Let me specify that my knowledge about Lie groups/algebras is reduced to bits and pieces I learned from various diff geometry textbooks. I could not find a reference for the following question (I am not sure if it is true to start with):

Suppose $G$ is connected compact Lie group. I know that $l=Lie(G)$ splits into a direct product of Lie algebras $[l,l]$ and $z(l)$. If $H \leq G$ is connected such that $Lie(H)=[l,l]$, since $[l,l]$ is semi-simple with negative definite Killing form, an argument using Myers theorem gives that $H$ is a closed(embedded) subgroup of $G$.

Now let $Z \leq G$ such that $Lie(Z)=z(l)$. Clearly $Z$ is Abelian. Can one deduce that $Z$ is also a closed (or embedded) subgroup? Is there a choice of Z that satisfies this? Is it true for all such $Z$?

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    $\begingroup$ Is that really the center of the group? If so, it is pretty easy to show the center of $G$ is closed. $\endgroup$ – Thomas Andrews Feb 1 '15 at 22:10
  • $\begingroup$ Indeed. I missed a simple fact. $\endgroup$ – Hammerhead Feb 1 '15 at 22:29

(Assuming the $Z$ you've defined is the same as the usual "center" of a group.)

Let $C_g=\{g_1\in G\mid g_1g=gg_1\}$. This set is closed, it is the inverse image of the closed set $\{e\}$ of the continuous function $G\to G$ defined as $g_1\mapsto gg_1g^{-1}g_1^{-1}$.

But $Z=\bigcap_{g\in G} C_g$ then also must be closed.


The center of any (Hausdorff) topological group is closed. The point is that the centrality condition, that $zg = gz$ for any $g \in G$, is a "closed condition." Intuitively, it is preserved under taking limits in $z$ (and this is a proof for any second-countable (Hausdorff) topological group, in particular any Lie group). More formally, for any $g \in G$ the centralizer

$$C(g) = \{ z \in G : zg = gz \}$$

is the preimage of the identity under the commutator map

$$G \ni z \mapsto [z, g] = z g z^{-1} g^{-1} \in G$$

and since $G$ is Hausdorff, the identity is closed. Since the preimage of a closed subset is closed, $C(g)$ is closed. Now the center is the intersection of the $C(g)$ for all $g \in G$, and the intersection of closed subsets is closed.

  • $\begingroup$ Can you verify for me if his last paragraph is actually defining the "usual" center? He seems to be defining something else, at least, a different definition than the usual center. $\endgroup$ – Thomas Andrews Feb 1 '15 at 22:16
  • $\begingroup$ @Thomas: I think it's right. If I've got my Lie theory straight, the point is that the Lie algebra $\mathfrak{g}$ of a compact connected Lie group is reductive, so decomposes as the direct sum of a semisimple part and an abelian part. The abelian part is then naturally isomorphic to both the center and to the abelianization of $\mathfrak{g}$. $\endgroup$ – Qiaochu Yuan Feb 1 '15 at 22:17
  • $\begingroup$ for me, Z was not quite the center but, your efforts already answer my question. I missed this simple fact. $\endgroup$ – Hammerhead Feb 1 '15 at 22:32

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