Real Projective Space has homeomorphic atlas

Question

I am trying to show that $\mathbb{RP}^{n}$ is a smooth manifold with atlas $({U_{i},\phi_{i}})$ where $\phi_{i}:U_{i}\to \mathbb{R}^{n}$ is defined by $[x_{0},\dots,x_{n}]\mapsto (x_{0}/x_{i},\dots,x_{n}/x_{i})$. I know that in the quotient topology to show a set $V$ is open, we need $\pi^{-1}(V)$ to be open, where $\pi$ maps $x$ to $[x]$ the equivalence class. Ultimately what this boils down to is showing that $$ \pi^{-1}(\phi^{-1}(V)) = \{x\in \mathbb{R}^{n+1} \mid \phi\circ\pi (x)\in V\} $$ is open however I have no idea how to go about that. Any help would be appreciated!

Answer 1

I don't know about your choice of chart since $U_i$ is not defined and I'm not exactly sure what $[x_0,\cdots,x_n]$ represents; a representative of an equivalence class, viewing this space as a quotient of $S^n$? Or of $\Bbb R^{n+1}\setminus\{0\}$?

You know $S^n$ is a manifold (stereographic projection) and you know that $\Bbb RP^n$ may be defined up to homeomorphism as $S^n$ quotient the antipodal relation.

Denote by $q$ the defining quotient map $S^n\twoheadrightarrow\Bbb RP^n$. Take any point $x\in\Bbb RP^n$ and choose a representative $y\in q^{-1}\{x\}$. Find a chart $(U;\varphi)$ at $y$ using the standard smooth structure on the sphere. Without loss of generality, we may choose $U$ so small that $q$ is injective on $U$. $V:=q(U)$ is open in $\Bbb RP^n$ (and a neighbourhood of $x$), realising that $q^{-1}(V)=U\sqcup(-U)$ is evidently open in $S^n$. In fact $q$ is an open map in general: this should address your main question. $q^{-1}(q(W))=W\cup(-W)$ is open for open $W$ hence $q(W)$ is open in the quotient space (by definition of quotient topology).

I claim that $q\circ\varphi^{-1}:\Bbb R^n\to U\to V$ is a homeomorphism. Why? Well, $q$ is an open map and is injective on $U$ hence a homeomorphism $U\cong V$. So, we have a chart! It follows $\Bbb RP^n$ is a topological manifold. To check smoothness (with respect to these charts) is not so hard. The transition maps are smooth because they are either of form $\psi\circ\varphi^{-1}$ (restricted to suitable sub-co/domains) or are of form $\psi\circ(-\varphi^{-1})$ where $\psi,\varphi$ come from the known-to-be-smooth structure of $S^n$. To be really pedantic, you can check $x\mapsto -x$ is smooth in the differential-manifold sense, going $S^n\to S^n$, by an easy explicit calculation with the stereographic charts. Or maybe there is some better method: I'm not experienced with differential topology.

Oh, and the second-countable criterion holds because $q$ is open here.

This argument was quite easy and I suspect a general result about "nice" quotients of compact smooth manifolds will apply. I just can't think of the right hypotheses yet. A key point was that $q$ identifies points in a smooth way e.g. $-W$ is also open if $W$ is, and $-f$ is also a smooth function if $f$ is.

This proof strategy should apply for your chart once you have a clear definition sorted out.

Answer 2

It seems that you want to have a proof based on the definition $$\mathbb{RP}^n = (\mathbb R^{n+1} \setminus \{0\})/ \sim$$ where $(x_0,\ldots,x_n) \sim (y_0,\ldots,y_n)$ iff there exists $\lambda \in \mathbb R$ such that $(x_0,\ldots,x_n) = \lambda(y_0,\ldots,y_n)$. Note that the number $\lambda$ occurring here cannot be $0$.

The equivalence class of $x = (x_0,\ldots,x_n)$ is denoted by $[x] = [x_0,\ldots,x_n)]$ and the quotient map by $\pi : \mathbb R^{n+1} \setminus \{0\} \to \mathbb{RP}^n$. We have $$[x] = \{\lambda x \mid \lambda \in \mathbb R \setminus \{0\} \}.$$

Your sets $U_i$ are defined by $$U_i = \pi(U'_i)$$ where $U'_i = \{(x_0,\ldots,x_n) \mid x_i \ne 0\}$ which is open in $\mathbb R^{n+1} \setminus \{0\}$. The $U'_i$ cover $\mathbb R^{n+1} \setminus \{0\}$, thus the $U_i$ cover $\mathbb{RP}^n$.

Let us prove that $\pi$ is an open map; this will show that the $U_i$ are open.

So let $V \subset \mathbb R^{n+1} \setminus \{0\}$ be open. We have to show that $\pi^{-1}(\pi(V))$ is open. We have $$\pi^{-1}(\pi(V)) = \{ \lambda x \mid x \in V, \lambda \in \mathbb R \setminus \{0\} \} = \bigcup_{\lambda \in \mathbb R \setminus \{0\} } \lambda V \tag{1}$$ where $\lambda V = \{ \lambda x \mid x \in V \}$. The multiplication maps $\mu_\lambda : \mathbb R^{n+1} \setminus \{0\} \to \mathbb R^{n+1} \setminus \{0\}, \mu_\lambda(x) = \lambda x$, are homeomorphisms, thus the $\lambda V = \mu_\lambda(V)$ are open. By $(1)$ $\pi^{-1}(\pi(V))$ is open.

The map $$\phi'_i : U'_i \to \mathbb R^n,\phi'_i(x_0,\ldots, x_n) = \left(\frac{x_0}{x_i}, \ldots, \frac{x_{i-1}}{x_i},\frac{x_{i+1}}{x_i}, \ldots \frac{x_n}{x_i}\right) $$ is compatible with $\sim$ and thus induces a unique function $\phi_i : U_i \to \mathbb R^n$ such that $\phi_i \circ \pi_i = \phi'_i$ where $\pi_i : U'_i \to U_i$ is the restriction of $\pi$. Clearly $\pi_i$ is an open surjection (since $\pi$ is an open map and $U'_i$ is open), thus it is a quotient map. Hence $\phi_i'$ is continuous.

The map $$\psi'_i : \mathbb R^n \to U'_i , \psi_i(w_1,\ldots w_n) = (w_1, \ldots, w_i,1,w_{i+1},\ldots, w_n)$$ is continuous and so is $\psi_i = \pi_i \circ \psi'_i : \mathbb R^n \to U_i$. It is easy to see that $\psi_i \circ \phi_i = id$ and $\phi_i \circ \psi_i = id$, thus $\phi_i$ is a homeomorphism.

This shows that $\mathbb {RP}^n$ is locally Euclidean.

Since $\mathbb {RP}^n$ is covered by the finitely many open $U_i$ all of which are second countable, it is itself second countable.