Graduate Compact Lie Group: Proof of Discrete Center & Finite Size?

[Calabi]

Hello, let be $G$ a connected Lie group. I suppose$Ad(G) \subset Gl(T_{e}G)$ is compact and the center $Z(G)$ of $G$ is discret (just to remember, forall $g \in G$, $Ad(g) = T_{e}i_{g}$ with $i_{g} : x \rightarrow gxg^{-1}$.).

I saw without any proof that in those hypothesis $G$ is compact and the center is finite.

Have you got a proof of this results please?

Here is the things I try to do : if we show $G$ compact, then since the center is closed and discret then it will be finite. But since $G$ is connected we show that $Z(G)$ is the kernel of
$Ad$ so as $Ad(G)$ is compact it's also a Lie group so that $Ad(G)$ is isomorph to $G / Z(G)$. Then if I show the commutator group is dense in $G / Z(G)$, I saw in a books that we get the result but I don't know how to do.

Thank you in advance and have a nice afternoon.

 

[fresh_42]

I've seen a proof, where it is also required, that there doesn't exist any non-trivial continuous homomorphism from $G$ to (Abelian) $\mathbb{R}^+$, which the density of $[G,G]$ is one possibility to get this. (The proof goes along the lines: If $Z(G)$ is infinite, then there is such a homomorphism from $Z(G)$ which can be extended to one of $G$. By ruling this out, we get a finite center.)

 

[Calabi]

Hello Fresh_42(that's been a long time, hope you're fine.).
Your argument is interessting. Have you got a reference?

I stump on this result which not seem trivial I'm curious to see how to show it.

 

[fresh_42]

I've found it here: https://www.amazon.com/dp/0387909699/?tag=pfamazon01-20, in section 4.11 theorem 4.11.5 to prepare Weyl's theorem. It doesn't look very complicated, but it's four lemmas on roughly two pages that deal with this homomorphism and it's extensibility to $G$. The result itself is proven for connected, locally compact Lie groups satisfying second countability with discrete center and $G/Z(G)$ compact and this non-existence requirement for mappings to $\mathbb{R}^+$.

 

[Calabi]

Hello fresh_42, I consult the book in my University's library. I see the theorem but how to link to the first result I try to show?

I remember I try to show $Ad(G) \subset Gl(T_{e}G)$ is compact and the center $Z(G)$ of $G$ is discret (just to remember, forall $g \in G$, $Ad(g) = T_{e}i_{g}$ with $i_{g} : x \rightarrow gxg^{-1}$.). implies $G$ is compact and the center is finite.
It's enough to show $G$ is compact as $Z(G)$ is finite as claim at the beginning.
Sorry for the late of the answer but I was perparing my class entry.

 

[Calabi]

Then, the non existence on continuous homomorphism is not garanti.

 

[fresh_42]

A closer look on Varadarajan's theorem showed me, that he only requires a central discrete subgroup $C$ (along with the conditions on $G$ and $G/C$ compact), which is weaker than the entire center to be discrete. The stronger condition on $Z(G)$ might already be sufficient, not sure.

As I see it, this somehow technical requirement is an attempt to be a) as general as possible (using central subgoups $C\,$) and b) get a hold on the center, i.e. guarantee it is finite, which you also used in your argument. The problem with the center is always, that you cannot factor it out, because $G/Z(G)$ can still have a center, and center elements always lead to degrees of freedom in any representation.

The proof Varadarajan gives shows that the conditions on $G$ (locally compact, connected, 2nd order countability) and the discreteness of $C$ with $G/C$ compact imply that $C$ is finitely generated. Then an infinite central subgroup $C$ would imply a non-trivial, continuous homomorphism to $\mathbb{R}^+$ and he proceeds to lift such a homomorphism $\varphi$ to $G$.

So if we can rule out $\varphi$ we can in return guarantee that $C=Z(G)$ is finite, which we really want. The trick here is that we search for a topological property (finite center) and the proof gives us an algebraic mean to test this (no $\varphi$). Therefore we have a sufficient condition. I don't know, whether it is also a necessary condition. But maybe the given conditions in case $C=Z(G)$, which imply that $Z(G)$ is finally generated (main result in Varadarajan's proof), is already sufficient to rule out the existence of $\varphi$.

To summarize it, we are looking for a way to rule out the situation $Z(G) \cong \{c_1,\ldots c_m\} \times \mathbb{Z}^n$ with $n \geq 1$. The non-existence of $\varphi$ is one possibility to rule it out.

 

[Calabi]

OK. I understand a few things. But I'm not sure to be able to show what you said.
Thanks a lot for your answer. If someone as idea I'll be pleased to listen it.

