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1. It is not clear to me that linear duals, and not just Hodge duals, can be represented in geometric algebra at all. See, for example, here.

Can linear duals (i.e. linear functionals) be represented using the geometric algebra formalism?

2. It also seems like most tensors cannot be represented, see for example here. This makes intuitive sense since any geometric algebra is a quotient of the corresponding tensor algebra. Also it seems like only some contravariant tensors (and no covariant tensors whatsoever) can be represented unless the answer to 1. is no.

Which types of tensors admit a representation using geometric algebra?

3. The exterior algebra under which differential forms operate can clearly be represented by geometric algebra and its outer product.

However, do objects "sufficiently isomorphic" to differential forms admit a representation in geometric algebra?

This would be preferred given how geometric algebra is more geometrically intuitive than differential forms. See also here.

4. This question seemingly depends on the answers to 1. and 3., since derivations=vector fields are the linear dual of differential forms.

Can vector fields=derivations be represented using geometric algebra?

This paper seems to suggest that the answer is yes, although it was unclear to me. It also listed as references Snygg's and Hestene's books for representing derivations=vector fields via geometric algebra. However, I quickly searched Snygg's book and could not even find the use of the word "bundle" once, which seems to cast doubt on the claim.

Moreover, derivations are just the Lie algebra of smooth functions between manifolds, correct? Since Lie algebras are non-associative, it seems doubtful to me that derivations could be represented effectively by the associative geometric algebra. On the other hand, quaternions are somehow also Lie algebras, and they can be represented in geometric algebra, so I am not sure.

5. This probably a duplicate of 4. but I am asking it anyway.

Do tangent/cotangent spaces/bundles admit a representation using geometric algebra?

This one is especially unclear to me, since using "ctrl-f" the word "bundle" is not used even once in Snygg's book "Differential Geometry via Geometric Algebra", which appears to be the most thorough treatment of the subject.

(Incidentally, the word "dual" also only appears once, in reference to Pyotr Kapitza's dual British and Russian citizenship.)


Basically I am wondering if differential geometry can be "translated" completely using the language of geometric algebra. I think the answer is no because Hestene's conjecture regarding smooth and vector manifolds has yet to be proved (see the comments here), but it seems like we would run up with barriers even sooner than that. Although I probably am misunderstanding the comment.

I have found differential geometry difficult to understand at times, and would like to learn it by translating it as much into geometric algebra and then back. The extent to which the two is "equivalent" obviously presents a barrier to how much this is possible. Still, I already feel like I understand the concepts and motivations of multilinear algebra and related fields much better after having just learned a little geometric algebra, and would like to apply this as much as possible to the rest of differential geometry.

These questions are also related: symmetric products are the inner product from geometric algebra, and wedge products are the outer product from geometric algebra; geometric algebra is a special type of Clifford algebra which contains the exterior algebra over the reals; and this question discusses derivations in algebras in detail.

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    $\begingroup$ Lie brackets aren't necessarily non-associative. Some are, some aren't. Quaternion multiplication is associative. $\endgroup$ – user137731 Jun 26 '16 at 18:09
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    $\begingroup$ @Bye_World To be explicit, the Jacobi identity says that a Lie bracket is associative if and only if the Lie algebra is 2-step nilpotent (and thus the products involved in writing the associativity relation are zero). Such things are rare, but infinitely many exist. As for your actual question, one should make sure to note that you can't use geometric algebra unless you've equipped yourself with a metric in the first place, which loses some of the metric-independent appeal of differential forms. $\endgroup$ – user98602 Jun 26 '16 at 19:46
  • $\begingroup$ @MikeMiller To the best of my knowledge though even using geometric-algebra the integral is still metric-indpendent, see:physicsforums.com/threads/… Reference physicsforums.com/threads/… $\endgroup$ – Chill2Macht Jun 26 '16 at 19:58
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    $\begingroup$ @MikeMiller That's true. But I was just trying to explain that there is no discrepancy between quaternions forming a Lie algebra and quaternion multiplication being associative. The fact that Lie brackets aren't defined to be associative doesn't mean they can't be (under suitable conditons). $\endgroup$ – user137731 Jun 26 '16 at 19:59
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    $\begingroup$ @William I've started a bounty on this question to see if someone with more differential geometry expertise than you or I takes notice. In particular, that last (sub)question in your post is interesting and I'd like an answer if there is one. We'll see if it works. $\endgroup$ – user137731 Jul 5 '16 at 12:26
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Can linear duals (i.e. linear functionals) be represented using the geometric algebra formalism?

Yes and no.

