I am trying to solve the following exercise, from do Carmo's Riemannian Geometry:
Show that the real projective plane $P^2(\Bbb{R})$ is not orientable.
Hint: Show that if the manifold $M$ is orientable, then every open set of $M$ is an orientable manifold. Observe that $P^2(\Bbb{R})$ contains an open set diffeomorphic to the Möbius strip, which is not orientable.
I am trying to follow the hint, and managed to prove the first part. Now, I have the intuition that if we take a neighborhood of the equator in the sphere $S^2$, this amounts to an open set in $P^2(\Bbb{R})$. On the other hand, it also amounts to the Möbius band, which I think of, following the book, as the quotient manifold $C/G$, where $C$ is a right cylinder and $G = \{ Id, A\}$, where $A$ is the antipodal map.
In other words, I am thinking of a map
$$
U \subset P^2(\Bbb{R}) \rightarrow V \subset S^2 \rightarrow C \xrightarrow{\pi} C/G \\
[p] \mapsto p \mapsto q \mapsto[q]
$$
where $V = \{p \in S^2 \ : \ |p_3| < \varepsilon\}$ is a neighborhood of the equator and $q$ can be obtained from $p$ by
$$
p = (p_1, p_2, p_3) \mapsto \left(\frac{p_1}{\sqrt{p_1^2 + p_2^2}}, \frac{p_2}{\sqrt{p_1^2 + p_2^2}}, p_3 \right) = q
$$
However, I am not able to write this map explicitly, nor show that it is a differomorphism. I am also havin difficulty determining the domain $U \subset P^2(\Bbb{R})$.
I will apreciate any hints or solutions.
Thanks in advance and kind regards.
Best Answer
do Carmo introduces the (open) Möbius band as the quotient manifold $C/G$, where $C = S^1 \times (-1,1)$ is an open circular cylinder and $G=\{Id,A\}$, where $A$ is the antipodal map.
$P^2(\mathbb R)$ is quotient manifold $S^2/G$. Let $N =(0,0,1)$ and $S = (0,0,-1)$ denote the north and south pole of $S^2$ and $M = S^2 \setminus \{N,S\}$. Then $M/G = P^2(\mathbb R) \setminus \{P\}$, where $P$ is the equivalence class consisting of the pair of antipodal points $\{N,S\}$. Clearly $P^2(\mathbb R) \setminus \{P\}$ is an open subset of $P^2(\mathbb R)$.
Obviously the map $$\phi : M \to S^1 \times (-1,1), \phi(x,y,z) = (x/\sqrt{x^2+y^2},y/\sqrt{x^2+y^2},z)$$ (which is a special case of the map you considered in your question) is a diffeomorphism which is compatible with the action of $G$. It therefore induces a diffeomorphism
$$\phi' : P^2(\mathbb R) \setminus \{P\} \to C/G .$$ This is what you have been looking for.
PS. The inverse of $\phi$ is $$\psi : S^1 \times (-1,1) \to M, \psi(u,v,w) = (\sqrt{1-w^2}u,\sqrt{1-w^2}v,w) .$$