Why is the cross product is alternating? In the book An introduction to manifolds by Tu, Loring W, it says that

The cross product $v\times w$ on $\mathbb R^3$ is alternating.

However in the definition of alternating, it says that

A $k$-linear function $f:V^k\to\mathbb R$ is symmetric, if
\begin{align*}
    f\left(v_{\sigma(1)}, \ldots, v_{\sigma(k)}\right)=f\left(v_{1}, \ldots, v_{k}\right)
\end{align*}
for all permutations $\sigma\in S_k$; it is alternating, if
\begin{align*}
    f\left(v_{\sigma(1)}, \ldots, v_{\sigma(k)}\right)=(\operatorname{sgn} \sigma) f\left(v_{1}, \ldots, v_{k}\right)
\end{align*}
for all $\sigma\in S_k$.


However, I can't understand why the cross product is alternating. I think that the cross product $\times$ can be viewed as a function $\mathbb R^3\times \mathbb R^3\to\mathbb R^3$, but not any function that maps to $\mathbb R$. So why this is alternating?
 A: If you expand the definition of alternating to linear functions from $\mathbb{R}^n$ to $\mathbb{R}^m$, one has the following.
If $u$ and $v$ are linearly independant vectors, then $u\times v$ is defined to be the unique vector such that

*

*$\|u\times v\| = \|u\|\|v\||\sin(u,v)|$

*$u\times v$ is orthogonal to the plane generated by $u$ and $v$

*$(u,v,u\times v)$ is a positively oriented basis

Thus, if $(u,v,u\times v)$ is positively oriented, then $(v,u,u\times v)$ is negatively oriented. Hence, $(v,u,-(u\times v))$ is again positively oriented. The norm of $-(u\times v)$ is the good one, and it is orthogonal to $u$ and $v$. By uniqueness, $-(u\times v) = v\times v$.
As the cross product is bilinear, and the only permutations on a set of two elements are the identity or the transposition, it follows that the cross product is alternating.
A: You're correct that using the given definition strictly, this doesn't make sense. However, there are a couple small changes in interpretation you could choose to make it work:

*

*Extend the definition of a symmetric and alternating functions to apply to $k-$linear functions $V^k \to W$ for any vector space $W$. (You can use the same defining formulae.) In the case of the cross product, it's alternating in this sense simply because $u \times v = - v \times u.$


*"Lower an index" on the cross product via the metric, i.e. consider $f(u,v,w) = (u \times v) \cdot w.$ This is then a 3-tensor on $\mathbb R^3$ that co-incides with the determinant.
