Coexistence of Tensor and Geometric Products in Multilinear Algebra

[Rasalhague]

They aren't equivalent in general, but do they ever coincide, and, if so, under what conditions? I've seen both denoted by juxtaposition. Is there a way to tell, in such cases, which is meant, or is it necessary to always use a different notation for the tensor product when the geometric product is denoted by juxtaposition? E.g. using juxtaposition for the geometric product, I think

$$\mathrm{d}x \, \mathrm{d}y = \mathrm{d}x \wedge \mathrm{d}y$$

since this is a 2-form, being the exterior derivative of $x \, \mathrm{d}y$, and therefore antisymmetric, but is this equivalent to

$$\mathrm{d}x \otimes \mathrm{d}y \enspace ?$$

And would I be right in thinking that

$$\mathrm{d}x^\alpha \otimes \mathrm{d}x^\beta \neq \mathrm{d}x^\alpha \wedge \mathrm{d}x^\beta \enspace$$

since

$$\mathrm{d}x^\alpha \otimes \mathrm{d}x^\alpha \neq \mathrm{d}x^\alpha \wedge \mathrm{d}x^\alpha \enspace = 0$$

 

[arkajad]

Wedge product normally applies to differential forms. Tensor product applies to all kind of tensors. Wedge product includes antisymmetrization.

You can also define wedge product to vectors and multivectors. Usually the associated name is "Grassmann algebra". Be careful with the use of "geometric product". Usually by geometric product one means Clifford algebra product (you need a scalar product to define it, Grassmann algebra can be thought of as a particular case of a Clifford algebra with zero scalar product.) Better use "exterior product" or "wedge product".

 

[Rasalhague]

My question was indeed about the "geometric product", in the Clifford alegabra sense. I'm aware that this is not the same thing as the wedge product, unless the result is antisymmetric. I mentioned the wedge product because of the relation $ab=a \cdot b + a \wedge b$. I was trying to use this to establish whether geometric and tensor products coincided in a particular case, and how to interpret the juxtaposition in expressions like $\mathrm{d}x \, \mathrm{d}y$ in integrals and descriptions of the exterior derivative (I think geometric product, happening here to coincide with the wedge product because of the antisymmetric property of forms), or in the definition of a metric tensor, $g = g_{\alpha \beta} \, \mathrm{d}x^\alpha \, \mathrm{d}x^\beta$ (I think tensor product, as the metric tensor has no antisymmetric part). I was wondering if there's any general rule about when juxtaposition means one thing, and when the other, in such expressions.

 

[arkajad]

Your formula for the Clifford product holds for vectors (or, by duality, 1-forms). For antisymmetric tensors it gets more complicated. For 1-vectors and 1-forms you can see that Clifford product coincides with wedge product if a and b ore orthogonal.

In your expression for the metric you evidently have tensor product (in fact, even symmetrized tensor product).

In multiple integrals wedge product is often omitted owing to conventions that some authors state explicitly and some not at all.

 

[Rasalhague]

Your formula for the Clifford product holds for vectors (or, by duality, 1-forms). For antisymmetric tensors it gets more complicated. For 1-vectors and 1-forms you can see that Clifford product coincides with wedge product if a and b ore orthogonal.

Okay, good.

In your expression for the metric you evidently have tensor product (in fact, even symmetrized tensor product).

Speaking of which, I was wondering if, rather than introduce the potentially confusing notation "juxtaposition = symmetrised tensor product" as the Wikipedia article "Metric tensor (general relativity)" does, it would be better to just state that a metric tensor must be symmetric, and let the components g_ij ensure that. Won't any specific formula for a metric tensor field, in any chart, give symmetric components without us having to also state that this is the symmetrisation of some other, presumably nonsymmetric tensor field?

In multiple integrals wedge product is often omitted owing to conventions that some authors state explicitly and some not at all.

Is this convention consistent with treating the justaposition of coordinate basis 1-forms in multiple integrals as a geometric product (i.e. can I always be sure that the geometric product will reduce to the wedge product in this case)? Is it consistant with treating their juxtaposition here as a tensor product, as well as or instead of a geometric product? When I see a bunch of 1-forms juxtaposed in this way, can I always put a wedge between them?

 

[arkajad]

Indeed, when it is clear that, in a give context, the metric tensor is symmetric, it is not necessary to symmetrize the tensor product. But some authors use non-symmetric tensors and still call it a "metric".

In integrals you do not have a "geometric product". You either integrate forms over manifolds or you integrate functions over domains of $$R^{n}$$. In good textbooks it is explained how integration of forms over manifolds is implemented by integrating functions over domains of $$R^{n}$$. What is what should follow from the context. But occasionally it may be somewhat confusing.

 

[Rasalhague]

In integrals you do not have a "geometric product". You either integrate forms over manifolds or you integrate functions over domains of $$R^{n}$$.

The latter being a special case of the former, right? (Scalar fields being defined as 0-form fields.)

In good textbooks it is explained how integration of forms over manifolds is implemented by integrating functions over domains of $$R^{n}$$. What is what should follow from the context. But occasionally it may be somewhat confusing.

Does the following argument work? When $\mathrm{d}x^i$ are exterior derivatives of the coordinate functions $x^i$ of $\mathbb{R}^n$, they're orthogonal, so the geometric product of any pair of them should coincide with their wedge product: $\mathrm{d}x^a \, \mathrm{d}x^b = \mathrm{d}x^a \wedge \mathrm{d}x^b$. By the Leibniz rule, and the fact that the exterior derivative of a scalar field is a 1-form field, and the fact that multiple application of the exterior derivative results in 0, for any values of a and b, $\mathrm{d}(x_a \mathrm{d}x^b) = \mathrm{d}x^a \otimes \mathrm{d}x^b$. Providing the coordinates are properly defined so as to be sufficiently differentiable, this 2-form field can be integrated over a region of $\mathbb{R}^n$, so the tensor product symbol shouldn't be out of place in an integral expression, and your statement in #4 that the wedge product is "often omitted" in multiple integrals suggests that the wedge product is also a correct notation to use in multiple integrals. Being a 2-form field, $\mathrm{d}x^a \otimes \mathrm{d}x^b$ this must be antisymmetric, which is consistent with the antisymmetry of the wedge product. So, in this context at least, we do have a geometric product that coincides with the wedge product and the tensor product: $\mathrm{d}x^a \, \mathrm{d}x^b = \mathrm{d}x^a \otimes \mathrm{d}x^b = \mathrm{d}x^a \wedge \mathrm{d}x^b$, where juxtaposition denotes the geometric product.

 

[arkajad]

You may like to read these online notes

http://www.nmt.edu/~iavramid/notes/diffforms.pdf

 

[Rasalhague]

Wow, thanks arkajad. This is great: something relating the geometric algebra concepts I've been reading about reecently to what I've learned about tensors. Although I don't think it mentions the geometric product explicity, it has lots of good and relevant stuff.