Let $G$ be a Lie group with left translation action:

$L: G \times G \rightarrow G$ $\quad$ $(g,h) \rightarrow L_g(h) := gh$

I want to show that this action is transitive,free, and proper.

Transitive and free follow from definition more or less.

However, I am having trouble showing it is proper.

This is the definition of a proper action that I am given:

The action $\phi:G \times M \rightarrow M$ of a group $G$ on a manifold $M$ is proper if, whenever the sequences $\{x_n\}$ and $\{g_nx_n\}$ converge in $M$, the sequence $\{g_n\}$ has a convergent subsequence in $G$.

It is clear to me that if $G$ is a compact Lie group, then the action is proper. This is so because in a compact group, every sequence $\{g_n\}$ has a convergent subsequence. But we are not given that the group $G$ is compact. My question is, how do we show this for a generic Lie group $G$?

Any hints/suggestions are most welcomed.


1 Answer 1


Hint : if $x_n\to x, g_nx_n \to y$, then $g_n = g_nx_nx_n^{-1}\to ?$

  • $\begingroup$ if $x_n\to x, g_nx_n \to y$, then $g_n = g_nx_nx_n^{-1}\to yx^{-1}$. But ok, can we just conclude that since we've found a convergent subsequence $g_n$ that works, we're done? $\endgroup$ Apr 29, 2019 at 18:17

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