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I need to decide if this statement is true or false:

" Every Lie subgroup of non-abelian compact Lie group is compact."

I think that it is false. I thought in a counterexample in which the center is discrete and infinite. Let $G=\prod_{n=1}^{\infty}SU(2)$. By Tychonoff theorem $G$ is compact (since $SU(2)$ is compact). But since $Z(SU(2))\cong \mathbb{Z}_{2}$ (cyclic group of order two) then $Z(G)\cong \prod_{n=1}^{\infty}\mathbb{Z}_{2}$ which is infinite and discrete. Therefore although $G$ is compact, its center (which is a closed Lie subgroup) is not compact. This example is right? Thanks!!

Kauê Orlando Pereira
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    In what sense is $G$ a Lie group? Lie groups are usually finite-dimensional by definition, and the various infinite-dimensional generalizations can lose some of the standard properties. – Kajelad Jul 15 '23 at 04:07
  • It was a doubt that I had. There is some example finite dimensional? – Kauê Orlando Pereira Jul 15 '23 at 12:15
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    See here for example. – Callum Jul 15 '23 at 15:17
  • In this example the torus is abelian. I was thinking that this statement is true but i don't know how prove it. – Kauê Orlando Pereira Jul 15 '23 at 15:30
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    @KauêOrlandoPereira Note the ambiguity of weather "Lie subgroup" includes closed as a criteria; you might want to clarify that in your question. If you aren't requiring closure, you can embed the torus in $SO(4)$ for a nonabelian counterexample. – Kajelad Jul 15 '23 at 17:47

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