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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.

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    $\begingroup$ What is your definition of $\mathbb{RP}^n$? $\endgroup$ – Peter Franek Dec 30 '14 at 15:39
  • $\begingroup$ We defined $\mathbb{R}P^n$ to be $(\mathbb{R}^{n+1} \setminus \{0\})/\sim$ with $x\sim \lambda x$ for all $\lambda\neq 0$ $\endgroup$ – iwriteonbananas Dec 30 '14 at 15:42
  • $\begingroup$ Hint: Examine the set $\{x_{i} = 1\} \subset U_{i}$. $\endgroup$ – Andrew D. Hwang Dec 30 '14 at 16:03
  • $\begingroup$ Boothby's An Introduction to Differentiable Manifolds and Riemannian Geometry-ChIII.2 pretty much says it all. $\endgroup$ – Troy Woo Dec 30 '14 at 16:26
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For the Hausdorff property,

enter image description here

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.

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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.

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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 $\hat{\pi}^{-1}(\hat{\pi}(U))$).

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