Suppose G is a connected semisimple Lie group with finite center, and A, B are one parameter subgroups of the same Cartan subgroup. If the connected components of the identity of the centralizers of A and B are equal does it follow that the centralizers of A and B are equal ?

$\begingroup$ I'm not sure about your terminology here, so mentioning a source would be useful. Can you give an example of such a one parameter subgroup whose centralizer fails to be connected? $\endgroup$– Jim HumphreysNov 8 '13 at 0:06

$\begingroup$ @Jim: the group $diag(e^t,e^t,e^{2t})$ in $SL_3$. The centralizer is $S(GL_2\times GL_1)$. In the reals the connected centralizer has index 2 (the determinant if each block has to be positive). $\endgroup$– YCorNov 8 '13 at 0:24

$\begingroup$ (given the context, the topology is clearly meant to be the transcendental one). Btw, the subgroup $\{diag(e^t,e^{t}),t\in\mathbf{R}\}$ in $SL_2(\mathbf{R})$ already has a nonconnected centralizer in $SL_2(\mathbf{R})$. $\endgroup$– YCorNov 8 '13 at 0:54

1$\begingroup$ @Yves: My questions were mostly rhetorical, since I felt a lack of background, examples, motivation in the question. I was hoping Daryl would write down a little more, rather than just stating something in exercise format. $\endgroup$– Jim HumphreysNov 8 '13 at 13:24

