# Is there a deeper meaning behind the “determinant” formula for the cross product?

We all know that for all vectors $\mathbf{a}, \mathbf{b} \in \mathbb{R^3}$, if $(a_x,a_y,a_z)^\top$ is the component form of $\mathbf{a}$ and similarly $(b_x, b_y, b_z)^\top$ is the component form of $\mathbf{b}$ then, the cross product can be evaluated by the formal determinant \begin{equation} \mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k}\\ a_x & a_y & a_z \\ b_x & b_y & b_z \end{vmatrix} \end{equation}

However, I'm not satisfied with such an understanding. I think the formula is too elegant to be a mere formal, memorisation aid. So is there a generalisation of the determinant, which has an obvious geometric meaning, the same way the determinant can be interpreted the area multiplication factor of a linear transformation?

In geometric algebra, one can generalise the determinant of the matrix using the wedge product. If $\mathbf{a,b,c}$ are vectors in $\mathbf{R}^3$ then the determinant of the matrix, which has the vectors $\mathbf{a,b,c}$ as columns is $|\mathbf{a} \wedge \mathbf{b} \wedge \mathbf{c}|.$ However, the wedge product is defined for general multivectors, so I tried evaluating $(\mathbf{ii} + a_x\mathbf{j} + b_x\mathbf{k}) \wedge (\mathbf{ji} + a_y\mathbf{j} + b_y\mathbf{k}) \wedge (\mathbf{ki} + a_z\mathbf{j} + b_z\mathbf{k})$ to obtain, $\left(a_y b_z-a_z b_y\right) \mathbf{jk} + \left(a_y-b_z\right) \mathbf{ijk}.$ However, the result doesn't have any obvious connection to the cross product.

## 2 Answers

The coordinate expansion of a wedge product is

\begin{aligned}\mathbf{a} \wedge \mathbf{b} &= (\mathbf{e}_i a_i) \wedge (\mathbf{e}_j b_j) \\ &= \sum_{i, j} a_i b_j \mathbf{e}_i \wedge \mathbf{e}_j \\ &= \sum_{i < j} \begin{vmatrix}a_i & a_j \\ b_i & b_j\end{vmatrix}\mathbf{e}_i \wedge \mathbf{e}_j.\end{aligned}

In $$\mathbb{R}^3$$, this can be related to the wedge product by a duality transformation (i.e. factoring out of a pseudoscalar for the space which such as $$\mathbf{e}_1 \mathbf{e}_2 \mathbf{e}_3$$), as in

\begin{aligned}\mathbf{a} \wedge \mathbf{b} &=\begin{vmatrix}a_1 & a_2 \\ b_1 & b_2\end{vmatrix}\mathbf{e}_1 \mathbf{e}_2+\begin{vmatrix}a_2 & a_3 \\ b_2 & b_3\end{vmatrix}\mathbf{e}_2 \mathbf{e}_3+\begin{vmatrix}a_3 & a_1 \\ b_3 & b_1\end{vmatrix}\mathbf{e}_3 \mathbf{e}_1 \\ &=\mathbf{e}_3\begin{vmatrix}a_1 & a_2 \\ b_1 & b_2\end{vmatrix}\mathbf{e}_1 \mathbf{e}_2 \mathbf{e}_3+\mathbf{e}_1\begin{vmatrix}a_2 & a_3 \\ b_2 & b_3\end{vmatrix}\mathbf{e}_1 \mathbf{e}_2 \mathbf{e}_3+\mathbf{e}_2\begin{vmatrix}a_3 & a_1 \\ b_3 & b_1\end{vmatrix}\mathbf{e}_1 \mathbf{e}_2 \mathbf{e}_3 \\ &=(\mathbf{a} \times \mathbf{b}) \mathbf{e}_1 \mathbf{e}_2 \mathbf{e}_3.\end{aligned}

The above factorization has made use of $$\mathbf{e}_i \mathbf{e}_i = 1$$, and $$\mathbf{e}_i \mathbf{e}_j = -\mathbf{e}_j \mathbf{e}_i, i \ne j$$.

Note that the wedge product has the geometric meaning of the area of the parallelopiped that are spanned by the two vectors, except that it is an oriented area, with a bivector orientation that equals the geometric product (or wedge product) of two perpendicular vectors that lie in the span of the plane.

In 3D you can have a mapping between the normal to that oriented area (i.e. the cross product). Which normal you get depends on the choice of the pseudoscalar used in the duality transformation.

The rationale for your choice of that multivector wedge product is not clear, so it is not suprising that it does not look anything like the cross product.

• you should specify that you are working in the the geometric algebra G^3 – user48672 Apr 10 '16 at 1:10
• @user48672 The OP's post specified that the vectors are in $\mathbb{R}^3$. After stating the coordinate expansion of the wedge product, I also then restricted attention to the $\mathbb{R}^3$ case. – Peeter Joot Feb 24 '20 at 15:22

The formula $A \cdot (B \times C) = \textrm{Det}(A,B,C)$ shows this the cross product can be thought of as the transpose of the linear map $\textrm{Det}(\cdot,B,C)$.

Using the notation of riemannian geometry (hodge star, sharps, and flats) another way to say this is that $A \times B=\star(A^\flat \wedge B^\flat)^\sharp$. This is the connection between the cross product and exterior product you were looking for.

Hmm. I actually just remembered that I answered a similar question here. I am thinking of the entries as columns rather than rows in the determinant here (because that is how my brain works), but I think you will see the connection.

You may also be interested in this question on MO.

• I upvoted your answer for now, because it isn't quite what I'm looking for. I understand that the cross product can be defined using the hodge star and sharps and flats. This isn't quite the generalisation I'm looking for, is something of the following form: Determinant ---> Generalised Determinant (i.e. something where the "determinant" form of the cross product makes sense) ---> obvious geometric equality between the cross product and the generalised determinant. – 11Kilobytes Jan 20 '15 at 17:30
• This might be asking for too much, so if in the next couple of days I don't receive a satisfactory answer, then I'll just accept your answer. BTW, M2O2C2 was awesome, when are you offering it again? – 11Kilobytes Jan 20 '15 at 17:32
• I am in the final stages of writing my thesis, so not this semester. I am working on a calculus 1,2,3 project in the same system now. I hope to also produce more videos, some 3d graphing interactives, and a differential forms section before running it again. – Steven Gubkin Jan 20 '15 at 17:34
• To ask for something specific, could you please elaborate on how to, derive the "determinant" form of the cross product from any of the notions you gave here? – 11Kilobytes Jan 20 '15 at 17:51
• Done! Thanks for your kind words about m2o2c2 btw: I think the course is a little wonky, but I really hope to improve it in the near future. – Steven Gubkin Jan 20 '15 at 18:21