In this answer the following is stated in the 2-dim case:

\begin{eqnarray} v\land w & = & \frac{1}{2!}(v\land w-w\land v) \\ & = & \frac{1}{2!}\epsilon_{\mu\nu}v^{\mu}\land w^{\nu} \\ & = & \frac{1}{2!}\epsilon_{\mu\nu}(v^{\mu}\otimes w^{\nu}-w^{\nu}\otimes v^{\mu}) \\ & = & \epsilon_{\mu\nu}v^{\mu}\otimes w^{\nu}. \end{eqnarray}


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I wanted to see the index mechanics at play replicating what was done in this answer, building a 2-vector from two vectors in $v, w \in\mathbb R^3, $ such as $v=1e_1+3e_2-2e_3$ and $w=5e_1+2e_2+8e_3:$

$$v\wedge w=(1\cdot e_1 + 3 \cdot e_2 - 2 \cdot e_3) \wedge (5\cdot e_1 + 2 \cdot e_2 + 8 \cdot e_3) = \\[2ex] 1\cdot 5 \cdot e_1 \wedge e_1 + 1\cdot 2 \cdot e_1 \wedge e_2 + 1\cdot 8 \cdot e_1 \wedge e_3 \\ +3\cdot 5 \cdot e_2\wedge e_1 +3\cdot 2 \cdot e_2\wedge e_2 +3\cdot 8 \cdot e_2\wedge e_3 \\ -2\cdot 5 \cdot e_3\wedge e_1 -2\cdot 2 \cdot e_3\wedge e_2 -2\cdot 8 \cdot e_3\wedge e_3 = \\[2ex] 5 \cdot \mathbb O + 2 \cdot e_1 \wedge e_2 - 8 \cdot e_3 \wedge e_1 \\ -15 \cdot e_1\wedge e_2 +6 \cdot \mathbb O +24 \cdot e_2\wedge e_3 \\ -10 \cdot e_3\wedge e_1 +4 \cdot e_2\wedge e_3 -16 \cdot \mathbb O = \\[2ex] \bbox[5px,border:2px solid red] { 28 \cdot e_2\wedge e_3-18 \cdot e_3\wedge e_1 - 13 \cdot e_1\wedge e_2}$$

starting off at the end, and trying to calculate $\epsilon_{\mu\nu}v^\mu\otimes w^\nu:$

$$\epsilon_{\mu\nu}v^\mu\otimes w^\nu= \\[2ex] \color{blue}{\epsilon_{11}} 1\cdot 5 \cdot e_1 \otimes e_1 + \color{blue}{\epsilon_{12}} 1\cdot 2 \cdot e_1 \otimes e_2 + \color{blue}{\epsilon_{13}} 1\cdot 8 \cdot e_1 \otimes e_3 + \\ \color{blue}{\epsilon_{21}} 3\cdot 5 \cdot e_2\otimes e_1 + \color{blue}{\epsilon_{22}} 3\cdot 2 \cdot e_2\otimes e_2 + \color{blue}{\epsilon_{23}} 3\cdot 8 \cdot e_2\otimes e_3 + \\ \color{blue}{\epsilon_{31}} (-2)\cdot 5 \cdot e_3\otimes e_1 +\color{blue}{\epsilon_{32}}(-2)\cdot 2 \cdot e_3\otimes e_2 +\color{blue}{\epsilon_{33}}(-2)\cdot 8 \cdot e_3\otimes e_3 = \\[2ex] \color{blue}0\cdot 1\cdot 5 \cdot e_1 \otimes e_1 + \color{blue}1\cdot 1\cdot 2 e_1 \otimes e_2 + \color{blue}1 \cdot 1\cdot 8 e_1 \otimes e_3 + \\ \color{blue}{(-1)}\cdot 3\cdot 5 e_2\otimes e_1 + \color{blue}0 \cdot 3\cdot 2 e_2\otimes e_2 + \color{blue}1 \cdot 3\cdot 8 e_2\otimes e_3 + \\ \color{blue}{(-1)}\cdot (-2)\cdot 5 e_3\otimes e_1 +\color{blue}{(-1)}\cdot(-2)\cdot 2 e_3\otimes e_2 +\color{blue}0\cdot (-2)\cdot 8 \cdot e_3\otimes e_3 = \\[2ex] \bbox[5px,border:2px solid red] { 2 e_1 \otimes e_2 + 8 e_1 \otimes e_3 - 15 e_2\otimes e_1 + 24 e_2\otimes e_3 + 10 e_3\otimes e_1 + 4 e_3\otimes e_2} $$

How do I reconcile these two results?

