Suppose $H$ is a closed subgroup of a Lie group $G$. Then in Lee's book Introduction to Smooth Manifolds (Ch. 9) he showed that the action $H\times G\to G$ $(h,g)\mapsto gh$ is a smooth, free, proper action. I have a small problem regarding showing the action is proper. Note that if $g_ih_i\to g_0$ and $g_i\to g_{00}$ then we must have that $h_i\to g_{00}^{-1}g_0\mathrel{:=}h_0$ and as $H$ is closed $h_i\to h_0\in H$. I want to understand why this would imply the action is proper. I actually need to show that $h_i\to h$ in the topology of $H$ but the above argument shows that the convergence happens in the subspace topology of $H$. I know that as $H$ is closed, $H$ must be embedded. But to this point in Lee's book this is not proved. Is there any way to see why the argument in the Lee's book makes sense?
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asked Dec 6, 2020 at 9:44
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3$\begingroup$ The topology of $H$ is the subspace topology. $\endgroup$– user130903Commented Dec 6, 2020 at 11:18
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$\begingroup$ @Zero. Please read the question clearly and then comment. $\endgroup$– A beginner mathmaticianCommented Dec 6, 2020 at 12:49
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2$\begingroup$ The question is a bit unclear. If your starting point is a closed subgroup and you're asking why the "topology on $H$" is the same as the subspace topology, you seem to assume that $H$ is given another topology... but this is not part of the setting. Maybe your starting point is that $H$ is a Lie group (with assumptions??) with an injective homomorphism with closed image into $G$ and you're asking whether this is a homeomorphism onto its image. Whether this holds, depends on the assumptions... $\endgroup$– YCorCommented Dec 6, 2020 at 13:07
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$\begingroup$ Try this with $H=\mathbb{Z}$ and $G=(\mathbb{R},+)$ $\endgroup$– Liviu NicolaescuCommented Dec 6, 2020 at 20:52
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3$\begingroup$ Lee defines Lie subgroups as subgroups endowed with some topology and smooth structure that makes them immersed submanifolds. So the argument in the book has a gap, as it is implicitly assuming that H is an embedded submanifold. It appears that the "closed subgroup theorem" should have been proved before the result that you are quoting, as it is done for instance in the book of Warner, Foundations of Differentiable Manifolds and Lie groups. $\endgroup$– Ramiro LafuenteCommented Dec 9, 2020 at 23:28
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1 Answer
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As @Ramiro Lafuente pointed out in a comment, there's a gap in this proof.
You're apparently using the first edition of my Smooth Manifolds book. The problem is fixed in the second edition, because the closed subgroup theorem is proved before this theorem.
Meanwhile, some years ago I posted a correction in my errata list for the first edition, which you can find here.
