The classical splitting theorem (Toponogov, Milka) says that if a smooth complete Riemannian manifold (more generally, Alexandrov space) $M^n$ of non-negative sectional curvature contains a line (i.e. infinite geodesic line minimizing distance between any pair of its points) then $M$ is isometric to $\mathbb{R}\times N^{n-1}$ where $N$ is non-negatively curved.

My question is whether there exists something in the spirit of the following quantitative version of the above theorem.

Let us fix $n\in \mathbb{N}$,$l>0$, $\delta>0$, and $r>1000$. Does there exist $\varepsilon>0$ and $L>0$ with the following properties. Let $M^n$ be a Riemannian manifold or Alexandrov space (not necessarily complete!) of curvature $\geq -\varepsilon$ such that it contains a compact ball $\bar B(x,r)$. Assume there exists a minimal geodesic of length $L$ such that $x$ is its middle point, and the whole space $M$ is contained in 1-neighborhood of this geodesic. Then there exists an open subset of the ball $\bar B(x,r)$ which is bi-Lipschitz homeomorphic to $(0,l)\times N^{n-1}$ where $N$ has curvature $\geq -\delta$.


It is not hard to show that the space is close to a subset in the product in Gromov--Hausdorff sense, but you should assume bit more, say $\bar B(x,L)$ is compact.

You can get some constant, by taking limit. (If you want effective estimates, assume $z$ and $z'$ are the ends of geodesic, you can apply gradient flow along the distance functions to these two points and show that they nearly inverse of each other. Hence the statement follows.)

This should imply bi-Lipschitz homeomorphism to the product space which is also Hausdorff approximation. (No one know the proof but it is conceivable that Perelman did know how to prove it.)

On the other hand, you should not expect both Lipschitz constant to be close to $1$. (I have a counterexample in my pocket.)

  • $\begingroup$ Thank you very much. So could you please formulate a precise statement which you know to be true. $\endgroup$ – MKO Oct 8 '15 at 11:29
  • $\begingroup$ In fact, I am surprised that both Lipschitz constants cannot be close to 1. $\endgroup$ – MKO Oct 8 '15 at 12:14

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