Let $(M^2,g)$ be a Riemannian manifold, with manifold boundary $\partial M$. We assume that the metric degenerates at the boundary, in the sense that the (Gauss) curvature diverges like $K \to +\infty$ as one approaches it.

Question. Is it true that 'nearly all' geodesics avoid the boundary? Perhaps those going to the boundary are meagre in some sense?

  • In general, it seems intuitively reasonable that 'most' geodesics would avoid regions where the curvature is 'large and positive'. (This is of course a bit vague—I was unable to come up with a precise formulation.)
  • The question is motivated by a comment made in passing by Robert Bryant in his answer to this question, where he discussed the geodesics of a very specific metric.
  • $\begingroup$ the thing that comes to mind is the Clairaut relation for surfaces of revolution such as $z = x^2 + y^2,$ with positive curvature near the origin. Every geodesic that does not pass through the origin (not a meridean) achieves a minimum distance from it, so every non-meridean geodesic can be produced by specifying a point where it is parallel to the $xy$ plane. en.wikipedia.org/wiki/… $\endgroup$
    – Will Jagy
    Oct 13, 2022 at 21:14
  • $\begingroup$ @WillJagy Sure, Clairaut's relation is essentially the heuristic which suggested to me that 'most' geodesics ought to avoid regions where the curvature is 'large'. However, I am not sure it gives much more here, as I don't want to impose rotational symmetry. $\endgroup$
    – Leo Moos
    Oct 13, 2022 at 22:31
  • $\begingroup$ I had never thought about this before, but am having trouble imagining a smooth 2D riemannian manifold M with boundary such that its gaussian curvature K(p) → ∞ as p → ∂M. Is that even possible? $\endgroup$ Jul 1 at 19:09
  • $\begingroup$ Now that I've thought about it, it seems possible (but very counterintuitive to me): $\endgroup$ Jul 1 at 20:32

1 Answer 1


It seems to fail for $(y^2+z^2-1)^2=x^3$, $x\geqslant 0.$


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