I was confused about the relationship between a set of basis vectors in 3D, $ \left\{\hat e_1, \hat e_2, \hat e_3 \right\} $ and their exterior products. In my head, it makes sense that the identity for the unit vector cross products would be true for the exterior products, an example of this would be:
$$ \hat e_1 \wedge \hat e_2 = \hat e_3 $$
My reasoning was that if I have two vectors $ \vec a = a_x\hat e_1+ a_y\hat e_2+a_z\hat e_3 $ and $ \vec b = b_x\hat e_1 + b_y \hat e_2 + b_z\hat e_3 $, their cross product would be:
$$ \vec a \times \vec b = (a_yb_z-a_zb_y)\hat e_1 + (a_zb_x-a_xb_z)\hat e_2 + (a_xb_y-a_yb_x)\hat e_3 $$
However, their exterior product would be:
$$ \vec a \wedge \vec b = (a_yb_z-a_zb_y) (\hat e_2\wedge \hat e_3) + (a_zb_x-a_xb_z)(\hat e_3\wedge \hat e_1) + (a_xb_y-a_yb_x)(\hat e_1\wedge \hat e_2) $$
Is the cross product just a special case of the exterior product, or is there something I'm missing? Of course, I'm basing this on the assumption that $\left[ \vec a\times \vec b = \vec a \wedge \vec b \right]$. Would it be completely wrong to say that $ \hat e_1 \wedge \hat e_2 = \hat e_3 $, or is that just a special case?
Thanks in advanced guys 🙂
Best Answer
They are not equal: $\vec a\wedge\vec b$ is a 2-vector, while $\vec a\times \vec b$ is just a vector. They are related by the Hodge dual operator: $$ \star: \vec e_i\wedge\vec e_j\mapsto \text{sgn}(\sigma)\vec e_k $$ where $\sigma$ is the permutation $(1,2,3)\mapsto(i,j,k)$.
In Clifford algebra $\mathcal{Cl}_3$, they are related by: $$ \vec a\wedge\vec b=(\vec a\times\vec b)\vec e_{123},\quad \vec e_{123}=\vec e_1\vec e_2\vec e_3. $$ Here, the Clifford product is defined by: $$ \vec e_i\vec e_j=\begin{cases} -\vec e_j\vec e_i & i\neq j\\ 1 & i=j \end{cases} $$ Knowledge of such can be easily picked up from Lounesto's Clifford Algebras and Spinors.