Understanding Dual Manifolds in General Relativity

[Phrak]

Background.

We define vectors in general relativity as the differential operators

$$\frac{\cdot}{d\lambda}=\frac{dx^\mu}{d\lambda}\frac{\cdot}{\partial x^\mu}$$

which act on infinitessimals--dual vectors,

$$df=\frac{\partial f}{\partial x^\mu} dx^\mu \ ,$$

as linear maps to reals.

However, both vectors and dual vectors are elements of their own vector spaces which
recognize no distinction between vectors and dual vectors.

So.

Call the manifold which does recognized vectors as vectors and dual vectors as dual vectors, M. It would seem natural to look for a manifold M* where the vectors on M are dual vectors on M* and the dual vectors on M are vectors on M*.

Is this sort of dual manifold definable?

Observations.

It may be rather difficult to come up M* all at once, if at all. However finding the relationships for a single point, M_p* from a point M_p may be the best place to start.

The units of the manifold M* would be inverted. That is, (t,x,y,z) on M would become (\omega, k_x, k_y, k_z) on M*.

 

[Phrak]

I'm afraid I may have introduced some confusion with the nonstandard notation, $\frac{\cdot}{d\lambda}$.

The action of a vector on a dual vector is a scalar.
$df/d\lambda = S$ , where S is a scalar.

The dot stands in place of whatever function will stand to the right of the derivative with respect to lambda. We can't really leave the 'd' on top of $d\lambda$, because d is part of the dual vector df.

Using the usual notation where a vector is represented as $d/d\lambda$, the action of the vector on df is completely wrong! You get $d^2f/d\lambda^2$.

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In any case, this "dual manifold" may not be such a stupid ideal after all.

And besides it's all pervect's fault for giving me the idea in the first place.

The idea is to put, in correspondence, vectors on a manifold M with dual vectors on a manifold M*. This would be a good thing to be able to do, because some things that cannot be done with tensors with upper indices on M without using connections could be done with their corresponding dual vectors on M*.

Since we want to have some sort of equivalence, we can set the components of a vector on M equal the the components of it's dual vector on M*--and the converse. There seems no obvious way to have the basis vectors "pointing in the same direction", when using differential operators as vectors and infinitessimals as dual vectors, so we can ignore them for now.

Prepend everything on the manifold M* with a star, and use x for coordinates on M, and y for coordinates on M*.

$$\frac{\partial f}{\partial x^\mu} = \frac{\partial y^\mu}{\partial \lambda ^*}$$

$$\frac{\partial x^\mu}{\partial \lambda} = \frac{\partial f^*}{\partial y^\mu}$$

This works out nicely. Vector contractions on M are equal to their equivalent vector contractions on M*.

$$\frac{df}{d\lambda} = \frac{df^*}{d\lambda^*}$$