I'm very unsure about all this business so please forgive any inaccuracies in my question.

Essentially what I'm having trouble understanding right now is how we "decide" the dimensionality of a manifold. For instance, take the 2-sphere, $S^2$. It's commonly used as an example of why more than one chart may be required to cover a manifold; in this example, a stereographic projection is used to map points on $S^2$ to a plane ($\mathbb{R}^2$). But couldn't we also just map points on $S^2$ to $\mathbb{R^3}$ using just one chart with the identity map, and say that $S^2$ is a three-dimensional manifold? Or, even map the points on the 2-sphere to $\mathbb{R}^1$, using an infinite number of charts, and thus call $S^2$ a one-dimensional manifold?

Basically I'm not sure if, in a manifold where we have charts $\phi_{\alpha}:U_{\alpha}\to \mathbb{R}^n$, the dimensionality of the manifold is given by $n$, since there is some ambiguity (at least I believe there is) as to what dimensionality of Euclidean space the charts are mapping points of the manifold to.


I think the concept of charts might be leading you astray here. A simple, practical and intuitive way to define the dimension of a manifold is the number of numbers you would need to locate a point on that manifold. For the case of a sphere, you need two, commonly written as $\theta, \phi$. For $\mathbb{R}^3$, you need 3, $x,y,z$.

Sure you can embed $S^2$ in $\mathbb{R}^3$, but this doesn't mean $S^2$ has somehow become 3-dimensional. If we use spherical coordinates in $\mathbb{R}^3$, the $S^2$ is embedded via the following equation: $r=\text{const}.$ So although there are 3 coordinates in $\mathbb{R}^3$, one must be fixed to a constant to embed the $S^2$.

There is no direct connection between the number of charts and the dimensionality of the space.

If you map the points of $S^2$ to $\mathbb{R}^1$, then this is not an isomorphism between two manifolds--it is a projection. Many points on the $S^2$ are being mapped to the same point on the $\mathbb{R}^1$.

  • $\begingroup$ In other words, the map $U_\alpha\subset S^2\rightarrow U'_\alpha\subset \mathbb{R}^1$ is not one-to-one, which is by the definition of the manifold required for all such maps. (@OP, not you, Commander.) $\endgroup$ – Ryan Unger Feb 10 '15 at 3:31
  • $\begingroup$ Doesn't the second paragraph only apply to a sphere centered on the origin? So you don't really need to fix anything constant for the general case of a sphere embedded in $\mathbb{R^3}$, and could end up needing to use 3 numbers to locate points on the sphere. $\endgroup$ – Physics Llama Feb 10 '15 at 3:34
  • $\begingroup$ So there's a distinction between embedding a sphere, and foliating the space with spheres. If I want to embed an $S^2$, I use the equation written above. I can foliate the space with spheres, which is essentially a fancy way of saying that spherical coordinates exist. But now I have 2 coordinates for where I am on the sphere, and another variable coordinate for "which sphere I'm on"--$r$. So this is a 3-dimensional manifold. $\endgroup$ – Surgical Commander Feb 10 '15 at 3:37
  • $\begingroup$ As for the map to $\mathbb{R^1}$ not being isomorphic, I think I see your point (or at least can't think of a way to construct a set of isomorphic maps that will do the trick), so I accept that you can't consider the sphere to be two-dimensional. $\endgroup$ – Physics Llama Feb 10 '15 at 3:37
  • $\begingroup$ @PhysicsLlama: The origin is arbitrary. And embedding does not change the dimension of the manifold. What what we're really interested in are the local homeomorphisms. Do we need more than $2$ coordinates locally to map the sphere? $\endgroup$ – Ryan Unger Feb 10 '15 at 3:38

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