 

[Calabi]

The things is that we use $Ad(G)$ compact only to show $G / Z(G)$ which is not enough.
And the discret properties is not use too much.

 

[fresh_42]

I'm not quite sure, if I understood you correctly. E.g. if we take $G=\mathbb{R}^n$ as connected, Abelian Lie group, then $\operatorname{Ad}G=\{1\}$ which is compact, but $Z(G)=\mathbb{R}^n$ is not discrete. On the other hand, if we take $G=SL_n(\mathbb{R})$ with odd $n$ as connected Lie group, then $\operatorname{Ad}(G) \cong G = SL_n(\mathbb{R})$ is not compact but $Z(G)=\{1\}$ is discrete. Of course these are extreme examples, but it shows that connectedness alone doesn't imply anything.

 

[Calabi]

Hello it's doesen't work. My goal is to show $G$ connected $Ad(G)$ compact and $Z(G)$ discret $\Rightarrow$
$G$ compact and $Z(G)$ finite.

 

[Calabi]

I saw in a book that if we add $T_{e}G$ semi simple it's work.

It beginn the same ways as me by remarking $Ad(G) \simeq G / Z(G)$ is compact and is a Lie group because $Z(G)$ is discret.
Then he claim that $G / Z(G)$ is equats to its derivate group and that it conclude.

Do you see why?

 

[Calabi]

The derivate $D$ group of $G' = G / Z(G)$ is not always closed(so it's not necessarly a Lie Group for the induce topolgie.). So we must show that $D$ is closed.
Then if we show that $[T_{e}G', T_{e}G'] = T_{e}G'$(the equality is true because of semi simplicity.). is the tangent bundle of
$D$ then $D = G'$.

But I don't see how conclude even with that.

 

[Calabi]

What do you think please?

 

[fresh_42]

Hello it's doesen't work. My goal is to show $G$ connected $Ad(G)$ compact and $Z(G)$ discret $\Rightarrow$
$G$ compact and $Z(G)$ finite.

In short we have an exact sequence $Z(G)_{discrete} \rightarrowtail G_{connected} \twoheadrightarrow \operatorname{Ad}(G)_{compact}$. What I don't see is how you could rule out that $Z(G)$ contains copies of $\mathbb{Z}$. Couldn't we even extend $G$ by any central lattice without affecting the other properties? It certainly doesn't affect $\operatorname{Ad}(G)$ and it should be doable in a connected way. So something more has to be required.

I saw in a book that if we add $T_{e}G$ semi simple it's work.

It beginn the same ways as me by remarking $Ad(G) \simeq G / Z(G)$ is compact and is a Lie group because $Z(G)$ is discret.
Then he claim that $G / Z(G)$ is equats to its derivate group and that it conclude.

Do you see why?

The derivate $D$ group of $G' = G / Z(G)$ is not always closed(so it's not necessarly a Lie Group for the induce topolgie.). So we must show that $D$ is closed.
Then if we show that $[T_{e}G', T_{e}G'] = T_{e}G'$(the equality is true because of semi simplicity.). is the tangent bundle of
$D$ then $D = G'$.

But I don't see how conclude even with that.

I'm not sure that I'm fit enough for the changes between group and algebra and back. However, $T_e(G)$ semisimple is quite a strong condition as we have then
$$
T_e(G)= \mathfrak{g} = [\mathfrak{g},\mathfrak{g}] \cong \operatorname{ad} (\mathfrak{g}) = T_e(\operatorname{Ad}(G)) \cong \operatorname{Der}(\mathfrak{g})
$$
We therefore have $[G/Z(G), G/Z(G)]=[G,G]/Z(G)$ and $[\mathfrak{g},\mathfrak{g}]=\mathfrak{g}$. Now the question is, what does the semisimplicity of $T_e(G)=\mathfrak{g}$ says about the connection component of $e \in G$.

Varadarajan proves (as a byproduct) that if $\mathfrak{g}$ is a real semisimple Lie algebra and $\operatorname{Ad}(G)$ compact, then all analytic groups $G$ with $T_e(G) \cong \mathfrak{g}$ are compact. This is all in the chapter I mentioned earlier and too long to type in here and I'm not sure, whether there is a shorter way to see it. His theorems are quite general, so maybe there is a shortcut in this special case. Varadarajan proves it with Weyl's theorem, and with this result your statements follow immediately (in case of semisimple $\mathfrak{g}\,$).

I still think that the general case without this assumption is not true and that connectedness alone is too weak to rule out e.g. $Z(G)=\mathbb{Z}$ but I admit that I have no counterexample.