Differentiable manifold not riemannian

In summary, a differentiable manifold without an associated metric is possible by not specifying a metric, or by embedding the space into a Riemannian space and making it path-connected. If the space is not second countable, the integral that lets us go from local to global metricity may diverge.
  • #1
redrzewski
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0
I'm looking for a simple example of a differentiable manifold that doesn't have an associated riemann metric.

thanks
 
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  • #2
Strictly speaking, "Riemannian" implies the metric is positive definite, so you can just use a metric with indefinite signature...

Or, you can simply not specify a metric. Voila! It's easy to define differentiable manifolds without metrics...you just don't give them a metric.

Or do you want an example of a differentiable manifold that is not metrizable? That might be more difficult. I can't think of any examples.

Edit: In fact, there can be no examples. Every differentiable manifold can be embedded in R^n for some n, and therefore can always be given a metric by taking the induced metric from R^n.
 
  • #3
It depends on your definition of differentiable manifold. If you only have that it is locally homeomorphic to Euclidean space and the overlapping coordinates give a differentiable function, then you can have weird things like the long line, which is defined as:
Pick an uncountable ordinal W. Take the set [0,1)xW (an uncountable number of copies of [0,1). This is essentially too long to be embedded into Euclidean space. I imagine it's not metrizable because if two copies of [0,1) are infinitely far apart the distance between them probably has to be infinite, but I can't think of a reason why so don't take that as fact
 
  • #4
The proof that every manifold has a metric (as well as the proof of Whitney's embedding theorem) relies on paracompactness. If you drop this requirement, you can have all sorts of aberrations.

In fact, if your space has a metric, it has to be second countable (delta-balls type argument).
 
  • #5
zhentil said:
The proof that every manifold has a metric (as well as the proof of Whitney's embedding theorem) relies on paracompactness. If you drop this requirement, you can have all sorts of aberrations.

In fact, if your space has a metric, it has to be second countable (delta-balls type argument).

We have to be careful with definitions here. In particular, it seems to me that a non-second-countable smooth manifold may be equipped with a local Riemann metric but not be a metric space.
 
  • #6
hamster143 said:
We have to be careful with definitions here. In particular, it seems to me that a non-second-countable smooth manifold may be equipped with a local Riemann metric but not be a metric space.
True, true. The space must be path-connected for what I said to hold.
 
  • #7
zhentil said:
True, true. The space must be path-connected for what I said to hold.

Not enough. If it's non-second-countable, the integral that let's us go from local Riemannian metric to global metricity may diverge.
 
  • #8
I don't follow. |c'(t)| is a continuous function on a compact set. How could its interval diverge?
 
  • #9
Nevermind, I was wrong.
 
  • #10
The real two dimensional vector space R2.
 
  • #11
I'm going thru Arnold's Math Methods of Classical Mechanics.

His definition of differentiable manifold looks to be as OfficeShredder says. Arnold assumes it is connected as well, but there doesn't appear to be the 2nd countable requirement (that Lee explicitly calls out for instance).

I was confused since Arnold calls out adding the additional structure of the riemann metric.

Thanks for all the clarifications.
 

1. What is a differentiable manifold that is not Riemannian?

A differentiable manifold that is not Riemannian is a type of mathematical structure that allows for smooth, continuous functions to be defined on it, but does not necessarily have a metric tensor that satisfies the laws of Riemannian geometry. This means that the concept of distance and angles may not be well-defined on the manifold, making it a more abstract object.

2. How is a differentiable manifold not Riemannian different from a Riemannian manifold?

A Riemannian manifold has a metric tensor that satisfies the laws of Riemannian geometry, meaning that distance and angles are well-defined. On the other hand, a differentiable manifold not Riemannian does not have this metric tensor, making it a more general and abstract mathematical structure.

3. What are some examples of differentiable manifolds that are not Riemannian?

Examples of differentiable manifolds that are not Riemannian include topological manifolds, which are manifolds that are locally Euclidean but do not have a metric tensor, and smooth manifolds with non-positive curvature, such as hyperbolic spaces.

4. What is the significance of studying differentiable manifolds that are not Riemannian?

Studying differentiable manifolds that are not Riemannian allows for a more abstract understanding of mathematical structures and can be useful in fields such as differential geometry, topology, and physics. These types of manifolds also have important implications in understanding spaces with non-Euclidean geometries.

5. Can a differentiable manifold not Riemannian be embedded in a higher dimensional space?

Yes, a differentiable manifold not Riemannian can be embedded in a higher dimensional space, just like a Riemannian manifold. However, the embedding may not preserve the metric or other geometric properties of the manifold, making it a non-isometric embedding.

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