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

Calabi
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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:oldbiggrin:.
 
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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.)
 
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.
 
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}^+##.
 
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.
 
Then, the non existence on continuous homomorphism is not garanti.
 
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.
 
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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.
 
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.
 
  • #10
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.
 
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  • #11
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.
 
  • #12
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?
 
  • #13
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.
 
  • #14
What do you think please?
 
  • #15
Calabi said:
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.
Calabi said:
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 said:
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.
 
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