# Closures of one-parameter subgroups of lie groups

I'm reading some basic facts of Lie groups. I meat with difficulties when I try to solve the following statement: "Prove the closure of a non-closed one-parameter subgroup of a Lie group is a torus."

I have proved that a compact connected abelian Lie group must be a torus. However, I have no idea how to show that the closure is compact. Thanks in advance.

• Could you provide a counter-example that the closure isn't compact? Thanks. – 貓貓吃狗狗 Dec 19 '15 at 5:19
• Ah, sorry, I confused myself. I guess the statement is true. – Qiaochu Yuan Dec 19 '15 at 6:06

Let $\gamma:\mathbb R \to G$ denote the one parameter sub-group, and $H$ its image in $G$. The closure $\overline H$ of $H$ is an abelian (connected) Lie group, and so can be written (non-canonically) in the form $\overline H = V \times T$, where $V$ is a vector-space and $T$ a torus. We want to show, if $H\not=\overline H$, that $V= 0$. Suppose $p=(v,t)\in\overline H \setminus H$. The projection of $\gamma$ on $V$ is of the form $x \mapsto xw$, for some non-zero $w\in V$ (non-zero, for otherwise $H$ is not dense in $\overline H$ (!), unless $V=0$), so there is an open neighborhood $B$ of $p$, with compact closure, and a bound $R$ such that $|x|>R$ implies $\gamma (x) \not \in \overline B$. The set $\gamma([-R,R])$ is compact, and contains $p$ in its closure, so contains $p$. Therefore $p\in H$. Contradiction! Therefore $V=0$.
• Why does such a $B$ exist? – Andre Gomes Jun 30 '18 at 21:45
• Because the linear map $t \to wt$ goes to infinity: one can take $B = A \times T$, for some bounded open ball $A$ which contains $w$. For $t$ large $wt$ cannot belong to $\overline A$. – peter a g Jul 1 '18 at 1:11