# $\mathbb{R}P^n$ is an $n$-manifold: how to show locally Euclidean and Hausdorff properties?

Show that the real projective space $\mathbb{R}P^n$ is an $n$-manifold.

We need to show that $\mathbb{R}P^n$ is second countable, locally Euclidean and Hausdorff.

Second countability simply follows from second countability of $\mathbb{R}^{n+1} \setminus \{0\}$.

To prove the locally Euclidean property, I follow a hint and consider the sets $U_i = \{ (x_0,...,x_n) \in \mathbb{R}^{n+1} : x_i \neq 0 \}$. Then we can construct the maps $$\varphi_i : U_i \to \mathbb{R}^n, (x_0,...,x_n) \mapsto \left(\frac{x_0}{x_i},...,\frac{x_{i-1}}{x_i}, \frac{x_{i+1}}{x_i},...,\frac{x_n}{x_i}\right).$$ I don't know how to proceed now. What can we use these maps for? Are they homeomorphisms?

I also haven't been able to prove the Hausdorff property.

Thank you for every hint.

• What is your definition of $\mathbb{RP}^n$? Dec 30, 2014 at 15:39
• We defined $\mathbb{R}P^n$ to be $(\mathbb{R}^{n+1} \setminus \{0\})/\sim$ with $x\sim \lambda x$ for all $\lambda\neq 0$ Dec 30, 2014 at 15:42
• Hint: Examine the set $\{x_{i} = 1\} \subset U_{i}$. Dec 30, 2014 at 16:03
• Boothby's An Introduction to Differentiable Manifolds and Riemannian Geometry-ChIII.2 pretty much says it all. Dec 30, 2014 at 16:26

For the Hausdorff property,

For your first question note that the images of the $U_i$ in the quotient are open sets, and that these form an open cover. Your maps $\varphi_i$ pass to the quotient and induce homeomorphisms onto their images (write down an inverse map!). If you need more details, please tell me so.

By definition, $$\mathbb{RP}^n = \left(\mathbb{R}^{n+1} \setminus \{0\}\right) \big/ \sim\, ,$$ where $$x \sim y \iff \exists\ \lambda \in \mathbb{R} \setminus \{ 0 \} \text{ such that } x = \lambda y.$$ The topology on $$\mathbb{RP}^n$$ is, by definition, the quotient topology induced by the canonical projection \begin{align} \pi &: \mathbb{R}^{n+1} \setminus \{ 0 \} \to \mathbb{RP}^n\\ &: (x_0,\dots,x_n) \mapsto [x_0,\dots,x_n] \end{align} where $$[x_0,\dots,x_n] \in\mathbb{RP}^n$$ denotes the equivalence class of $$(x_0,\dots,x_n) \in \mathbb{R}^{n+1} \setminus \{0 \}$$. This makes $$\pi$$ a quotient map.

To show that $$\mathbb{RP}^n$$ is locally Euclidean, we need to exhibit a cover for $$\mathbb{RP}^n$$ by coordinate charts. For each $$0 \leq i \leq n$$, define $$U_i \subset \mathbb{R}^{n+1} \setminus \{ 0 \}$$ by $$U_i = \left\{ (x_0,\dots,x_n) \in \mathbb{R}^{n+1} \setminus \{ 0 \} : x_i \neq 0 \right\}.$$ One can check that $$U_i$$ is an open subset of $$\mathbb{R}^{n+1} \setminus \{ 0 \}$$. Define $$V_i \subset \mathbb{RP}^n$$ to be $$\pi(U_i)$$. Then, one can check that $$V_i$$ is an open subset of $$\mathbb{RP}^n$$ and $$\pi_i = \pi |_{U_i} : U_i \to V_i$$ is also a quotient map. The sets $$V_i$$, $$0 \leq i \leq n$$, form an open cover of $$\mathbb{RP}^n$$.

We show that each $$V_i$$ is homeomorphic to $$\mathbb{R}^n$$ as follows. For each $$0 \leq i \leq n$$, define the map $$\psi_i : V_i \to \mathbb{R}^n$$ by $$\psi_i[x_0,\dots,x_n] = \left( \frac{x_0}{x_i},\dots,\frac{x_{i-1}}{x_i},\frac{x_{i+1}}{x_i},\dots,\frac{x_n}{x_i} \right).$$

Continuity of $$\psi_i$$:
The map $$\varphi_i = \psi_i \circ \pi_i : U_i \to \mathbb{R}^n$$ is given by $$\varphi_i(x_0,\dots,x_n) = \left( \frac{x_0}{x_i},\dots,\frac{x_{i-1}}{x_i},\frac{x_{i+1}}{x_i},\dots,\frac{x_n}{x_i} \right).$$ Since $$\varphi_i$$ is continuous, by the characteristic property of quotient maps $$\psi_i$$ is also continuous.

Bijectivity of $$\psi_i$$:
Note that $$\psi_i$$ is surjective because for every $$(u_1,\dots,u_n) \in \mathbb{R}^n$$, $$[u_1,\dots,u_i,1,u_{i+1},\dots,u_n] \in V_i$$ and $$\psi_i([u_1,\dots,u_i,1,u_{i+1},\dots,u_n]) = (u_1,\dots,u_n)$$. Note that every element in $$V_i$$ has a unique representative whose $$i$$th coordinate equals $$1$$. This fact easily implies that $$\psi_i$$ is injective.

