Can we replace our use of cross products with the BAC-CAB rule?

Question

A trending question here asks a question about cross products which are related to the existence of an orientation on 3D space.

At some level what we are trying to do with cross products seems to involve trying to take the orthogonal component of one vector relative to another, $$\operatorname {orth}_{\mathbf u}\mathbf v = \mathbf v - \mathbf u \frac{\mathbf u \cdot \mathbf v}{\mathbf u \cdot \mathbf u},$$ e.g. taking the component of a force that points normal to some displacement vector from the center, and calling that the “torque” of the force. Playing around with it a bit, it was clear that this takes a more symmetric form when multiplied through by $u^2$ and indeed seemed to suggest something which is indeed a formulation of the BAC-CAB rule, $$\{\mathbf u, \mathbf v|\mathbf w\} = \mathbf u (\mathbf v \cdot \mathbf w) - \mathbf v (\mathbf u \cdot \mathbf w),$$where in 3D space $\{\mathbf u, \mathbf v|\mathbf w\} = \mathbf w\times(\mathbf u \times \mathbf v).$ But the above expression is cyclic, multilinear, and does not require the space to have an orientation.

Indeed viewing $\{\mathbf u, \mathbf v|\bullet\}$ as a more generic “antiproduct” we have it as a [1, 1]-tensor, so that in $\mathbb R^2$, $$ \left\{\begin{bmatrix}a\\b\end{bmatrix},\begin{bmatrix}c\\d\end{bmatrix}\middle|~\bullet~\right\} = \begin{bmatrix}0&ad-bc\\bc-ad&0\end{bmatrix}$$and in $\mathbb R^3$, $$ \left\{\begin{bmatrix}a\\b\\c\end{bmatrix},\begin{bmatrix}d\\e\\f\end{bmatrix}\middle|~\bullet~\right\} = \begin{bmatrix}0&ae-bd&af-cd\\bd-ae&0&bf-ce\\cd-af&ce-bf&0\end{bmatrix},$$ and so on, containing the elements of the cross product $\mathbf u \times \mathbf v,$ just not packaged into one vector in $\mathbb R^3$. Similarly it looks like one could always take an $N$-dimensional vector field and construct a curl-field as a [1, 1]-tensor field.

Given these sorts of examples it seems to me that the above constitutes some sort of isomorphism between the exterior square that the wedge product generates, and the antisymmetric [1, 1]-tensors of the space, with no need to refer to any sort of orientation of the space itself. Where it exists the orientation really just says that this antiproduct of the two vectors can, in the 3D case, be regarded as itself a vector.

Is this some sort of well-known seed for a general framework for replacing the cross product in rotation and magnetism and the like? If so, is there a “triple-wedge-product“ analogue? Or is this ultimately flawed to only work in 2D and 3D systems because it lacks the full set of indices for the orientation tensor and that tensor really is essential to understanding something about rotation/magnetism?

Answer 1

We're not usually "trying to take the orthogonal vector" when we're taking the cross product. We're actually looking for the plane of the parallelogram defined by the two vectors we're taking the cross product of, and it just so happens that in 3d a plane can be equivalently defined by the unique direction orthogonal to it (the cross product is a vector in that direction, with its length the area of the parallelogram spanned by the two vectors in the place). Mathematically, the dimension-agnostic definition of e.g. angular momentum is not via a cross product, but as the bivector that is the exterior product of position and momentum. In 3D, the pseudovector of angular momentum we know and love is the Hodge dual of this bivector, see also this answer of mine and this answer of mine.

Your "antisymmetric [1,1]-tensors" are particular ways to write the second degree of the exterior algebra, i.e. you're just writing bivectors in a non-standard way. Given the framework of the exterior algebra, it's also obvious that there are "triple-wedge-products", i.e. third-degree-elements of the exterior algebra, but in 3D these are isomorphic to scalars (up to reflection due to the properties of the Hodge dual), so 3D pseudoscalars are a non-empty class of examples for these objects.