80
$\begingroup$

There is a fascinating open problem in Riemannian Geometry which I would like to advertise here because I do not think that it is as well-known as it deserves to be. Euclid's famous fifth postulate, or more precisely Playfair's version of it, states that, in the Euclidean plane, through every point outside a line $\ell$ there passes one and only one line which does not intersect $\ell$.

Question: Is the Euclidean plane the only complete Riemannian manifold homeomorphic to $R^2$ which satisfies the fifth postulate, i.e., through any point outside a complete geodesic $\gamma$ there passes one and only one complete geodesic which does not intersect $\gamma$?

In other words, does the fifth postulate force the curvature to be zero? The reason this is interesting, or historically significant, is that it was the attempts to reduce the fifth postulate to the other axioms of Euclid, throughout the middle ages, which eventually led to the discovery of non Euclidean geometries and the notion of curvature by Gauss and Riemann.

This problem appears to be originally due to Burns and Knieper in 1991: see the survey paper by Burns and Matveev, which also includes other nice problems on geometry of geodesics. The problem is also mentioned in papers of Croke, and Bangert and Emmerich, and has been studied most recently by Ge, Guijarro, and Solorzano.

Improve this question
$\endgroup$
7
  • 1
    $\begingroup$ The answer is yes for those Riemannian steucture which satisfy the equivalency of Pythagoras theorem and 5th postulate. Please see this mathforum post: mathforum.org/kb/message.jspa?messageID=3545911 $\endgroup$
    – Ali Taghavi
    Commented Oct 15, 2017 at 14:16
  • $\begingroup$ I thank you and Wojowu for your comment on my answer to this question. Your comment helped me to realize my answer was not relevant. So I deleted it. $\endgroup$
    – Ali Taghavi
    Commented Oct 15, 2017 at 14:24
  • 2
    $\begingroup$ Dear Mohammad: I take it you do not want to assume any of the other four postulates -- thus the geometry need not be homogeneous or isotropic. Correct? $\endgroup$
    – Richard Montgomery
    Commented Oct 18, 2017 at 4:57
  • 1
    $\begingroup$ Cones over circles of radius 1/2 or less come close to doing what you want -but the complete geodesics must be stopped when they hit the cone point since they fail to minimize past the cone point. Have you thought of trying to make Riemannian metric counterexamples by smoothing out the cone point of such metrics? $\endgroup$
    – Richard Montgomery
    Commented Oct 18, 2017 at 5:04
  • $\begingroup$ Hi Richard. No assumption is made other than those mentioned above. I do not think that smoothing the cone is going to work, since all asymptotically flat surfaces appear to be OK according to the recent paper of Ge, Guijarro, and Solorzano mentioned above. $\endgroup$
    – Mohammad Ghomi
    Commented Oct 18, 2017 at 12:36

1 Answer 1

Reset to default
6
$\begingroup$

It is a very nice question.

I am under the impression that a Riemannian metric on $\mathbb R^2$ which has conjugate points cannot satisfy the fifth postulate.

Now, there are various results showing that, under some conditions, metrics without conjugate points must be flat. For a $\mathbb Z^2$-periodic metric, for instance, this is a theorem of Hopf. For asymptotically Euclidean metrics, it seems to be a very recent result of Guillarmou, Mazzucchelli and Zuo: https://arxiv.org/pdf/1909.01488.pdf

Improve this answer
$\endgroup$
3
  • 1
    $\begingroup$ Welcome to mathoverflow. $\endgroup$
    – Ben McKay
    Commented Jan 27, 2021 at 17:30
  • $\begingroup$ In talking about the limits of their result, these authors point to section 2.3 of GLT2019, i.e. arxiv.org/abs/1910.09631, page 10. Do those examples provide a negative answer to the question here? $\endgroup$
    – user44143
    Commented Jan 27, 2021 at 19:20
  • 2
    $\begingroup$ They do not because they have negative curvature, hence do not have conjugate points. Heuristically, the existence of parallels in the fifth postulate should rule out positive curvature while the uniqueness should rule out negative curvature. $\endgroup$
    – Nicolast
    Commented Jan 27, 2021 at 19:47

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .