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I can't understand the red line well. Seemly, if $\tilde \gamma_1\cap \tilde \gamma_2$ has at least two points, it will contradict with the simple connectivity. I want to know why it is ?

PS(2024-3-26): As the hint of Moishe Kohan, the right way is to use the Hadamard Theorem.

Hadamard Theorem: Let $M$ be a complete Riemannian manifold, simply connected, with sectional curvature $K\le 0$. Then $M$ is diffeomorphic to $\mathbb R^n$, $n=\dim M$; more precisely $\exp_p:T_pM\rightarrow M$ is a diffeomorphism.

If there are at least two points, assuming two of them are $p,q$, then, there are two different vectors $u,v\in T_pM$ such that $\exp_pu=\exp_pv =q$. It contradict with $\exp_p$ is diffeomorphism.

Picture below is from the 260th page of do Carmo's Riemannian Geometry. enter image description here

Enhao Lan
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1 Answers1

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Suppose the intersection contains just one point, $P$. Where must that point go after the translation, i.e., what is $f(P)$? It must lie on both geodesics, because we're assuming both are (setwise) invariant under $f$. But the only point on both geodesics is $P$. So we get $f(P) = P$, i.e., $P$ is a fixed-point of $f$. But $f$ has no fixed points (by the opening clause of that sentence), so that's a contradiction.

John Hughes
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  • I know why one point is impossible. But I don't know why two point or more is impossible. Maybe, I didn't make my question clear. – Enhao Lan Mar 21 '24 at 12:19
  • If you have two points, $P$ and $Q$, presumably the idea is that a path from $P$ to $Q$ along $\gamma_1$ and then back to $P$ along $\gamma_2$ is, for some reason, a non-contractible path, so that it becomes a nonzero element in $\pi_1$. That's probably proven in an earlier lemma. – John Hughes Mar 21 '24 at 12:37
  • See also Moishe's comment ... which presumably relates to an earlier lemma as well. – John Hughes Mar 21 '24 at 12:38