# Where does the right hand rule appear in the “tensor” definition of the cross product?

When doing the cross product of two vectors according to the usual geometric definition of $\mathbf{A}\times\mathbf{B}$ being perpendicular to both $\mathbf{A}$ and $\mathbf{B}$, it's pretty clear that some kind of convention has to be made, because there are, in fact, two directions perpendicular to $\mathbf{A}$ and $\mathbf{B}$. The right hand rule tells us which one to pick, but we could just as well have picked the left hand rule and everything would still be fine as long as we replaced "right" by "left" everywhere.

Recently I learned a different (but equivalent, obviously) way of calculating the cross product: $(\mathbf{A}\times\mathbf{B})_i = \epsilon_{ijk}A_jB_k$, where $\epsilon_{ijk}$ is the Levi-Civita symbol, antysimmetric in all its indices and satisfying $\epsilon_{123} = 1$. The thing is that I'm having trouble seeing where the right hand rule comes into play here; the right hand side of the equation looks like a perfectly natural formula, no arbitrary conventions involved. What am I missing?

• The right hand rule appears in $\epsilon_{123} = 1$. The left hand rule would corresponds to $\epsilon_{123} = -1$. Note that $\epsilon_{i,j,k}=det(e_i,e_j,e_k)$ where $(e_1,e_2,e_3)$ is the canonical basis of $\mathbb R^3$, and that $e_3$ is the unit vector orthogonal to both $e_1$ and $e_2$ obtained by the right hand rule. – Taladris Aug 24 '14 at 15:29

## 3 Answers

From the algebraic definition it follows that for linearly independent vectors ${\bf a}$, ${\bf b}$ the triple $({\bf a},{\bf b},{\bf a}\times{\bf b})$ is positively oriented with respect to the triple $({\bf e}_1,{\bf e}_2,{\bf e}_3)$. This means that the matrix expressing $({\bf a},{\bf b},{\bf a}\times{\bf b})$ in terms of $({\bf e}_1,{\bf e}_2,{\bf e}_3)$ is positive.

Since we draw our figures with a "right handed" orientation of the basis vectors ${\bf e}_i$ $\>(1\leq i\leq 3)$ the right hand rule for the cross product is a simple mnemotechnic device to assure correct drawing of ${\bf a}\times{\bf b}$.

The right hand rule comes in because the Levi-Civita symbol is antisymmetric, so if you switch the indices $j$ and $k$ you need to multiply by $-1$. Think of that $-1$ as "flipping your hand" when you switch the vectors.

Recall how you get the components of a vector: given a vector $v$, you can extract the components $v^i$ by using the basis vectors $e_i$: $v^i = v \cdot e_i$.

The components $\epsilon^{ijk}$ can be extracted in a similar manner.

But Muphrid, what are the $\epsilon^{ijk}$ components of? Clearly they're not components of a vector or covector.

Indeed, they're components of a trivector instead. In clifford algebra, we can form bivectors, trivectors, and the like using the wedge product ($\wedge$) of vectors. Clifford algebra also makes meaningful the dot product of trivectors and trivectors.

So when we write the Levi-Civita components, we can say we're doing this:

$$\epsilon^{ijk} = \epsilon \cdot (e_k \wedge e_j \wedge e_i)$$

(Yes, the ordering is backwards. This is a necessary convention so that $\epsilon^{123} = +1$ and not $-1$.)

The quantity $\epsilon$ is not just any trivector: it is, in particular, $e^1 \wedge e^2 \wedge e^3$. This ordering is what makes $\epsilon$ "right-handed". You could write cross products in terms of a different trivector $\sigma = e^3 \wedge e^2 \wedge e^1$ and get left-handed cross products instead.