A Exterior Algebra Dual for Cross Product & Rank 2 Tensor Det

[MisterX]

The determinant of some rank 2 tensor can be expressed via the exterior product.
$$T = \sum \mathbf{v}_i \otimes \mathbf{e}_i \;\;\; \text{or}\sum \mathbf{v}_i \otimes \mathbf{e}^T_i $$
$$ \mathbf{v}_1\wedge \dots \wedge \mathbf{v}_N = det(T) \;\mathbf{e}_1\wedge \dots \wedge\mathbf{e}_N$$
The determinant is sometimes used to describe the cross product.
$$\mathbf{a} \times \mathbf{b} =det\, \begin{pmatrix} \mathbf{e}_1 & a_1 & b_1 \\\mathbf{e}_2 & a_2 & b_2 \\ \mathbf{e}_3 & a_3 & b_3\end{pmatrix}$$
I am wondering if we may represent this as
$$det\, \begin{pmatrix} \mathbf{e}_1 & a_1 & b_1 \\\mathbf{e}_2 & a_2 & b_2 \\ \mathbf{e}_3 & a_3 & b_3\end{pmatrix}\mathbf{e}_1\wedge \mathbf{e}_2 \wedge\mathbf{e}_N = \left( \sum_{i=1}^3 \mathbf{e}_i \otimes \mathbf{e}_i \right) \wedge \mathbf{a} \wedge \mathbf{b} $$
or something along those lines. Does this describe the cross product and how does it generalize compared to the Hodge dual from $\wedge^{N-1} V \to V$ or other such duals?

 

[jambaugh]

A couple of notes, the determinate form of the matrix is more formal than definitive since you are creating the oddity of a matrix of scalars and basis vectors. It is more a calculating convention using the formal determinant. Your use of it in that last bit is a bit unclear.

But in n-dimensions a determinant is the Hodge dual (up to sign) of the exterior product of the row vectors of the matrix in question (or the column vectors since it's transpose invariant).

In three dimensions you have duality between rank 2 anti-symmetric tensors (bivectors) and the rank 1 tensors (vectors). The cross product of two vectors is the Hodge dual of the exterior product (again subject to sign convention). In higher (as in N-)dimensions the rank 2 anti-symmetric tensors will be dual to rank N-2 anti-symmetric tensors so we cannot define a cross product except as the exterior product itself. You would rather have to design a multi-product combining N-1 vectors together to yield the dual of a vector. I.e. you could define a "triple cross product" in 4 dim or "quad cross product" in 5.

Now as to your questions, note that the form of T you have is not the most general operator form and it is specifically not the same as the matrix written in the cross product's determinate formula. So I'm not clear on whether the result you used will still hold in your application.

[I'd say more but it's 3:15AM and my eyes are starting to blur. Let me look at this in the morning.]

 

[chiro]

Hey MisterX.

Do you understand how to relate the determinant to a wedge product similar to how you can express a normal cross product as A X B = |A||B|sin(a,b)N where N is unit length?

It's basically the same thing except you have a wedge product with more rows and columns (so a sort of "generalization" of the cross product).