$\begingroup$ This question arises in the classification of limits of conjugates of certain subgroups of G. The subgroups are those fixed by an involution. This is part of a program to study deformations of homogeneous spaces from an algebraic point of view. It is related to the idea of Wigner Inönü contraction of a Lie algebra. $\endgroup$– Daryl CooperNov 8 '13 at 15:50
The answer is affirmative, and it is not necessary to assume that the center of $G$ is finite; discreteness (which is a consequence of semisimplicity of the Lie algebra) is sufficient. The crux of the matter is to carefully turn the analytic problem into an algebraic one (keeping track of connectedness issues on the analytic and algebraic sides of the story), even though the given Lie group and certainly the 1parameter subgroups need not have any algebraic structure at all. The merit of the algebraic setting is that one has very general connectedness results for torus centralizers in connected linear algebraic groups over any field (such as $\mathbf{R}$).
To explain this, we first review the close connection between linear algebraic groups and the image of the adjoint representation of connected semisimple Lie groups, as well as a related algebraicity feature for Cartan subgroups.
Let $\mathfrak{g} = {\rm{Lie}}(G)$. The natural map ${\rm{Ad}}_G:G \rightarrow {\rm{GL}}(\mathfrak{g})$ induces ${\rm{ad}}_{\mathfrak{g}}$ on Lie algebras. This makes $\mathfrak{g}$ a perfect Lie subalgebra of $\mathfrak{gl}(\mathfrak{g})$. Bringing in the theory of linear algebraic groups with the Zariski topology, there is a unique (smooth) connected closed $\mathbf{R}$subgroup $j:\mathbf{G} \hookrightarrow {\rm{GL}}(\mathfrak{g})$ such that ${\rm{Lie}}(\mathbf{G}) = \mathfrak{g}$ inside $\mathfrak{gl}(\mathfrak{g})$ (inclusion of $\mathfrak{g}$ via ${\rm{ad}}_{\mathfrak{g}}$); see Corollary 7.9 in Ch. II of Borel's textbook on algebraic groups. Since ${\rm{Lie}}(\mathbf{G})$ is semisimple, it follows by various means that $\mathbf{G}$ is semisimple in the sense of linear algebraic groups.
Consider the $\mathbf{R}$homomorphism ${\rm{Ad}}_{\mathbf{G}}:\mathbf{G} \rightarrow {\rm{GL}}(\mathfrak{g})$. (Here we abuse notation in the usual manner, letting ${\rm{GL}}(V)$ to denote either the linear algebraic $\mathbf{R}$group or the associated disconnected Lie group of $\mathbf{R}$points, with the context making the intended meaning clear.) On Lie algebras this induces ${\rm{ad}}_{\mathfrak{g}}$, so ${\rm{Ad}}_{\mathbf{G}} = j$ due to Zariski connectedness of $\mathbf{G}$. In particular, $\ker {\rm{Ad}}_{\mathbf{G}} = 1$, which is to say $\mathbf{G}$ is of adjoint type. The map ${\rm{Ad}}_G$ factors through the closed subgroup $\mathbf{G}(\mathbf{R})^0$ because it does so on Lie algebras (due to the compatibility of formation of Lie algebras with respect to passage from algebraic $\mathbf{R}$groups to Lie groups of $\mathbf{R}$points).
The resulting Lie group homomorphism $G \rightarrow \mathbf{G}(\mathbf{R})^0$ with discrete kernel $Z := Z_G$ (center of $G$) is an isomorphism on Lie algebras and hence identifies $\mathbf{G}(\mathbf{R})^0$ with $G/Z$. (Note also that $\mathbf{G}(\mathbf{R})^0$ always has finite index inside $\mathbf{G}(\mathbf{R})$; we will not use it). This is the "algebraicity" (up to disconnectedness!) of the image $G/Z$ of ${\rm{Ad}}_G$ inside ${\rm{GL}}(\mathfrak{g})$.
Likewise, if $\mathfrak{a} \subset \mathfrak{g}$ is a Cartan subalgebra then there is a (unique) maximal $\mathbf{R}$torus $\mathbf{T} \subset \mathbf{G}$ with associated Lie algebra $\mathfrak{a}$ inside $\mathfrak{g}$ (as for connected reductive groups over any field of characteristic 0), ultimately because the groupcentralizer of a semisimple element in the Lie algebra of a connected reductive group has reductive identity component (so we can perform dimension induction, using that the center of a connected reductive group has identity component that is a torus). Thus, $\mathbf{T}(\mathbf{R})$ has preimage in $G$ whose identity component $C$ maps onto $\mathbf{T}(\mathbf{R})^0$ with discrete kernel and hence is the Cartan subgroup ${\rm{exp}}_G(\mathfrak{a}) \subset G$ associated to $\mathfrak{a}$. This establishes a bijection between the sets of Cartan subgroups of $G$ and maximal $\mathbf{R}$tori in $\mathbf{G}$.
Now we address the question at hand. Let $f:\mathbf{R} \rightarrow G$ be a 1parameter subgroup, and consider the induced homomorphism $\overline{f}:\mathbf{R} \rightarrow G/Z = \mathbf{G}(\mathbf{R})^0 \subset \mathbf{G}(\mathbf{R})$. Since $Z$ is discrete and central in $G$, $f$ is uniquely determined by $\overline{f}$. Hence, if $g \in G$ then $g$ centralizes $f$ if and only if its image $\overline{g} \in \mathbf{G}(\mathbf{R})$ centralizes $\overline{f}$. Thus, the centralizer $Z(f)$ of $f$ in $G$ is the full preimage under $G \rightarrow \mathbf{G}(\mathbf{R})$ of the centralizer $Z(\overline{f})$ of $\overline{f}$ in $\mathbf{G}(\mathbf{R})$, with $Z(f)$ mapping onto $Z(\overline{f}) \cap \mathbf{G}(\mathbf{R})^0$ (with discrete kernel). In particular, $Z(f)^0$ maps onto $Z(\overline{f})^0$.
Let $h:\mathbf{R} \rightarrow G$ be a second 1parameter subgroup and assume $Z(h)^0 = Z(f)^0$, so $Z(\overline{h})^0 = Z(\overline{f})^0$. To prove that $Z(h) = Z(f)$ (centralizers in $G$) it suffices to prove that $Z(\overline{h}) = Z(\overline{f})$ (centralizers in $\mathbf{G}(\mathbf{R})$, which may of course be disconnected).
Assume $h$ and $f$ are valued inside a common Cartan subgroup $C$ of $G$. Clearly $\overline{h}$ and $\overline{f}$ are 1parameter subgroups of $\mathbf{G}(\mathbf{R})$ valued in the group of $\mathbf{R}$points of the associated maximal $\mathbf{R}$torus $\mathbf{T}$ of $\mathbf{G}$.
Now comes the key idea: we consider Zariski closures of analyticallydefined and typically nonalgebraic subgroups of $\mathbf{G}(\mathbf{R})$. Let $\mathbf{S}, \mathbf{S}' \subset \mathbf{G}$ be the respective Zariski closures of the images of $\overline{h}$ and $\overline{f}$. In view of the "algebraicity" of the group law on $\mathbf{G}(\mathbf{R})$, $Z(\overline{h})$ is the group of $\mathbf{R}$points of the algebraic group centralizer $Z_{\mathbf{G}}(\mathbf{S})$, and likewise $Z(\overline{f})$ is the group of $\mathbf{R}$points of $Z_{\mathbf{G}}(\mathbf{S}')$. Thus, it suffices to show that $Z_{\mathbf{G}}(\mathbf{S}) = Z_{\mathbf{G}}(\mathbf{S}')$ inside $\mathbf{G}$.
The hypothesis that $Z(\overline{h})^0 = Z(\overline{f})^0$ implies that the closed $\mathbf{R}$subgroups $Z_{\mathbf{G}}(\mathbf{S})$ and $Z_{\mathbf{G}}(\mathbf{S}')$ in $\mathbf{G}$ have the same Lie algebra inside $\mathfrak{g}$ because the Lie algebra of any algebraic $\mathbf{R}$group is naturally identified with the analyticallydefined Lie algebra of the identity component of its group of $\mathbf{R}$points. But in characteristic 0, a connected (for Zariki topology!) smooth closed subgroup of a linear algebraic group is uniquely determined by its associated Lie algebra as a subalgebra of the ambient Lie algebra, so it suffices to prove that $Z_{\mathbf{G}}(\mathbf{S})$ and $Z_{\mathbf{G}}(\mathbf{S}')$ are connected for the Zariski topology.
By construction, $\mathbf{S}$ and $\mathbf{S}'$ are (smooth) closed $\mathbf{R}$subgroups of the $\mathbf{R}$torus $\mathbf{T}$, so they are tori provided that they are Zariskiconnected. Moreover, it is a wellknown fact in the algebraic theory that the centralizer of a torus in a (Zariski)connected linear algebraic group over any field whatsoever is always (Zariski)connected. This is really the whole point of this entire argument: in the algebraic setting, the disconnectedness problems that plague the consideration of centralizers in the analytic theory often do not arise.
So it suffices to prove that $\mathbf{S}$ and $\mathbf{S}'$ are connected for the Zariski topology. More generally, if $\mathbf{G}$ is an arbitrary linear algebraic group over $\mathbf{R}$ (not assumed to be connected for the Zariski topology) and if $H \subset \mathbf{G}(\mathbf{R})$ is an arbitrary subgroup (not necessarily closed!) that is connected for the subspace topology defined via the analytic topology on $\mathbf{R}$ (e.g., $\overline{h}(\mathbf{R})$ and $\overline{f}(\mathbf{R})$, with $\mathbf{G}$ the connected semisimple $\mathbf{R}$group of adjoint type considered above) then we claim that the Zariski closure $\mathbf{H}$ of $H$ in $\mathbf{G}$ is Zariskiconnected. In view of the generality of this setup, since $\mathbf{H}$ is certainly itself a linear algebraic group we may rename it as $\mathbf{G}$ to reduce to proving that if $\mathbf{G}(\mathbf{R})$ contains a subgroup that is connected for the analytic topology (though not necessarily closed for the analytic topology) and is also Zariskidense in $\mathbf{G}$ then in fact $\mathbf{G}$ is Zariskiconnected. We can replace $\mathbf{G}$ with $\mathbf{G}/\mathbf{G}^0$ to reduce to the case when $\mathbf{G}$ is finite etale, and then the only connected subgroup of the finite discrete group $\mathbf{G}(\mathbf{R})$ is the trivial one, so the identity point is Zariskidense in the finite etale $\mathbf{R}$group $\mathbf{G}$. This forces $\mathbf{G} = 1$, so we are done. QED

