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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?

noot
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1 Answers1

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$F_*$ 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.

Rogelio Molina
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  • Why is $F*$ surjective? – noot Oct 13 '13 at 09:48
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    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 09:52
  • 'the tangent space at a point in P can be seen as the set of equivalence classes of curves.' could you explain using the definition equation of velocity vector in my original post?to me P is not in any Euclidean space so I do not know what the tangent space at a point in P is like. – noot Oct 13 '13 at 09:56
  • 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
  • Are curves $\alpha$ and $\beta$ lying in P or in the unit sphere? How can you define their preimage in the sphere? – noot Oct 13 '13 at 10:22
  • The definition I gave is valid for any differentiable manifold, so it applies in particular to both $S^2$ and $P$, you may want to use it explicitely in both cases to show in detail that $F_*$ is surjective. – Rogelio Molina Oct 13 '13 at 10:24
  • for a curve $\alpha$ in P, how to define its preimage in the unit sphere? – noot Oct 13 '13 at 10:25
  • For your problem you really need just the curve in a small neighbourhood of the point in $P$, you may try to construct it explicitely but you don't have to. That is because $F$ being surjective tells you that a preimage exists for sure, so you actually work the way around. – Rogelio Molina Oct 13 '13 at 10:33
  • Could you write out the construction explicitly? How can you decide which antipodal point to choose and why the resulting curve is differentiable> – noot Oct 13 '13 at 10:38
  • I can give a geometrical picture of how a preimage could be found. You could also try to work out the details in a coordinate system. Picture a model for $P$: an $S^2$ with antipodal points identified. You just need a very small piece of the curve to find the derivative and hence the tangent vector. Draw a piece of the curve small enough on your model: It will produce two pieces of curves (one being the antipodal image of the other). Now $F: S^2 \to P$, any of the two pieces will serve as a preimage to your small curve in $P$. – Rogelio Molina Oct 13 '13 at 10:46
  • Thanks.Could you also tell me how to prove that the tangent space at a point of P is two dimensional, not one? – noot Oct 13 '13 at 12:35
  • The dimension of the tangent space of a manifold is equal to the dimension of the manifold, so all you have to do is observe that $P$ is two dimensional (because it is locally homeomorphic to $S^2$ and hence to $\mathbb{R}^2$) – Rogelio Molina Oct 14 '13 at 06:30