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 $A - G$. 



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> 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?)
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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](http://en.wikipedia.org/wiki/File:Root_system_G2.svg)). 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.. 



Simply connected algebraic groups and reductive subgroups of maximal rank
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 $F\_4$.


