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I am having difficulty showing the following relation with respect to orthonormal basis $\{{\bf{e}}_{1},{\bf{e}}_{2},{\bf{e}}_{3}\}$:

$$ \epsilon_{ijk}{\bf{S}}_{jk} = \epsilon_{ijk}{\bf{W}}_{jk}\tag{a}\label{a} $$

where $\bf{S}$ is a tensor and $\bf{W}$ is the skew tensor component of $\bf{S}$. It is also given that the axial vector of $\bf{S}$ is ${\bf{w}} = \text{axl}\bf{S}$ and is defined by:

$$ {\bf{W}\bf{u} = \bf{w} \times \bf{u}} \tag{b}\label{b}$$

I believe I understand it at a qualitative level. I know the following:

  1. The cross product of two vectors is an axial vector. The scalar components of the cross product are the scalar components of an axial vector.
  2. The scalar components of the cross product can be expressed as the product of a skew matrix (i.e. the scalar components of a skew tensor) and a vector.
  3. The scalar components of the cross product can be expressed in the following ways using indicial notation \begin{align} ({\bf{a}} \times {\bf{b}})_i &= \epsilon_{jki}a_jb_k \quad \text{(3x1 matrix of the scalar components of the cross product)}\tag{1}\label{eq1}\\ ({\bf{a}} \times {\bf{b}})_{ij} &= a_ib_j - a_jb_i\tag{2}\label{eq2} \end{align}

where Eq.~\eqref{eq2} is a 3x1 matrix of cross product scalar components represented as the product of a skew matrix (3x3) and a vector (3x1).

Given all the above, Expression \eqref{a} makes sense when I think of it in terms of the scalar components of each side of the equation. The left side is the product of a skew matrix and a vector and the right side is simply the scalar components of the cross product of $\bf{w}$ and $\bf{u}$.

Where I struggle is finding a starting point for Expression \eqref{a}. Since I don't know where to start I will begin by stating what the left side, $\epsilon_{ijk}{\bf{S}}_{jk}$, represents. This quantity is a compact way of stating the scalar components of a vector, represented by the following 3x1 matrix: \begin{bmatrix} S_{23} - S_{32}\\ S_{31} - S_{13}\\ S_{12} - S_{21} \end{bmatrix}

This looks awfully familiar to expressions I have seen before of the scalar components of the cross product. In fact, if $S_{jk} = w_j u_k$ I would say this is equal to the scalar components of the cross product of $\bf{w} \times \bf{u}$. However, I do not know why $S_{jk}$ would be equal to $w_j u_k$ and I am failing to see how this helps me show my original objective.

I would appreciate any hints or comments on what I have stated thus far.

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I believe you are overthinking this; your starting point should simply be to take your matrix and express it as a sum of symmetric and antisymmetric parts: $$\begin{align}S_{ij} &= \frac12 S_{ij} + \frac12 S_{ij}\\ &=\frac12 S_{ij} + \frac12 S_{ji} + \frac12 S_{ij} - \frac12 S_{ji}\\ &=S_{(ij)} + S_{[ij]}, \end{align}$$ where the parentheses represent those first two terms and symmetrize over the indices, and the square braces represent those last two terms and antisymmetrize over the indices.

Operating on this with any antisymmetric tensor, including $\epsilon_{ijk}$, gives $$\sum_{ij} S_{ij} ~\epsilon_{ijk}=\sum_{ij}S_{(ij)}~\epsilon_{ijk} + \sum_{ij} S_{[ij]}~\epsilon_{ijk},$$where we can recognize that the first term on the right is zero. In general for symmetric $H$ we have $\sum_{ij}H_{ij}~\epsilon_{ijk} = \sum_{ji} H_{ji}~\epsilon_{jik}$ by simple renaming, but by index swapping we find that this term equals $\sum_{ij} (+H_{ij})~(-\epsilon_{ijk})$. So this number (for any $k$) is equal to its own negative, which means it is zero.

Defining $K_{ij} = S_{[ij]}$ one gets quite straightforwardly after this vanishing that $$S_{ij}~\epsilon_{ijk} = K_{ij}~\epsilon_{ijk},$$ with the Einstein summation convention.

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  • $\begingroup$ Wow - I really was overthinking it. I appreciate the tip. I was able to follow your reasoning and came to the same conclusion. The key was to recognize the two expressions (you had them in terms of the tensor H) are equivalent and that you picked up a negative sign from the alternating tensor when you swapped the two indices. $\endgroup$ – Nukesub Sep 19 '17 at 0:56

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