Let $(M,g)$ be a complete simply connected Riemannian manifold with non-positive curvature. Because of the Hopf-Rinow theorem, any two points are connected by a geodesic segment.

Pick three distinct points $o$, $a$ and $b$. Let $L_a(t)$ and $L_b(t)$ be the geodesic segments joining $o$ and $a$, and $o$ and $b$, parametrised with unit speed.

For any $t$, let $m(t)$ be the midpoint between $L_a(t)$ and $L_b(t)$.

Is $m(t)$ a geodesic segment?


1) I think we can assume that $M$ is two dimensional.

2) The answer in the case of the Euclidean plane is affirmative.

  • 2
    $\begingroup$ You don't need to assume that any two points are connected by a geodesic segment; this follows from the Hopf-Rinow theorem. $\endgroup$
    – Ben McKay
    Jul 29, 2016 at 8:55
  • $\begingroup$ Do you assume that there is a unique midpoint between any two points? That is equivalent to $M$ being simply connected, since geodesic loops give multiple midpoints. $\endgroup$
    – Ben McKay
    Jul 29, 2016 at 8:57
  • $\begingroup$ yes! edited the question $\endgroup$
    – Giulio
    Jul 29, 2016 at 8:59
  • 5
    $\begingroup$ Also for the hyperbolic plane the answer is affirmative: given two geodesic starting at $o$ there is a symmetry fixing $o$ which exchange them. Then since the construction is geometric the middle point $m(t)$ belong to the fixed set of the symmetry hence $m(t)$ is a point in the same geodesic for all $t$. $\endgroup$
    – Holonomia
    Jul 29, 2016 at 9:21

1 Answer 1


The answer is no.

Define the manifold

Let $f$ be a smooth function on $\mathbb{R}$ satisfying

  1. $f(r) = |r|$ for $|r| > 3$
  2. $f(r) > 0$
  3. $f''(r) \geq 0$.

The corresponding warped product metric on $\mathbb{R}^2$

$$ \mathrm{d}s^2 = \mathrm{d}r^2 + f(r)^2 \mathrm{d}\theta^2 $$

is complete and has non-positive curvature. The exterior regions $|r| > 3$ are flat. It is in fact the simply connected cover of a corresponding warped metric on $\mathbb{R}\times\mathbb{S}^1$, which when $|r| > 3$ are two copies of the Euclidean plane with a ball removed.

Since geodesics (lines) in the Euclidean plane are confined to a half plane, we know that geodesics satisfying $|r(\gamma)| > 3$ must have bounded $\theta$ within some $[\theta_0, \theta_0+\pi]$. In other words, for a lot of what we are going to say we can just work in the quotient picture, without worrying too much about the lift to upstairs.


In the quotient picture fix, in rectangular coordinates on $\mathbb{R}^2 \setminus B_3(0)$ a point $(x,y)$ in the first quadrant, such that

  1. $x> y > 3$
  2. the line $\tilde{\gamma}: t\mapsto (x-t/\sqrt{2}, y-t/\sqrt{2})$ does not enter the ball of radius 3.

Let $o = (x,y)$ (or rather its lift to our manifold defined above), and $a = (x,0)$, $b = (y,0)$.

Notice that $\tilde{\gamma}$ is geodesic. Note also that we don't have to worry about the quotient/lift since $L_a, L_b$ and $\tilde{\gamma}$ all avoid the "cut" at $\{x = y \wedge x,y < 0\}$.

Note further that for some $\epsilon$ time that $\tilde{\gamma}(t)|_{t\in (-\epsilon,\epsilon)} = m(t)$.

Note also that $\tilde{\gamma}(t)$ always stay on the "$L_a$ side" of the "hole"

But this does not hold for all times: once the geodesic segment joining $L_a(t)$ to $L_b(t)$ starts to enter the ball of radius 3, its midpoint would start to deviate from $\tilde{\gamma}$, and move closer toward $L_b(t)$.

This is easiest to see when $t$ is very very large. When $t$ is very very large, on the original manifold, the geodesic connecting $L_b(t)$ to $L_a(t)$ must first go into the hole, which requires travelling distance approximately $t - y + O(1/t)$, and exit the hole, and reach $L_a(t)$, traveling another distance that is approximately $t-x + O(1/t)$. The distance traveled in the hole is $\leq 6\pi$. So asymptotically the midpoint $m(t)$ remains in fact in a compact region of our manifold. If initially $y$ is much smaller compared to $x$ (in the sense that $x-y > 6\pi$), in fact we expect $m(t)$ to appear "on the other side of the hole" eventually.

Since $m(t)$ and $\tilde{\gamma}(t)$ coincide for some time interval, but their image do not coincide globally, we conclude that $m(t)$ cannot be always a local geodesic.

  • $\begingroup$ @what about if the scalar curvature is constant? Thanks $\endgroup$
    – Giulio
    Sep 27, 2016 at 10:25
  • $\begingroup$ In two dimensions the answer is trivial if you make the scalar curvature constant. Generally as dimension increase, scalar curvature becomes weaker and weaker as restriction on geometry, so if you ask me to bet I would say that for sufficiently high dimensions there may be counterexamples. I don't have a proof one way or another. $\endgroup$ Sep 27, 2016 at 13:02
  • $\begingroup$ One thing to try is to take a warped product of what I outlined above with a copy of $\mathbb{H}^2$; this way the geodesics orthogonal to the $\mathbb{H}^2$ factor remains unchanged. To ensure constant scalar curvature you will have to solve an second order ODE in $r$ for the warping factor; if this ODE can be solved to have a global positive solution you will get an example. There's a one parameter freedom in choosing the constant for the scalar curvature, perhaps the ODE only has a good ground state for a good choice of this constant. $\endgroup$ Sep 27, 2016 at 13:11

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