I recently discovered Clifford's geometric algebra and its application to differential geometry. Some claim that this conceptual framework subsumes and generalizes the more traditional approach based on differential forms. Is this true?

More generally, are these frameworks strictly equivalent? I have heard that geometric algebra is only a suitable approach once a metric tensor has been chosen. Thus, it would seem that this approach is in fact less powerful.

Putting aside all personal and aesthetics preferences, which framework is more general (if any)?

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    $\begingroup$ Hestenes sure seems to think they are. I admit, I find his insistence on this point a bit unconvincing. Perhaps it's true that many (all?) manifolds can be embedded in some Euclidean space and that embedded manifold can be considered a vector manifold, but it's clumsy to have to embed everything under the sun, in my opinion. Still, the metric in GA is no more meaningful than the metric of that Euclidean embedding--it could have strong geometric meaning, but it might not, depending on the problem. I don't think that's a weakness of GA. $\endgroup$
    – Muphrid
    Feb 28, 2013 at 12:05
  • $\begingroup$ Thanks for your input! I found this question to be enlightening: math.stackexchange.com/questions/233110/…. Edit: I just realized you provided one of the answers to this question. $\endgroup$ Feb 28, 2013 at 12:22
  • $\begingroup$ I don't know enough about differential forms to talk about them, but isn't the geometric calculus version just a reexpression of differential forms? I could swear I ran across something like that while reading something recently... A large part of "geometric algebra" is reexpression of things in the notation of Clifford algebra. $\endgroup$
    – rschwieb
    Feb 28, 2013 at 14:50

2 Answers 2


Part 1

The differential forms approach is indeed very powerful. What Hestenes points out in his From Clifford Algebra to Geometric Calculus is that to give a complete treatment of differential geometry of manifolds, you need various structures. In the book, you will find an alternative. The starting point (as was pointed out above) is the notion of a vector manifold.

A vector manifold is an abstract set contained within an infinite dimensional abstract algebra. This can be given different interpretations. There are special elements of this algebra called vectors and others called pseudoscalars. The set of elements that define a vector manifold is a set of vectors. These vectors generate tangent space that are not part of the set. This is customary: one often invokes tangent spaces on manifolds. Notice that the vectors which define the vector manifold and the vectors which form the tangent space are very different. For visualization purposes, the elements of a vector manifold are called "points." The space tangent to a point generates a tangent algebra; this algebra contains ONE element called its pseudoscalar.

The function $I_m(x)$, which is pseudoscalar-valued and takes as arguments "points" of the vector manifold, characterizes the vector manifold. If this function is differentiable then one says that the vector manifold is differentiable. If so, all the differential geometry of a vector manifold can be carried out using only $I_m(x)$. Here, $m$ is the dimension of the tangent spaces, and this also defines the dimension of the vector manifold.

So the idea behind the definition is the following. If you have a manifold, its points lack algebraic structure, and so one imposes other structures as needed. If, on the other hand, you start with a vector manifold which is algebraically rich, then no further structures are needed later on. A manifold can be defined as a space which is isomorphic to a vector manifold. This isomorphism can be thought of as a mere stripping of the algebraic structure. A manifold can be treated as a completely abstract object, but for the purposes for which it was constructed, a geometrical interpretation is rewarding. This is also true for vector manifolds.

Some may think that vector manifolds are embedded in a Euclidean space or that one needs a metric, but this is not true. The main reason could be (this is my opinion) the nomenclature of GA. For example a vector manifold sounds like a special kind of manifold. Another case is the inner product. This product is defined algebraically and does not need a metric. It can be interpreted metrically and thus lead to very important developments, but it is only an interpretation. I use "only" in the sense that a metric does not define the abstract algebra or anything about it.

Part 2

Differential forms can live in the GA framework as follows. There is a unique $k$-vector for every $k$-form. A one-to-one correspondence.

