Manifold question?

1. Mar 2, 2005

waht

Is the manifold a space defined by the metric tensor or is it a completetly different thing. I'm new to tensor analysis though.

Thanks.

2. Mar 2, 2005

3. Mar 2, 2005

jcsd

No, infact a manifold doesn't necessarily need to have a (global) metric dfeined in order to be a manifold! A manifold is basically anything that can be continously parameterized or more formally, is a set of points where every point has an open neighbourhood that is homeomorphic to Rn (where n is the dimension of the manifold). The metric tensor field defines the scalar product at each point on a manifold, it doesn't define the manifold as a manifold is still a manifold whether the scalar product is defined or not.

4. Mar 2, 2005

gvk

For comfort, think about manifold as a curve or surface embedded in 3D space. And work hard to know everything about those stuff including n-D hypersurfaces, Gauss curvature, Riemannian curvature, and parallel transport, you be ready to step up to the manifold's world without metric.

5. Mar 2, 2005

jcsd

Actually gvk things like curvature and parallel transport dpend on the properties of the metric tensor, a mainofld at it's most basic is really just a set o fpoitns where for each point we have set of other points which are 'near to it'.

6. Mar 2, 2005

waht

re

that makes more sense now, so basically the parametric equations define a manifold.

"Actually gvk things like curvature and parallel transport dpend on the properties of the metric tensor, a mainofld at it's most basic is really just a set o fpoitns where for each point we have set of other points which are 'near to it "

and the wolfram describes the manifold as a topological space that is locally Euclidean.

I don't get this part.

7. Mar 2, 2005

jcsd

Okay by locally Eudlidean what is meant that for every point on the mainfold there is also a neighbourhood (if you like all the points that are less than x distance away) which is Euclidean, i.e. we can if we like treat this neighbourhood just like normal Euclidean space. So for example in general relativty the laws of special relativity are not true in a general sense but thanks to the fact that spavcetime is represnted by a manifold they are always true locally (i.e. they are true as long as we only talk about a small region of spacetime).

8. Mar 3, 2005

Peterdevis

I agree that a manifold is a topological space (by introducing open sets) that locally lokes like the euclidian space.
In my opinion you introduce (if you want it ar not) the standard metric with this last restriction(lokes like the euclidian space) in the topological space. I agree that at this level of the manifold you don' t use it. But when you define a calculus on that manifold you use the fact that the euclidian space is equiped with the standard metric.

9. Mar 3, 2005

mathwonk

an example of a manifold is a sphere.

If you include also the family of tangent planes to the sphere and a smoothly varying dot product on all these planes, you have a (Riemannian) metric.

a family of velocity vectors, v(p), one at each point p of the sphere, is an example of a "vector field".

The family of linear functionals, <v(p), > on tangent vectors defined by a vector field and a dot product, is an example of a "covector" field.

The family of dot products itself < , >(p), is an example of a "tensor field".

so naturally if you view your original sphere as embedded in three space, then the planes and dot product come along for free, and you do not notice they are extra structure.

And by the way, you do not need a metric to do calculus on a manifold, at least not to define derivatives, velocity vectors, and integrate differential forms. Only to measure arc lengths and curvature, volume, etc...

Last edited: Mar 3, 2005