This is a problem from the book Gallot, Hulin, Lafontaine: Riemannian geometry (3rd edition).

Exercise 2.118: For a compact Riemannian manifold, let $p,q$ two points such that $d(p,q) = \text{diam}(M)$. Show that there are at least two geodesic segments from $p$ to $q$.

Question: It's quite easy to prove it under the additional assumption that $q$ and $p$ are not conjugate (I provide a proof below). Is this assumption really required, or can be dropped?

The proof (assuming $p$ and $q$ not conjugate) goes as follows (sorry if I am a bit verbose):


Let $\gamma:[0,1] \to M$ be a minimizing geodesic joining $p$ with $q$. Let $v \in T_p M$ such that $\gamma(t)=\exp_p(tv)$. Indeed, since $d(p,q) = \text{diam}(M)$, this geodesic is no longer minimizing when prolonged past time $1$. Thus, for any point $:q_\epsilon = \gamma(1+\epsilon)$ there exists another minimizing geodesic $\gamma_\epsilon:[0,1]\to M$ connecting $p$ with $q_\epsilon$.

Let $v_\epsilon$ the initial vector of such a geodesic, i.e. $\gamma_\epsilon(t) = \exp_p(t v_\epsilon)$. In particular we have, by construction

$$\exp_p(v_\epsilon) = \exp_p((1+\epsilon)v)$$

Indeed $v_\epsilon \neq v(1+\epsilon)$ for all $\epsilon > 0$. (otherwise the original geodesic would be miniminzing past its final point).

By compactness, for $\epsilon \to 0$, these geodesics converge to a limit geodesic $\gamma_0$, with initial vector $v_0$. It remains to prove that $v_0 \neq v$ (thus giving a truly different geodesic).

By absurdum, assume $v_0 = v$. Neglecting terms of higher order, we can assume $v_\epsilon = v+ \epsilon w$. By the discussion above $w \neq v$. Now consider the two Jacobi fields along the original geodesic $\gamma(t)$:

$$J_1(t) = \left.\frac{d}{d\epsilon}\right|_{\epsilon =0} \exp_p(tv(1+\epsilon))$$

$$J_2(t) =\left.\frac{d}{d\epsilon}\right|_{\epsilon =0} \exp_p(tv_\epsilon)$$

By construction $J_1(0) = J_2(0) = 0$. By construction $J_2(1) = J_1(1)$. Moreover the two fields are independent, in fact by computing the covariant derivative w.r.t.

$$D_t J_1(0) = v$$

$$D_t J_2(0) = w$$

Then the Jacobi field $J(t) :=J_1(t) - J_2(t)$ is a non-trivial Jacobi field (recall that $v\neq w$) vanishing at its endpoints $p$ and $q$, which is absurdum. Thus $v_0$ must be different from $v$.

Indeed the last part of the proof, namely to show that the limit geodesic $\gamma_0(t) = \exp_p(t v_0)$ is a truly different geodesic, requires the non-conjugacy assumption.


Assume there exists precisely one geodesic segment from $p$ to $q$, say $\gamma : [0,l] \to M$, parametrized by arc length. By compactness we can extend $\gamma$ to the intervall $[0,\infty[$. Now consider a sequence $\gamma_n : [0,l_n] \to M$ of minimal arc length geodesics from $p$ to $\gamma (l + \frac 1 n)$. Since limits of minimal geodesics are minimal geodesics and $\gamma$ is the unique minimal arc length geodesic from $p$ to $q$ it follows that $\gamma_n$ converges to $\gamma$. Therefore, the angle between $\dot \gamma(l + \frac 1 n)$ and $\dot \gamma_n (l_n)$ converges to $0$.

Now one can use the first variation formula to calculate that $l_n$, the length of $\gamma_n$ is bigger than $l$ for large $n$, contradicting the fact that $l$ equals the diameter of $M$.

