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I'm reading some notes about integration of differential forms and at the begining the author claims:

A $1$-manifold in $n$ dimensions is just a curve parametrized as $X: (a, b) \subseteq \mathbb{R} \rightarrow \mathbb{R}^{n}$ (Me: plus other conditions of smoothness, etc.). In general a $k$-manifold in $n$ dimensions is just the image of a function $X: D \subseteq \mathbb{R}^{k} \rightarrow \mathbb{R}^{n}$ (Me: with the aditional conditions).

My questions are: $(1)$ how can you come to such description of a manifold from the definiton of manifolds by charts, atlases and transition maps, and $(2)$ by the definiton I know, a manifold $M$ is of dimension $n$ (and is called a $n$-manifold) if for some chart (and hence for all) $(U \subseteq M, \phi)$, $\phi (U) \subseteq \mathbb{R}^{n}$. So how would I determine the dimension of a manifold with the definition of the notes (i.e. is the dimension $k$ or $n$)?

Antonio Velez
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  • Old fashioned, compare I §2 of Riemann's inaugural lecture: http://www.maths.tcd.ie/pub/HistMath/People/Riemann/Geom/WKCGeom.html – Michael Hoppe Nov 22 '13 at 14:13

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It appears the author is using a restricted definition, so all manifolds under consideration have a single chart. Thus the two definitions are not the same; the one from the notes is a special case of the more general one. (Assuming you have added enough conditions; you need to rule out figure-8s and so on).

However, the map $X$ is still a chart, and so the definition of dimension is the usual one; the dimension is $k$. This fits the intuition of the object being parameterised by $k$ independent parameters.

mdp
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    One should maybe add that there is also the definition of an abstract manifold, i.e. a topological space not necessarily embedded in euclidean space as $X:(a,b)\subseteq\mathbb{R}\to\mathbb{R}^n$ would suggest. These two notions are equivalent by Whitney's embedding theorem. – gofvonx Nov 22 '13 at 17:56