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.


3 Answers 3


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.

  • $\begingroup$ What is considered the "traditional abstract approach"? $\endgroup$
    – rschwieb
    Nov 8, 2012 at 21:44
  • $\begingroup$ The one you reference as on p 93 of the notes: Defining a manifold as a topological space equipped with an atlas of coordinate charts, etc. $\endgroup$
    – Neal
    Nov 8, 2012 at 22:12
  • $\begingroup$ Ah, I didn't see that: I'm not the OP. At first I suspected you meant the atlas definition, and now I'm realizing that the "coordinate map Atlas" definition is "coordinate free," somewhat ironically. It makes specific coordinates irrelevant by talking about all possible coordinatizations... $\endgroup$
    – rschwieb
    Nov 9, 2012 at 0:57
  • $\begingroup$ @rschwieb Whoops, my bad on the misidentification. Yes, I meant exactly what you suspected and then realized :) $\endgroup$
    – Neal
    Nov 9, 2012 at 4:06
  • $\begingroup$ The Nash embedding theorem guarantees that any Riemannian manifold can be isometrically embedded into Euclidean space of sufficiently high dimension. This is especially relevant because the metric is part of the algebra. $\endgroup$
    – mr_e_man
    Jul 15, 2020 at 0:41

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.

  • $\begingroup$ Do you know any concrete examples if this, manifestly corrdinste free manifolds that are not embedded. For example how would a sphere be defined in that framework. I'm thinking along the lines of some set with spinors as tangent vectors. Is there anything more obvious $\endgroup$
    – Kugutsu-o
    Jan 11, 2022 at 20:21

To answer your question I will give some context.

Firstly, the concept of Universal Geometric Algebra (UGA). The elements of UGA are called multivectors, among them are the special elements called vectors, pseudoscalars and scalars. The scalars are said to be identical to the real numbers- complex numbers are unnecessary. UGA is generated from an infinite dimensional vector space. One can generate every finite dimensional Geometric Algebras by the selection of it's pseudoscalar - hence the name UGA.

for precise mathematical details see chapter 1 of "from Clifford Algebras to Geometric Calculus" where the concept was first introduced.

For an up-to-date quick explanation see this tutorial

Secondly, the concept of vector manifold. A set of points (endowed with an algebraic structure) such that at each point there is a pseudoscalar of definite dimension. The dimension of the pseudoscalar is the dimension of the vector manifold. The algebraic structure imposed on the points is that of the vectors in UGA. By which I mean the points of the vector manifold are vectors in UGA.

The points in the vector manifold are NOT closed under addition and multiplication by scalars. They do NOT generate a vector space of finite dimension.

A manifold is then defined as a set of points isomorphic to a vector manifold. So in this way the "shape" of the manifold is given by the vector manifold and a manifold is the set of points which have the same "shape" but do not have the algebraic structure imposed on them. If you have a manifold which is locally like $\mathbb R^n$, $n$ is what I was calling the dimension of the pseudoscalar.

The way in which one constructs a vector manifold is similar to the way one constructs a GA, by the choosing of a psuedoscalar. The difference is that in the vector manifold the pseudoscalar is not constant, it is a function of points on the vector manifold. If the pseudoscalar is differentiable then so is the vector manifold. From this pseudoscalar one can calculate all quantities relevant to differential geometry i.e. the curvature.

See chapter 5 of the same book referred to above for precise mathematical details.

For an up-to-date explanation see chapter 19 of "Guide to Geometric Algebra in Practice". This chapter is available here

The name is appropriate as the points are abstract vectors in UGA, however it might be unfortunate as some misguidedly believe that vector manifolds are less general. This might be because GA has a metrical interpretation and so concepts such as inner product and pedusoscalar which do have very important metrical interpretations are confused at the abstract level. GA also has a projective interpretation where the inner product and the pseudoscalar are given different interpretations. These can be found in the article which Muphrid provided. (the definition of inner product in GA is the symmetrical part of the geometric product, the product which defines the algebra)

  • 2
    $\begingroup$ So you think that, by using the universal geometric algebra, the problem of the arbitrariness of an embedding is neatly ducked, since all finite spaces can be seen as embedded in the UGA? I think this is an interesting approach to the problem. I'm a little concerned, however, that by choosing a pseudoscalar, we may be confining ourselves to orientable manifolds, unless we're saying that this pseudoscalar is not necessarily single-valued. $\endgroup$
    – Muphrid
    May 22, 2013 at 16:30

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