NB: This is impossible to reconcile as per the comments: A change of signs cannot relate $v\otimes w$ to $w\otimes v$ - the initial equations are not correct.

The second issue is reflected on my extended comment / "answer" below, and makes reference to the use of the Levi-Civita symbols seemingly undoing the usual change of signs when permuting wedge products...

The essential issue is that the LeviCivita symbols don't seem to naturally "handle" the basis of the wedge product, as well as they do, say, in the case of the cross product. How should the LC symbols be applied in the wedge product?

  • $\begingroup$ What is "LC symbol"? $\endgroup$ – lisyarus Aug 21 at 11:56
  • $\begingroup$ @lisyarus Levi Civita. Probably not standard. $\endgroup$ – Antoni Parellada Aug 21 at 11:58
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    $\begingroup$ The formula stated in that answer is just plain wrong, as there is no way to use a sign to relate $v\otimes w$ and $w\otimes v$. $\endgroup$ – Ted Shifrin Aug 23 at 23:41
  • $\begingroup$ @TedShifrin I gather from your comment that this would be the problem step in the Physics.SE answer referenced above: $\begin{align}&\small{\frac{1}{2!}}\epsilon_{\mu\nu}(v^{\mu}\otimes w^{\nu}-w^{\nu}\otimes v^{\mu}) \\ =& \epsilon_{\mu\nu}v^{\mu}\otimes w^{\nu}\end{align}.$ $\endgroup$ – Antoni Parellada Aug 24 at 1:53
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    $\begingroup$ @AntoniParellada Yup, precisely. $\endgroup$ – Ted Shifrin Aug 24 at 3:04

Let me dive a bit into two ways of looking at exterior products.

The first one, which is how I prefer it, is to view $v \wedge w$ as an element of the exterior square $\Lambda^2 V$ of the original vector space $V$ that contains vectors $v$ and $w$. It is a vector space of dimension ${\dim V \choose 2}=\frac{\dim V \cdot(\dim V-1)}{2}$ crafted specifically as the place where exterior products of 2 vectors live.

The second way, that is more common in classical differential geometry & physics, is to embed $\Lambda^2 V$ as a subspace of $V \otimes V$, namely the space of alternating (antisymmetric) tensors. The embedding looks like this:

$$v \wedge w \mapsto \frac{1}{2!}(v\otimes w - w \otimes v)$$

or like this

$$v \wedge w \mapsto v\otimes w - w \otimes v$$

Using these embeddings implicitly, one can view this as the definition of the wedge product, taking values in the space of alternating tensors, completely skipping the exterior square part.

Both embeddings seem to be in use in literature; the difference, as I see it, is only a matter of taste: some calculations get easier with the first embedding, and some with the second (unless we work over a field of scalars that has $\operatorname{char}\neq 0$).

It is not entirely clear to me whether you intend to use the first or the second embedding, and indeed this is precisely the reason I tend not to like the idea of identifying wedge products with alternating tensors. Once we work solely in $\Lambda^2 V$, everything is precisely defined.

As for calculating the wedge product of $v=1e_1+3e_2−2e_3$ and $w=5e_1+2e_2+8e_3$, using, say, the second embedding, one could go like this:

$$v \wedge w = v \otimes w - w \otimes v = \\ = (1e_1+3e_2−2e_3) \otimes (5e_1+2e_2+8e_3) - (5e_1+2e_2+8e_3) \otimes (1e_1+3e_2−2e_3) = \\ = \big[5e_1\otimes e_1 + 2e_1\otimes e_2+8e_1\otimes e_3+15e_2\otimes e_1+6e_2\otimes e_2+24e_2\otimes e_3-10e_3\otimes e_1-4e_3\otimes e_2-16e_3\otimes e_3\big] - \big[5e_1\otimes e_1+15e_1\otimes e_2-10e_1\otimes e_3+2e_2\otimes e_1+6e_2\otimes e_2-4e_2\otimes e_3+8e_3\otimes e_1+24e_3\otimes e_2-16e_3\otimes e_3\big] = \\ = -13e_1\otimes e_2+18e_1\otimes e_3+13e_2\otimes e_1+28e_2\otimes e_3-18e_3\otimes e_1-28e_3\otimes e_2 = \\ = -13(e_1\otimes e_2-e_2\otimes e_1)+28(e_2\otimes e_3-e_3\otimes e_2)-18(e_3\otimes e_1-e_1\otimes e_3) = \\ = -13e_1\wedge e_2 +28e_2\wedge e_3 - 18 e_3\wedge e_1$$

