## (1)

The real projective plane \(\mathbb{R}P^2\) is defined the be the set of all lines in \(\mathbb{R}^3\) that pass through the origin. We must show that this forms a 2-dimensional manifold. We can cover this set with three subsets:

$$\begin{align}U_x &:= \text{all lines which do not lie on the yz-plane}, \\

U_ y &:= \text{all lines which do not lie on the xz-plane}, \\

U_z &:= \text{all lines which do not lie on the xy-plane}.\end{align}$$

For each of these we can define a coordinate system by associating each line is the set with a point in the plane. That is, a map \(\phi_i(U_i) : \mathbb{R}P^2 \to \mathbb{R}^2, \text { for } i = x,y,z\). To do this for \(U_z\), chose any point \((x,y,z)\) other than the origin on the line and construct

$$\phi_z(U_z) = \left(\frac{x}{z},\frac{y}{z}\right).$$

So a line in \(U_z\) gets mapped to a point in \(\mathbb{R}^2\) which is the intersection of that line and the \(z=1\) plane. The other coordinate patches are defined similarity.

To show that \(\mathbb{R}P^2\) forms a manifold, we must show that for any line existing in more than one coordinate patch, the coordinates in one patch can be differentiably expressed in terms of the coordinates in another patch.

$$\phi_x(\phi_z^{-1}(a,b)) = \phi_x(x,y,1) = \left(\frac{y}{x},\frac{1}{x}\right).$$

This is \(C^\infty\) over \(U_z \cap U_x\) (\(x \neq 0\) in \(U_x\)). Likewise for all other patch relationships. Therefore, the real projective plane is a 2-dimensional manifold.

## (2)

The real projective space \(\mathbb{R}P^3\) is a 3-dimensional manifold. A point \(p = (x^1,x^2,x^3,x^4)\) is identified with \((\lambda x^1, \lambda x^2, \lambda x^3,\lambda x^4)\) for all \(\lambda \neq 0 \in \mathbb{R}\). Following the example of \(\mathbb{R}P^2\), we cover \(\mathbb{R}P^3\) with the 4 coordinate patches

$$U_i = \{(\lambda x^1, \lambda x^2, \lambda x^3,\lambda x^4) \, | \, x^i \neq 0\}, \,\,\, i = 1,2,3,4.$$

Then the coordinate maps \(\phi_i : U_i \to \mathbb{R}^3\) are clear, e.g.

$$\phi_1(U_1) = \left(\frac{x^2}{x^1}, \frac{x^3}{x^1}, \frac{x^4}{x^1}\right).$$

Thus,

$$\phi_2(\phi_1^{-1}(a,b,c,d)) = \phi_2(1,x^2,x^3,x^4) = \left(\frac{1}{x^2}, \frac{x^3}{x^2}, \frac{x^4}{x^2}\right),$$

which is \(C^\infty\). Hence \(\mathbb{R}P^3\) is a 3-dimensional manifold.