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I have come across a question, which I am not convinced I have the right answer to.

Let (M,g) be a connected, non-compact, complete Riemannian manifold, $p\in M$.

a) Show that there exists a sequence $(x_i)_{i\in \mathbb{N}}$ with $d(p,x_i)\overset{i\to \infty}{\longrightarrow} \infty$

b) Show that there exists $X_i\in T_pM$ with $|| X_i ||=1$ so that $x_i = exp_p(d(p,x_i)X_i)$

where $d:M\times M \to \mathbb{R}$ is the distance function.

For a) I thought, that since $M$ is non-compact, there exists a diverging geodesic $\gamma$ with $\gamma(0) = p$. Because $M$ is complete its length is $\infty$. Is that correct reasoning? In class we have not talked about non-compact manifolds.

For $b)$ I thought about using the fact that $exp_p$ is a radial isometry and sends straight lines through $0\in T_pM$ to geodesics through $p\in M$.

Thanks in advance

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  • $\begingroup$ I have seen it, but I don't know where it states that. This is a question from an exam. $\endgroup$
    – Jonas W.
    Commented Sep 19, 2018 at 11:08
  • $\begingroup$ Ok. It makes the second part easier, and my statement is wrong anyway : $\mathbb R$ is a counterexample. The right way to state it is that like $\mathbb R^n$, every closed and bounded set is compact in a complete Riemannian manifold. $\endgroup$ Commented Sep 19, 2018 at 12:44

1 Answer 1

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Unfortunately no, you can't use that for (a) because it is circular reasoning (how do you propose to prove the existence of the ray $\gamma$ otherwise?). But (b) follows from standard equivalent definitions of completeness (e.g. Hopf-Rinow).

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  • $\begingroup$ thanks. So how does a) follow? $\endgroup$
    – Jonas W.
    Commented Sep 19, 2018 at 11:04
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    $\begingroup$ Argue by contradiction. If the distance from p is all at most some finite R then $M$ is the image of the closed ball $\overline{B(0;R)}\subset T_pM$ under $\exp_p$ so would be compact. $\endgroup$ Commented Sep 19, 2018 at 11:08
  • $\begingroup$ nice, thanks! This solves b) too right? $\endgroup$
    – Jonas W.
    Commented Sep 19, 2018 at 11:11
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    $\begingroup$ Not quite. You want to use some version of Hopf-Rinow to say there is a minimizing geodesic connecting $p$ to $x_i$ then look at the derivative (i.e. direction) at $p$. $\endgroup$ Commented Sep 19, 2018 at 11:14

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