# What's manifold?

1. May 5, 2012

### KFC

Hi there,
I find that the term 'manifold' appears in many book of statistical physics or classical mechanics while talking about phase space. I try to search the explanation on online but it is quite abstract and hard to understand what's manifold really refers to. Can anyway explain this a bit with a simplest picture? Thanks

2. May 5, 2012

### K^2

An n-dimensional manifold, in simplest terms, is a representation of non-Euclidean (in general) n-dimensional space that is locally Euclidean* at any point. In other words, you can define a local coordinate system anywhere, but not necessarily a global coordinate system.

Keep in mind, I am oversimplifying a bit. For formal definition, look up pretty much any topology text.

3. May 5, 2012

### Matterwave

One can think of a Manifold as any set of objects which can be described by a set of coordinate charts (called an atlas). A coordinate chart is a mapping between your set of objects to some Euclidean space R^n (n is then the dimension of your manifold). There's really nothing much more to it than that. If you can describe your set of objects in such a way, then it's a manifold.

4. May 5, 2012

### agostino981

I am a bit confused. Isn't it the "definition" of differentiable manifold instead of manifold?

5. May 6, 2012

### K^2

Could be. Did I make an assumption that mapping is differentiable when I said that coordinate system could be defined? I think you might be right about that. I'm sure the OP would be dealing with such, but I probably shouldn't have defined it so narrowly anyways. Matterwave's explanation is a lot closer to a formal, general definition of a manifold.

6. May 6, 2012

### HallsofIvy

A "manifold" is a geometric object that is locally Euclidean. More precisely, an n dimensional manifold is a topological space, together with a set of "pairs", $\{(M_\alpha, \phi_\alpha)\}$, in which one member of each pair, $M_\alpha$, is an open set and the other member, $\phi_\alpha$, is a continuous function from $M_\alpha$ to Rn.

We require, further, that if p is any point in the manifold there exist at least one $\alpha$ such that p is in $M_\alpha$ ( the open sets cover the manifold). We require, further, that in the intersection of two such sets, $M_\alpha$ and $M_\beta$, $\phi_\alpha o\phi_\beta$ be a homeomorphism from Rn to itself.

In order to have a differentiable manifold, we require that, in the intersection of $M_\alpha$ and $M_\beta$, both $\phi_\alpha o \phi_\beta$ and $\phi_\beta o\phi_\alpha$ be differentiable.

7. May 6, 2012

### K^2

Halls, OP states that he looked at former definitions. I think he was looking for a simpler illustration.

8. May 6, 2012

### haruspex

If 'o' here just means the usual composition of functions, there's some inversion missing. But then the functions must be 1-1 onto.
As I read it, the individual functions are required to be homeomorphisms. The composition of one with the inverse of another then is automatically a homeomorphism.

Going back to the original question, maybe some examples help. The mapping of the surface of the earth to flat maps by carving it into overlapping regions is an obvious one.