In the book Riemannian geometry - modern introduction by Isaac Chavel, three different definitions of convexity are introduced. I am looking for an example of a set which is convex but not strongly convex according to the following definitions (cited from the book):

Let $M$ be a complete Riemannian manifold, and $A \subset M$. $A$ is:

  1. convex if for any $p,q\in A$ there exists a geodesic $\gamma_{pq} \subset A$ such that $\gamma_{pq}$ is the unique minimizer in $M$ connecting $p$ to $q$.
  2. strongly convex if for any $p,q\in A$ there exists a geodesic $\gamma_{pq}\subset A$ such that $\gamma_{pq}$ is the unique minimizer in $M$ connecting $p$ to $q$, and $\gamma_{pq}$ is the only geodesic contained in $A$ joining $p$ and $q$.

I believe that this picture gives an example. This is not an accurate figure! The set I'm referring to is the one centered at around a geodesic segment with end points conjugate one to the other along it, and bounded symmetrically by two close non minimal geodesics.

Is my example correct? If not, then can someone provide a valid example?

Edit: Caution I know of something like 7-8 different definitions of convexity in the case of a Riemannian manifold. Try to refer to the definitions I gave above.

  • $\begingroup$ It's a good thing you added the edit. I deleted my non-answer, I was too hasty and didn't pay attention to the difference in the definition. $\endgroup$ Commented Aug 23, 2010 at 13:09

1 Answer 1


I believe your example is correct. As you are allowing $A$ closed the following is pretty similar. Choose a real number $ N > 1.$ Take the smooth and analytic curve (write it as a level curve and check the gradient) $$ y^2 = x^3 - 3 N^2 x^2 + 3 N^4 x = \left( \frac{x}{4} \right) \left( (2 x - 3 N^2)^2 + 3 N^4 \right) $$ Note that $$ 2 y y' = 3 x^2 - 6 N^2 x + 3 N^4 = 3 (x - N^2)^2.$$ Also $$ 4 y^3 y'' = 3 (x^4 - 4 N^2 x^3 + 6 N^4 x^2 - 3 N^8 ) = 3\left((x-N^2)^4 + 4N^6(x-N^2)\right). $$ Furthermore, when $$x = N^2, y = \pm N^3.$$ Revolve this around the $x$-axis, making a simply connected surface. There is now a closed geodesic along $ x = N^2,$ of circumference $ 2 \pi N^3.$ The minimizing geodesic between $( N^2, N^3, 0)$ and $( N^2, -N^3, 0)$ is the original curve in the plane $z=0,$ of length no larger than $ 2 N^2 + 2 N^3.$ The length of half the closed geodesic is $ \pi N^3,$ which is larger for large enough $N.$ So, as I did not say, we are taking $N > 2$ and $$ A = \left\{(x,y,z) \in \mathbb R^3 : y^2 + z^2 = x^3 - 3 N^2 x^2 + 3 N^4 x \; \mbox{and} \; x \leq N^2.\right\} $$

  • $\begingroup$ @Will: It seems that this is a correct example. However, I don't see how can we assure here that every two points are connected by a unique minimal geodesic. For some $0<x<N^2$ two "antipodal" points may be connected by two different (symmetric) minimal geodesics? It's true for the points on the "equator" you mentioned, but what about other points on the surface? Is this an original example, or did you cite it? $\endgroup$ Commented Aug 24, 2010 at 9:49
  • $\begingroup$ Dear Dror, I just sent you an email. I have come to believe that for some $ 0 < x < N^2$ and two "antipodal" points, the original elliptic curve gives the only geodesic. $$ $$ Any geodesic meeting the interior of $A$ has a "vertex," a point with minimal $x,$ similar idea to the "vertex" of a parabola. I believe that every point in the interior of $A$ is the vertex of a (minimizing) geodesic arc with two endpoints on the "equator" $ x = N^2.$ $$ $$ This may follow from Clairaut's theorem. $\endgroup$
    – Will Jagy
    Commented Aug 24, 2010 at 19:47
  • $\begingroup$ Let's see, the example is original, I made it up. I am convinced now that the behavior you are worried about is precisely what happens when cutting an ellipsoid of revolution in half when it is a cigar-shaped "prolate spheroid." en.wikipedia.org/wiki/Prolate_spheroid $$ $$ But the behavior we get here is that of a pancake shaped "oblate spheroid," en.wikipedia.org/wiki/Oblate_spheroid $$ $$ Note that there is a bifurcation in behavior in passing between the two, for the perfect sphere all geodesics pass through antipodal points. $\endgroup$
    – Will Jagy
    Commented Aug 25, 2010 at 0:47
  • 1
    $\begingroup$ Here you can find a visualization of Will's surface and set: page.mi.fu-berlin.de/atariah/gallery.html#convex_example_2 $\endgroup$ Commented Aug 25, 2010 at 8:34

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