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To demonstrate this do I just have to show that $P^2$ is Hausdorff and locally Euclidean? I can show that the space is Hausdorff but I'm having a little trouble demonstrating that it is locally Euclidean.

user62931
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1 Answers1

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The real projective plane, $\mathbb{R}P^2$, is the space of lines through the origin in $\mathbb{R}^3$, so it's natural to use homogeneous coordinates, where $(x, y, z) \sim (tx, ty, tz)$ for any $t \in \mathbb{R} \setminus \{0\}$.

We can manage to cover the projective plane with three charts, one for each coordinate in $\mathbb{R}^3$. Here's one of them. Let $U_x \subset \mathbb{R}P^2$ denote the subset of points in $\mathbb{R}^3$ with nonzero $x$-coordinate.

$$U_x = \{ (x, y, z) \in \mathbb{R}^3 \; | \; x \ne 0\}$$

Considering these as homogeneous coordinates, though, $$(x, y, z) \sim \left(1, \frac{y}{x}, \frac{z}{x}\right).$$

Geometrically, this projects any line that pierces the plane $x = 1$ onto its intersection point with that plane. Voilà, this is our first chart.

$$\begin{align} \psi_x: U_x &\to \mathbb{R}^2 \\ (x,y,z) &\mapsto \left(\frac{y}{x}, \frac{z}{x}\right) \end{align}$$

Notice that the inverse, $\psi_x^{-1}: \mathbb{R}^2 \to U_x$ is given by $(u, v) \mapsto (1, u, v)$.

In a strictly analogous fashion, we can construct charts for open sets $U_y$ and $U_z$. To stitch these charts together into an atlas, we have to look at the various compositions that arise where the charts overlap, mapping open subsets of Euclidean space into itself.

$$\begin{array}{*{5}{c}} \psi_x(U_x \cap U_y) &\to& U_x \cap U_y &\to& \psi_y(U_x \cap U_y) \\ (u, v) &\mapsto& (1, u, v) &\mapsto& \left( \frac{1}{u}, \frac{v}{u}\right) \end{array}$$

This composition is a diffeomorphism, which is to say that its smooth and so is its inverse. This is what makes $\mathbb{R}P^2$ into a smooth manifold. (It may be that you're only interested in the projective plane as a topological manifold, in which case you are only interested in these maps being homeomorphisms.)

Sammy Black
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