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Let $G$ be a connected Lie group of dimension one and let $U=\{u_t\}_{t \in \mathbb R }$ be a one-parameter subgroup. I wonder if it is true that $U=G$?

It is easy to see that $U$ is closed. So by a theorem in Lie theory, $U$ is a Lie subgroup. $U$ is not discrete and thus it must have dimension one. But I don't know how to go further

No One
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  • Up to isomorphism there are only two Lie groups of dimension $1$, and you can compute the one-parameter subgroups of both them explicitly. (In fact, one of the two groups is the universal covering group of the other, so you really only need to compute for one of them.) – Travis Willse Aug 21 '20 at 22:37
  • Correction : two connected Lie groups of dim 1: the circle and the line. – markvs Aug 21 '20 at 22:39
  • What are those two types of lie groups explicitly? – No One Aug 21 '20 at 22:40
  • The two groups are $\mathbb{R}/\mathbb{Z}$ (circle) and $\mathbb{R}$ (line). – markvs Aug 21 '20 at 22:50

1 Answers1

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Since $\dim U=\dim G$, $U$ is open in $G$, and thus clopen. Since $G$ is connected, this implies $U=G$.

Eric Wofsey
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