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My question arises from Chapter 21 of Lee's book, Introduction to Smooth Manifolds, 2nd edition.


Let $G$ be a Lie group acting smoothly, freely and properly on a manifold $M$ on the left, denoted by $g\cdot p$, $g\in G, p\in M$. Let $H$ be a closed normal subgroup of $G$. Then we have the following conclusions by some theorems in Lee's book

By another question on topological group action, we also know that

  • The group action of $G$ on $M$ induces a (left) group action of $G/H$ on $M/H$ by: \begin{equation}\tag{1} gH\cdot (H\cdot p) := H\cdot(g\cdot p),\quad p\in M, g\in G. \end{equation}
  • The induced action of $G/H$ on $M/H$ is also smooth and free.
  • The quotient spaces $(M/H)/(G/H)$ and $M/G$ are identical as sets, and have same quotient topology.

The natural question is, whether the induced action defined in $(1)$ is proper? If so, then the quotient manifold $(M/H)/(G/H)$ is well defined, and we can easily check that $(M/H)/(G/H)$ and $M/G$ have same smooth structure (by Quotient Manifold Theorem). But I do not know how to check this properness...

I can say more in the very special case that $H$ and $G/H$ are both compact subgroups. By Exercise 13 in Section 26 of Munkres' topology book, $G$ is also compact. Then by Corollary 21.6 in Lee's book, the two group actions of $G$ on $M$ and $G/H$ on $M/H$ are both proper.

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