Continuity of $$\psi_i^{-1}$$:
For each $$0 \leq i \leq n$$, consider the map $$\theta_i : \mathbb{R}^n \to \mathbb{R}^{n+1} \setminus \{ 0 \}$$ given by $$\theta_i(u_1,\dots,u_n) = (u_1,\dots,u_i,1,u_{i+1},\dots,u_n).$$ Then, $$\theta_i$$ is continuous and its image is contained in $$V_i$$. One now checks that $$\pi_i \circ \theta_i = \psi_i^{-1}$$. So, $$\psi_i^{-1}$$ is continuous.

Hence, $$\psi_i$$ is a homeomorphism for each $$0 \leq i \leq n$$.

To show that $$\mathbb{RP}^n$$ is Hausdorff, choose $$\tilde{x}$$ and $$\tilde{y}$$, two distinct points in $$\mathbb{RP}^n$$.

If there exists $$0 \leq i \leq n$$ such that both points lie in $$V_i$$, then $$\psi_i(\tilde{x})$$ and $$\psi_i(\tilde{y})$$ are two distinct points in $$\mathbb{R}^n$$. Since $$\mathbb{R}^n$$ is Hausdorff, there exists a pair of disjoint open sets $$A$$ and $$B$$ with $$\psi_i(\tilde{x}) \in A$$ and $$\psi_i(\tilde{y}) \in B$$. Hence, $$\psi_i^{-1}(A)$$ and $$\psi_i^{-1}(B)$$ are disjoint open subsets of $$V_i$$ (and hence of $$\mathbb{RP}^n$$) such that $$\tilde{x} \in \psi_i^{-1}(A)$$ and $$\tilde{y} \in \psi_i^{-1}(B)$$.

On the other hand, suppose there is no $$i$$, $$0 \leq i \leq n$$, such that $$\tilde{x}$$ and $$\tilde{y}$$ both lie in $$V_i$$. Let $$(x_0,\dots,x_n)$$ and $$(y_0,\dots,y_n)$$ be representatives of $$\tilde{x}$$ and $$\tilde{y}$$, respectively. There exists $$i \neq j$$, $$0 \leq i,j \leq n$$, such that \begin{align} &x_i \neq 0, y_i = 0, \quad \text{ and}\\ &x_j = 0, y_j \neq 0. \end{align} Fix the representatives so that $$x_i = 1 = y_j$$. WLOG, let $$i < j$$. Choose $$0 < \epsilon < 1$$. The sets \begin{align} A &= \{ [a_0,\dots,a_{i-1},1,a_{i+1},\dots,a_n] : |a_k - x_k| < \epsilon\ \forall\ k \neq i \} \subset V_i, \quad \text{ and}\\ B &= \{ [b_0,\dots,b_{j-1},1,b_{j+1},\dots,b_n] : |b_k - y_k| < \epsilon\ \forall\ k \neq j \} \subset V_j \end{align} are open sets containing $$\tilde{x}$$ and $$\tilde{y}$$, respectively. This is because $$\psi_i(A)$$ is an open rectangle in $$\mathbb{R}^n$$ centered on $$\psi_i(\tilde{x})$$ having side length $$2 \epsilon$$, and similarly $$\psi_j(B)$$ is an open rectangle in $$\mathbb{R}^n$$ centered on $$\psi_j(\tilde{y})$$ having side length $$2 \epsilon$$. They are disjoint because if $$[a_0,\dots,a_{i-1},1,a_{i+1},\dots,a_n] = [b_0,\dots,b_{j-1},1,b_{j+1},\dots,b_n]$$, then we must have $$a_j \neq 0$$, $$b_i \neq 0$$, and $$a_j b_i = 1$$. But, $$|a_j| < 1$$ and $$|b_i| < 1$$, so this is not possible.

Hence, $$\mathbb{RP}^n$$ is Hausdorff, and so $$\mathbb{RP}^n$$ is an $$n$$-manifold.

Let $\pi : \mathbb{R}^{n+1} \backslash \{ 0 \} \to \mathbb{R}P^n$ denote the quotient map. In my answer to Real projective space is Hausdorff: is this proof correct? you will find a proof that

a) $\mathbb{R}P^n$ is Hausdorff.

b) The restriction $\hat{\pi} = \pi \mid_{S^n} : S^n \to \mathbb{R}P^n$ is also a quotient map (where $S^n \subset \mathbb{R}^{n+1}$ denotes the standard $n$-sphere).

I guess you know that that $S^n$ is an $n$-manifold. To show that also $\mathbb{R}P^n$ is one, it therefore suffices to show that each sufficiently small $U \subset S^n$ is mapped by $\hat{\pi}$ homeomorphically onto an open $V \subset \mathbb{R}P^n$.

Let $x \in S^n$ and $\varepsilon > 0$. Then $U = \{ y \in S^n \mid \lVert y - x \rVert < \varepsilon \}$ is open in $S^n$ and for sufficiently small $\varepsilon$ we see that $\hat{\pi}$ maps the closure $\overline{U}$ bijectively onto $\hat{\pi}(\overline{U})$. But $\overline{U}$ is compact so that this mapping is a homeomorphism. Hence also $U$ is mapped homeomorphically onto $\hat{\pi}(U)$. Clearly $\hat{\pi}(U)$ is open in $\mathbb{R}P^n$ (consider $\hat{\pi}^{-1}(\hat{\pi}(U))$).