Let $(M,g)$ be a complete riemanian manifold with $\text{diam}(M,g) < \infty$. If there are two points $p,q$ s.t. $d(p,q) = \text{diam}(M,g)$ then there are at least two shortest paths between them.

So my idea is to take some geodesic (exists because of completeness) between p and q, lets call it $\gamma: [0,1] \to M$. Then find another geodesic from p to $\gamma(1 + \epsilon)$ which has to exists because of completeness and can not be the same as $\gamma$ since the length has to be $\le \text{diam}(M,g)$.

Then what I want to do is look back at the tangent space $T_pM$ at $p$ and find something like a Cauchy-Sequence of the $v_\epsilon$, where $\exp(tv_\epsilon)$ is the geodesic from $p$ to $\gamma(1+\epsilon)$ and see that the limit of these $v_\epsilon$ gives me another geodesic. But this is where I am stuck. It is not clear that a Cauchy-Sequence like that will exist to me and that the limit does the right thing.


I will tell you how to solve the part where you are stuck, but not the original problem. Your approach is, in fact correct. If $\gamma_\varepsilon$ is the minimizer with endpoint $\gamma(1+\epsilon)$ (starting at that point), parametrized by arclength (!), then the preimage of $\gamma_{\varepsilon}(0)$ under the exponential will converge tothe origin of $T_{\gamma(1)}M$, which corresponds to $\gamma(1)$. Now note that the initial vectors have all length $1$, and that the set of these vectors is compact. This means that a subsequence will converge to some $v\in T_{\gamma(1)}M$ .

If you then look at the geodesic which starts in $\gamma(1)$ in direction $v$ this will be a minimizing (why?) geodesic between the two points you started out with.

Now the tricky part (which I will not solve for you) is to show that the limit $v$ can be chosen in such a way that this limiting geodesic is different from $\gamma$.

  • $\begingroup$ Does this work? Instead of parametrizing by arclength, I choose the vectors in $T_pM$ s.t. $\exp_p(v_\epsilon) = \gamma(1+\epsilon)$ and minimizing. They all lie in the compact ball of radius diam(M,g) and thus also converge. Also they now obviously converge to a $\tilde{v}$ s.t. the exponential map maps $\tilde{v}$ to a minimizing geodesic (where in your approach it was not clear to me why it was minimizing). Now assume $\tilde{v} = v$. For each $v_\epsilon$ I know make a variation. $\beta(s,t) = \exp_p(t(v + s (v_\epsilon - v)))$. $\endgroup$ – kave Aug 4 '17 at 12:04
  • $\begingroup$ Then the angle between $\gamma'(1)$ and $\frac{\partial}{\partial s} \beta(0, 1)$ converges to one and plugging this into the variation formula we get that our original geodesic wasn't a critical point $\endgroup$ – kave Aug 4 '17 at 12:04
  • $\begingroup$ @kave I'm not sure what you are doing with the $v_\varepsilon$. In your question you have chosen a minimizing geodesic $\gamma_\varepsilon$ from $\gamma(1+\varepsilon)$ to $\gamma(0)$. This geodesic cannot coincide with a part of $\gamma([0,1])$ since by assumption $\gamma$ cannot be a global minimizer on $[0,t]$ for any $t>1$. It is this sequence/set of geodesics $\gamma_\varepsilon$ for which you have to show that it contains a convergent subsequence, and this is done by a compactness argument like the one I proposed. Such a limit will be minimizing since $\endgroup$ – Thomas Aug 4 '17 at 12:28
  • $\begingroup$ ...all $\gamma_\varepsilon$ will have length $< $ diam $M$ and because the length funcional is well behaved with regard to this convergence. You cannot show that the limit is different from $\gamma $ in general. If it is equal to $\gamma$ you have shown that $\gamma(0)$ is conjugate to $\gamma(1)$, which may well happen. Just have a look at the sphere. $\endgroup$ – Thomas Aug 4 '17 at 12:30
  • $\begingroup$ I am trying to show convergence for the tangential vectors and not the whole geodesics (since that seems pretty complicated to me and I have never looked at convergence of curves on an manifold). I am choosing a sequence of tangent vectors s.t. $\exp_p(v_\epsilon) = \gamma(1 + \epsilon)$ and then showing convergence of these. Then I am using the variation formula to prove that they could not converge to the original $v$ (with $\exp_p(v) = \gamma(1)$). $\endgroup$ – kave Aug 4 '17 at 15:08

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