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Recall that a connected semisimple algebraic group $G$ over an algebraically closed field $K$ of arbitrary characteristic was defined by Chevalley to be simply connected if the character group $X(T)$ of a maximal torus $T$ in $G$ is "as large as possible": equal to the abstract weight lattice associated to the root lattice $\mathbb{Z} \Phi \subset X(T)$ (here $\Phi$ denotes the root system of $G$ relative to $T$). Since all maximal tori are conjugate, this is independent of the choice of $T$. Fortunately this notion of "simply connected" agrees with the usual topological notion when $K = \mathbb{C}$. A typical example is $\mathrm{SL}_n$.

The closed reductive subgroups $H$ of $G$ containing $T$ have been well studied, but one loose end still bothers me. The connected component of the identity $H^\circ$ is generated by $T$ and certain pairs of opposite root groups for roots forming a subsystem $\Psi$ of $\Phi$ (Borel-Tits). Typical examples are the derived groups of Levi subgroups in parabolic subgroups of $G$ (complementary there to the unipoent radical), which are always connected. Imitating work of Borel and de Siebenthal (1949) on compact Lie groups, one gets these and sometimes more examples by taking root subsystems corresponding to proper subsets of vertices in the extended Dynkin diagram.

Equally natural are the groups $H = C_G(s)$, centralizers of semisimple elements (not necessarily connected, unless $G$ is simply connected). These have especially been studied in prime characteristic and have identity components which turn out to be of the type described above: work of Springer and Steinberg, Carter and Deriziotis, McNinch and Sommers.

Assume for convenience that $G$ is both simply connected and almost-simple, thus of Lie type $A - G$.

Is the semisimple derived group $H'$ of a connected reductive subgroup of $G$ including $T$ always simply connected? (If so, is there a uniform proof?)

When $H$ is a Levi subgroup of some parabolic subgroup of $G$, this was proved by Borel-Tits (in the more complicated context of a field of definition). Some experimental work with subgroups of Borel and deSiebenthal type suggests that the derived groups are also simply connected, but if so the reason is obscure. Maybe the topological case in characteristic 0 is suggestive?

Example: Take $G$ to be of type $G_2$ (so it is both simply connected and adjoint), with short simple root $\alpha$ and long simple root $\beta$ (picture here). The six long roots form a subsystem of type $A_2$. Since the original highest root is $3\alpha +2\beta$, its negative (call it $\gamma$) along with $\delta := \beta$ correspond to two vertices of the extended Dynkin diagram and span a subsystem $\Psi$ of type $A_2$ belonging to a subgroup $H$. Here $H$ is already connected and almost-simple. In fact it is simply connected: direct calculation shows that the two fundamental weights for $H$ are $-2\alpha -\beta$ and $-\alpha$, thus lie in the weight lattice (= root lattice) of $G$.

ADDED: My formulation (and example) probably oversimplify the computations involved in most cases. When $H'$, with a maximal torus $S := T \cap H'$, has lower rank than $G$ or has a root system with multiple irreducible components, the study of $X(S)$ tends to be more complicated. This already shows up in the usual Levi subgroups. For instance, the group $G_2$ has a standard Levi subgroup with derived group of type $\mathrm{SL}_2$ corresponding to either $\alpha$ or $\beta$, but the fundamental weight in $X(S)$ won't lie in $X(T)$. I hoped one might see whether or not a uniform pattern exists without too much case-by-case work, but that may be unrealistic.

UPDATE: The comments (especially by Paul) help to clarify what is going on, though I'm left with the partial question: Is there any predictable pattern to the occurrence or non-occurrence of simply connected groups $H'$? I was motivated to look more closely at an extreme case involving the simple (both adjoint and simply connected) group $G$ of type $E_8$. By removing the vertex of the extended Dynkin diagram corresponding to $\alpha_2$ (Bourbaki numbering), one gets a subgroup $H= H'$ of type $A_8$ with the same maximal torus $T$ and character group $X(T)$. Using Chevalley's classification, $H$ will be one of three nontrivial quotients of $\mathrm{SL}_9$ (as a group scheme). Some tedious bookkeeping with roots and weights shows that $H$ is the "intermediate" group, whose group of rational points has a center of order 3 (when the characteristic is not 3). Thus $H$ fails to be simply connected or adjoint, the former not obvious a priori but certainly consistent with results of McNinch-Sommers..

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Jim: I do not know an answer but one approach would be to look at Dynkin et al classification of maximal subgroups of G. Either you would find a counter example or you can induct. This is a bit unsatisfactory since it requires case-by-case analysis especially in the exceptional cases. –  Misha Apr 4 '13 at 16:15
    
P.S. I finally got around to looking again at the 1949 paper by Borel and de Siebenthal on compact Lie groups. Though I remembered their algorithm and table, which carry over to root systems in algebraic settings, I had completely forgotten their useful Remarque II. I'm not quite sure whether it applies readily to algebraic groups, especially in prime characteristic, but it shortcuts some fundamental group questions (still leaving perhaps a case-free question). –  Jim Humphreys Apr 13 '13 at 23:03

4 Answers 4

up vote 3 down vote accepted

Surely it's easier to check whether $H'$ is simply-connected by inspecting the co-root lattice...? For the example of $G_2$ containing $SO(4)$ that Allen mentions, we have a pseudo-Levi subalgebra of type $A_1\times \tilde{A}_1$ where $\{(3\alpha+2\beta),\alpha\}$ is a basis of simple roots. Now the cocharacter $(3\alpha+2\beta)^\vee=\alpha^\vee+2\beta^\vee$, so the lattice:

${\mathbb Z}(3\alpha+2\beta)^\vee+{\mathbb Z}\alpha^\vee = {\mathbb Z}\alpha^\vee+{\mathbb Z}(2\beta^\vee)$

is of index two in the cocharacter lattice ${\mathbb Z}\alpha^\vee+{\mathbb Z}\beta^\vee$ for $T$.

