Exterior Algebra Dual for Cross Product & Rank 2 Tensor Det

In summary, the determinant of a rank 2 tensor can be expressed through the exterior product, and in n-dimensions it is the Hodge dual of the exterior product of the row vectors of the matrix. In three dimensions, there is duality between rank 2 anti-symmetric tensors and rank 1 tensors, allowing for the cross product to be expressed as the Hodge dual of the exterior product. However, in higher dimensions, a multi-product would be needed. The form of T used in the conversation is not the most general operator form and may not yield the same result as the matrix in the cross product's determinate formula.
  • #1
MisterX
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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?
 
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  • #2
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.]
 
  • #3
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).
 

1. What is exterior algebra dual for cross product?

The exterior algebra dual for cross product is a mathematical concept that extends the traditional cross product operation in three-dimensional space to higher dimensions. It is used to compute the cross product of two vectors in a higher dimensional space, such as four-dimensional space.

2. How is exterior algebra dual for cross product related to rank 2 tensor det?

The exterior algebra dual for cross product is closely related to the rank 2 tensor determinant, as both involve the calculation of a determinant. In fact, the rank 2 tensor determinant can be seen as a special case of the exterior algebra dual for cross product.

3. Why is the exterior algebra dual for cross product useful?

The exterior algebra dual for cross product is useful in many areas of mathematics and science, including physics, engineering, and computer graphics. It allows for the calculation of cross products in higher dimensional spaces, which is necessary for many advanced calculations and applications.

4. How is the exterior algebra dual for cross product calculated?

The exterior algebra dual for cross product is calculated using the wedge product, which is a mathematical operation that combines two vectors to form a new vector. The resulting vector is known as the exterior product and is equivalent to the cross product in three-dimensional space.

5. Can the exterior algebra dual for cross product be extended to even higher dimensions?

Yes, the exterior algebra dual for cross product can be extended to any number of dimensions. In fact, it is a crucial concept in the field of differential forms, which deals with objects that behave like vectors in higher dimensional spaces.

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