# Differential manifolds

1. Feb 6, 2012

### hedipaldi

Hi everebody,
I want to clear something.An n-dimentional differential manifoled is locally endowed by topologies defined by the metrices from the local parametrisations.I suppose that these topologies may all be different.Am i right?If i am mistaken ,then why?
thank's

2. Feb 6, 2012

### mathwonk

i think you are mistaken. the local coordinate systems not only define the same topology on overlaps but also the same differentiable structure.

3. Feb 7, 2012

### Bacle2

I'm not sure I understand your question, but the manifold as a whole is a metric space, i.e., it has a global/general topology generated by ( with topological basis ) the open balls in the metric, where the distance is given by the length of the shortest path between points.

4. Feb 7, 2012

### Alesak

It follows from definition on manifold. It must be locally homeomorphic to ℝn, so it inherits topology from there. A subset O of manifold is open iff for every chart (U; γ) of M a coordinate representation of intersection of O and U is open in ℝn.

I might have not put it in the best way, but in fact its quite simple.

You don't even have to define manifold as a topological space, but as a set with an atlas, and topology follows from that.

Ordinary manifold shouldn't be a metric space I believe. You cannot measure lenghts on it.

5. Feb 7, 2012

### Bacle2

The standard definition/conditions for a topological space to be a manifold I know of, guarantees its metrizability: 2nd countable, paracompact, Hausdorff, etc. and then use some variants of Urysohn metrization theorem to guarantee its metrizable. I'll look up some refs. for a more detailed argument.

6. Feb 7, 2012

### Bacle2

I just checked Wiki and they state that 2nd countable manifolds are metrizable. Urysohn's metrization says that Hausdorff +2nd countable +regular => metrizable. In my use, these conditions are assumed..

Still, I may be off in stating that in all manifolds the topology is determined by path length,(so that open balls B(p,r) are given by path length) only when we have a Riemannian metric, so that the manifold becomes a(n) intrinsic length space. Maybe someone can double-check on that. See the 'examples' section in:
http://en.wikipedia.org/wiki/Length_space , for more, and a better explanation.

Basically, a Riemannian structure on M allows us to define the length of paths
between points. Then we define , for x,y in M , d_L(x,y) := inf {L(γ): γ:I→M; γ(0)=a,
γ(1)=b} , where L is the length of the path. One can show this d_L is a metric, and the topology determined by this metric ( i.e., base elements are open balls B(p,r)) agrees with the intrinsic topology of the manifold.

And I think we only need the manifold to be C^1 , for it to allow a Riemannian structure.

Last edited: Feb 7, 2012
7. Feb 7, 2012

### Alesak

I capitulate, since I don't understand this area, but I think you mistook metric and metrizable space.

8. Feb 7, 2012

### Bacle2

Well, I am making a distinction between the Riemannian metric --more accurately metric tensor-- and a metric as a distance function in the manifold, if that is what you meant. My point is that a Riemannian metric gives rise to a distance function metric on the manifold by the method I stated in my previous post.

9. Feb 8, 2012

### hedipaldi

Thank's all for your constructive replies