Relation between volume form and cross product

Question

Euclidean three-dimensional space (it's simpler). Defining $\eta={e^*}^1 \wedge {e^*}^2 \wedge {e^*}^3$, with $\{{e^*}^1,{e^*}^2,{e^*}^3\}$ dual of the orthonormal basis, and indicating the classic cross product with $\times$, is the relation $$\eta_{ijk} X^j Y^k = \frac{1}{6} (X \times Y)^i \quad i,j,k=1,2,3$$ correct?

I know it's trivial but I need to understand some mechanisms of volume forms, wedge product, Hodge operator, etc.

Answer 1

Since $$X\times Y=(X^2Y^3-X^3Y^2)e_1+(X^3Y^1-X^1Y^3)e_2+(X^1Y^2-X^2Y^1)e_3$$ then $$(X\times Y)^i=\eta_{ijk}X^jY^k,$$ is the correct formula for components.

In the other hand if we agree the volume form be $e^{*1}\wedge e^{*2}\wedge e^{*3}$ and defined by $$e^{*1}\wedge e^{*2}\wedge e^{*3}=\sum_{\sigma\in S_3}(-1)^{\sigma} e^{*\sigma(1)}\otimes e^{*\sigma(2)}\otimes e^{*\sigma(3)}$$ where $S_3$ is the symmetric group on three labels, then $$e^{*1}\wedge e^{*2}\wedge e^{*3}(X,Y,Z)=X\bullet(Y\times Z)=\det\left( \begin{array}{ccc} X^1&Y^1&Z^1\\ X^2&Y^2&Z^2\\ X^3&Y^3&Z^3 \end{array} \right) .$$ In the literature $X\bullet(Y\times Z)$ is called the triple product.

Answer 2

If you're already using exterior algebra, clifford algebra isn't that much of a stretch to use. It just relies on an associative product called the geometric product. In an orthonormal basis, its properties are

$$e_i e_j = \begin{cases} 1 & i = j \\ -e_j e_i & i \neq j \end{cases}$$

Elements of the clifford algebra are members of a $2^n$ dimensional, graded vector space. Vectors are the grade-1 elements, bivectors are grade-2 elements, and in the sense that all of these are elements of the $2^n$ dimensional clifford vector space, you can add vectors, bivectors, and so on.

With this product in mind, you can define $\eta = e_1 e_2 e_3$. You can also look at the geometric product of two vectors and see that

$$ab = a \cdot b + a \wedge b$$

The wedge product $a \wedge b$ can be written as a linear combination of $e_1 e_2, e_2 e_3, e_3 e_1$, so you can multiply it by $\eta$:

$$(\eta )(a \wedge b) = \text{vector}$$

Typically, though, the vector returned by the cross product is the negative of this.

$$a \times b = -\eta (a \wedge b)$$

You can verify this yourself by directly computing the components. For instance, $\eta (e_1 \wedge e_2) = e_1 e_2 e_3 (e_1 e_2) = -e_3$. You'll find that clifford multiplication by $\eta$ is conceptually the same as finding the Hodge dual, and one only needs to keep track of which minus signs are conventional.