Take the following example. Suppose you want to manipulate an expression before integrating. You do so and right before you integrate, you "multiply" by a differential. Then you integrate. You can "multiply" by the differential from the very beginning, but you do not need to do so and would probably not do so. The difference between a $k$-vector and a $k$-form is somewhat similar. If you already know what a $k$-form is, you could think of a $k$-vector as a $k$-form without the differential. If you don't know what a $k$-form is, you can study $k$-vectors and some of their mathematics and right before you need to integrate a $k$-vector, you turn it into a $k$-form. (I point this out because it is more or less how Hestenes proceeds in the book I mentioned above: forms are only strictly needed when integration is to be carried out.)

What I am pictorially trying to answer with the above example is "are these frameworks strictly equivalent?" When scalar-valued integrals are considered, the answer is YES. GA can also handle vector-valued integrals or in general multivector integrals. And integrals are Riemannian! Outside of integration theory, GA is more general. For example, the exterior derivative of differential forms has a counterpart in GA. Both are grade raising. GA also has a grade lowering derivative, but in differential forms, this can only be achieved with a metric. When comparing the two, one might get confused and think GA also needs a metric for this. This is not true. Furthermore, the sum of these grade raising and lowering operators is in actuality the fundamental derivative of GA. This derivative is invertible while the exterior derivative is in general not.

The overall answer to your question is geometric algebra (technically geometric calculus) is more general.

Comment. There are also other alternatives to geometric calculus which do not use the concept of a vector manifold. They are also more general by treating any manifold and use Clifford algebra structures and do not use coordinates. Some other approaches exist which use coordinates here and there but also use this algebra structure and are more general.

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    $\begingroup$ Can you provide references to the alternatives to Geometric Calculus that you're talking about? $\endgroup$
    – Luke Burns
    Feb 24, 2014 at 20:14
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    $\begingroup$ Geometric algebra inherently depends on a metric, because $v^2 = |v|^2$. Differential forms transform under any diffeomorphism, geometric algebra only under an isometry. The relationship between differential forms and GA is analogous to the relationship between vector spaces and inner product spaces. Of course you can choose an arbitrary metric/inner product, but that's usually not a good thing. $\endgroup$
    – Jules
    Jan 12, 2018 at 12:02

I think that two things must be distinguished here. One is the "traditional"/mainstream Clifford algebras, and the other, Hestenes'.

The traditional definition (which you'll find in pretty much any book) takes a vector space V, and builds an algebra out of its tensor algebra. In particular, it takes a quotient of the tensor agebra by the relation $v\otimes v = |v|^2 I$. Now, from looking at this definition, it is obvious that the vector space must come with a chosen quadratic form. A global version of this is the requirement of a metric on the manifold (if you're considering the tangent bundle, of course). On the other hand, because of the dependence of the algebraic structure on the metric, pretty much every "tangential" object lives in this algebra. For example, vectors and one-forms are naturally expressed as representations of the same element in different bases (cf. Snygg: Clifford Algebra, a Computational Tool for Physicists). Computationally, this is very useful: look, for example, at the definition of the Hodge dual in both the differential form formalism, and the Clifford formalism.

Mor formally, while the alternating and Clifford algebras are not isomorphic themselves, the alternating algebra is isomorphic to the graded algebra associated with a filtration of the Clifford algebra. This means that they are indeed very close to each other, the main difference being this: take vectors $v\in V$. While in $\Lambda^*(V)$ we necessarily have $v\wedge v=0$ (identifying $V$ with its dual with the metric), in $CL(V)$ this is not so. What happens is that taking the gradation implies to quotient out lower order terms. Another formulation of this fact relates to the multiplication of orthogonal vs non-orthogonal elements of $V$.

Now, Hestenes, in his book, has stripped the Clifford algebras to their bare algebraic properties, taking an axiomatic, rather than constructive approach. It is true that from this point of view, no assumption is made on the existence of a metric. But it is also true that it is not clear how to build such an object without choosing one (i.e., a construction different from the one above). As long as you ignore this point, his formalism can be very interesting.


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