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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?

flag	asked May 19 2012 at 17:18 Michael Kunzinger 33●2

Anton Petrunin · Accepted Answer · 2012-05-19 18:13:14Z UTC

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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.

answered May 19 2012 at 18:13

Anton Petrunin
16.4k●41●110

	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 2012 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 2012 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 2012 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 2012 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 2012 at 7:49

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

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