0
$\begingroup$

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 $Z$?

$\endgroup$
  • 1
    $\begingroup$ 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
5
$\begingroup$

(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.

$\endgroup$
5
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.