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In Milnor/Stasheff they give the definition of smooth manifold as follows (page 4):

A subset $M \subset \mathbb R^A$ is a smooth manifold of dimension $n \ge 0$ if, for each $x \in M$ there exists a smooth function $$ h: U \to \mathbb R^A$$ defined on an open set $U \subset \mathbb R^n$ such that

1) $h$ maps $U$ homeomorphically onto an open neighborhood $V$ of $x$ in $M$ and

2) for each $u \in U$ the matrix $[\partial h_\alpha (u) / \partial u_j]$ has rank $n$.

This differs from the definition I know: $M$ is a smooth $n$-manifold if for any two charts $\varphi, \psi$ the transition map $\varphi \circ \psi^{-1}$ is smooth.

Are these two definitions equivalent?

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  • $\begingroup$ If you're asking about only equivalence — yes, they're equivalent. As I understand the Milnor/Stasheff definition is just based on the concept of parameterized hypersurface and rephrases usual conditions for manifold in case of Euclidean spaces. $\endgroup$ – Evgeny Jul 29 '13 at 10:20
  • $\begingroup$ @user84127 it seems worthwhile to note that the Whitney embedding theorem(s) indicate that any finite-dimensional manifold can be diffeomorphically embedded into a euclidean space $\mathbb{R}^n$ with sufficiently high $n$. The theorem(s) even give an idea of what $n$ will work. So, in this sense, the definitions are not so far removed. However, I prefer the chart definition since it reflects the real abstraction which proper manifold theory intends. $\endgroup$ – James S. Cook Jul 31 '13 at 18:29
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The definition in Milnor and Stasheff is a bit of a hybrid between a purely "coordinate chart" definition you alluded to (requiring transition functions to be smooth), and a purely Euclidean space definition, which runs as follows:

An $n$-dimensional manifold is a subset of $\mathbb{R}^A$ (here $A$ may be much bigger than $n$) such that each point has a neighborhood which is the graph of a differentiable function over a suitable coordinate subspace $\mathbb{R}^n\subset\mathbb{R}^A$.

Note that in this definition we only need the standard coordinate planes (it is not necessary to take arbitrary subspaces), by the implicit function theorem.

For example, the circle in the plane is the graph of a function of type $\sqrt{1-t^2}$ near every point, either over the $x$-axis or over the $y$-axis.

The "coordinate chart" definition has the advantage that no apriori structure is assumed on $M$ (other than being a set). In particular, the topology results from the smooth structure imposed by the transition functions.

From this point of view, the Milnor-Stasheff definion has a disadvantage that we must already know about topological spaces and the notion of a homeomorphism.

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  • $\begingroup$ Thank you for this great answer. This purely Euclidean definition though doesn't seem to be equivalent to the one with charts: if I'm not mistaken the sphere $S^2$ would not be a manifold because we can't have a function with two values. From this I deduce that the definition in Milnor/Stasheff is less general than the definition with charts. $\endgroup$ – student Jul 30 '13 at 6:18
  • $\begingroup$ One question I have: where you write by the implicit function theorem -- could you elaborate a bit? I don't understand what you mean by the sentence. Thank you. $\endgroup$ – student Jul 30 '13 at 6:19
  • $\begingroup$ The sphere is certainly a manifold according to all definitions. Each point on the unit sphere in 3-space has a neighborhood where it is a graph of a suitable function of 2 variables over either the $xy$-plane, the $xz$-plane, or the $yz$-plane. The version of the implicit function theorem I have in mind is the following, in the 2-dimensional case. If the $y$-partial derivative of $f(x,y)$ is nonzero, then the level curve of $f$ is locally a graph over the $x$-axis. $\endgroup$ – Mikhail Katz Jul 30 '13 at 7:50
  • $\begingroup$ I think I understand it. If all the derivatives of $f$ are linearly independent vectors we may apply the inverse function theorem to write $(f_x,f_y)$ as $A\vec{v}=A(x,y)$ with $A$ invertible. Considering the level curve reduces it by one degree of freedom so that one may solve the system of equations $c=(f_x,f_y)=A(x,y)$ to obtain a function of $x$. $\endgroup$ – student Jul 30 '13 at 9:26

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