$\begingroup$ Many thanks. I owe you a bottle of wine ! $\endgroup$ Nov 8 '13 at 16:13


1$\begingroup$ @Daryl: I've heard that Syrah is more reductive than Pinot Noir. $\endgroup$– MarguaxNov 9 '13 at 2:24
This is only a very partial answer but is a short argument to show that this is true in $\mathrm{SL}_d(\mathbf{R})$, namely that if any two subsets $X_1,X_2$ of the set $D$ of diagonal matrices have the same connected centralizers in $\mathrm{SL}_d(\mathbf{R})$ then they have the same centralizer. Indeed let $A_i$ be the subalgebra generated by $X_i$. A simple observation is that the (unital) $\mathbf{R}$subalgebras of $D\simeq\mathbf{R}^d$ (product of rings) are classified by equivalence relations on $\{1,\dots,d\}$, namely if $R$ is such an equivalence relation, then $D_R=\{(x_1,\dots,x_d):x_i=x_j$ whenever $i\,R\,j\}$ is a subalgebra and these are the only ones. Since the centralizer of $X_i$ is the same as the centralizer of $A_i$, to conclude it is enough to check that for any two distinct equivalence relations $R,R'$ the connected centralizers of $D_R$ and $D_{R'}$ are distinct and this is easy: for $R$, the centralizer is the set of determinant one matrices withs diagonal blocks along the $R$decomposition, and the connected centralizer is the set for which all diagonal blocks have positive determinant, and this clearly remembers $R$.
Maybe a similar approach can be done with other groups, at least for linear ones, using a linear embedding (the above argument always shows that there are only finitely many subgroups occurring as a centralizer of a subset of a connected Cartan subgroup!). Note that the case of $\mathrm{SL}_d$ does not imply the case of groups locally isomorphic to $\mathrm{SL}_d$, so for example $\mathrm{PSL}_d(\mathbf{R})$ for even $d$ should be checked.