A complete, simply connected Riemannian manifold has no conjugate points if and only if every geodesic is length-minimizing. I just realized that I don't know whether the same is true for a locally CAT(k) space (i.e. a geodesic space with curvature bounded above in the Alexandrov sense).

Thanks to Alexander and Bishop, there is a developed "geodesic analysis" in these spaces, including Jacobi fields and conjugate points. And there is a Cartan-Hadamard theorem for spaces without conjugate points: if it is simply connected, then every pair of points is connected by a unique geodesic.

In the Riemannian case, the converse statement follows from the basic fact that a geodesic beyond a conjugate point is no longer minimizing. This is proved by constructing a length-decreasing variation, or something similar, from a vanishing Jacobi field. Unfortunately, this argument uses a lower curvature bound. Well, not quite that, because it also works in Finsler geometry, but anyway it fails for CAT(k): on a bouquet of two spheres there are geodesics that remain minimizing beyond a conjugate point.

However this does not disprove the converse Cartan-Hadamard theorem. Hence the question:

Let $X$ be a space with curvature locally bounded above. Let's not talk about monsters: the space is complete, locally compact, all geodesics are extensible (otherwise one can play dirty tricks with a boundary). Suppose that every geodesic in $X$ is minimizing. Or even better: every pair of points is connected by a unique geodesic. Does this imply that the geodesics have no conjugate points?

UPDATE. Thanks to Henry Wilton, I've found that there is no standard definition of a conjugate point. In fact, some definitions are designed so as to imply the affirmative answer to my question immediately. When I asked the question, I meant the following (maybe not the best possible) definition.

Fix a point $p\in X$ and consider the space $X_p$ of geodesic segments emanating from $p$. The segments are parametrized by $[0,1]$ proportionally to arc length. The space $X_p$ is regarded with the $C^0$ metric. The exponential map $\exp_p:X_p\to X$ is defined by $\exp_p(\gamma)=\gamma(1)$. A point $q=\gamma(1)$ is conjugate to $p$ along $\gamma$ iff $\exp_p$ is not bi-Lipschitz near $\gamma$.

  • $\begingroup$ What are conjugate points? $\endgroup$
    – HJRW
    Mar 31, 2010 at 1:44
  • $\begingroup$ I'm confused by the word "locally" in your hypothesis. If it's, say, locally CAT(0) but not CAT(0) then it has non-trivial fundamental group. (Every locally CAT(0) space has CAT(0) universal cover, by the Cartan--Hadamard theorem for CAT(0) spaces.) So, in particular, geodesics are never unique. $\endgroup$
    – HJRW
    Mar 31, 2010 at 2:00
  • $\begingroup$ I guess I'm just observing that in a (nice) locally CAT(0) space, uniqueness of geodesics implies that it's actually CAT(0). $\endgroup$
    – HJRW
    Mar 31, 2010 at 2:07
  • $\begingroup$ @Henry Wilton: uniqueness of geodesics implies that the space is simply connected (in fact, contractible). Simply connected and locally CAT(0) implies CAT(0). $\endgroup$ Mar 31, 2010 at 11:40

1 Answer 1


Consider a surface of revolution with an equator $\ell$ of lenght $2{\cdot}\pi$ such that its Gauss curvature $$K=1/\left(1+\sqrt[5]{\mathrm{dist}_ \ell}\right).$$ Choose $z\in \ell$ and let $\Sigma=B_{\pi/2}(z)$. Clearly $\Sigma$ is a $\mathrm{CAT}(1)$-space it has just one pair of conjugate points (say $p$ and $q$ --- the ends of $\Sigma\cap\ell$) and it has unique geodesics between each pair.

It remains to make geodesics extensible. To do this, we take $\Lambda=(S^1\times [0,\infty), d)$ with flat metric and concave boundary $\partial \Lambda=\partial\Sigma$. Then we glue $\Lambda$ and $\Sigma$ along the boundary.

The metric on $\Lambda$ is completely described by curvature $k(u)$ of its boundary [$u\in \partial \Lambda=\partial \Sigma$]. We only need to choose a function $k$ which is

  • on one had is large enough so that the glued surface still has unique geodesics between each pair (in particular $k(p)=k(q)=\infty$).
  • on the other hand is $\int_{S^1} k<\infty$, so that glued space is locally compact.

I believe it is possible...

  • $\begingroup$ I don't see how to make $\int k<\infty$. Consider a geodesic $p'q'$ where $p'$ and $q'$ are $\delta$-close to $p$ and $q$ and $q'$ is on the boundary of $\Sigma$. A Jacobi field $J$ starting at $p'$ with $J=0$ and $J'=1$ seems to have $J'\approx-1$ and $|J|<C\delta$ at $q'$. To compensate, you need to insert curvature of order $1/\delta$ at $q'$. And if you do so, the integral diverges. $\endgroup$ Apr 4, 2010 at 10:05
  • $\begingroup$ I have updated the answer. $\endgroup$ Apr 5, 2010 at 3:13
  • $\begingroup$ Do you confirm that your example is a complete, locally compact, intrinsic metric and uniquely geodesic space, with conjugate points ? So that, the lack of conjugate points is not necessary for a (complete, locally compact, intrinsic metric) space to be uniquely geodesic. I ask you this question because in this post I ask about a (as general as possible) condition for a complete, locally compact, intrinsic metric space to be uniquely geodesic. $\endgroup$ Aug 23, 2013 at 8:29
  • $\begingroup$ @SébastienPalcoux, One has to perform some calculations to make sure it works, but I believe it is works. $\endgroup$ Aug 23, 2013 at 19:35

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