# Compact manifolds locally bi-Lipschitz to Euclidean space

I have a compact manifold $M$, and I am allowed to choose some Riemannian metric on it, exactly which I don't care. But I would love it if I could choose the metric $g$ such that every point has an open neighbourhood $U$ bi-Lipschitz, (Edit: as a metric space, equipped with the geodesic metric) to an open ball in $\mathbb{R}^n$ with the Euclidean metric.

Is it possible to always choose some metric such that this is true? Or else do we know the/a class of manifolds where this is possible?

One could also ask if something slightly weaker is always possible, such as choosing a metric locally quasi-isometric to an open ball in Euclidean space.

I suspect these conditions are too strong, but identifying what doesn't work may help me shape my proof around the lack of these conditions.

• Now it seems that every Riemannian metric (and you always have such on a compact manifold) will do. Am I missing something? Apr 20, 2016 at 7:31
• David, I don't know what you (or the references you mention) mean by "regular" or "homogeneously regular" but the fact that for every point in every Riemannian manifold every small enough neighborhood is bi-Lip to the unit ball in the Euclidean space seems quite basic to me. Apr 20, 2016 at 10:04
• I don't quite understand the relevance. In the link above $M$ is non-compact and uniform bi-Lipschitz constants are discussed. In any case, I am not an expert, so maybe I am missing something obvious, moreover the MO system starts shouting that I am being chatty, so I'll stop it here. Apr 20, 2016 at 10:43
• The relevant notion here is that of the Riemannian exponential map. Most Riemannian geometry textbooks show that for sufficiently small $r$, the $r$-ball in the tangent space is bilipschitz in this sense to the corresponding ball in the manifold. To know more you need estimates on curvature and Jacobi fields. Apr 20, 2016 at 12:50
• You don't need the exponential map. It's true even if the metric is only continuous and is analogous to the fact that any continuous function on a compact manifold is uniformly continuous. Just take any (finite) cover of $M$ of balls. Since $g$ is positive definite on the closure of each ball, there exists a positive constant $a > 0$ such that $a^{-1} \le g \le a$ and therefore the ball with the Riemannian metric structure is bilipschitz to the Euclidean ball. Apr 20, 2016 at 14:33

This question could be seen by some as too basic, but I think it is precisely the kind of situation MO is for (question is not trivial to an outsider, but easy to answer to an insider of the field), so let me give a complete argument to show that the property you seek is true for any metric (only adding some flesh to the comments).

Edit Here is a modified argument, applying to continuous metrics (and not resorting to the exponential map) even when one wants to control the induced distance on the neighborhood, rather than its length metric.

Let $g$ be any continuous Riemannian metric on $M$ and fix $p\in M$. Consider any chart $\varphi: B\to M$ where $B$ is the unit Euclidean ball, such that $\varphi(0)=p$. Compare the metrics $g$ and $\tilde g:=\varphi_* g_0$ (where $g_0$ is the Euclidean metric) on the image $D:=\varphi(\frac12 B)$ of the ball of radius $1/2$. Since both are continuous (here we need the chart to be $C^1$, but not more), there are $k,K$ such that $k\tilde g\le g \le K\tilde g$ on the closure of $D$.

Denote by $\ell(\gamma)$ the length of a curve $\gamma$ on $D$ with respect to the metric $g$, and $\tilde\ell(\gamma)$ its length with respect to the metric $\tilde g$. Then the formula for length of curves immediately gives $$k\tilde\ell(\gamma) \le \ell(\gamma)\le K\tilde\ell(\gamma).$$ By taking the infimum, we obtain that the length metrics induced by $g$ and $\tilde g$ on $D$ are bi-Lipschitz.

Now, there is a small enough $r>0$ such that $k(1-2r)\ge 2K r$; consider the smaller Euclidean ball $rB$ and its image $D_r:=\varphi(rB)\subset D$. I claim that the restriction to $D_r$ of the length metric of $(M,g)$ is bi-Lipschitz to the length metric induced on $D_r$ by $\tilde g$.

