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Let $M$ be a 2-dimensional Riemannian manifold of positive curvature and $A, B$ two closed geodesics. Show that $A$ and $B$ must intersect.

I know that this is a Frankel-Hadamard type of proof and I also looked at this post here. But I don't know how to hit the problem without the assumption of $M$ being compact and connected.

I think that what I can do for now is to assume $A$ and $B$ don't intersect. Then if I can find a shortest geodesic c from $A$ to B, and show that this geodesic must be perpendicular to each closed geodesic, and then find a parallel field along $c$ that is tangent to $A$ and $B$ at the endpoints of $c$, using the second variation formula to get a shorter curve from $A$ to $B$ I can get it done. But I don't really know how to define the geodesic and the parallel field.

I'd appreciate any help.

  • It's quite clear that $M$ has to be connected. And I also do believe that $M$ should be compact for the statement to hold (but would not know by heart). – Thomas Dec 02 '22 at 17:34
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    If $M$ is not compact, there are counterexamples if you assume $M$ is not complete, so let's add completeness as an assumption. Even if $M$ is non-compact, the images of the two closed geodesics still form a compact subset of $M$, so the distance between them is realized. Hopf-Rinow gives you a geodesic connecting the two curves, as desired. – Jason DeVito - on hiatus Dec 02 '22 at 17:51
  • @JasonDeVito Thanks so much. So according to the comments, the manifold has to be complete so using the Hopf-Rinow theorem, it has to be geodesically complete too. So, this actually gives me the geodesic I'm looking for and this part is done. Now, I think I have to use the second variation formula to prove that it's not possible for this geodesic not to be perpendicular to A and B, but I don't know how to do that exactly. I'd really appreciate it if you could help me with that. –  Dec 07 '22 at 18:27
  • That particular question is a duplicate of https://math.stackexchange.com/questions/2793667/let-n-subset-m-be-a-submanifold-a-minimizing-geodesic-which-joins-a-point-q – Jason DeVito - on hiatus Dec 07 '22 at 18:41

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EDIT: In my original answer, I assumed that $\Gamma$ was a family of geodesics, but it can be any family of curves such that $\Gamma(s,\cdot) = \gamma$. The answer has been edited to reflect this.

Here is a sketch of one way to see this:

First, recall that if a curve $\gamma: [0,1] \rightarrow M$ is a constant speed geodesic whose length is the distance from $A$ to $B$, it minimizes the energy functional $$ E[c] \int_{t=0}^{t=1} g(c',c')\, dt, $$ where $c: [0,1] \rightarrow M$ is any smooth curve such that $c(0) \in A$ and $c(1) \in B$.

Let $\Gamma: (-\delta,\delta)\times [0,1] \rightarrow M$ be a variation of $\gamma$, where $\Gamma(s,0) \in A$ and $\Gamma(s,1) \in B$. Let $S = \partial_s\Gamma$ and $T = \partial_t\Gamma$. The energy of each curve $\Gamma(s,\cdot)$ is $$ E[\Gamma(s,\cdot)] = \int_{t=0}^{t=1} |T|^2\,dt. $$ Since $\Gamma(0,\cdot) = \gamma$ minimizes energy, \begin{align*} 0 &= \frac{1}{2}\left.\frac{d}{ds}E[\Gamma(s,\cdot)]\right|_{s=0}\\ &= \int_{t=0}^{t=1} g(T,\nabla_ST)\,dt\\ &= \int_{t=0}^{t=1} g(T,\nabla_TS)\,dt\\ &= g(T(0,1),S(0,1)) - g(T(0,0),S(0,0)) - \int_{t=0}^{t=1} g(\nabla_TT, S)\,dt\\ &= g(T(0,1),S(0,1)) - g(T(0,0),S(0,0)) \end{align*} Since this holds for any variation $\Gamma$ such that $S(0,0)$ is tangent to $A$ and $S(0,1)$ is tangent to $B = S(0,0)$, it follows that $\gamma$ is orthogonal to both $A$ and $B$.

Deane
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