Product of Two Levi-Civita Tensors
Product of Two Levi-Civita Tensors
Harun Basmacı, Ankara University
July 2022
Abstract
In this teeny-weeny piece of paper, I will be trying to derive a formula for the product of two Levi-Civita tensors by using basic trigonometric properties- basically expressing them as four Kronecker delta. Also, I will demonstrate the expression I finally get, whether it works or not.
1 Derivation
Suppose $\hat{e}_i, \hat{e}_j,$ and $\hat{e}_k$ are unit vectors, therefore
\begin{equation} \hat{e}_i = \langle 1,0,0 \rangle \end{equation}
\begin{equation} \hat{e}_j = \langle 0,1,0 \rangle \end{equation}
\begin{equation} \hat{e}_k = \langle 0,0,1 \rangle. \end{equation}
We know that
\begin{equation}\epsilon_{ijk} = \hat{e}_i \cdot (\hat{e}_j \times \hat{e}_k)\end{equation}
\begin{equation} =\hat{e}_i \cdot (\lVert{\hat{e}_j}\rVert \lVert{\hat{e}_k}\rVert \sin{\theta_{jk}}\hat{e}_i)\end{equation}
($\theta_{jk}$ represents the angle between the vectors $\hat{e}_j$ and $\hat{e}_k$)
\begin{equation}= (\hat{e}_i \cdot \hat{e}_i)(\sin{\theta_{jk}})\end{equation}
(the norm of both vectors $\hat{e}_j$ and $\hat{e}_k$ equal to 1, since they are unit vectors)
\begin{equation} \hat{e}_i \cdot \hat{e}_i = \delta_{ii} = 1 \end{equation}
thus,
\begin{equation} \epsilon_{ijk} = \sin{\theta_{jk}} \end{equation}
(notice that the expression is independent of $i$).
If we write $\theta_{jk}$ as the difference of two angles $\theta_j$ and $\theta_k$, we get
\begin{equation}\sin{(\theta_j-\theta_k)} = \sin{\theta_j}\cos{\theta_k} - \sin{\theta_k}\cos{\theta_j}\end{equation}
(by using the sine difference formula).
Hence, we get the Levi-Civita tensor as
\begin{equation}\epsilon_{ijk} = \sin{\theta_j}\cos{\theta_k} - \sin{\theta_k}\cos{\theta_j}\end{equation}
and (by replacing indices with $l, m, n$)
\begin{equation}\epsilon_{lmn} = \sin{\theta_m}\cos{\theta_n} - \sin{\theta_n}\cos{\theta_m}.\end{equation}
Consequently, the product of two Levi-Civita tensors must be
\begin{equation}\epsilon_{ijk} \epsilon_{lmn} = (\sin{\theta_j}\cos{\theta_k} - \sin{\theta_k}\cos{\theta_j})(\sin{\theta_m}\cos{\theta_n} - \sin{\theta_n}\cos{\theta_m})\end{equation}
\begin{equation}= \sin{\theta_j}\sin{\theta_m}\cos{\theta_k}\cos{\theta_n} - \sin{\theta_j}\sin{\theta_n}\cos{\theta_k}\cos{\theta_m}\end{equation}
$$- \sin{\theta_k}\sin{\theta_m}\cos{\theta_j}\cos{\theta_n} + \sin{\theta_k}\sin{\theta_n}\cos{\theta_j}\cos{\theta_m}$$
\begin{equation}= \cos{\theta_{jm}} \cos{\theta_{kn}} - \cos{\theta_{km}} \cos{\theta_{jn}}\end{equation}
(by using the cosine difference formula)
\begin{equation}= \delta_{jm}\delta_{kn} - \delta_{jn}\delta_{km}\end{equation}
(since, any pair $x$ and $y$, $\delta_{xy} = \hat{e}_x \cdot \hat{e}_y = \lVert{\hat{e}_x}\rVert\lVert{\hat{e}_y}\rVert\cos{\theta_{xy}} = \cos{\theta_{xy}}$).
