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In a recent question, I initially doubted that $\mathbb{C}^\times\cong S^1$, my intuition being that $\mathbb{C}^\times$ has one more "dimension" than $S^1$ - in rigorous terms, $S^1$ is (or rather, can be given the structure of) a 1-dimensional Lie group, and $\mathbb{C}^\times$ is (can be given the structure of) a 2-dimensional Lie group.

My question is rather naive: Given a Lie group $G$ of dimension $n\in\mathbb{N}\cup\{\infty\}$, for which $m\in\mathbb{N}\cup\{\infty\}$ does there exist a Lie group $H$ of dimension $m$ such that $G\cong H$ as groups (forgetting about the manifold structure)? Surely there is at least one $G$ such that the answer isn't all $m$?

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Zev Chonoles
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  • I guess your question is motivated by the fact that for all $n,m \in \mathbb{N}$, you have a group ($\mathbb{Q}$-vector space) isomorphism $\mathbb{R}^n \simeq \mathbb{R}^m$, right ? – Joel Cohen Jun 10 '11 at 01:23
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    I had also thought about that while composing the question, but I was mainly motivated by my incorrect intuition about $\mathbb{C}^\times\cong S^1$. – Zev Chonoles Jun 10 '11 at 02:16
  • Still informally, re the dimension issue, $C^x$ is homotopic to $S^1$, by retraction. – gary Jun 10 '11 at 16:52

2 Answers2

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I believe all 1-dimensional (connected) real lie groups are abelian, so given a non-abelian lie group there exists no lie group of dimension 1 isomorphic to it.

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jspecter
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A related question was asked at MathOverflow. The author argues that there is a unique Lie group structure on $SU(2)$.

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Cheerful Parsnip
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