# Using the quotient manifold theorem to show real projective spaces are smooth manifolds

The goal of this post is, as stated in the title,

Use the quotient manifold theorem to show real projective spaces are smooth manifolds.

I know this is an overkill, but I am just curious about this theorem so that I want to use it on some examples. Btw, I just learnt that we can form products of manifolds quite naturally so that I am curious under what conditions we can do the same thing for quotients. Then that brought me to the post and hence Thm 7.10 in Lee's book (well, I didn't know there was such a good book before lol):

(Quotient manifold theorem) If $$G$$ is a Lie group acting smoothly, properly and freely on a smooth manifold $$M$$, then $$M/G$$ is a topological manifold with dimension $$\dim M - \dim G$$ and has a unique smooth structure such that the canonical surjection is a smooth submersion.

I went through the relevant definitions and I have shown the following (these are relatively easy):

1. The multiplicative group $$\Bbb{R}^\times$$ is a Lie group.
2. The map $$\phi = (\lambda ,x )\mapsto \lambda x$$ is a group action by $$\Bbb{R}^\times$$ on $$\Bbb{R}^n \backslash \{0\}$$.
3. The action is smooth as a map from the manifold $$\Bbb{R}^\times \times \Bbb{R}^n \backslash \{0\}$$ to $$\Bbb{R}^n \backslash \{0\}$$.
4. The map $$f_\phi = (\lambda, x) \mapsto (\lambda x, x)$$ is injective, hence the action $$\phi$$ is free.

Then it remains to show the group action is proper (i.e if $$K \subseteq M \times M$$ is compact, $$f_\phi^{-1}(K) \subseteq G \times M$$ is compact). It has been shown in Lemma 7.1 that

(Criterion of properness) A continuous group action $$\phi$$ by $$G$$ on Hausdorff space $$M$$ is proper iff for all compact subset $$K \subseteq M$$, $$G_K = \{g \in G: gk \in K \text{ for some } k \in K \} = \pi_G( f_\phi^{-1}(K \times K))$$ is a compact subset of $$G$$.

Attempt: Let $$K$$ be compact subset of $$\Bbb{R}^n \backslash \{0\}$$. Although $$\Bbb{R}^\times$$ is no longer complete, it still has the property that bounded $$\Rightarrow$$ totally bounded. So it suffices to show $$G_K$$ is bounded and complete.

Bounded: The norm function $$f(x) = \|x\|$$ is real valued and continuous on $$\Bbb{R}^n\backslash \{0\}$$, so $$f(K)$$ is compact and it has minimum $$a$$ and maximum $$b$$. Then $$G_K$$ is bounded subset of $$\Bbb{R}^\times$$ since for each $$g \in G_K$$, there are $$k_1, k_2 \in K$$ such that $$gk_1 = k_2 \Rightarrow |g|\|k_1\| = \|k_2\| \Rightarrow |g| \leq \|k_1\|/\|k_2\| \leq b/a$$.

Complete: Let $$\{g_n\}$$ be a Cauchy sequence in $$G_K$$. Then for each $$n$$, there is $$x_n$$ such that $$g_n x_n \in K$$. Since $$K$$ is compact, there is some subsequence $$\{x_{n_k}\}$$ and $$x \in K$$ such that $$x_{n_k} \to x$$. $$g_{n_k} x_{n_k}$$ is again a Cauchy sequence in $$K$$ so by completeness, $$g_{n_k} x_{n_k} \to y$$ for some $$y \in K$$. It remains to show that $$y = g x$$ for some $$g \neq 0$$ and $$g_{n_k} \to g$$, then the proof is complete. I am still was thinking of a way of showing that...

Edit: so the specific questions are

1. Since the proof was finished below, can you give another (shorter/more elegant/more direct) proof to show that the group action $$(\lambda , x) \mapsto \lambda x$$ is proper?
2. Is there any problem in this proof?
• In the statement of the theorem you wrote "canonical injection" whereas you probably meant "canonical projection" (you can inject a subspace into a superspace, here you project a space onto its quotient). Mar 2, 2019 at 8:23
• ah surjection lol Mar 2, 2019 at 8:31
• The version of my book that you seem to be referring to is an illegally pirated early draft of the first edition, which somebody posted on the internet without permission. It’s full of errors, and I don’t recommend that anyone use it to try to learn about smooth manifolds. You'll have a much easier time working with the published second edition. Mar 2, 2019 at 19:34
• Do you spot that out by the theorem number? :P Mar 3, 2019 at 1:42

So I realize that the remaining part of the proof can be finished with linear algebra.