In geometric algebra, dual vectors can be computed through Hodge duality. Let $\{u_1, u_2, \ldots, u_n\}$ be an orthogonal basis set for an $n$-dimensional vector space. Let $I$ be their geometric product, which is grade-$n$ due to orthogonality. Then $u^i = I u_i$ is, within a scale factor, a unique vector such that $u^i \cdot u_j = 0$ for $i\neq j$ but $u^i \cdot u_i \neq 0$ for nonzero $u_i$. Do a little more work normalizing $I$, and you would get the correct vector that corresponds to the element of the dual space that is dual to $u_i$.

So, geometric algebra lets you compute those vectors, but linear functionals themselves--as functions--have no place in the algebra. The algebra has elements and functions of elements, but I would hesitate to say that linear functionals are elements of the algebra.

That said, you can also construct a geometric algebra over the dual space.

Which types of tensors admit a representation using geometric algebra?

Any that you can suggest an isomorphism between tensors of that form and the algebra itself.

...yes, I know that borders on a non-answer, but let me give an example.

For instance, the linear map $T(a) = B \cdot a$ for vector $a$ and bivector $B$ is a tensor, moreover a linear operator. It's clear that this tensor directly, and uniquely, corresponds to $B$. $B$ entirely determines the action of the tensor.

Contrast this against the form of a general linear operator, $T(a) = \sum_i^n (a \cdot u^i) v_i$ for a basis set $u_i$ and some other set $v_i$, and you see that there is no such direct correspondence in the general case.

However, do objects "sufficiently isomorphic" to differential forms admit a representation in geometric algebra?

That's an easy one. You can write a $k$-form as a $k$-covector field. Any differential form can be written in terms of the algebra--perhaps with exception of "vector-valued forms" and other such things, but these are no more complicated in geometric calculus than they are in traditional differential forms. Doran and Lasenby or Hestene and Sobczyk both have extensive chapters on calculus with GA.

Can vector fields=derivations be represented using geometric algebra?

No, with a caveat: the geometric algebra is merely an algebra. It does not care what the underlying vector space is that it is built upon. It does not care whether vector fields are actually derivations.

So, GA can't represent vector fields being derivations because such a consideration is wholly separate from it.

In other words, if you want to take the wedge product of two vector fields and interpret that as meaning something in terms of derivations, that's on you. All GA says is that, if there is a meaningful metric you can impose on the vectors in this vector space, you can build a geometric algebra on it.

Do tangent/cotangent spaces/bundles admit a representation using geometric algebra?

The geometric algebra and its calculus can represent vector fields, but I'm not aware of any construction that allows it to invert things and recover the tangent bundle.

However, if I had to guess, I would say such a thing is probably the inverse of the unit pseudoscalar function on a manifold. Such a function is from $M$ to a grade-$n$ multivector, where $n$ is the dimension of $M$. Inverting this map would yield a map from a pseudoscalar to the manifold, which seems almost exactly like the tangent bundle. Such a function, however, would rely on the pseudoscalar admitting an inverse, which it might not do globally, and I can only imagine this making sense in terms of an embedding.

So where do we stand?

In my opinion, geometric algebra and calculus is more than capable of serving as a full foundation for someone studying differential geometry. Even if you throw away the notion of Hestenes' vector manifolds, you can still use geometric algebra and calculus to compute relations between vector fields or between differential forms. You can translate any differential forms expression into geometric algebra, and general tensors that don't correspond to GA elements can still be represented as linear functions on those elements instead.

There's already been considerable work on the relationship between GA/GC and differential geometry. I recommend Doran and Lasenby for this; they have an in-depth chapter building on Hestenes' vector manifold theory, in which they develop and expand on much of the calculus of GA. But moreover, they also have a chapter on general relativity, in which they develop an alternative to curved spaces for differential geometry, preferring "gauge fields" on flat manifolds instead. This method is superficially very similar to moving frames, and they use it to generate GA equivalents of the Cartan structure equations.

