Are all manifolds in the usual sense also "vector manifolds"?

Question

In geometric calculus, there is a concept of a vector manifold where the points are considered vectors in a general geometric algebra (a vector space with vector multiplication) which can then be shown to have the properties of a manifold (tangent spaces etc.). For a more precise definition of a vector manifold, see page 65 of http://montgomerycollege.edu/Departments/planet/planet/Numerical_Relativity/bookGA.pdf For what I believe to be the standard definition of a manifold see page 93 of the same book. This question doesn't seem to be answered inside the above book.

It seems that this approach is simpler than the traditional coordinate approach to manifolds since it is automatically independent of coordinates, except it might be less general. So, my question is: Are all manifolds to be vector manifolds or at least for every usual manifold to be isomorphic to some vector manifold? If not, are there simple additional criteria necessary to add to a manifold in order for it to be isomorphic to a vector manifold?

Note: I'm new to differential geometry, so feel free to correct anything above which may be incorrect. Also, thank you for any ideas/advice/answers you give.

Answer 1

The notes' definition of "vector manifold" in a geometric algebra $V$ is just as a subset $W\subset V$, but since he goes on to talk about smooth paths and tangent vectors, I expect that he means a smoothly embedded submanifold of $V$.

The Whitney embedding theorem guarantees that any smooth manifold can be smoothly embedded into Euclidean space of sufficiently high dimension. There are geometric algebras of arbitrarily high dimension (the Clifford algebra $Cl_n(\mathbb{R})$ is a vector space of dimension $2^n$). So any smooth manifold can be realized as a vector manifold.

The traditional abstract approach to defining manifolds is already automatically coordinate-free, and makes no reference to any ambient space. However, defining manifolds in ambient Euclidean space is equivalent to the coordinate-free approach - for example, standard textbooks such as Differential Topology (Guillemin-Pollack) define and work with manifolds entirely in ambient space.

Answer 2

I'm a few months late to the party, but I think Neal is somewhat off base. This is not Neal's fault, as the reference does not really emphasize the point. Hestenes often repeats the point that vector manifolds need not be embedded. (See, for example, this article.)

After some thought, I think I agree with him. Indeed, the point is almost trivial to see. Often, we take a point $x = x(x^\mu)$ to be a mapping from $\mathbb R^n$ to a manifold $M$. All Hestenes is adding on is that $x$ is a vector in some geometric algebra, as opposed to merely a coordinate tuple. This geometric algebra need not have a higher dimension than the manifold. Rather, assigning this structure automatically makes it such that the tangent space is the base space for the geometric algebra. The basis vectors $e_\mu = \partial_\mu x$ automatically fall out.

Can all points on the manifold be assigned vectors in this way, essentially using vectors in the tangent space to globally cover the manifold? Probably not. But this is no different than how most manifolds require more than one coordinate chart to be covered (see, for example, a sphere).

It is, admittedly, hard to be very very sure of all these things without a comprehensive text on vector manifold theory to prove all the results you would want, but it seems to work pretty well. Hestenes certainly claims that any manifold is isomorphic to a vector manifold, and I absolutely find the usual notion that tangent vectors are $\partial_\mu$ to be little more than voodoo. At the least, Hestenes is right to point out that doing so makes it impossible to use the algebraic structure of GA without tacking it back on. GA is very powerful, so such a loss is quite significant.

I do think Hestenes overplays how coordinate free vector manifolds are. Most important results in differential geometry are proven to be independent of coordinate chart. Rather, I think Hestenes is trying to say that in differential geometry we often say we're working in some arbitrary chart (even if we don't choose it) to do abstract calculations while vector manifolds allow you to cut out that step and not even talk about a chart at all.

At any rate, geometric algebra and geometric calculus are very powerful frameworks. Hestenes' definition of the vector derivative the limit of a surface integral as the volume shrinks to zero is, to me, particularly ingenious, and that requires no reference to embedding at all.