Eventually,
\begin{equation} \epsilon_{ijk}\epsilon_{lmn}=\delta_{jm}\delta_{kn}-\delta_{jn}\delta_{km} \end{equation}
2 Demonstration
We obtain the expression
\begin{equation}\epsilon_{ijk}\epsilon_{lmn} = \delta_{jm}\delta_{kn} - \delta_{jn}\delta_{km}.\end{equation}
Notice that it is independent of $i$ and $j$, therefore we can let the first indices of Levi-Civita tensors be the same
\begin{equation}\epsilon_{ijk}\epsilon_{imn} = \delta_{jm}\delta_{kn} - \delta_{jn}\delta_{km}. \end{equation}
Also, from previous knowledge
\begin{equation}\epsilon_{ijk}\epsilon_{imn} = [\hat{e}_i \cdot (\hat{e}_j \times \hat{e}_k)][\hat{e}_i \cdot (\hat{e}_m \times \hat{e}_n)]. \end{equation}
Now, it is time to examine the unit vectors in 3-D space
This is because from the fundamental property of the Levi-Civita tensor,
\begin{eqnarray} \begin{array}{rcl} \epsilon_{ijk} & = \begin{cases} +1 & \text{if $(i,j,k)$ is $(1,2,3)$, $(2,3,1)$, or $(3,1,2)$}\\-1 & \text{if $(i,j,k)$ is $(1,3,2)$, $(2,1,3)$, or $(3,2,1)$} \\0 & \text{if $i=j$, $i=k$, or $j=k$}\end{cases} \end{array} \end{eqnarray}
if we let $i\neq j$, $i\neq k$, $j\neq k$, $i\neq m$, $i\neq n$, and $m\neq n$, thus
\begin{equation}\epsilon_{ijk}=1\end{equation}
\begin{equation} = \hat{e}_i \cdot (\hat{e}_j \times \hat{e}_k)\end{equation}
\begin{equation} \hat{e}_j \times \hat{e}_k = \hat{e}_i \end{equation}
\begin{equation}\epsilon_{imn}=1\end{equation}
\begin{equation} = \hat{e}_i \cdot (\hat{e}_m \times \hat{e}_n) \end{equation}
\begin{equation} \hat{e}_m \times \hat{e}_n = \hat{e}_i \end{equation}
and this gives us the previous figure.
From the equations (21) and (24), we see that product of two Levi-Civita tensors
\begin{equation}\epsilon_{ijk} \epsilon_{imn} = 1.\end{equation}
Now, let's look at the right-hand side of the equation (18),
\begin{equation}\delta_{jm}\delta_{kn} - \delta_{jn}\delta_{km}.\end{equation}
If we turn back to equation (14)
\begin{equation}\delta_{jm}\delta_{kn} - \delta_{jn}\delta_{km} = \cos{\theta_{jm}}\cos{\theta_{kn}}-\cos{\theta_{jn}}\cos{\theta_{km}}\end{equation}
since we know the placement of the unit vectors in space,
\begin{equation}\theta_{jm} = 0^\circ \end{equation}
\begin{equation}\theta_{kn} = 0^\circ \end{equation}
\begin{equation}\theta_{jn} = 90^\circ \end{equation}
\begin{equation}\theta_{jn} = 90^\circ \end{equation}
we can determine the angles.
Therefore the equation (29) will be
\begin{equation}\cos{\theta_{jm}}\cos{\theta_{kn}}-\cos{\theta_{jn}}\cos{\theta_{km}} = \cos{0^\circ}\cos{0^\circ}-\cos{90^\circ}\cos{90^\circ}\end{equation}
\begin{equation}= 1 - 0\end{equation}
\begin{equation}= 1.\end{equation}
So we are done,
\begin{equation}\epsilon_{ijk}\epsilon_{imn} = \delta_{jm}\delta_{kn} - \delta_{jn}\delta_{km} \end{equation}
\begin{equation}1 = 1.\end{equation}