Can we define this alternate version of the cross product?

Question

I've been playing with the definition of the cross product and am trying to grasp the atomic algebraic assumptions needed to define the unique cross product. I remember seeing a post that was saying that the cross product is necessarily the only vector valued vector multiplication that satisfies the distributive property with scalar multiplication and addition (1)-(2) below and some other simple assumptions. One such requirement would be that the product is orthogonal to both arguments (4) below. In trying to find these assumptions I have arrived at the following investigation.

The cross product is defined as an operation $\times : \mathbf{R^3}\times\mathbf{R^3}\rightarrow\mathbf{R^3}$ with the following algebraic properties.

(1) $c\mathbf{v}\times\mathbf{w} = \mathbf{v}\times c\mathbf{w} = c(\mathbf{v}\times\mathbf{w})$

(2a) $(\mathbf{v} + \mathbf{u})\times \mathbf{w} = \mathbf{v}\times \mathbf{w} + \mathbf{u} + \mathbf{w}$

(2b) $\mathbf{v} \times (\mathbf{u} + \mathbf{w}) = \mathbf{v}\times \mathbf{u} + \mathbf{v} + \mathbf{w}$

(3) $\mathbf{v} \times \mathbf{w} = -(\mathbf{w} \times \mathbf{v})$.

With these properties along with the assumptions

(i) $\hat{i}\times \hat{j} = \hat{k}$

(ii) $\hat{j}\times \hat{k} = \hat{i}$

(iii) $\hat{k}\times \hat{i} = \hat{j}$

we can derive the definition for such a product by computing $\mathbf{v} \times \mathbf{w} = (v_1 \hat{i} + v_2 \hat{j} + v_3 \hat{k}) \times (w_1 \hat{i} + w_2 \hat{j} + w_3 \hat{k})$.

My question is, is it possible to replace rule (3) with $\mathbf{v} \times \mathbf{w} = \mathbf{w} \times \mathbf{v}$ and assume only (i) $\hat{i}\times \hat{j} = \hat{k}$ to define a consistent multiplication? It seems like this shouldn't work but I haven't been able to find a contradiction yet.

An alternate question is: can we assume (1)-(2) with (i) along with

(4) $\mathbf{v} \cdot(\mathbf{v}\times \mathbf{w}) = \mathbf{w} \cdot(\mathbf{v}\times \mathbf{w}) =\mathbf{0}$

and derive (3) by contradiction?

Answer 1

No, if you require $v\times w=w\times v$, then even specifying (i)-(iii) alone is not enough to completely define the multiplication. So, if you specify (i) alone, then you definitely have too little.

Your properties (1),(2a),(2b) together say that $\times$ is a bilinear map $\Bbb{R}^3\times\Bbb{R}^3\to\Bbb{R}^3$. So, let us first understand linear and bilinear maps more carefully first. Say you have vector spaces $X,Y,Z$ over the same field $\Bbb{F}$ with $X$ and $Y$ having finite dimension, say $n,m$ respectively (in the above example, $\Bbb{F}=\Bbb{R}$ is the field of real numbers and $n=m=3$ and $X=Y=Z=\Bbb{R}^3$). Now, in order to completely specify a bilinear mapping $T:X\times Y\to Z$, it suffices to fix a basis $\alpha=\{v_1,\dots, v_n\}$ for $V$ and a basis $\beta=\{w_1,\dots, w_m\}$ for $W$, and to specify $T(v_i,w_j)$ for all $i,j$ (see here for the proof of this claim in the case of linear maps; I leave it to you to prove the analogue for bilinear, and more generally, multilinear maps).

So, in order to fully specify a bilinear map $T:X\times Y\to Z$, you have to specify a total of $nm$ pieces of information (i.e the values $T(v_i,w_j)\in Z$ for all $i,j$).

Now, let us come to the special case where $X=Y$, so we’re interested in bilinear maps $T:X\times X\to Z$. Then of course one has to specify a total of $n^2$ pieces of information. There are two special cases of interest:

• anti-symmetric $T$ (assuming further that $\Bbb{F}$ doesn’t have characteristic $2$, which is certainly the case for $\Bbb{R}$):, i.e a bilinear $T$ such that for all $x,y\in X$, $T(x,y)=-T(y,x)$. In this case, you only have to specify $\frac{n^2-n}{2}=\frac{n(n-1)}{2}$ pieces of information, namely $T(v_i,v_j)$ for all $i,j\in\{1,\dots, n\}$ such that $i<j$ (if you imagine storing the values $T(v_i,v_j)$ in an $n\times n$ matrix, then you have to specify the values strict upper triangle; from here, the entries on the main diagonal must be zero due to anti-symetry, and those on the strict lower triangle are the additive inverses of those on the upper triangle). In the case of the cross product, $n=3$, so $\frac{n(n-1)}{2}=3$, which is why you only needed to tell me what $i\times j, i\times k, j\times k$ are. From here I can figure out the rest (6 others) based on anti-symmetry.
• symmetric $T$: here, you need to specify a total of $\frac{n^2-n}{2}+n=\frac{n(n+1)}{2}$ pieces of information, i.e $T(v_i,v_j)$ for all $i,j\in\{1,\dots, n\}$ such that $i\leq j$ (notice the difference the $<$ vs $\leq $ makes in how much you need to specify). In matrix language, you have to specify the values on the upper triangle and the main diagonal; then those on the strict lower triangle are known. So, in the case of $n=3$, you need to specify $\frac{n(n+1)}{2}=6$ pieces of information. However, properties (i)-(iii) only give us 3, so this is insufficient to get a unique “cross product” (there are infinitely many symmetric bilinear maps which satisfy (i)-(iii)).