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  • $\begingroup$ Thanks so much for this detailed answer! I get that geometric algebra is just an "algebra", hence we only need to choose a construction that satisfies its properties. But in practice, if we want to say that certain formulations are equivalent, don't we need to specify some preferred construction? Or have some theorem like for the real numbers saying that all constructions of structures with the desired properties are equivalent in some sense? $\endgroup$ – Chill2Macht Jul 6 '16 at 3:35
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    $\begingroup$ @William When you construct an algebra (including the real numbers), you're really just constructing a model of the (abstract) algebra. The model is designed to have all of the properties associated with the algebra. Thus this answer wouldn't change at all if we choose some preferred model over another because all of the properties Muphrid mentions would be shared by each one. $\endgroup$ – user137731 Jul 6 '16 at 12:26
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    $\begingroup$ Yeah, I agree with @Bye_World here. It may be one would want to prove these consequences with one particular construction, but if one also proves that all these constructions are equivalent, then no construction is preferred in any canonical sense. $\endgroup$ – Muphrid Jul 6 '16 at 16:41
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    $\begingroup$ @William I'd just generalize from the multivector elements of a GA to sufficiently smooth/differentiable multivector-valued functions on a manifold, and I think that takes care of it, really. $\endgroup$ – Muphrid Jul 6 '16 at 18:07
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    $\begingroup$ To be honest, it sounds like you have enough material for a separate question. I know the moderators here would prefer not to have a discussion of much more length in comments, so that might be the best thing to do. $\endgroup$ – Muphrid Jul 6 '16 at 20:31
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2. Which types of tensors admit a representation using geometric algebra?

On p.4 of this document we can see that multivectors over a given Euclidean space $\mathbb{R}^n$ do not have arbitrarily high grade/order; instead the highest order possible is $n$ (the determinant/volume element). This is because of the two products available in geometric algebra, the inner (symmetric) algebra and the outer (exterior) algebra, the inner product is grade-reducing, while the outer product is always degenerate for $k>n$.

However, general tensors can have arbitrarily high grade, so it therefore follows that the tensor algebra on $\mathbb{R}^n$ cannot be embedded into the geometric algebra on the same space. And algebraically we would have expected this result anyway, since as a Clifford algebra on $\mathbb{R}^n$, the geometric algebra is isomorphic to a quotient of the tensor algebra of Euclidean space. From this is doesn't necessarily follow that the cardinality is strictly lesser and thus that no injection exists, but from this fact alone it would seem implausible that any multilinear embedding could exist, and indeed, as the first paragraph of this answer shows, it is impossible.

For grades $1-n$, it seems clear to me that the tensors which are elements of the symmetric and exterior algebras on $\mathbb{R}^n$ should all be representable in the geometric algebra, and those tensors which can be represented as linear combinations or geometric products of those elements, but otherwise I don't believe (although I am not 100% certain) that any other elements of the tensor algebra could exist inside of the geometric algebra.

Here is a similar question on Math.SE: what's the relationship of tensor and multivector. Basically all multivectors are tensors, but not all tensors are multivectors. These documents (1) (2) outline how to represent multivectors of the geometric algebra for $\mathbb{R}^3$ and the regular inner product as tensors.

3. Do objects "sufficiently isomorphic" to differential forms admit a representation in geometric algebra?

Differential forms are tensors on $\mathbb{R}^n$, but not just any tensors, they are alternating tensors, i.e. the exterior algebra on $\mathbb{R}^n$. Since the geometric algebra contains the exterior algebra as a sub-algebra, it follows that geometric algebra completely subsumes differential forms (perhaps not their vector space duals, but most certainly differential forms themselves). A detailed explanation of how that can be done is to be found here.

1. Can linear duals (i.e. linear functionals) be represented using the geometric algebra formalism?

The dual operation for geometric algebras is equivalent to what is called the Hodge dual in the language of differential forms. See page 13 of this document. In particular, like the simpler exterior algebra on $\mathbb{R}^n$, the geometric algebra is not an involutive algebra under its dual operation, in contrast to the vector space/tensor dual on the corresponding tensor algebra of $\mathbb{R}^n$. See here.

However, the tensor algebra on $\mathbb{R}^n$ is involutive with respect to its dual operation (transposition, the vector space dual operation). Thus, the Hodge/geometric algebra dual and the vector space dual are clearly not equivalent.

Since the geometric algebra is isomorphic to a subset of the tensor algebra, it should be possible to define dual (transpose/adjoint) objects for all of its elements in the vector space sense. However, I believe that we should only expect these dual/transpose/adjoint objects to exist inside the geometric algebra for even-grade bivectors, or at least for all dimensions greater than or equal to two. This is for at least two reasons.

1: Column vectors (tensors/multivectors of rank 1) are clearly not closed under transposition, whereas square matrices (tensors/multivectors of rank 2) clearly are. I am not familiar enough with higher order tensors to make any definitive statements about them.

2: The spinor algebra on $\mathbb{R}^n$, which is a sub-algebra of the geometric algebra consisting of all multivectors of even grade, appears to be closed under transposition (i.e. vector space duals), at least as far as I can tell. See here.

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