Tangent bundles

[Quetzalcoatl]

What is the correct thing to say when speaking of a curved Reimannian geometry:

 

1. The base manifold is curved, not the tangent bundle over it.

2. The tangent bundle is what's curved, not the base manifold.

3. The curvature of the base manifold is "induced" onto the tangent bundle over it. As a result, both are curved in the same way.

 

Huh?

[ajb]

Really, technically it is the tangent bundle that is curved. The (vector bundle) connection used in general relativity is on the tangent bundle. (You can also think of connections in a principle bundle, in this case we need the frame bundle. I don't think we need to think about this at the moment.)

 

 

Now, as the tangent bundle is an example of a natural bundle it is fine to make reference only to the base manifold. This is what happens in practice in both general relativity (physics) and differential geometry (mathematics).

 

A natural bundle is "a functor from the category of smooth manifolds to the category of (smooth) vector bundles such that local diffeomorphisms become vector bundle automorphisms".

 

In plain language given a smooth manifold you get for free all natural bundles over the manifold. The tangent bundle is a great example of this.

 

Let us employ local coordinates [math]\{x^{a}\}[/math] on a manifold [math]M[/math]. Furthermore, let us employ natural coordinates (adapted to the vector bundle structure) [math]\{x^{a}, y^{a} \}[/math] on the tangent bundle [math]TM[/math].

 

(Local) Diffeomorphisms on the manifold are of the form

 

[math]\overline{x}^{a} = \overline{x}^{a}(x)[/math]. (usual abuse of notation)

 

These transformations "induce" the transformations on the tangent bundle. The fibre coordinate transforms as

 

[math]\overline{y}^{a} = y^{b}\frac{\partial \overline{x}^{a}}{\partial x^{b}}[/math].

Edited July 11, 2010 by ajb

[Quetzalcoatl]

Thanks for the quick reply, I'm trying to generalize on this a little:

 

In a sense, what you've shown here is that the "obstruction" to defining a global coordinate system on a manifold finds itself manifested as the tangent bundle's curvature, makes sense. I didn't really understand what a natural bundle is, but I'm guessing the cotangent bundle is another example, and so are all tensor bundles.

 

What about the bundle of scalars over M (zero-rank tensors)? Is it also a natural bundle?

Do all bundles over a manifold inherit the manifold's curvature through the coordinate obstruction?

[ajb]

In a sense, what you've shown here is that the "obstruction" to defining a global coordinate system on a manifold finds itself manifested as the tangent bundle's curvature, makes sense.

 

In a sense you can think of the curvature as an obstruction to a global trivialisation of a bundle. Look up Characteristic classes. They measure the deviation from triviality of principle bundles and their associated vector bundles. Most of these are written in terms of curvatures.

 

A principle bundle is trivial if it admits a global section. This does not carry over to the associated vector bundles as their is always the zero section.

 

For the case of the tangent bundle, which is a vector bundle the thing to look at is the (tangent) frame bundle. The thing you are interested in is the triviality or not of this.

 

 

I didn't really understand what a natural bundle is, but I'm guessing the cotangent bundle is another example, and so are all tensor bundles.

 

Yes indeed. Also line bundles of densities, that is one can also include the Jacobian as part of the transformation rules for geometric objects.

 

What about the bundle of scalars over M (zero-rank tensors)? Is it also a natural bundle?

 

I have never thought about functions on a smooth manifold as sections of a bundle. Usually you think of them as sections of a sheaf as they are global objects. This is the structure sheaf of a smooth manifold. All the coordinate changes, the notion of smoothness etc is all encoded in this sheaf.

 

You could then consider the stalk of the sheaf. I don't know if this is really a useful notion here.

 

Slightly more generally one can think of sections of line bundles. In general a line bundle is a vector bundle of rank one. They are not natural bundles as the transition functions need to be specified.

 

Do all bundles over a manifold inherit the manifold's curvature through the coordinate obstruction?

 

There is in general no notion of the curvature of a manifold. Curvature usually refers to the curvature of a specified connection in a principle bundle or on a vector bundle.

 

If we denote a connection on a vector bundle as [math]\nabla[/math] (covariant derivative, I'll be lax about what this acts on and suppress any indices ) the curvature is

 

[math]F = \pm [\nabla, \nabla] [/math]

 

In Riemannian geometry you have a unique symmetric connection on the tangent bundle completely specified by the metric. This is the Levi-Civita connection. It's existence and uniqueness is known as the "fundamental theorem of differential geometry".

 

You must keep telling yourself that curvature refers to a connection, this can then tell you something about the bundles via characteristic classes.

[Quetzalcoatl]

Thanks a lot man, these questions have been bugging me for a while. I'm reading a book about the subject ("The Geometry of Physics" by Frankel) and your answers helped clarifying some subtle points.

[ajb]

An important theorem you must be aware of is the "Homotopic theorem" and the "triviality of bundles over contractable spaces".

 

Theorem (Homoptopy theorem of sometimes Axiom)

Let [math]E \rightarrow M[/math] be a fibre bundle. Let [math]f[/math] and [math]g[/math] be homotopic maps from [math]M[/math] to [math]M'[/math]. Then the fibre bundles [math]f^{*}E[/math] and [math]g^{*}E[/math] are isomorphic bundles over [math] M'[/math].

 

 

A direct corollary of this is that all fibre bundles over a contractable space are trivial. Intuitively, if the space can be shrunk to a point then all the fibre bundles must be trivial.

 

A simple example are line bundles over an interval of the real line. Any fibre bundle over a disk. Any fibre bundle over a sphere with a point removed.

 

A counter example is the circle and line bundles over it. We have the cylinder and the Möbius band. The first is trivial, the second has a twist.

 

This theorem can be very useful when breaking up spaces in to coordinate patches. You can sometimes "push" the non-triviality to a specific coordinate patch. For example, think about the a sphere. You can cut it up into three pieces, the north hemisphere, the south hemisphere and an equatorial band.

 

The hemispheres are all isomorphic to a disk and thus contractable, bundles over them are trivial. The equatorial band is homotopic to a circle, so it is here that the triviality or not is a non-trivial question.

 

I hope I have helped you think about bundles, rather than confuse you!