Consider the plane $\mathbb{R}^2$ with a complete metric $g$ without conjugate points. We will denote by $K$ the Gaussian curvature.
Question: Is there a constant $C >0$ such that $K(p) \leq C$, for every $p \in \mathbb{R}^2$?
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4$\begingroup$ Just an idea for a counterexample: take the hyperbolic plane and put infinitely many spherical caps on it with radii tending to 0... $\endgroup$– Ivan IzmestievCommented Jan 26, 2017 at 15:45
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3$\begingroup$ It seems to me that a counterexample, as described by Ivan, can be constructed by making sure there are only finitely many caps of appropriate size along any geodesic. $\endgroup$– Deane YangCommented Jan 26, 2017 at 16:10
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2$\begingroup$ @IvanIzmestiev Why don't you write it as an answer, so the question will be closed? $\endgroup$– Anton PetruninCommented Jan 27, 2017 at 0:39
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$\begingroup$ @AntonPetrunin I'm no more sure it works. A geodesic going "straight through" a bubble is not minimizing; the shortest one with the same endpoints "goes across a slope". So it seems, for an appropriate choice of endpoints, the minimizing geodesic will have a Jacobi field vanishing at the endpoints. Right? $\endgroup$– Ivan IzmestievCommented Jan 27, 2017 at 18:10
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2$\begingroup$ @IvanIzmestiev No, not if the bubble has small diameter. $\endgroup$– Anton PetruninCommented Jan 27, 2017 at 22:11
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1 Answer
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It boils down to the following statement.
There a rotationally symmetric $g$ on $\mathbb{R}^2$ without conjugate points, with huge positive curvature at $0$ and curvature $\equiv -1$ outside of the unit ball.
If such metrics are constructed, then we can cut pieces from it and glue them together along domains of curvature $\equiv -1$. This way we construct a Riemannian manifold without conjugate points and no upper curvature bounds. (We need to spread the unit balls far apart, so globally the glued space looks like Lobachevsky plane, and yet make sure no geodesic pass thru 3 balls — this is easy to achieve.)
To construct such examples, prescribe the curvature depending on the distance to the origin, so that at $0$ it is huge positive, but quickly becomes $-1$ and stay constant. The no conjugate points condition follows from the Jacobi equation.
answered Jan 27, 2017 at 22:09