Why is the tangent space to a real projective plane two dimensional?

Let P be the projective plane obtained by identifying antipode points on the unit sphere.

How to prove that the tangent space at $q \in P$ to the projective plane P is 2 dimensional?

My questions are

1, P is not a submanifold of the Euclidean space and its tangent vectors are defined in terms of equivalence classes. How to show that there exist two linearly independent tangent vectors?

2 I hope someone can give me detailed, elementary proof without using more advanced facts--I am just starting out. Thanks

• Is the statement intuitively clear to you? If yes, ignore my comment. If not, then maybe try to look at it this way. Your question is local (as you're asking about the tangent space at a point), and locally, the projective plane "looks like" (i.e. is homeomorphic to an open subset of) $\Bbb R^2$. Now non-empty open subsets of $\Bbb R^2$ have two dimensional tangent space ("on a small open disk, you can go into two directions"), hence the same should hold for the projective plane. – Nils Matthes Oct 13 '13 at 13:04

3 Answers

Although Nils's comment is correct and general, I'll give you a very explicit description of the tangent space $T_q\mathbb P^2$ to the projective plane $\mathbb P^2$ at $q$.
The projective plane consists of pairs $q=\{a,-a\}$ of antipodal points $a,-a\in S^2 \; (a\cdot a=1)$, and a tangent vector $V\in T_q\mathbb P^2$ consists of pairs of couples $V=\{(a,v),(-a,-v)\}$with $v\in \mathbb R^3$ orthogonal to $a\in S^2$ : $\: a\cdot v=0$ .

A basis of $T_q\mathbb P^2$ consists in two linearly independant tangent vectors $V, V'\in T_q\mathbb P^2$ described as $V=\{(a,v),(-a,-v)\}, V'=\{(a,v'),(-a,-v')\}$ with $v,v'\in \mathbb R^3$ linearly independant but both orthogonal to $a$.
This makes it clear that the dimension of $T_q\mathbb P^2$ is $2$.

Reality check
Can you compute $2V-3V'\in T_q\mathbb P^2$ ?

• +1 Georges, I'd never such a concrete interpretation before! – user38268 Oct 13 '13 at 13:42
• Dear user38268, your sense of fair play is much more praiseworthy than all the descriptions of all the tangent spaces in the world: you have my genuine admiration for that. I learned that description from the book Characteristic Classes by Milnor and Stasheff. – Georges Elencwajg Oct 13 '13 at 13:56
• Dear @Georges, I posted a recent question here. Would you be able to help answer it? Thanks. – user38268 Oct 13 '13 at 14:11

The real projective plane can be identified with tuples $(a_0,a_1,a_3) \in \Bbb{R}^3$ modulo the equivalence $(a_0,a_1,a_2) \simeq (ka_0,ka_1,ka_2)$ with $k \neq 0$ and $a_0,a_1,a_2$ not all simultaneously zero. Think about why this is the same as identifying antipodal points on $S^2$. With these, we see immediately that $\Bbb{R}P^2$ can be covered by charts $U_1,U_2,U_3$ where $U_i$ is points in $\Bbb{R}P^2$ such that the $i$-th coordinate is not zero. It is clear that $U_i$ is homeomorphic to $\Bbb{R}^2$ and so we see that each point has a neighbourhood that is homeomorphic to $\Bbb{R}^2$. So the dimension of the tangent space is $2$.

You have a map $p:S^2\longrightarrow \mathbb{R}P^2$ that identifies antipodal points. Given an element $x\in S^2$, there is an open set small enough so that the restriction of $p$ to $U$ is a diffeommorphism. This means that the induced map on tangent spaces is bijective. And $S^2$ certainly has a two-dimensional tangent space.