# velocity vector on the projective plane

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

Let $\alpha$ be a curve in P.

The book I am reading says that $\alpha'(t)$ is the function such that

$\alpha'(t)[f]=\frac{d}{dt}f(\alpha(t))$

for every differentiable real valued function $f$ on P.

Let $F$ be the projection of the unit sphere to P and $F*$ the tangent map.

How to prove that two distinct tangent vectors to the unit sphere with the same point of application result in distinct tangent vectors to the projective plane with the same point of application under the tangent map? In other words, how to prove

If $v\neq w$, $v ,w$ are both tangent vectors to the unit sphere at point say $q$. Then $F*(v)\neq F*(w)$? How to write detailed proof?

• How to prove that F is surjective so $dF(T_qS^2)$ is 2 dimensional? – noot Oct 13 '13 at 9:45

$F_*$ is an isomorphism (it maps between two vector spaces of the same dimension, and it is clearly surjective, this implies injectiviy as the dimension is finite). Assume $F_*(v) = F_*(w)$ then this implies $F_*(v-w)=0$, which then implies $u-v=0$ hence $u=v$. So if the images are equal, the preimages are also equal.
• Why is $F*$ surjective? – noot Oct 13 '13 at 9:48
• because $F$ is surjective: the tangent space at a point in $P$ can be seen as the set of equivalence classes of curves. Since $F$ is surjective any curve has a preimage in $S^2$ and so each vector in $F(q) \in P$ has a preimage in $T_qS^2$. – Rogelio Molina Oct 13 '13 at 9:52
• A tangent vector in a point $p$ can be defined as the equivalence class, $[\alpha]$, of all curves $\alpha(t)$ such that two curves are equivalent when $\alpha(0) = \beta (0) p$ and with the same action on $f$ you have defined. Such equivalence class is a vector on $p$. By the way, the tangent space of $P$ is just a real plane, even though $P$ is non-euclidean (the same statement is true for all finite dimensional real differentiable manifolds I know). – Rogelio Molina Oct 13 '13 at 10:10
• I meant to write $\alpha(0)=\beta(0)=p$ above, sorry about that. – Rogelio Molina Oct 13 '13 at 10:17