## differentiable manifold not riemannian

I'm looking for a simple example of a differentiable manifold that doesn't have an associated riemann metric.

thanks
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 Recognitions: Science Advisor 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.
 Blog Entries: 1 Recognitions: Homework Help 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

## differentiable manifold not riemannian

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).

 Quote by zhentil 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.

 Quote by hamster143 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.

 Quote by zhentil 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 lets us go from local Riemannian metric to global metricity may diverge.
 I don't follow. |c'(t)| is a continuous function on a compact set. How could its interval diverge?
 Nevermind, I was wrong.
 Blog Entries: 1 Recognitions: Gold Member Homework Help The real two dimensional vector space R2.
 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.