The only thing to check is that we can restrict to curves on $D$, i.e. that given $x,y\in D_r$, there is no curve $\gamma$ on $M$ between them which is shorter than $\ell(\tilde\gamma)$, where $\tilde\gamma$ is the $\tilde g$ geodesic from $x$ to $y$ (which is the $\varphi$-image of a Euclidean segment). If there where, the curve could not stay in $D$, so that $\gamma$ would cross twice the annulus $D\setminus D_r$. Its $\tilde g$ length would be at least $1-2r$, hence its $g$ length would be at least $$k(1-2r)\ge 2Kr\ge K\tilde\ell(\tilde\gamma)\ge \ell(\tilde \gamma)$$ and we are done.

Of course, by compactness of $M$, one can take the constants $k$ and $K$ uniform over $M$. For a non-compact $M$, the above still holds non-uniformly.

The previous argument

Cover your manifold with finitely many charts $(\varphi_i)$ defined on Euclidean balls $(B_i)$, such that the images of the balls with half radius $(\frac12 B_i)$ still cover $M$. Each point has $D_i:=\varphi(\frac12 B_i)$ as a neighborhood. Compare for that chart $\varphi_i$ the metric $g$ on $\varphi_i(B_i)$ and the pushforward of the Euclidean metric, $\tilde g=\varphi_*g_0$. This second metric is isometric to a Euclidean ball by definition.

Since both metrics are smooth (in fact we only need them to be continuous, as mentioned by Deane Yang), there are constants $k$ and $K$ such that $k\tilde g\le g \le K\tilde g$ on the closure of $D_i$ (here is why we took balls of half radius, to use compactness).

Denote by $\ell(\gamma)$ the length of a curve $\gamma$ on $D_i$ with respect to the metric $g$, and $\tilde\ell(\gamma)$ its length with respect to the metric $\tilde g$. Then the formula for length of curves immediately gives $$k\tilde\ell(\gamma) \le \ell(\gamma)\le K\tilde\ell(\gamma)$$ and passing to the infimum, you get that $\varphi$ is bi-Lipschitz (in fact, all of this is a lengthy spelling-out of the fact that diffeomorphisms between Riemannian manifolds are always locally Lipschitz!)

Now, if you want to compare with a Euclidean metric the restriction to $D_i$ of the length metric on $M$ (as opposed to the length metric of the restriction of $g$ to $D_i$), then there is a small issue left: the geodesic of $M$ between two points of $D_i$ might go outside $D_i$. Then you can use the exponential map instead of any chart, which has the additional property that image of small balls are convex, taking care of this point.

• Nice answer. Is there a way to do the last step without the exponential map and valid for a continuous metric? Would a smoothing argument work? Apr 21, 2016 at 12:40
• @DeaneYang: I think you can avoid the exponential and deal with a continuous metric by controlling the length of a curve that goes out of $D_i$: replace $\frac12B_i$ by two balls, one small and the other large; define $k$ and $K$ using the latter, and say that the $g$ geodesic connecting two points of the image of the small ball either stays in the image of the large one (in which case we are done), or must be quite long to go out and back; choosing (a posterori) the small ball small enough, you can ensure that this can never happen. This may need some care on the order of quantifiers. Apr 21, 2016 at 13:28
• Thanks, Benoît! This is just the sort of thing I was after. I'm glad this wasn't entirely trivial. Apr 21, 2016 at 14:47
• Benoit, thanks. I had tried for a few minutes the approach you suggest but wasn't successful. Of course that says more about me than the idea. Apr 21, 2016 at 18:28
• @DavidRoberts: This was written some time ago, but yes I think I claimed this. And the needed regularity for the manifold is the needed regularity for the charts, i.e. $C^1$ (we only need to be able to pull back the continuous metric on $M$ to a continuous metric on the domain of the chart). Dec 9, 2017 at 16:08