4
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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 $\gamma$.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.