# Another manifold definition deficiency?

## Main Question or Discussion Point

Conventional manifold definition refers to the neighbor of every point having a Euclidean space description. http://en.wikipedia.org/wiki/Manifold" [Broken] But if most manifolds have additional property of some curvature, then won't such manifold definition actually be describing a tangent space i.e. not part of curved manifold?

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## Answers and Replies

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Hurkyl
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Conventional manifold definition refers to the neighbor of every point having a Euclidean space description.
Only the topology -- none of the other properties of Euclidean space are used. Note the fact that each chart is merely a homeomorphism -- an isomorphism of topologies -- as opposed to anything stronger like a diffeomorphism (isomorphism of differential calculus) or an isometry (isomorphism of geometry).

Manifolds are purely topological in nature -- they don't have any geometry or differential structure. Those are extra structure we might add in addition to being a manifold (e.g. differential manifolds and Riemannian manifolds)

Conventional manifold definition refers to the neighbor of every point having a Euclidean space description. http://en.wikipedia.org/wiki/Manifold" [Broken] But if most manifolds have additional property of some curvature, then won't such manifold definition actually be describing a tangent space i.e. not part of curved manifold?
Not withstanding homeomorphism definition of manifold, wherein one has 1:1 mapping and bicontinuity (attention to inbetweenness); might one also consider the conjecture (?) that perhaps most manifolds are non-smooth;that is, non-differential. Hence one could not consider the neighborhood of a 'point' nor a patch of manifold. So the concept of tangent space would not even be realizable. Also the word Euclidean, to most, implies the additional property of flatness (zero curvature). Such implied additional property of flatness is probably a special case; that is most manifolds probably have additional property of non-zero curvature. Hence even a limited use of the word Euclidean in a definition of manifold, would seem highly misleading, to other than mathematicians of course. http://en.wikipedia.org/wiki/Manifold#Differentiable_manifolds"

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Hurkyl
Staff Emeritus
Gold Member
most manifolds are non-smooth;that is, non-differential. Hence one could not consider the neighborhood of a 'point' nor a patch of manifold.
The former has absolutely nothing to do with the latter.

The definition of "topological space" says that every point has a neighborhood. The definition of "manifold" says that every point has a neighborhood with a specific property.

So the concept of tangent space would not even be realizable.
Putting a tangent bundle on a manifold is equivalent to putting a differential structure on the manifold, so those bits are equivalent. But this has nothing to do with neighborhoods or patches.

most manifolds probably have additional property of non-zero curvature.
Manifolds don't have curvature, be it zero or otherwise. Curvature is a property of Riemannian manifolds, and other similar structures.

To the OP: are you aware that a hemisphere and a disk are homeomorphic? That a square and a circle are homeomorphic? Not agreeing with your intuition <> deficient.

What you seem to be looking for is a local isometry condition: but by Cartan's theorem, this would imply that every manifold has euclidean space as a universal cover. I think you would agree this is a bit restrictive.

Conventional manifold definition refers to the neighbor of every point having a Euclidean space description. http://en.wikipedia.org/wiki/Manifold" [Broken] But if most manifolds have additional property of some curvature, then won't such manifold definition actually be describing a tangent space i.e. not part of curved manifold?
The answer to your question is yes. In order to have curvature one needs a tangent space and that requires more than the space being just locally Euclidean. It requires that the coordinate transformations on overlapping charts be at least twice differentiable(I think) since curvature is defined in terms of second derivatives.

I guess there is a question of the degree of differentiability of the coordinate transformations. If they are for instance, only C1, are they compatible with a C2 structure? I think Whitney did the research on this. But C2 should be all you need for curvature.

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