# I How is a manifold locally Euclidean?

1. Dec 23, 2016

### FallenApple

So if I pick any 2 points on a 2d manifold, say x1 and x2, then the distance between these two points is a secant line that passes through 3 space that isn't part of the manifold. So no matter what, there doesn't exist an point epsilon, e , where ||e ||>||0 || and ||x2-x1||<|| e ||

No matter how small we shrink the neighborhood by decreasing the length of e, the distance, ||x2-x1|| is the distance of a secant line that does not lie on the curved surface itself. So it seems that there is no neighborhood on a surface that is euclidian.

2. Dec 23, 2016

### FactChecker

A 2d manifold will not be an open set in the 3d space, but it can have it's own metric and topology where there are neighborhoods similar to 2d Euclidean space.

3. Dec 23, 2016

### Staff: Mentor

... which is the main reason to consider manifolds! With them mathematicians got rid of the embedding.

4. Dec 23, 2016

### lavinia

A surface that is completely flat in some region will be Euclidean in that region in your sense.

However surfaces that can not be embedded in 3 space can be completely flat in the sense that locally they look exactly like flat pieces of paper.

No closed surface in3 space can be completely flat.

More generally a n dimensional manifold can look locally exactly like a piece of flat Euclidean space.

Last edited: Dec 25, 2016
5. Dec 24, 2016

### FallenApple

I think the common idea is the if you zoom in enough, the space looks euclidian. Kinda like how if you zoom in to an tangent line on a curve, the difference in y values approaches 0. Is that what you mean by similar in metric?

Topologically, it makes sense in that a surface and plane can be deformed into each one another.

6. Dec 24, 2016

### FallenApple

Oh ok. So that means that we need another notion resembling that of a neighborhood. Something that isn't the distance between an epsilion ball. Maybe the same definition but having the epsilion be something like a geodesic.

7. Dec 24, 2016

### FactChecker

And it can be of a lower dimension than the space it is embedded in.

8. Dec 25, 2016

### lavinia

This is true but it does not mean that the space is flat. It just means that the deviation from flat becomes small. In a truly flat space the deviation is exactly zero. The deviation from flatness can be detected with a quantity called the Gauss curvature. For a flat space the Gauss curvature is zero everywhere.

A topological deformation can create any shape at all. The process can erase the geometry and flatten out a curved space. Local flatness is not a topological property. It is a geometric property. For instance a sphere is nowhere locally flat.

You can not deform a closed surface into a region of a plane with out destroying its topology.

Last edited: Dec 26, 2016
9. Jan 3, 2017

### FallenApple

Do you mean that the gaussian curvature approaches 0 for small patches? Does this mean locally euclidian? Or does the curvature need to be exactly 0?

10. Jan 4, 2017

### lavinia

No. The Gauss curvature is defined at each point. A space that is Euclidean has exactly zero Gauss curvature at every point of the surface.

Last edited: Jan 4, 2017
11. Jan 7, 2017

### Lukas Juhrich

One should be careful in not mixing up the concepts of a manifold and a riemannian manifold. Your question is entirely focused on the definition of a manifold, which is, as you correctly seem to know, a locally euclidean hausdorff space, i.e. a space with some nice enough topology.
Note that this definition doesn't impose anything like a metric, metric tensor, or a concept of differentiability! Your analogy with the tangent line completely breaks down, since it is not ensured that a tangent line exists. A simple example would be a cube in $ℝ³$.
To be precise, “locally euclidean” means “locally homeomorphic to $ℝ^n$”, or equivalently to an open subset of the latter. Now, recall that homeomorphism denotes a continuous bijection, and does not introduce any more restrictions such as being differentiable (which wouldn't even be well defined in general).
Also, I discourage the use of the term ”flat euclidean space”, since it usually means “$ℝ^n$ equipped with the flat metric tensor”. Everything we need from $ℝ^n$ in this context is only its topology and no additional structure.

Considering your last quote, similarly, the notion of curvature is neither defined nor relevant in this realm.