The determinant of some rank 2 tensor can be expressed via the exterior product.(adsbygoogle = window.adsbygoogle || []).push({});

$$T = \sum \mathbf{v}_i \otimes \mathbf{e}_i \;\;\; \text{or}\sum \mathbf{v}_i \otimes \mathbf{e}^T_i $$

$$ \mathbf{v}_1\wedge \dots \wedge \mathbf{v}_N = det(T) \;\mathbf{e}_1\wedge \dots \wedge\mathbf{e}_N$$

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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# A Exterior Algebra Dual

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