As for the linked physics.se answer, the crucial thing is that it works in 2 dimensions. Using the second embedding, we get

$$v \wedge w = (v^1w^2-v^2w^1) e_1 \wedge e_2$$

and since the wedge product of any two vectors is proportional to $e_1 \wedge e_2$, it is common to identify 2-vectors with numbers (see Hodge dual). In this case, dropping the $e_1 \wedge e_2$ part, we get

$$v \wedge w = \epsilon_{ij}v^iw^j$$

In general, the n-fold wedge product of n vectors is a multiple of $e_1 \wedge \dots \wedge e_n$ and is commonly identified with numbers; the wedge product can be computed using the Levi-Civita with n indices:

$$v_1 \wedge \dots \wedge v_n = \epsilon_{i_1\dots i_n}v_1^{i_1}v_2^{i_2}\dots v_n^{i_n}$$

which is actually the same as the determinant.

  • $\begingroup$ Your explanation is wonderful. I know you are being tactful or respectful, but it sounds as though you went to the Physics.SE post, and are a bit skeptical of that answer... Or is it that my attempt to mirror your answer to a similar question failing at some fundamental level? $\endgroup$ – Antoni Parellada Aug 21 at 13:49
  • $\begingroup$ @AntoniParellada Thank you! Yes, I failed to understand what the physics.se post means. I have some ideas, though; I'll update the answer. $\endgroup$ – lisyarus Aug 21 at 13:53
  • $\begingroup$ Would you go about calculating the RHS of the equation the same way I did in the OP, resulting in $2 e_1 \otimes e_2 + 8 e_1 \otimes e_3 - 15 e_2\otimes e_1 + 24 e_2\otimes e_3 + 10 e_3\otimes e_1 + 4 e_3\otimes e_2$? $\endgroup$ – Antoni Parellada Aug 21 at 23:02
  • $\begingroup$ @AntoniParellada Sure. I've updated the answer. $\endgroup$ – lisyarus Aug 22 at 12:43
  • $\begingroup$ I really appreciate your effort to help with this problem, and I wonder if you could take it a bit farther, and look into the calculations in the equations on the Physics.SE referenced at the top of the OP, assuming that on the RHS of the equation the basis vectors are implied, as in this post. $\endgroup$ – Antoni Parellada Aug 23 at 0:42

More like an extended comment...

If we assume that the RHS of the equation is really meant to signify

$$\frac{1}{2!}\epsilon_{\mu\nu}v^\mu w^\nu \, e_\mu\wedge e_\nu,$$

the only way I can see a way to use LeviCivita symbols is to establish basis vectors for the wedge product ahead of time, as in $\{e_1 \wedge e_2, \;e_2 \wedge e_3, \; e_1 \wedge e_3\}.$

Remembering that $v=1e_1+3e_2-2e_3$ and $w=5e_1+2e_2+8e_3,$

We can establish a parallel with the use of LC symbols in the cross product - and algebraically identical operation in 3-dim:

$$\begin{align} v \times w &= \begin{vmatrix}3 &- 2\\2&8 \end{vmatrix} e_1 - \begin{vmatrix}1 &- 2\\5 &8 \end{vmatrix} e_2 + \begin{vmatrix}1 & 3\\5& 2 \end{vmatrix} e_3\\[2ex] &= \epsilon_{ijk}\;v_i\,w_j\; e_k \\[2ex] &= \epsilon_{123} \; 1\cdot 2\;e_3 + \epsilon_{213}\; 3\cdot 5 \; e_3\\ &+ \epsilon_{132} \; 1\cdot 8\; e_2 + \epsilon_{312}\; (-2)\cdot 5 \; e_2 \\ &+ \epsilon_{231} \; 3\cdot 8\; e_1 + \epsilon_{321}\; (-2)\cdot 2 \; e_1 \\[2ex] &= 1 \;\cdot 2\;e_3 + (-1)\;\cdot 15 \; e_3\\ &-1 \; \cdot 8\; e_2 + 1\; \cdot (-10) \; e_2 \\ &+ 1 \; \cdot 24\; e_1 -1\; \cdot (-4) \; e_1 \\[2ex] &=28 \,e_1 -18 \, e_2 -13 \,e_3 \end{align}$$

If we can replace the basis vectors above with bivector basis $e_1\wedge e_2$ instead of $e_3;$ $e_2\wedge e_3$ for $e_1;$ and $e_1\wedge e_3,$ or even better, $e_3 \wedge e_1$ for $e_2, $ we end up with a strict correspondence of coefficients with correct sign. However, in the use of LC symbols for the crossproduct, we didn't have to arrange the basis vectors just so - the symbols took care of matching coefficients with the corresponding $e_1,$ $e_2$ or $e_3.$

I just don't see how this can be extrapolated to the wedge product...


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