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Assume that there is a parametrized smooth curve $c$ on the manifold $M$, mapping from $[a,b]$ to $M$. Also assume that there is a tangent vector on $M$ in the form $(p,v)$. Tu's text states that it is assumed that the curve $c$ is starting at $p$ if $c(0)=p$. Is there a way to show that $c(0)=p$ in any way? Any help would be appreciated!

Thanks in advance!

Marion Crane
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  • Without any conditions on $c$, how do we know it even passes through $p$? – J126 Mar 04 '15 at 02:18
  • @JoeJohnson126 If $p$ is some point on $M$, then can we show that a parametrized curve $c$ passes through that point $p$ at $c(0)$. That's my intention – Marion Crane Mar 04 '15 at 02:21
  • As long as $M$ is path connected, there will always exist some curve that passes through $p$. And you can always arrange the curve to passes through $p$ at time $0$, as long as $a\leq 0\leq b$. On top of that you can always have it so that $c'(0)=v$ for any tangent vector $v$. – J126 Mar 04 '15 at 02:27
  • @JoeJohnson126 i really thought that. But why does the book give an exercise to show it then? – Marion Crane Mar 04 '15 at 02:30
  • Are you stating the whole exercise as printed in the book? I think you are missing some details. – J126 Mar 04 '15 at 02:34
  • @JoeJohnson126 The exercise asks that if there is a tangent vector $(p,v)$, then show that there is a smooth curve $c$ mapping (a,b) to $M$ such that $c(0)=p$ – Marion Crane Mar 04 '15 at 02:37
  • @JoeJohnson126 I did know from the reading that $c(0)=p$. But I do not know how to show that – Marion Crane Mar 04 '15 at 02:38

2 Answers2

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If there is a tangent vector $v$ at $p$, then $v \in T_p(M)$. Since $M$ is a manifold, there exists a parametrization $\phi: U \rightarrow V$ on a neighborhood $p \in V$ such that $\phi(0) = p$, and $d\phi_0(u) = v$ for some $u \in U$. Consider the line given by $L = \{ x\cdot u|x\in\mathbb{R}\}$ and consider $L \cap U = \{ x\cdot u | x\in (a,b) \}$. Then the curve $c:(a,b)\rightarrow V\subset M$ given by $c(x) = \phi(x\cdot u)$ is a smooth curve to the manifold with the required tangent vector at the given point.

Machine Caliber
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If $M$ is a manifold, then there is some coordinate chart $\phi:\mathbb{R}^n\rightarrow M$ that is a smooth homeomorphism onto its image such that $\phi(0)=p$. Restrict $\phi$ to a small portion of the $x$-axis that includes $0$. This will be a smooth curve such that $c(0)=p$. Further, if you want $c'(0)=v$, you know that $\phi_*:T\mathbb{R}^n\rightarrow T_pM$ is surjective.

J126
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