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 :)

  • $\begingroup$ The first thing to notice, as Troy alludes to, is that the set of vectors is closed under the cross product, but it is not closed under the exterior product. $\endgroup$ – rschwieb Jan 17 '15 at 12:16

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.

  • $\begingroup$ Thanks for your fast response, and how would you define $\vec e_1 \vec e_2 \vec e_3 $? Haven't seen that done. Thanks for the awesome reference too :) $\endgroup$ – user2662833 Jan 17 '15 at 11:37
  • $\begingroup$ @user2662833 I added one paragraph. But you are really supposed to read the book I recommended. $\endgroup$ – Troy Woo Jan 17 '15 at 12:11
  • $\begingroup$ I'll be sure to check it out. Thanks again! $\endgroup$ – user2662833 Jan 17 '15 at 15:08

These two products are definitely not the same, but they are dual to each other in $3$-D.

There are a couple giveaways. For one thing, for every $a,b\in \Bbb R^3$, $a\times b\in \Bbb R^3$, whereas $e_1\wedge e_2\notin \Bbb R^3$. Instead, this wedge product is in the exterior algebra, but outside $\Bbb R^3$.

The second giveaway is that the wedge is associative, whereas the cross product is not associative.

Thirdly you can just stumble across some different behaviors like this: $e_1\times(e_2\times e_1)=e_1\times(-e_3)=e_2$.

But $e_1\wedge(e_2\wedge e_1)=-e_1\wedge e_1\wedge e_2=0$.


$ e_1 \wedge e_2 $ is a 2 vector also called a 2 blade. It represent a signed area of the parallelogram spaned by the two vectors. For three vectors it represens a volume, and so on for n≥2 vectors.

It is characteristic by that $ e_i \wedge e_i = 0 $. The volumes (area) are also calculated by $ \det(e_1,e_2,e_3, \cdots ) $ ( the determinant composed of the column vectors).


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