Equivalence of wedge and tensor product with Levi-Civita symbol

Question

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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NB:

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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.

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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?

Answer 1

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...

Answer 2

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.

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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:

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

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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.

Answer 3

I feel that the answer you copied, in order to point out some problems with it, on physics stack exchange, is not as bad as you think. That answerer simply got mixed up between the notation of the O.P., $x^0$ and $x^1$ (passing to two Euclidean dimensions, as he did, which is a good pedagogic strategy), and his own, which was $v$ and $w$. So, in context, it seems to me he means $x^0$ = v and $x^1$ =w. This is another good pedagogic strategy since the superscripts on $dx^0$ do not indicate components at all. But then he fell back into the superscripts, putting them on the v and w, which was just carelessness. Since $w^1$ really means $v$, and $w^2$ means w, and $v^1$ means v, and $v^2$ means w, due to carelessness, the manipulation is justified by the usual splitting the sum into two sums and making a "change of variable" in one of them, which is what I think he intended, instead of "commuting", which as you and Prof. Shifrin point out, is invalid.

Indeed, you must see that he did not mean components, since you can't take the tensor product of components, so he couldn't have possibly really meant what he carelessly wrote.

I do feel that such carelessness is made less avoidable by the usual sloppy use of indices for two purposes: one, to enumerate the elements of a set, like the four elements of a basis set, and two, to enumerate the components of a vector, like the four components of a four-vector. But this ambiguity is common, and one must get used to it, at least in Physics.