Thread: Vector fields, tensor fields etc

1. So, in order to get to some problems with Gauge Theory I have been having over the last few months, let's do this.......

I suppose an n-manifold - for now, I insist on no particular properties, other that it is a manifold.

I suppose that for each point I can find find a vector space each of whose elements are tangent to at that point, and call this vector space as .

I now form the set-theoretic (disjoint) union of all such vector spaces, and call this the "tangent bundle over " - one usually writes for this beast (somewhat confusingly) Ask me why I insist on non-disjunction if you dare!

Recalling that I have placed no restrictions on my manifold, and accepting the fact that therefore I have not excluded the possibility of what is called an "embedding space", I may visualize my tangent vectors as "sticking out" into this embedding space. This is not good (which we may get to later), but first this....

Consider the 1-manifold which is usually called the real circle. Now consider the vector space at each point which is, essentially by definition, the real line - a vector space - , and remember we are not at present excluding the possible existence of an embedding space.

For visualisation, let us take our tangent vector under this (false) scenario to be, not exactly tangent to our circle but perpendicular to it, again a false scenario. Then it is not hard to see intuitively that the bundle is a "cylinder" of sorts. This is of course a manifold of dimension 2. We would be well advised to use this for intuition only, but note this......

The vector bundle over any n-manifold is also a manifold of dimension 2n.

Agreed? Good. So, again without excluding the existence of an embedding space, I want to define a vector field over any manifold as a section of its tangent bundle, where "section" can be taken more-or-less literally, that is to each and every point I may assign a vector, a point without insisting on any sort of uniqueness.

So this is where our problem starts - our embedding space. How do we know whether our section is a set of elements from our embedding space or whether - as we require - it is a set of elements from the bundle itself? Specifically we need to know what is meant by a tangent vector on an n-manifold where we deny the existence of an embedding space

More later (possibly), as this is more than enough for now  2. Originally Posted by Guitarist So, in order to get to some problems with Gauge Theory I have been having over the last few months, let's do this.......

I suppose an n-manifold - for now, I insist on no particular properties, other that it is a manifold.

I suppose that for each point I can find find a vector space each of whose elements are tangent to at that point, and call this vector space as .
You are already assuming a LOT more than just being a manifold.

To even start to talk about tangent vectors you need to assume that the manifold had some differentiable structure. It you assume anything less than then the definition of a tangent vector becomes a bit tricky.

Moreover, by assuming that you have some vector space each of whose elements are tangent to the manifold at some given point, you appear to be assuming that the manifold is embedded in some Euclidean space. Now, there are embedding theorems that show that one can embed a differentiable manifold in a Euclidean space of suitably high dimension, but the theorems that show this are non-trivial. Originally Posted by Guitarist I now form the set-theoretic (disjoint) union of all such vector spaces, and call this the "tangent bundle over " - one usually writes for this beast (somewhat confusingly) Ask me why I insist on non-disjunction if you dare!
You need not only the disjoint union, but also some means of gluing them together smoothly. There is a hell of a lot more to a bundle than just the disjoint union of a bunch of fibers. The relationship among the fibers is critical to the character of the bundle. One line bundle over the circle (note your example below) is the cylinder. But the Mobius strip can also be realized as a line bundle over the circle, and the cylinder and Mobius strip are completely different animals. Originally Posted by Guitarist Recalling that I have placed no restrictions on my manifold, and accepting the fact that therefore I have not excluded the possibility of what is called an "embedding space", I may visualize my tangent vectors as "sticking out" into this embedding space. This is not good (which we may get to later), but first this....

Consider the 1-manifold which is usually called the real circle. Now consider the vector space at each point which is, essentially by definition, the real line - a vector space - , and remember we are not at present excluding the possible existence of an embedding space.
Note the comments above. One has to be a bit careful with your visualization. Consider the circle as being just the unit circle in the plane. Then your visualization of a tangent vector as a vector lying along some line that is tangent to the circle is a valid way to visualize a tangent vector. But note that it is not a vector in the vector space (Euclidean 2-space) in which the circle is embedded, if for no other reason than the tail of the vector does not lie at the origin of the plane. Originally Posted by Guitarist For visualisation, let us take our tangent vector under this (false) scenario to be, not exactly tangent to our circle but perpendicular to it, again a false scenario. Then it is not hard to see intuitively that the bundle is a "cylinder" of sorts. This is of course a manifold of dimension 2. We would be well advised to use this for intuition only, but note this......

The vector bundle over any n-manifold is also a manifold of dimension 2n.
The vector bundle for a circle is a cylinder. That is the way mathematicians think of it.

The vector space that is the tangent space at a point of manifold of dimension n is also of dimension [if you don't assume a smooth manifold then this is done by fiat]. A vector bundle is a special case of a fiber bundle, and one requirement of a fiber bundle is that it is "locally trivial". Local triviality means that if one restricts the bundle to a sufficiently small neighborhood of a point then the bundle is homeomorphic to a Cartesian product of the fiber of the bundle (in the case of the tangent bundle the fiber is just n-space) with the neighborhood. Since every point has a neighborhood that is homeomorphic to n-space then local triviality shows that the bundle has dimension 2n. Originally Posted by Guitarist Agreed? Good. So, again without excluding the existence of an embedding space, I want to define a vector field over any manifold as a section of its tangent bundle, where "section" can be taken more-or-less literally, that is to each and every point I may assign a vector, a point without insisting on any sort of uniqueness.
A vector field is usually taken to be a smooth section. I assume that you intend that the field be smooth. If you are willing to allow non-smooth or even discontinuous vector fields then you can do that, but since such field are so general you will not be able to say much about them. Originally Posted by Guitarist So this is where our problem starts - our embedding space. How do we know whether our section is a set of elements from our embedding space or whether - as we require - it is a set of elements from the bundle itself? Specifically we need to know what is meant by a tangent vector on an n-manifold where we deny the existence of an embedding space

More later (possibly), as this is more than enough for now
Ah.

It is probably easiest to first define what is mean by a "tangent vector" to a manifold without recourse to any embedding space and only later deal with the relationship to that space.

There are several ways to define the notion of a tangent vector. See the various references on differential geometry in the "brief survey of mathematics" thread. In all cases the idea is to replace the naïve idea of a tangent vector with a partial differential operator that is essentially the directional derivative in the direction of what you think of as the tangent vector. That can be done as an equivalence class of smooth curves on the manifold, as derivations on the sheaf of germs of smooth functions at the point, or one can proceed globally with derivations on the algebra of smooth functions on the manifold and then localize. But the key idea is that tangent vectors at a point m on the manifold operate on smooth functions to produce what one would want to call the directional derivative of the functions in the direction of the vector at the point m.

A section of a bundle is always a function from the manifold to the bundle that, at a point m, has as its value an element from the fiber over m.

The way to relate the tangent space at a point m to vectors in the embedding space (if you have an embedding space) is to take a smooth curve, say mapping (-1,1) to the manifold that carries 0 to m and then take the derivative at m. That vector will be tangent to the manifold at m and can be thought of as an element in the tangent space at m. Two curves that produce the same tangent vector are equivalent, and that is how one identifies equivalence classes of (smooth) curves with tangent vectors.

Aside. If you are trying to learn this stuff from a physics book and are having trouble, then I suggest that you find a mathematics text on differentiable manifolds and try reading that. The precision will help.  Posting Permissions
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