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I have a doubt about the domain of the chart on a manifold. Suppose $M$ is a smooth manifold and that $(U, \varphi)$ is a chart on $M$, then $\varphi : U \to \mathbb{R}^n$ has $U$ as it's domain. That's fine to me. My doubt has to do with the comparison of this to the case of regular surfaces in $\mathbb{R}^3$. In that case, the surface $S$ is a subset of $\mathbb{R}^3$ and so, if $(V, \psi)$ is a chart on $S$, then I know that $\psi$ takes points in $\mathbb{R}^3$ and takes to $\mathbb{R}^2$. In other, words, I know that I can write:

$$(u,v)=\psi(x_1,x_2,x_3)$$

And so, I know many things, for instance, I know that to build the chart I'll need to find one expression of $x_1, x_2,x_3$ which gives $(u,v)$ respecting the properties required. On general Manifolds, however, I feel a little confused. I mean, the manifold isn't inside another higher dimensional ambient space, so, if $(U, \varphi)$ is a chart, I would have:

$$(x^1, \dots, x^n) = \varphi(\text{what goes here ?})$$

And the $\text{what goes here ?}$ is because if I plug there some $m$-tuple of numbers, I'm supposing that $M$ is a subset of $\mathbb{R}^m$ and that supposition shouldn't be necessary. I've read for instance, that to find charts for the $n$-sphere, we can consider it as a subset of $\mathbb{R}^{n+1}$, but that isn't really necessary.

My point is, I know that elements of the codomain of a map of a chart will be $n$-tuples of numbers, that's fine, I know how to work with such objects. But how will be the elements of the domain, if we do not express the manifold as a subset of a higher dimensional ambient space?

I think I've made clear my point. If I've failed to explain my doubt, please ask and I'll try to explain better. Thanks very much in advance!

EDIT: The definition of Manifold I'm working with is the definition as presented by Manfredo Do Carmo: A smooth manifold of dimension $n$ is a set $M$ with a family of bijective maps $\varphi_\alpha : U_\alpha \to M$ from open sets $U_\alpha\subset \mathbb{R}^n$ to $M$ such that:

  1. $\bigcup_\alpha\varphi_\alpha(U_\alpha)=M$
  2. For each pair $\alpha, \beta$ with $\varphi_\alpha(U_\alpha)\cap\varphi_\beta(U_\beta)=W\neq\emptyset$ we have $\varphi_\alpha^{-1}(W)$, $\varphi_\beta^{-1}(W)$ open in $\mathbb{R}^n$ and $\varphi_\beta^{-1}\circ\varphi_\alpha$, $\varphi_\alpha^{-1}\circ\varphi_\beta$ are differentiable.
  3. The family $\left\{U_\alpha, \varphi_\alpha\right\}$ is maximum with respect to conditions 1 and 2.

The only point is that on my text above, I've decided to change the direction of the maps, but the definition is yet that one.

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What is your definition of a manifold? I believe that the answer to your question lies essentially in that definition. –  Nils Matthes Apr 3 '13 at 15:02
    
Hi @NilsMatthes, I've posted the definition of manifold I'm working with. –  user1620696 Apr 3 '13 at 17:29

1 Answer 1

up vote 2 down vote accepted

A manifold is a priori an abstract space, so it consists of a set together with a topology and a maximal atlas of compatible charts. The charts are pairs of domains and maps from that domain to some Euclidean space with fixed dimension. The domains are open subsets of the underlying topological space, so they are elements of the topology that comes with the manifold. If you want to have a connection with the case of regular surfaces, think about the surface as a set of points in space without any coordinate system attached to it.

Lastly, every manifold can be embedded into some high-dimensional space, so if you really must, you can think of it that way, but this view usually obscures things.

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