Curiosity about the wedge product and the Levi-Civita symbol

Let us use the definition of the wedge product of two vectors:

$$\vec{u}\wedge\vec{v} = \vec{u}\otimes\vec{v} - \vec{v}\otimes\vec{u}$$

writing $\vec{u}$ and $\vec{v}$ in dyadic form as $\vec{u} = u_a\vec{e}^a$ and $\vec{v} = v_a\vec{e}^a$ the wedge product above yields:

$$\vec{u}\wedge\vec{v} = u_av_b\vec{e}^a \otimes \vec{e}^b - u_av_b\vec{e}^b \otimes \vec{e}^a$$

$$= u_av_b(\delta_l^a\delta_m^b - \delta_l^b\delta_m^a)\vec{e}^l \otimes \vec{e}^m$$

$$= u_av_b\epsilon_{lm}^{ab}\vec{e}^l \otimes \vec{e}^m = u_av_b\epsilon^{ab}\epsilon_{lm}\vec{e}^l \otimes \vec{e}^m$$

Assuming all the above is correct, expanding the $a,b$ contraction above in say $\Bbb{R}^2$ does indeed give $u_av_b\epsilon^{ab} = u_1v_2 - u_2v_1$ and on the other hand the second contraction on the basis gives $\epsilon_{lm}\vec{e}^l \otimes \vec{e}^m = \vec{e}^1\otimes\vec{e}^2 - \vec{e}^2\otimes\vec{e}^1 = \vec{e}^1\wedge\vec{e}^2$. And I believe the above still holds even when the indices are in $\{1,2,3\}$

My question(s) is(are): following a similar analogy and trying to stick to as little 'new' notation as possible, extending the above to the wedge product of multiple vectors is proving problematic for me. And I cant seem to find any text that connects the wedge product to the Levi-Civita symbol intimately. So what is the definition for the product $\vec{u}\wedge\vec{v}\wedge\vec{w}$ were the vectors are in $\Bbb{R}^3$ in similar notation to above if correct?

• instead of doing the wedge product of vectors, physicists and mathematicians take wedge product of linear functionals (a.k.a covectors) as useful algebraic objects Sep 19, 2015 at 1:09
• @janmarqz Ive noticed that. But there is a correspondence between vectors and forms yes? my question is purely a curiousity trying to connect classical tensor calculus and the differential form language in my head. I understand covectors are the standard. Sep 19, 2015 at 1:15

$$u \wedge v \wedge w = u^a v^b w^c \epsilon_{abc} e_1 \wedge e_2 \wedge e_3$$
The Levi-Civita in general describes volume forms, or pseudoscalar quantities--any $n$-vector built from an $n$-dimensional vector space is such, so components of such an $n$-vector can be described in terms of the Levi-Civita for that space. This should make some intuitive sense: the Levi-Civita is antisymmetric on all its indices, just as wedge products are antisymmetric on all their vectors.
• Thanks for your answer @Muphrid. So it follows then for the wedge product of n vectors: $v_1\wedge v_2\wedge ...\wedge v_n = \mathbf{skew}(v_1\otimes v_2\otimes ...\otimes v_n)$ where the skew operator is the skew-symetric part of the n-fold tensor product? Which is just formed by contraction with the appropriate levi-civita symbol? Sep 19, 2015 at 12:55