# Doubt regarding the dimension of a manifold

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.

## 1 Answer

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 $\mathbb{R}^1$.

• 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.) – Ryan Unger Feb 10 '15 at 3:31
• 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. – Physics Llama Feb 10 '15 at 3:34
• 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. – Surgical Commander Feb 10 '15 at 3:37
• 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. – Physics Llama Feb 10 '15 at 3:37
• @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? – Ryan Unger Feb 10 '15 at 3:38