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
Best Answer
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 $2V-3V'\in T_q\mathbb P^2$ ?