Showing that $\mathbb{RP}^n$ is a manifold

Question

I would like to show that the real projective space $\mathbb{RP}^n$ is a manifold. This seems like a standard problem, and the explanation can be found in Lee's book on smooth manifolds as well as numerous proofs being available on this site. For the most part I understand the proof, but I am missing some intuition.

The idea, as I understand it, is to construct coordinate domains $U_i$ whose union cover our set, where $U_i = \pi(\tilde{U}_i)$, $\pi: \mathbb{R}^{n+1}\backslash\{0\} \rightarrow \mathbb{RP}^n$ is the quotient map, and each $\tilde{U_i} \subset \mathbb{R}^{n+1}$ is defined to be the set of all points such that $x_i \neq 0$.

The union of all such $U_i$ clearly cover $\mathbb{RP}^n$, so all that is left is to construct the coordinate maps $\varphi_i$ such that $\varphi_i(U_i) \subset \mathbb{R}^n$ and $\varphi_i$ is a homeomorphism. Lee defines this map as $$\varphi_i[x^1,\ldots, x^{n+1}] = \Big(\frac{x^1}{x^i}, \ldots, \frac{x^{i-1}}{x^i}, \frac{x^{i+1}}{x^i}, \ldots, \frac{x^{n+1}}{x^i}\Big)$$ and so the corresponding inverse is given by $$\varphi_i^{-1}(u_1,u_2,.....,u_n)=[u_1,u_2,.....,u_{i-1},1,u_{i+1},.....,u_n].$$

What is the motivation behind $\varphi_i$ (and its inverse), and why is the coordinate $x^i$ omitted in its image? From what I have gathered it has to do with some kind of slope of a hyperplane, but I do not see this or why we are scaling by $x^i$. Also, why is the inverse necessarily continuous? My guess is that it has to do with some property of quotient maps, but I am not sure.

As a side question, is the argument of $\varphi_i$ a single equivalence class and each $x^i$ is a component of $[x]$? If so, how come there are $n+1$ entries and not $n$?

Answer 1

It seems you wish to intuit the definition of the charts $\varphi_i:U_i\subseteq \mathbb{R}\mathbb{P}^n\to\mathbb{R}^n.$

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Consider the simple case for $n=1$. Then look at the open set: $$B_1(0)\backslash\{0\}\subseteq \mathbb{R}^2\backslash\{0\},$$ which is just the punctured open disc of radius 1. The "natural projection" of this set can be described by a portion of $S^1$ (the circle), which is known to be locally 1-dimensional. Similar to the way the charts are constructed for the circle, we may project onto an axis to get only one coordinate out. However, we already have the other coordinate at our disposal for going backwards. So, instead of using the square root function to define the second coordinate, we can just encode it into the axial-projection by division (provided it is nonzero). For example: $$[(x^1,x^2)]\mapsto \frac{1}{x^2}\cdot x^1$$ provided $x^2\neq 0$. Then, going backwards, we say $y\mapsto [(y,1)] = [x^2\cdot(y,1)] = [(x^2\cdot y,x^2\cdot1)]$, which upon composition gives the identity both ways. If each of these are continuous, we get a homeomorphism. Note as well that different representatives have the scalar cancel out, so this is well-defined.

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Note that: $$\pi(B_1(0)\backslash\{0\}) = \pi(\mathbb{R}^2\backslash\{0\})$$ and the only technicality we encountered was needing $x^2\neq 0$ to give the chart described above. Reaching into higher dimensions, this prompts the more general definitions of $(U_i,\varphi_i)$. I think this suffices for motivation.

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The notation that is used for the equivalence class of a point is: $$[x^1,...,x^{n+1}] := [(x^1,...,x^{n+1})].$$

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Lastly, we want to prove $\varphi_i$ and $\varphi_i^{-1}$ are continuous maps.

If $[x_j]\to [x]$ in $U_i$ then $\forall k\in\{1,...,n+1\}$ we have: $$\lim\limits_{j\to\infty}x_j^k = x^k.$$ Hence: $$\varphi_i([x]) = \frac{1}{x^i}(x^1,...,\hat{x^i},...,x^{n+1}) = \lim\limits_{j\to\infty}\frac{1}{x_j^i}(x_j^1,...,\hat{x_j^i},...,x_j^{n+1})=\lim\limits_{j\to\infty}\varphi_i([x_j]).$$ Likewise, if $y_j\to y$ in $\varphi_i(U_i)\subseteq \mathbb{R}^n$, we have: $$\varphi_i^{-1}(y) = [(y^1,...,1,...,y^{n})] = \lim\limits_{j\to\infty}[(y_j^1,...,1,...,y_j^{n})] = \lim\limits_{j\to\infty}\varphi_i^{-1}(y_j).$$ Thus, both $\varphi_i$ and $\varphi_i^{-1}$ preserve limits of sequences and are hence continuous. $\square$

