# Frankel 1.2 Problems

## (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.