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](https://en.wikipedia.org/wiki/Lipschitz_continuity#Definitions), (*Edit*: **as a metric space, equipped with the geodesic metric**) to an open ball in $\mathbb{R}^n$ with the Euclidean metric.



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> 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?
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One could also ask if something slightly weaker is always possible, such as choosing a metric locally [quasi-isometric](https://en.wikipedia.org/wiki/Quasi-isometry) 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.



Compact manifolds locally bi-Lipschitz to Euclidean space
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.




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*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.