(There is $$g \neq 0$$ such that $$y = gx$$) For each $$k$$, write $$y_k = g_{n_k} x_{n_k}$$ and let $$e_k = y_k - \text{proj}_x y_k$$. So the task is to show $$e_k \to 0$$. Since $$y_k \to y$$, that shows $$y = \text{proj}_x y$$ and the desired $$g$$ would be $$\frac{x \cdot y}{x \cdot x}$$. Now

\begin{align*}e_k & = y_k - \frac{y_k\cdot x}{x \cdot x} x = \frac{g_{n_k}x_{n_k}(x \cdot x)-(g_{n_k} x_{n_k} \cdot x) x}{x \cdot x} \\ & = g_{n_k}\frac{(x \cdot x)x_{n_k}-(x_{n_k} \cdot x) x}{x \cdot x} \end{align*}

But $$\{g_{n_k}\}$$ is bounded so it suffices to show $$\|(x \cdot x) x_{n_k} - (x_{n_k} \cdot x) x \| \to 0$$. \begin{align*} \|x_{n_k} (x \cdot x) - (x_{n_k} \cdot x) x\| &= \|x_{n_k} (x \cdot x) - (x\cdot x) x + (x \cdot x) x - (x_{n_k} \cdot x) x\|\\ & = \|(x \cdot x)( x_{n_k} - x ) + ((x -x_{n_k}) \cdot x) x\|\\ & \leq \|x\|^2 \| x_{n_k} - x \| + |(x -x_{n_k}) \cdot x| \|x\|\\ & \leq \|x\|^2 \| x_{n_k} - x \| + \|x -x_{n_k}\| \|x\| \|x\|\\ & = 2\|x\|^2 \| x_{n_k} - x \|\to 0 \end{align*} , where we used Cauchy Schwartz in the second inequality.

($$g_{n_k} \to g$$) \begin{align*} g_{n_k} - g & = \frac{g_{n_k} (x \cdot x) - y \cdot x}{x \cdot x} \end{align*} Again, we show $$|g_{n_k} (x \cdot x) - y \cdot x| \to 0$$. \begin{align*} |g_{n_k} (x \cdot x) - y \cdot x| & = |g_{n_k} (x \cdot x) - y_k \cdot x + y_k \cdot x - y \cdot x|\\ & = |(g_{n_k}x - g_{n_k} x_{n_k}) \cdot x + (y_k - y) \cdot x|\\ & \leq \underbrace{|g_{n_k}|}_{{\text{bounded}}}\|x - x_{n_k}\|\|x\|+ \|y_k - y \|\|x\| \to 0 \end{align*} So $$g_{n_k} \to g$$. Since $$\{g_n\}$$ is Cauchy, the whole sequence will converge to $$g$$. Recall that $$x,y = gx \in K$$, so that $$g \in G_K$$. So that the Cauchy sequence $$\{g_n\}$$ has limit $$g$$ in $$G_K$$ and we can conclude $$G_K$$ is complete.

(Another answer) This shows $$G_K$$ is complete with a different approach: Observe that

$$A \subseteq \Bbb{R}\backslash \{0\}$$ is complete iff $$A$$ is closed and $$d(0,A) > 0$$.

($$d(0,A) > 0$$) As in the original proof, let $$a = \min_{x \in K} \|x\|$$ and $$b = \max_{x\in K} \|x\|$$. For all $$g \in G_K$$, there are $$k_1, k_2 \in K$$ such that $$gk_1 = k_2$$. Then $$|g| = \|k_2\|/|k_1| \geq a /b$$. So $$d(0,G_K) \geq a/b > 0$$.

($$A$$ is closed) Notice that $$G_K = \{g \in G: gK \cap K \neq \emptyset\}$$. We now show $$\Bbb{R}^\times \backslash G_K$$ is open: Let $$g \neq 0$$ such that $$g K \cap K = \emptyset$$. But $$x \mapsto gx$$ is continuous, so $$gK$$ is compact. In metric spaces, disjoint closed sets and compact sets maintain a finite distance with each other and compact sets are closed. Say $$d(gK, K) = \epsilon > 0$$, take $$\delta = \epsilon/b$$. Then $$B_\delta(g) \subseteq \Bbb{R}^\times \backslash G_K$$.

Okay, the second answer already reduced most of the length in the original proof, but what I really want is the argument presented here by Mr. Kato, which I come across just now. TLDR, he proved this proposition:

Let $$G$$ be a compact Hausdorff group and $$M$$ be a Hausdorff space. Suppose $$G$$ acts continuously on $$M$$. Then $$G$$ acts properly on $$M$$ (and this will imply that the orbit space $$M/G$$ is Hausdorff).

His argument is nice in a sense that it can be used for any Hausdorff $$M$$ and compact Hausdorff group $$G$$. To use his argument, we shall collaspe our $$M_\text{old} = \Bbb{R}^{n+1}\backslash \{0\}$$ and $$G_\text{old} = \Bbb{R}\backslash \{0\}$$ to $$M = \Bbb{S}^n$$ and $$G = \{-1,1\}$$. This third version of answer will be completed if I can rigorously prove that.

PS. Silly me…isn't this written in the book as a lemma, right after stated the properness criterion? Lol

Notice that each element $$g \in \mathbb R^{\times}$$ acts as a homeomorphism on the pointed Euclidean space. Therefore for any compact $$K \subseteq R^n \setminus \{ 0 \}$$ we have that $$g^{-1}(K)$$ is the image of the compact set $$K$$ through a continuous function $$g^{-1}$$. Hence $$g^{-1}(K)$$ is compact for any compact $$K$$, i.e. $$g$$ is a proper map.

Oops, the above is correct but not really what OP asked for. Condition of properness in this context means that the map $$G \times M \to M \times M$$ is proper, whereas I proved that after fixing an element $$g \in G$$ the corresponding map $$M \to M$$ is proper. I will try to correct this answer later.