In a course on differential manifolds and Lie groups, the following theorem was stated, though never proven:

Let $M$ and $N$ be smooth manifolds, and suppose $G$ is a Lie group acting on $M$. If the group action is free and proper, then $M/G$ has a manifold structure so that the quotient map $\pi:M\to M/G$ is smooth. Additionally, if $f:M/G\to N$ and $f\circ\pi$ is smooth, then $f$ is smooth.

The definition of a proper group action was given as follows: given any compact set $K \subset M$, the action of $G$ on $M$ is proper if and only if the set $\{g\in G:gK\cap K\neq\emptyset\}$ has compact closure in $G$.

I am confused about two things:

1) The definition of proper group action differs from the one I got from Wikipedia and the books I have, where the acting group is assumed to be discrete. Moreover, I found it highly contrived due to a lack of understanding on my part. What exactly is this definition helping us to see? The need for the closure to be compact, for instance, is lost on me.

2) Removing the context of manifolds and Lie groups, if one were to look at arbitrary group actions, then how would the proper action be defined? What kind of topology on G makes sense?

I apologise for any naiveté; I am new to group actions, even in the algebraic case. Thank you for your patience!


1 Answer 1


If $G$ is a topological group acting on a topological space $M$, the usual definition is that the action is proper if the map $G\times M\to M\times M$ defined by $(g,x)\mapsto (g\cdot x,x)$ is a proper map, which means that the preimage of every compact set is compact. This definition works both in the topological category and in the smooth category.

For sufficiently nice spaces, there are other characterizations that are often useful. For example:

  • If $M$ is Hausdorff, then properness is equivalent to the condition you described.
  • If $M$ and $G$ are Hausdorff and first-countable, then properness is equivalent to the following condition: If $(x_i)$ is a sequence in $M$ and $(g_i)$ is a sequence in $G$ such that both $(x_i)$ and $(g_i\cdot x_i)$ converge, then a subsequence of $(g_i)$ converges.

Actions of compact groups (on Hausdorff spaces) are always proper, so properness really has meaning only for actions of noncompact groups. In this case, properness has the intuitive meaning that "most" of $G$ (i.e., all but a compact subset) moves compact sets of $M$ far away from themselves.

You can find more information on group actions in the topological category in my Introduction to Topological Manifolds (2nd ed.), and in the smooth category in Introduction to Smooth Manifolds (2nd ed.), among other places.

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    $\begingroup$ Gosh!! I was just suggested the same book by a friend. Thanks a lot. Just one clarification. Why the big fuss about moving compact sets and not any other kind??Does bringing in compactness facilitate anything??Thanks again!! $\endgroup$
    – Vishesh
    Commented Oct 24, 2014 at 17:09
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    $\begingroup$ Maybe I'm confused. The trivial group acting upon a space always has a proper action, but its element does not move compact sets away from themselves at all. $\endgroup$ Commented Oct 24, 2014 at 18:06
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    $\begingroup$ @Robin: When I said "most" of $G$ moves comact sets far away, the word "most" is an imprecise way of saying "all of $G$ except a compact subset." Actions of compact groups (at least on Hausdorff spaces) are always proper, so properness really has meaning only for noncompact groups. I'll edit my answer to make this clearer. $\endgroup$
    – Jack Lee
    Commented Oct 24, 2014 at 18:26
  • $\begingroup$ @JackLee -- Aha! I see. Thank you for clearing that up. $\endgroup$ Commented Oct 24, 2014 at 18:28
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    $\begingroup$ Of course when you write "equivalent to the condition you described", you meant equivalent to the condition the OP surely intended to describe, which is that for each compact $K$ the set of $g$ such that ... has compact closure. $\endgroup$ Commented Jun 8, 2021 at 14:09

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