Alternatively one can apply toponogovs theorem: Let $M$ satisfy the lower curvature bound $k < 0$. For large $n$ consider the comparison triangle $\Delta_n$ in $M_2^k$ with sides $\tilde \gamma, c_n$ and $\tilde \gamma_n$ of respective lengths $l, \frac 1 n$ and $l_n$. Then the angle between $c_n$ and $\tilde \gamma_n$ is smaller than the corresponding angle in $M$, which is arbitrary close to $0$ for large $n$. From the geometry of $M_2^k$ it is then clear that $l_n$ is striclty bigger than $l$ for large $n$, again contradicting the fact that $l$ equals the diameter of $M$.


Thanks wspin. However, let me elaborate a bit, some detail is still not clear to me.

For any $s \in (-\epsilon,\epsilon)$, let $\gamma_s(t)$ be a minimal geodesic connecting $\gamma(0)$ with $\gamma(t+s)$. Let $V(t) :=\partial_s \gamma_s(t)|_{s=0}$ be the associated vector field along the original geodesic $\gamma$.

By construction $V(0) = 0$ and $V(1) = \dot\gamma(1)$. We apply the first variation formula and we get:

$$ \left.\frac{d}{ds}\right|_{s=0} L(\gamma_s) = \langle V(t),\dot\gamma(t)\rangle|_0^1 - \int_0^1 \langle V(t),\nabla_t \dot\gamma(t)\rangle dt = \|\dot\gamma(1)\|^2 > 0$$

contradicting the fact that $\gamma$ realizes the maximum of the distance between its endpoints.

Question: the argument above works if you are able to find such a family $\gamma_s$ of geodesics, at least $C^1$ in $s$. This is indeed possible if the final point $\gamma(1)$ is not-conjugate, but without this assumption the minimal geodesic connecting $\gamma(0)$ with $\gamma(1+s)$ is not even unique. And even if it were, the its initial vector $v_s$ may be not $C^1$ in $s$.

I believe you are considering some sort of non-smooth family $\gamma_n$ as one (of the many possible) geodesic joining $\gamma(0)$ with $\gamma(1+\frac{1}{n})$. How can you apply the variation formula to this? As far as I know, the first variation formula works for piecewise $C^1$ variations.

  • $\begingroup$ Since $\gamma_n$ converges to $\gamma$, for $n$ big enough the $\gamma_n$'s should all be contained in some tubular neighbourhood of $\gamma$. This lets us interpolate between the $\gamma_n$'s with some smooth family of curves $\tilde{\gamma}_s$ (say with $s \in [0, \epsilon)$ and $\tilde{\gamma}_n = \gamma_n$ for those $1/n < s$). This would give us a smooth variation $\tilde{\gamma}_s$ of $\gamma$ with $V(1) = \dot{\gamma}(1)$, so that we can apply the first variation formula. $\endgroup$ – jef808 Aug 19 '14 at 17:52
  • 1
    $\begingroup$ Actually we must take some subsequence of the $\gamma_n$'s to interpolate with a smooth family in the tubular neighbourhood and I'm not sure how to choose this subsequence correctly... For example the $\gamma_n$'s could lie on alternating sides of $\gamma$ and then $\gamma_s$ won't be differentiable anywhere in the $\partial_s$ direction at $s=0$. Also in my above comment I meant $\tilde{\gamma}_{1/n} = \gamma_n$. $\endgroup$ – jef808 Aug 19 '14 at 18:04
  • $\begingroup$ That's precisely the kind of problem I have in mind, jef808. How to "interpolate" in some smooth way the sequence $\gamma_n$ (or some subsequence) $\endgroup$ – Raziel Aug 19 '14 at 18:46
  • $\begingroup$ Indeed i also do not see a nice, say smooth, way to do this. However, a first variation formula holds in more general context than smooth variations. See for example section 4.5. in the book Metric Geometry; math.psu.edu/petrunin/papers/alexandrov/bbi.pdf (If one is not familiar with Alexandrov spaces note that Riemannian Manifolds are examples of such spaces) Anyways, as this question is an exercise it should be ment to be solved by simpler means. Maybe the Toponogov argument in my first answer. $\endgroup$ – wspin Aug 20 '14 at 8:38

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