# Is the exponential map of a $C^{1,1}$ Riemannian metric a local homeomorphism?

Suppose that $g$ is a $C^{1,1}$ (i.e., continuously differentiable with locally Lipschitz first derivative) Riemannian metric on a smooth manifold $M$. It seems to be known that locally the exponential map $\exp_p$ corresponding to $g$ at some point $p$ of the manifold is still a local homeomorphism (and Lipschitz). I am looking for a proof of this result (if it is true). The standard proof for $C^2$ metrics uses the fact that the tangent map of $exp_p$ at $0\in T_pM$ is the identity, so that the inverse function theorem gives that $exp_p$ is a local diffeomorphism. However, for a $C^{1,1}$ metric, $\exp_p$ is only (locally) Lipschitz, so it is not clear that there even exists a tangent map of $exp_p$ at $0$ and the inverse function theorem does not apply. Is there some argument that can replace the use of the inverse function theorem in this situation?

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Note that $C^{1,1}$ metric admits an approximation by $C^2$-metrics with uniformly bounded $C^2$-norm. In particular the curvature is bounded, hence we get a bounds for the distortion of the exponential map depending on the size of the neighborhood of $0$. It remains to pass to the limit.

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Thanks a lot, I see how to approximate by $C^2$-metrics with bounded curvature. But could you give a bit more details on the remaining part of your argument (or some reference)? – Michael Kunzinger May 20 '12 at 8:06
Our manifold is a limit of manifolds with uniformly bounded curvature. The injectivity radius of such manifolds is bounded from below in terms of volume of some ball centered at p and bounds on curvature. This should be in Cheeger--Ebin. Is that what you asking? – Anton Petrunin May 20 '12 at 13:52
As to your suggestion (if I understand correctly), the closest result I was able to find is the following (taken from Petersen): (J. Cheeger, 1967) Given $n ≥ 2$ and $v, K \in (0, ∞)$ and a compact $n$-manifold $(M, g)$ with $|sec|≤K$, $volB(p,1) ≥ v$, for all $p ∈ M$, then $inj(M) ≥ i_0$, where $i_0$ depends only on $n$, $K$, and $v$. Now suppose we can use this to obtain a uniform lower bound for the injectivity radius of the approximating metrics. How does this carry over to the limit? (uniform limits of injective maps need not be injective) – Michael Kunzinger May 20 '12 at 15:43
In addition you get that differential of $\exp_p$ is almost isometry. The later follows from the standard estimates on Jacobi fields. In particular the length of curves after the mapping changes by a coefficient between $1\mp\epsilon$; it implies that the map is bi-Lipschitz in say $(i_0/2)$-neighborhood. – Anton Petrunin May 20 '12 at 16:29
I guess the standard estimates you refer to are given by the Rauch comparison theorem. This then gives uniform bounds on the derivative of $\exp^{g_m}$ (with $g_m\to g$ the approximating $C^2$-metrics) from above and below. This allows to control lengths of curves under mapping them with $\exp^{g_m}$ or its inverse, giving a bi-Lipschitz property for $\exp^{g_m}$. For this to carry over to $\exp^g$ one probably needs that $g$-geodesics cannot leave some fixed set - this should follow since it holds for the $g_m$ and curve lengths w.r.t. $g$ are close to those w.r.t. $g$. Is this what you mean? – Michael Kunzinger May 21 '12 at 7:49

More generally the exponential map of any Lipschitz connection or spray is a Lipeomorphism (bi-Lipschitz homeomorphism) in a neighborhood of the origin. The proof uses Leach's inverse function theorem which holds under strong differentiability. The non-trivial part is to show that the exponential map is strongly differentiable at the origin. Details are given here http://arxiv.org/abs/1308.6675

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