User:Harold/Charting RP^n

$\newcommand{\R}{\mathbb{R}} \newcommand{\RPn}{\R P^n} \newcommand{\Ztwo}{\mathbb{Z} / 2 \mathbb{Z}} \newcommand{\id}{\mathrm{Id}}$ This article contains information on possible charts for the real projective space of dimension $n$, denoted by $\RPn$.

Definition of $\RPn$

We shall first define $\RPn$. Let $S^n = \left\{ (x_0, \dotsc, x_n) \middle\vert \sum_{i = 0}^n x_i^2 = 1 \right\}$ be the $n$-sphere. Define a group action $\{-1, 1\} \cong \Ztwo$ on $S^n$ by mapping $(\epsilon, x) \mapsto \epsilon x$ with $\epsilon \in \{-1, 1\}$ and $x \in S^n$. This group action is "nice enough" so that the quotient space $S^n / \left( \Ztwo \right)$ is actually a real smooth compact Hausdorff manifold.

Construction of the charts

We now construct (the) (smooth) charts on $\RPn$. First we introduce some notation: if $x \in \RPn$, we write $x = [x_0 : \dotsc : x_n]$ if $(x_0, \dotsc, x_n)$ is a representative of the equivalence class $x$. Define the subsets $U_i \subset \RPn$ for $0 \leq i \leq n$ as

$$U_i := \{ [x_0 : \dotsc : x_n] \in \RPn \vert x_i \neq 0 \}.$$ This is well-defined, because the choice of representative only depends on a sign or a non-zero scalar multiple (if the definition of lines in $\R^{n+1}$ is chosen; see Real projective space). Now introduce maps

$$\begin{align*} \phi_i: U_i & \to \R^n \\ [x_0 : \dotsc : x_{i - 1} : 1 : x_{i+1} : \dotsc : x_n] & \mapsto (x_0, \dotsc, \widehat{x_i}, \dotsc, x_n) \end{align*}$$

where $\widehat{x_i}$ denotes that the $i$-th coordinate is omitted. These maps are well-defined, and homeomorphisms if one takes the quotient topology on $\RPn$, and actually define a smooth structure on $\RPn$, as the transition maps $\phi_j \circ \phi_i^{-1}$ are diffeomorphisms (where defined).

On the transition maps

We obtain the following transition maps:

• Let $i,j \in\{1,\ldots,n\}\subseteq\mathbb{N}$ be given.
  • In general the transition maps have the form: $(\phi_j \circ \phi_i^{-1}) : \phi_i(U_i \cap U_j) \to \phi_j(U_i \cap U_j)$.
  • We obtain the transition map by case analysis, as follows:
    1. $i\eq j$ - in this case, the transition map is the identity map $(\phi_i \circ \phi_i^{-1}): \underbrace{\phi_i(U_i \cap U_i)}_{\eq \R^n} \to \underbrace{\phi_i(U_i \cap U_i)}_{\eq \R^n}$ given by $(\phi_i \circ \phi_i^{-1}) : x \mapsto x$. Note that this is just the identity, i.e., we have $(\phi_i \circ \phi_i^{-1}) \eq \id_{\R^n}$.
    2. $i<j$ - we obtain the following map:
       • $$\phi_j \circ \phi_i^{-1} : \phi_i(U_i \cap U_j) \to \phi_j(U_i \cap U_j)$$ given by $$(\phi_j \circ \phi_i^{-1})(x_0, \dotsc, x_{i-1}, x_{i+1}, \dotsc, x_n) \eq \frac{(x_0, \dotsc, x_{i - 1}, 1, x_{i+1}, \dotsc, x_{j-1}, x_{j+1}, \dotsc, x_n)}{x_j}$$ .
    3. $i>j$ - we obtain the following map:
       • $$\phi_j \circ \phi_i^{-1} : \phi_i(U_i \cap U_j) \to \phi_j(U_i \cap U_j)$$ given by $$(\phi_j \circ \phi_i^{-1})(x_0, \dotsc, x_{i-1}, x_{i+1}, \dotsc, x_n) \eq \frac{(x_0, \dotsc, x_{j-1}, x_{j+1}, \dotsc, x_{i - 1}, 1, x_{i+1}, \dotsc, x_n)}{x_j}$$ .
  • This completes our case analysis
• Since $i,j$ were arbitrary we have shown this for all.
• These explicit expressions makes it obvious that the transition maps are smooth, and they are obviously invertible, with smooth inverse. As such, they are diffeomorphisms from $\phi_i(U_i \cap U_j) \to \phi_j(U_i \cap U_j)$.