I am having troubles with an exercise (Ex 12.19) in Lee's book Introduction to Topological Manifolds. The text of the exercise is as follows:

Suppose we are given a continuous action of a topological group $G$ on a second countable, locally compact Hausdorff space $E$. Show that the action is proper if and only the following condition is satisfied: whenever $(e_i)$ is a sequence in $E$ and $(g_i)$ is a sequence in $G$ such that both $(e_i)$ and $(g_i \cdot e_i)$ converge in $E$, a subsequence of $(g_i)$ converges in $G$.

Denote by $\theta: G \times E \rightarrow E$ the $G$ action, $\theta(g,e)= g\cdot e$, and by $\Theta: G\times E \rightarrow E\times E$ the map $(g,e)\mapsto(g\cdot e, e)$. By definition a map is proper if the preimage of a compact set is compact and the action of $G$ is proper if the map $\Theta$ (not $\theta$) is proper.

Actually, J. Lee gives a proof of the only if part in the book Introduction to Smooth Manifolds under the hypotheses that $G$ is a Lie group and $M$ a smooth manifold. I have tried to adapt that proof to the case of the exercise, see below, but I have not succeeded. Indications on how to modify the proof, or how to prove the result with other methods, would be welcome.

The argument is as follows: denote by $p$ the limit of $(e_i)$ and by $q$ the limit of $(g_i \cdot e_i)$. Let $U$, $V$ be precompact (have compact closure) open neighbourhoods of $p$, $q$. It is possible to have such $U$, $V$ since $E$ is locally compact Hausdorff. By dropping if necessary a finite number of terms, we can assume that the sequence $(\Theta(g_i,e_i))$ is contained in the compact set $\bar{ V}\times \bar{U}$. Then \begin{equation*} ((g_i,e_i))\subseteq (\Theta ^{-1}(\Theta(g_i,e_i)))\subseteq \Theta^{-1}(\bar{V}\times\bar{U}). \end{equation*} Since $\Theta $ is proper, $\Theta^{-1}(\bar{V}\times\bar{U})$ is compact. If we can assume that it is also sequentially compact we are done since then $(g_i, e_i)$ has a convergent subsequence. Because of the second countable Hausdorff hypothesis, on $E$ compactness is equivalent to sequential compactness, however not in $G$ (but it would for $G$ a Lie group). Therefore I do not see why $\Theta^{-1}(\bar{V}\times\bar{U})$ should be sequentially compact.

  • 1
    $\begingroup$ I think it is pretty clear that he wants compactness for subsets in G to be equivalent to sequential compactness. (Metrizable would suffice.) $\endgroup$ – Moishe Kohan Mar 21 '17 at 0:22
  • $\begingroup$ In his answer here: math.stackexchange.com/questions/987038/… Lee adds the assumption that G is 1st countable and Hausdorff (for topological groups, this is equivalent to being metrizable (Birkhoff-Kakutani theorem). $\endgroup$ – Moishe Kohan Mar 21 '17 at 0:36
  • 2
    $\begingroup$ To clarify: Problem 12-19 is incorrect as stated. The correction that @MoisheCohen mentioned is listed in my online correction list. $\endgroup$ – Jack Lee Mar 21 '17 at 17:18
  • $\begingroup$ That settles it then. Thanks to both for the help! $\endgroup$ – GFR Mar 21 '17 at 20:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.