cong SL_2$ (when $G \neq PGL_2$), when does the corresponding map $SL_2 \rightarrow G$ factor through $PGL_2$? Some condition on $\alpha$? $\endgroup$ May 19, 2017 at 17:33
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    $\begingroup$ I can't agree with nfdc23 - for example, for any short root in the root system of type $B_n$, the group we obtain this way is ${\rm SO}_3={\rm PGL}_2$. I think your reasoning is fine for simply-connected groups, but $\alpha^\vee(-1)$ can be contained in the kernel of the universal cover - only in type $B$, though, I think. $\endgroup$
    – Paul Levy
    May 19, 2017 at 19:12
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    $\begingroup$ @nfdc23. Will you give us the cheat sheet? $\endgroup$ May 20, 2017 at 12:08
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    $\begingroup$ Dear all, nfdc23 shared with me the cheat sheet. Here is a LaTeX version: math.stonybrook.edu/~jstarr/papers/Centers.pdf $\endgroup$ May 21, 2017 at 10:34

3 Answers 3

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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 ${\rm SL}_2$.

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    $\begingroup$ Is this method based on Lie algebras applicable in all positive characteristics? $\endgroup$
    – nfdc23
    May 20, 2017 at 13:12
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    $\begingroup$ The existence of the ${\rm SL}_2$-subgroup fails in small characteristics, even in type $A$. But if the characteristic is good then we still get a cocharacter $\lambda:k^\times\rightarrow G$ (unique up to $G^e$-conjugacy) which is directly analogous to the maximal torus ${\rm diag}(t,t^{-1})$ in $H$. This cocharacter is called an associated cocharacter for $e$, and the kernel is either trivial or $\mu_2$. Then all of the remarks above should hold (in good characteristic) on systematically replacing "$H={\rm PGL}_2$'' by "${\rm ker}\, \lambda=\mu_2$". $\endgroup$
    – Paul Levy
    May 20, 2017 at 13:32
  • $\begingroup$ OK, thanks for clarifying. I have now given an answer that isn't as comprehensive in terms of general nilpotents in the Lie algebra but focuses on the more limited framework of the question posed to allow all characteristics on equal footing. $\endgroup$
    – nfdc23
    May 20, 2017 at 14:36
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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.

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As Paul Levy's answer suggests, your question probably needs some case-by-case work to be answered completely. The most general perspective may come from older work of Borel-Tits in their 1972 IHES paper here, where they develop a uniform criterion for simple connectedness of various subgroups. Here it's easiest to start with a semisimple group having the split properties you indicate.

Now you have for each root $\alpha$ a minimal parabolic subgroup $P = LV$, where $L$ is connected reductive and $V$ is unipotent. The assumption is that $L$ has semisimple rank 1, permitting either of the two rank 1 groups to occur as an almost-direct factor along with a central torus. Assuming $G$ is simply connected, the criterion of Borel-Tits in $\S4$ shows that the derived group of $L$ is also simply connected, thus in this situation of type SL$_2$.

But for example when $G$ is assumed to be of adjoint (or intermediate) type, it's trickier to predict which rank 1 group you get. As far as I can tell, the Borel-Tits refinements for the action on $V$ here only tell you that the center of $L$ must be connected when $G$ is adjoint. This still leaves case-by-case study to determine how the almost-product of a rank 1 simple group and a torus actually looks. I think Paul has covered this aspect well enough to sort out the cases.

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    $\begingroup$ One can arrange the case-work to require closer inspection only for type ${\rm{B}}_2={\rm{C}}_2$; please see the answer I have posted. $\endgroup$
    – nfdc23
    May 20, 2017 at 14:39
  • $\begingroup$ @PaulLevy's answer referenced in the post. $\endgroup$
    – LSpice
    May 1, 2021 at 16:10
  • $\begingroup$ @nfdc23's answer referenced in the comments. $\endgroup$
    – LSpice
    May 1, 2021 at 16:10

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