Locally euclidean 2D-surfaces In most places, a Riemannian manifold is defined to be a locally euclidean surface (with other restrictions). When thinking of 2D-surfaces, this raises two questions:


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*Does locally euclidean mean locally flat (angles of triangle sum to 180 degrees), or just that it can be represented by a 2D coordinate system?
Taking the example of a sphere without poles, both the sphere without poles and a tiny patch on it have a coordinate representation in R2. But only the small patch is (almost) flat. So which of the two properties are really locally euclidean? Or does one is a consequence of the other?

*In 2D, a regular surface with a metric can be considered a Reimannian manifold. So what property of a regular surface makes it locally euclidean?
 A: *

*$\newcommand{\Reals}{\mathbf{R}}$"Just that the manifold can be represented (locally) by an $n$-dimensional coordinate system."
In the context of the Mathworld definition of smooth manifold, "locally Euclidean" might better be phrased "locally Cartesian": locally, the space is modeled on the smooth structure of $\Reals^{n}$.
As Arthur says, a smooth manifold comes with no natural definitions of length, angle, or curvature.
If a differential geometer speaks of a locally-Euclidean manifold, by contrast, it may well mean a flat Riemannian manifold, in which small geodesic triangles have total interior angle $\pi$. It's therefore potentially ambiguous to speak of "locally-Euclidean" in the context of Riemannian geometry.
Incidentally, a Riemannian manifold is "pointwise Euclidean" in the sense that each tangent space comes equipped with a positive-definite inner product. Flatness, however, is a differential condition on the metric components.

*If $U \subset \Reals^{n}$ is a non-empty open set, a regular mapping $\Phi:U \to \Reals^{N}$ is an immersion, and each point $x$ in $U$ has a neighborhood $V$ such that


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*$\Phi(V) \subset \Reals^{N}$ is a smooth $n$-manifold, and

*$\Phi:V \to \Phi(V)$ is a diffeomorphism.
The flat (Euclidean) metric on $\Reals^{N}$ pulls back to a Riemannian metric on $U$, but generally this metric is not flat.
