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I am fairly new to differential geometry and approaching it with a physics background (in the study of general relativity), as a result I'm having a few struggles with terminology etc, so please bear with me.

As far as I understand the statement that a manifold $M$ is locally homeomorphic to $\mathbb{R}^{n}$ means that any point $p$ in a given subset $U$ of a manifold can be mapped to a point in a subset $V$ of $\mathbb{R}^{n}$ by a homeomorphism, $\phi :U\subset M \rightarrow V\subset\mathbb{R}^{n}$. As these "coordinate maps" just label the points $p\in M$ we are free to choose any coordinates we like to suit the problem we are considering. Now, if we introduce a metric on the manifold such that there is an intrinsic notion of distance between two points in a given patch on the manifold, then the homeomorphisms from the manifold to $\mathbb{R}^{n}$ do not in general preserve the distance between points on the manifold, such that the "coordinate distance" in $\mathbb{R}^{n}$ does not correspond to the intrinsic distance between them on the manifold $M$. Is this all correct so far?

Now, my issue is that I've read that in general one cannot construct a Cartesian coordinate map (i.e. one that directly relates coordinate distance in $\mathbb{R}^{n}$ and intrinsic distance on $M$. The standard coordinate system for Euclidean, flat space) from a given patch on a manifold to a subset of $\mathbb{R}^{n}$. Is this because as, in general, the geometry of a given patch on a manifold will be non-Euclidean and so it will not be possible to construct a homeomorphism between this patch and a subset of $\mathbb{R}^{n}$ such the intrinsic distance (i.e. the metric) is preserved, which is what is required for a coordinate system to be Cartesian? I understand that given a sufficiently small neighbourhood around each point the geometry of a manifold is flat and thus Cartesian coordinate maps can presumably be constructed, however, in general one will want to consider much larger patches of a manifold (particularly in general relativity), so is the reasoning a gave above correct (at least heuristically)?

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(1) Yes, it is more or less correct, just note that a metric is usually defined on a smooth manifold, which is a slightly richer structure.

(2) The simplest example is the standard 2-sphere (boundary of a $3$-ball with the usual metric). For any point of a sphere, some neighborhood is homeomorphic to a piece of $\mathbb{R}^2$ but this homeomorphisms never preserves distances. The reason is that, on the piece of $\mathbb{R}^2$, you have Pythagorean theorem that holds exactly, no matter how small the piece is. On the sphere, however, the Pythagorean theorem does not hold and the sum of angles in a triangle is never 180 degrees, no matter how small neighborhood of a point you take.

So, the statement that "given a sufficiently small neighbourhood around each point the geometry of a manifold is flat" is actually wrong, although it is a common mistake. It is only true if the Riemannian curvature is zero in every point of the given neighborhood. The topology is locally Euclidean but the geometry is not.

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  • $\begingroup$ thanks for taking a look. So is it often the case that a manifold is nowhere flat? Does this mean when authors put "locally homeomorphic to Euclidean space" they mean so in the loosest sense in that each point in a given local patch on a manifold can be mapped to a unique element in $\mathbb{R}^{n}$, and is not at all a statement about the manifold being locally flat?! ... $\endgroup$ – Will Jun 20 '15 at 16:34
  • $\begingroup$ ... Also, is what I said about Cartesian coordinate systems correct? (This has been a big sticking point for me as I immediately associated "locally homeomorphic to Euclidean space" with being able to construct the standard Cartesian coordinate system in that patch, which I know isn't correct). $\endgroup$ – Will Jun 20 '15 at 16:35
  • $\begingroup$ Homeomorphic means "continuous, bijection, and inverse is also continuous". Locally flat is something much much stronger. Homeomorphism says about the topology but nothing about distances/angles etc. Locally flat means that you really have triangles, distances, angles etc, locally with all the properties that you are used to from elementary school. $\endgroup$ – Peter Franek Jun 20 '15 at 16:37
  • $\begingroup$ So the homomorphism locally deforms the manifold to $\mathbb{R}^{n}$ such that it is locally topologically equivalent to $\mathbb{R}^{n}$? Is the reason why one can't generally construct a Cartesian coordinate map because the map will not be invertible (i.e. one maybe able to map from M to $\mathbb{R}^{n}$, but an inverse to this map will not exist), or is what I put in my original post a better way to look at it? $\endgroup$ – Will Jun 20 '15 at 16:48
  • $\begingroup$ It will always be invertible. Maybe look up the definition of homeomorphism in your books. But that has nothing to do with the metric, that's the point. The metric on the manifold may well be undefined. $\endgroup$ – Peter Franek Jun 20 '15 at 17:44

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