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I observed some naive examples. Spheres, for example, when we cut out one point, can be embedded into $\mathbb{R}^n$. And if we cut out a measure zero set of a projective space, it can be embedded into the Euclidean space of the same dimension. So I wonder if all manifolds can be embedded into a same dimensional Euclidean space when we cut out a measure zero set? Can anyone prove it or disprove it by giving me some counterexamples?

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+1, interesting question! Another way of phrasing it is, how big can we make a chart? I tried to make a title that was easier to parse, but feel free to change it to something else if you want. – Zev Chonoles Mar 6 '13 at 1:35
@Zev: In your phrasing, the question is a duplicate of both… and also of…. I'm not sure what the policy is on closing because something is provable equivalent to a duplicate ;-). – Jason DeVito Mar 6 '13 at 1:46
One way to get at this question is to put a Riemannian metric on the manifold and then to consider how big the cut locus of a point can be. I believe it is known that the cut-locus of a point has measure zero, although I forget the proof. If this is the case, you could always remove a measure zero set and then use the exponential map at a point to get the required embedding. – treble Mar 6 '13 at 1:46
What is cut locus? – lee Mar 6 '13 at 13:16
Another idea would be to take a Morse function with a unique minimum. Then, take the union of all the ascending manifolds from higher index critical points. This is of codimension 1 and thus of measure 0. I'm pretty sure the complement is the ascending manifold of the minimum and thus diffeomorphic to a ball. Oh, wait, that's the link that Jason gave. Sorry. – Sam Lisi Mar 20 '13 at 13:39
up vote 5 down vote accepted

This question was essentially answered here, namely, that the cut-locus has measure zero (see the references provided in the link). I assume that your manifold $M$ is smooth, otherwise, I am not sure what notion of measure zero you would be using. I will also assume that $M$ is connected. (If not, apply this argument to each connected component.) Then, put a complete Riemannian metric on $M$, consider the cut-locus $C(p)$ of a point $p\in M$ and the open subset $U(p)=M\setminus C(p)$. The exponential map $\exp_p: T_pM\to M$ restricts to a diffeomorphism $V(p)\to U(p)$, where $V(p)\subset T_pM$ is a certain open subset (diffeomorphic to the open $n$-ball, where $n$ is the dimension of $M$).

One can ask a similar question in the context of connected $n$-dimensional topological manifolds $M$. Instead of removing a measure zero set, one can remove a closed nowhere dense subset. Then in all dimensions but 4 it follows from the results in the book by Kirby and Siebenmann that $M$ contains an open subset $U$ homeomorphic to $R^n$. (Maybe it was also known prior to their work, I am not sure.) I do not know what to say about dimension 4.

One last thing, Sullivan proved that every topological manifold $M$ of dimension $\ne 4$ admits a unique Lipschitz structure, i.e., an atlas where transition maps are locally Lipshitz. This implies that topological $n$-manifolds ($n\ne 4$) have a well-defined notion of measure zero sets (very indirect though). However, I do not know if the residual set in Kirby-Sibenmann will have measure zero in this sense, this goes well beyond my understanding of the work of their work and the one by Sullivan.

Edit. A better reference for topological manifolds (instead of Kirby-Siebenmann) is in my answer here, to papers by Brown (compact case) and Berlanga (noncompact case).

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I know, that every manifold can be equipped with a Riemannian metric. But what ensures, that we can put a complete Riemannian metric on every manifold? – hase_olaf May 18 at 9:00
@hase_olaf It is a theorem, and not a difficult one, that every Riemannian metric conformal to a complete one. Alternatively, take a pull back of the flat metrics using a proper embedding in Euclidean space (Whitney embedding theorem). The pull back will be complete. – studiosus May 18 at 13:16

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