Suppose we have a Riemannian manifold $M^n$ with families $F_i$ $1\leq i \leq n$ of geodesics. Each family consists of geodesics which cover all manifold and don't intersect (probably we should also assume that they vary "smoothly" in some sense). Moreover, any geodesic from one family is perpendicular to any geodesic in the other family if the intersection of two occurs. Is it true that $M^n$ is flat? I would be interested to know any comments even for modified versions of the statement. Thank you.

Some comments: clearly in $R^n$ you can find such families. Looking at the sphere, I also failed to find such families (which doesn't prove that they don't exist). For me, the problem to prove the above statement lies in the fact that it is hard to start the calculation — it is hard to transfer geometric data into data about vector fields to show anything about curvature.


In dimension 2, having arbitrarily small (convex) quadrilaterals of geodesics, with four right angles, unless curvature is zero. This is the Gauss-Bonnet theorem, in a form sometimes called the theorem of turning tangents.

If there were any point with nonzero curvature, by continuity there would be a small neighborhood with curvature all of the same sign. Take a small quadrilateral inside that neighborhood with four right angles. The theorem says that the integral of $K dA$ on the "rectangle" is zero. However, the sign condition says that integral is nonzero. This contradicts the assumption that there was a point with nonzero curvature.

  • $\begingroup$ Nice application of Gauss Bonnet! Hope there exist higher dimensional version. It seems that in similar way we might be able to prove that sectional curvature of subspaces generated by any two families is zero everywhere. This might imply that all sectional curvatures of manifold are zero which implies that curvature vanishes. $\endgroup$ – Vadim Mar 8 '18 at 18:33
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    $\begingroup$ In fact, it seems that we have a bunch of planes perpendicular to each other. I have to work out details, thank you! $\endgroup$ – Vadim Mar 8 '18 at 19:06

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