# Definition of manifold

From Wikipedia:

The broadest common definition of manifold is a topological space locally homeomorphic to a topological vector space over the reals.

A topological manifold is a topological space locally homeomorphic to a Euclidean space.

In both concepts, a topological space is homeomorphic to another topological space with richer structure than just topology. On the other hand, the homeomorphic mapping is only in the sense of topology without referring to the richer structure.

I was wondering what purpose it is to map from a set to another with richer structure, while the mapping preserves the less rich structure shared by both domain and codomain? How is the extra structure on the codomain going to be used? Is it to induce the extra structure from the codomain to the domain via the inverse of the mapping? How is the induction like for a manifold and for a topological manifold?

Thanks!

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Yes, exactly: the extra structure on the codomain is pulled back to give additional local structure on the domain. –  Zhen Lin Aug 14 '11 at 1:52
Thanks! How is the pull back like for a manifold and a topological manifold? –  Tim Aug 14 '11 at 1:59
I find your question confusing. What else is there to do? If you're going to say some object $X$ is locally like object $Y$, how can you do that in a way that gives $X$ more structure than $Y$? –  Ryan Budney Aug 14 '11 at 2:15
@Ryan: That is because I am confused. "If you're going to say some object X is locally like object Y, how can you do that in a way that gives X more structure than Y?" is also my question. –  Tim Aug 14 '11 at 2:25
–  Willie Wong Aug 14 '11 at 14:12

The reason to use topological vector spaces as model spaces (for differential manifolds, that is) is that you can define the differential of a curve in a topological vector space. And you can use this to define the differential of curves in your manifold, i.e. do differential geometry.

For more details see my answer here.

All finite dimensional topological vector spaces of dimension n are isomorph to $\mathbb{R}^n$ with its canonical topology, so there is not much choice. But in infinite dimensions things get really interesting :-)

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Yes, $\mathbb{R}^n$ has a richer structure than its topology, but we are not interested in those other structures. The topology of $\mathbb{R}^n$ is very special and corresponds to our most intuitive way of thinking of a "space". So, we are only asking that our manifold is a reasonable geometric object.

In other words, when you define a manifold you DO NOT want to use any other structure of an Wuclidian space, just the topological structure. That's why it's defined by a homeomorphism (a map that preserves the topological structure).

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Thanks! "when you define a manifold you DO NOT want to use any other structure of an euclidian space, just the topological structure." Then if replacing "Euclidean space" with a topological space in the definition of a topological manifold, will the definition still be the same? Similar question for the definition of a manifold, if replacing "a topological vector space over the reals" with a topological space. –  Tim Aug 14 '11 at 3:59
@Tim: No, the definition will definitely not be the same. A general topological space can have properties that are very different from those of Euclidean space, and the point is that you don't want your manifold to have any such strange topology. It should topologically look like Euclidean space. –  Hans Lundmark Aug 14 '11 at 8:11
@Hans: Thanks! Is there another purpose that the domain is endowed with the extra structure of the codomain via the inverse of the mapping? –  Tim Aug 14 '11 at 14:36
@Tim: I don't think I understand that question, I'm afraid... –  Hans Lundmark Aug 14 '11 at 14:55
@Hans: To make an analogy, in measure theory, a mapping from a measure space to a measurable space can induce a measure on the codomain. –  Tim Aug 14 '11 at 15:01