Two distinct geodesics joining two points on a compact manifold

Question

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):

Proof:

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.

Answer 1

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.

Answer 2

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 $M$.