enter image description here![enter image description here] From Clifford algebra to geometric calculus by d. Hestenes


The attempt above is to have the base manifold like its tangent bundle be a "vector manifold". This seems very practical since you don't need a function angebra, coordinates or large expressions to define tangent vectors since simply by subtracting two points you get a meaningful value because your points already are vectors. I'm curious why exactly and how this would work The way you get geometry of your manifold is by looking how the psudoscalar slides along it. In this particular case both on wiki and in the book string emphasis is made on intrinsic nature of vector manifolds and to me it doesn't seem that this is justified.

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    $\begingroup$ To justify the intrinsic nature of vector manifolds means, as far as I can see, to demonstrate that every manifold, in the sense of the usual definition in differential geometry, arises as a vector manifold, i.e. admits a smooth embedding into Euclidean space, i.e. the Whitney embedding theorem. $\endgroup$ – Ben McKay Feb 10 '20 at 13:55
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    $\begingroup$ What Hestenes calls a vector manifold is more commonly known as affine space. en.wikipedia.org/wiki/Affine_space $\endgroup$ – Deane Yang Feb 10 '20 at 14:13
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    $\begingroup$ @ben mckey if vector manifolds are not intrinsic than I don't consider them a coordinate free alternative to atlases and charts formulation of manifolds. $\endgroup$ – Ezio Feb 10 '20 at 14:42
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    $\begingroup$ @DeaneYang: Hestenes's definition of a vector manifold is: an embedded submanifold of a Clifford algebra, strangely. It is not merely an affine space. Hestenes has many followers who apparently feel that this definition is the best way forward for differential geometry, and that more abstract notions of pseudo-Riemannian metric are unnecessarily complicated. $\endgroup$ – Ben McKay Feb 10 '20 at 14:44
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    $\begingroup$ @BenMcKay: in regarding to the first comment, maybe we want Nash instead? A lot of the "power" of the vector manifold stuff seems to be using the Clifford structure and this requires an ambient pseudo-Riemannian structure, so probably the correct embedding theorem to think about is the isometric embedding. (Related: mathoverflow.net/a/127735/3948 ) $\endgroup$ – Willie Wong Feb 10 '20 at 14:49

The book of Hestenes does not give a definition of vector manifold. It says that a vector manifold is a set of vectors in a Clifford algebra, with some additional properties. It defines those properties in terms of a notion of interior, and a notion of boundary, neither of which are defined. So it relies on intuitive notions of interior and boundary. It defines tangent space at a point (of any set of vectors), as the set of velocities of curves through those vectors. Presumably the curves are required to be smooth enough to have velocity vectors. It states that at interior points, the tangent space is a vector space. But that can't be enough to define a manifold in the usual sense, because it doesn't say what an interior point is. A set of vectors much worse than a submanifold can have points at which all velocities of differential curves in the set are zero. So the property of having a vector space as tangent space does not decide for us how to define the notion of interior point. In the end, the authors are relying on prior experience with manifolds, especially surfaces, as is often the case for authors working close to classical physics, engineering or statistics.

So to answer the questions: there is no coordinate free definition of manifold, because every definition we currently have relies on some charts, or on being a submanifold of some other previously defined manifold, ending us up with Euclidean space. There is no way to fully justify Hestenes's definition, because he doesn't really have one, but we can say that Nash's embedding theorem proves the existence of isometric embeddings of Riemannian manifolds, and subsequent authors have generalized to pseudo-Riemannian manifolds.

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    $\begingroup$ You just said in a comment some people say this is the alternative way to do differential-geometry, using vector manifolds instead of pseudo riemann metrics. If hestenes doesn't have a precise definition of what he means who has? $\endgroup$ – Ezio Feb 10 '20 at 15:28
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    $\begingroup$ @Ezio: I am only really guessing that Hestenes's intension is to study manifolds embedded into Clifford algebras (subject to a nondegeneracy hypothesis which he explains in that book). You might ask Hestenes for a precise definition. $\endgroup$ – Ben McKay Feb 11 '20 at 13:28
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    $\begingroup$ Yes that's what's he's doing, but I don't see how that's different than an embedding into an Euclidean space. Embedding is still embedding. When you talk about standard charts and coordinates manifolds there is no embedding whatsoever. So a priori his approach can't be an alternative to the standard way no matter how clumsy he says the standard coordinate approach is. $\endgroup$ – Ezio Feb 11 '20 at 14:05

There seems to be quite a bit of confusion regarding what Hestenes means exactly by "geometric algebra". The subject matter of "geometric algebra" can be defined and understood precisely in the standard language of spin geometry as that part of the theory of Clifford bundles which relies on their isomorphic presentation as Kahler-Atiyah bundles, a presentation which follows from the classical Chevalley-Riesz isomorphism.

This definition of "geometric algebra" makes perfect sense in the ordinary language of intrinsic differential geometry and does not require an embedding into an ambient space -- in particular, the entire subject can be defined without using the notion of "vector manifold". In this formulation, "geometric algebra" is merely that part of spin geometry which relies on systematic application of the Chevalley-Riesz isomorphism in order to translate a certain class of problems in spin geometry into equivalent problems for differential forms and polyvector fields --- an approach which can be very useful in certain situations.

See also the answers to the following questions:

What's "geometric algebra"?

Clifford algebras as deformations of exterior algebras

  • $\begingroup$ I don't see how this answers the problem. The main point hestenes tried to make is that exactly what you seem to be talking about is just a redundancy ridden formalism that he tries to simplify in what he calls geometric algebra. Also geometric algebra is coordinate less, so it doesn't use coordinates. Thaugh his calculations are coordinate less, they seem to unequivocally imply an imbedding. I don't see how this is not the case, that is the question $\endgroup$ – Ezio Jun 29 '20 at 13:21
  • $\begingroup$ @Ezio I disagree that intrinsic differential geometry would be redundant. Since the 19th century, it has been the general opinion of mainstream mathematicians that extrinsic formulations based on embedding a manifold into an ambient space should be avoided if one is interested in the intrinsic content of differential geometry. The formulation of Hestenes appears to be coordinate independent only because it relies on an embedding of a manifold into a Clifford algebra, but such embeddings are highly non-unique. That's why his followers appear to be disconnected from the mathematics mainstream. $\endgroup$ – amathematician Jun 29 '20 at 13:31
  • $\begingroup$ This is exactly what I'm saying. He explicitly states his formulation is intrinsic yet I can't see how. At no poit did I say intrinsic geometry is redundant if anything, sometimes it's the other way around. I said that the mainstream intrinsic formulation has a lot of redundancy. Hestenes managed to remove this redundancy of structure in which he failed to retain the intrinsic formulation despite claiming otherwise. $\endgroup$ – Ezio Jun 29 '20 at 13:43
  • $\begingroup$ @Enzio I think that Hestenes's formulation is not intrinsic and that his claims to the contrary do not satisfy basic standards of mathematical clarity and rigor. Can you clarify what you mean when you state that the standard intrinsic formulation of differential geometry is redundant ? $\endgroup$ – amathematician Jun 29 '20 at 13:47
  • $\begingroup$ So I would like a simple, coordinate free manifold geometry theory without embedding. Hestenes has the coordinate free but not intrinsic , and mainstream formulation has a lot of redundancy and coordinates though it's intrinsic. $\endgroup$ – Ezio Jun 29 '20 at 13:48

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