# Abstract manifolds or do we need an ambient space?

I'm currently studying on J.Lee's "Introduction to smooth manifolds", but several other sources I consulted present the same line of thought.

The most natural description of the $n$-dimensional sphere $S^n$ follows from imagining the sphere sitting in $\mathbb{R}^{n+1}$, as the locus of points $x=(x^1,\dots,x^{n+1})$ such that $(x^1)^2+\dots+(x^{n+1})^2=1$.

It is then possible to introduce (for example) the stereographic projection of the points of the sphere and to construct the homeomorphism $\sigma: S^n \setminus \{N\} \mapsto \mathbb{R}^n$. Writing the analogous map $\tilde{\sigma}: S^n \setminus \{S\} \mapsto \mathbb{R}^n$, I get an atlas for $S^n$ and I can prove that it is indeed a smooth manifold.

What puzzles me is that the coordinates of the ambient space have been used to describe the homeomorphisms, while it is always said that manifolds exist in their own right, without the need of embedding them in a bigger space (and I'm aware - thanks to this past discussion - of the difficulties and limitations of the "embed everything"-approach). But then I don't know how to describe smooth charts, i.e. how to specify the point of the manifold that gets a certain coordinate representation under the homeomorphism.

Moreover, when studying the differential geometry of curves and surfaces in $\mathbb{R}^3$, I used to parametrize the object with functions from $\mathbb{R}$ or $\mathbb{R}^2$. This looks as defining the homeomorphisms of my local charts in the opposite direction, but always imagining the curve/surface living in $\mathbb{R}^3$.

So, the questions are:

• Speaking about $S^n$, is there an "intrinsic" way to associate its points with those of $\mathbb{R}^n$ to build the local charts?
• How to proceed with a generic manifold? [Wild guess: As long as I consider surfaces and their higher dimensional generalizations, it is useful and perhaps unavoidable working with their parametrization; what I'm looking for comes out when I study stranger spaces.]

 Surfaces and their higher-dimensional generalizations are generic manifolds! What do you mean "stranger spaces"? – Aaron Mazel-Gee Sep 5 '12 at 22:08 @Aaron: Excuse my poor lexical choice! :) I mean things like the projective space, grassmannians, manifolds obtained by particular identification of points of more familiar objects... Even if they should become quite standard, I'm not used to them yet and always have difficulties in building some sort of visual intuition of what's going on. – Andrea Orta Sep 5 '12 at 22:49 Ah. Well for starters, $\mathbb{R}P^n$ has $S^n$ as a covering space, and the condition "being a manifold" is a local condition. For the Grassmannian $Gr(n,N)$, IIRC the quickest way to show it's a manifold is the "Schubert cell decomposition". You need to be careful with gluing manifolds together though, even in a nice, manifoldy way -- otherwise you might end up with something like the "line with two origins". (Most people require their manifolds to be Hausdorff.) – Aaron Mazel-Gee Sep 6 '12 at 11:15 As you very well know, a manifold is a topological space that is locally homeomorphic (or diffemorphic) to Euclidean space. The way I see the above description of the sphere, is that we define a topological space: when we say that the $n$-sphere is the set of unit vectors in $\mathbb R^{n+1}$, we are specifying a topological space. It just turns out that this construction gives us an "obvious" choice of charts. But in some sense, this definition could be set in any Euclidean space of dimension at least $n+1$; the embedding is irrelevant. – M Turgeon Sep 9 '12 at 21:35
Start with two copies $A, B$ of $\mathbb R^2$ and consider the map $f:A\setminus\{(0,0)\}\to B\setminus\{(0,0)\}$, $(x,y)\mapsto(\frac x{x^2+y^2},\frac y{x^2+y^2})$. By glueing $A$ and $B$ together using $f$, you obtain a sphere without mentioning $\mathbb R^3$.
 Thanks for your interesting example. Could you please explain what is glueing? I see your $f$ function is an inversion of $\mathbb{R}^2\setminus{(0,0)}$, with the circle as the set of fixed points: does glueing mean to identify all the points of $A\setminus{(0,0)}$ with their image in $B\setminus{(0,0)}$ under $f$? Then the identification of points along any ray of $\mathbb{R}^2$ gives a circle and the number of rays is bijectively related to the points of another circle, so intuitively I get $S^1\times S^1=S^2$, but this doesn't sound too rigorous to me! :) – Andrea Orta Sep 9 '12 at 15:39 Yes, glueing involves such an identification (and requires a few technicalities). However, we don't have $S^1\times S^1=S^2$. Note that $S^1\times S^1$ is a torus. – Hagen von Eitzen Sep 9 '12 at 15:51 Oh, sure. I made a stupid mistake. – Andrea Orta Sep 9 '12 at 16:19