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P.S. To prove the Hausdorff property, for $[x],[y]\in \mathbb{R}\mathbb{P}^n$, define the following neighborhoods: $$N_{[x]}:= \pi\bigg(B_r\bigg(\frac{x}{|x|}\bigg)\bigg)\text{ }\text{ }\text{ and }\text{ }\text{ }N_{[y]}:= \pi\bigg(B_r\bigg(\frac{y}{|y|}\bigg)\bigg),$$ where $r:= \frac{1}{2}d\big(\frac{x}{|x|},\frac{y}{|y|}\big)$.

And for the Second-Countable property, note that $\pi$ preserves unions and that $\mathbb{R}^{n+1}\backslash\{0\}$ satisfies the Second-Countable property. So any open set in the quotient topology pulls back to a countable union of basis elements that can then be projected into a countable union of projected basis element. [Exercise: Details!]

Answer 2

Let us first check that $\varphi_i : U_i \to \mathbb R^n$ and $\varphi_i^{-1}: \mathbb R^n \to U_i$ are continuous for $i = 1,\ldots,n+1$.

1. $\varphi_i$ is continuous.
   Define $$\tilde \varphi_i : \tilde U_i \to \mathbb R^n, \tilde \varphi_i(x^1,\ldots, x^{n+1}) = \Big(\frac{x^1}{x^i}, \ldots, \frac{x^{i-1}}{x^i}, \frac{x^{i+1}}{x^i}, \ldots, \frac{x^{n+1}}{x^i}\Big) .$$ The quotient map $p : \mathbb R^{n+1} \setminus \{0\} \to \mathbb{RP}^n$ retricts to a quotient map $p_i : \tilde U_i \to p(\tilde U_i) = U_i$ since $U_i$ is open in $\mathbb{RP}^n$ and $\tilde U_i = p^{-1}(U_i)$. We have $\varphi_i \circ p_i = \tilde \varphi_i$, thus $\varphi_i$ is continuous. Here we used the universal property of quotient maps.

2. $\varphi_i^{-1}$ is continuous.
   This is trivial since $$\psi_i : \mathbb R^n \to \mathbb R^{n+1} \setminus \{0\}, \psi_i(u_1,\ldots,u_n) = (u_1,\ldots,u_{i-1}, 1, u_i,\ldots, u_n)$$ is continuous and $\varphi_i^{-1} = p \circ \psi_i$. Note that you have a typo in your definition of $\varphi_i^{-1}$.

Let us now discuss motivation. The elements of $\mathbb{RP}^n$ are "pointed lines" $L^* = L \setminus \{0\} \subset \mathbb R^{n+1} \setminus \{0\}$, where $L$ is a line in $\mathbb R^{n+1}$ going through the origin (i.e. $L$ is a $1$-dimensional linear suspace of $\mathbb R^{n+1}$). Thus the elements of $U_i$ are precisely the pointed lines $L^*$ not intersecting $E_i$ = coordinate hyperplane $x_i =0$. To specify a map $\varphi_i^{-1}: \mathbb R^n \to U_i$ we must therefore associate to each point $u = (u_1,\ldots, u_n) \in \mathbb R^n$ a pointed line $L_u^*$ not intersecting $E_i$. This is easily done by taking for $L_u$ the line through $0$ and $(u_1,\ldots,u_{i-1}, 1, u_i,\ldots, u_n)$. Then $L_u^*$ is the equivalence class $[u_1,\ldots,u_{i-1}, 1, u_i,\ldots, u_n] \in \mathbb{RP}^n$.

Once we have found $\varphi_i^{-1}$, it is easy to determine $\varphi_i = (\varphi_i^{-1})^{-1}$. In fact, each point in $U_i$ has a unique representative of the form $[u_1,\ldots,u_{i-1}, 1, u_i,\ldots, u_n]$ with $(u_1,\ldots,u_n) \in \mathbb R^n$. Now send $[u_1,\ldots,u_{i-1}, 1, u_i,\ldots, u_n]$ to $(u_1,\ldots,u_n)$. Since for $[x^1,\ldots,x^{n+1}] \ \in U_i$ we have $[x^1,\ldots,x^{n+1}] = \Big[\frac{x^1}{x^i}, \ldots, \frac{x^{i-1}}{x^i}, 1,\frac{x^{i+1}}{x^i}, \ldots, \frac{x^{n+1}}{x^i}\Big]$, we get the above formula for $\varphi_i$.