In fact this allows you to determine exactly what the pseudo-Levi subgroup is in each case.

For the maximal pseudo-Levis there's an easier trick to find non-simply-connected ones: if $s\in T$ and $L=Z_G(s)$ then $Z(L)/Z(L)^\circ$ is generated by $s$, by a result of Eric Sommers. So we can see almost immediately that hardly any maximal pseudo-Levi subgroups are simply-connected. For example, the pseudo-Levi of $F_4$ which is of type $C_3\times A_1$ has a cyclic centre, so it can't be isomorphic to $Sp_6\times SL_2$. Specifically, it is isomorphic to $(Sp_6\times SL_2)/\{ \pm (I,I)\}$.

EDIT: A mistake with this is that Sommers' result only holds for adjoint type groups. More generally we have $Z(L)/(Z(L)^\circ Z(G))$ is generated by $s$. Of course this makes no difference for type $F_4$.

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@Paul: This analysis looks reasonable. I was probably focused too much on cases where the root system of $H'$ is irreducible. I wonder if it's true in general to say that the derived group of $H$ is an almost-direct product of simply connected groups? I was looking for a good analogy of the Borel-Tits result on Levi subgroups, but that may be elusive. –  Jim Humphreys Apr 5 '13 at 21:12
    
The $A_7$ (standard) pseudo-Levi subgroup of $E_7$ is not simply-connected. To see this, note that it has basis of simple roots $-\hat\alpha$ and $\alpha_i$ with $i\neq 2$ and the corresponding cocharacters do not generate all of $Y(T)$. In particular we have $-\hat\alpha^\vee(-1)\alpha_3^\vee(-1)\alpha_5^\vee(-1)\alpha_7^\vee(-1)=1$ so that in fact this subgroup is $\SL_8/\{ \pm I\}$. Similarly, the subgroups of $E_8$ of type $A_8$ and $D_8$ are not simply-connected. This is related to the fact that the coefficient in the highest root of the "missing" simple root is $>1$. –  Paul Levy Apr 6 '13 at 10:40
    
For another example, the pseudo-Levi subgroup of $E_8$ of type $A_7\times A_1$ is not an almost direct product of $SL_2$ and $SL_8$ - the $A_7$ subgroup is actually $SL_8/\{\pm I\}$, as can be seen from considering the $\hat\alpha^\vee(-1)$. –  Paul Levy Apr 6 '13 at 10:47
    
@Paul: I guess you refer to the 2003 J. Algebra paper by George McNinch and Eric Sommers: front.math.ucdavis.edu/0204.5275 Note that they redefine "pseudo-Levi subgroup", differing a little from Eric's original definition. Anyway it's hard to decide on terminology here. Eric and George apply the "pseudo" label to actual Levi subgroups of parabolic subgroups, while I'd prefer to reserve the term for other reductive subgroups arising from the algorithm of Borel and de Siebenthal. . –  Jim Humphreys Apr 7 '13 at 13:39

$G_2 \supset SO(4)$

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If $H$ is the fixed points of an involution of $G$ the answer is known completely, and $H$ is "almost always" not simply connected. This includes many subgroups of maximal rank (whenever the involution is inner).

Suppose $G$ (complex) is simple and simply connected, $H$ is the fixed points of an involution of $G$, and $H$ is semisimple. Then the fundamental group of $H$ is $\mathbb Z/2\mathbb Z$ except in the following cases: $Sp(2n)\subset SL(2n)$, $Spin(n)\subset Spin(n+1)$ ($n\ge 3$), $Sp(2p)\times Sp(2q)\subset Sp(2(p+q))$, $F_4\subset E_6$, and $Spin(9)\subset F_4$. The proof is a fairly simple, case-free, root/weight lattice argument, along the lines of some of the preceding discussion. See http://www.ams.org/mathscinet-getitem?mr=2112326.

A closely related question is When is a finite dimensional real or complex Lie Group not a matrix group

It would be interesting, and probably not too difficult, to extend this to the case when $H$ is the fixed points of an automorphism of finite order, and is maximal.

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I don't think this is true. In type $D_4$ you have a closed root subsystem of type $A_1^4$ (the simple roots for the leaves of the diagram, the longest root and their opposites), for which the corresponding coroots generate a strict sublattice of the lattice generated by the simple coroots; in fact the index is $2$. Then their fundamental weights cannot lie in the the character lattice for $G$.

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@Marc: Thanks for pointing out this example, which I had skipped over too quickly. I'm not sure whether any predictable pattern exists, especially for these non-irreducible subsystems. In a way that's simpler, since there seems to be no conceptual reason for uniformity in thuis indirect method of constructing reductive subgroups of maximal rank. –  Jim Humphreys Apr 4 '13 at 20:28
    
P.S. Looking further at your example, I'm probably more confused than before. The root lattices for $G$ and for $H=H'$ are not the same even though the maximal torus is the same for both. What is the index of each root lattice in the full weight lattice for $G$, assumed to be $X(T)$? For a simply connected $D_4$ it's certainly 4. I have to check Bourbaki's table again, but my computation doesn't seem to agree with yours. It's confusing to me because of all the factors 1/2 that occur. –  Jim Humphreys Apr 4 '13 at 22:48

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