Notes_EN

# Contraction of completely anti-symmetric tensor

Contraction formula of the completely antisymmetric tensor (or Levi-Civita symbol) is indispensable for vector calculus.

If you think it is hard to remember the formula, it is wrong.
Actually, the derivation is very easy.

$$\epsilon_{ijk}\epsilon^{ilm}=?$$

### Preparations

Completely anti-symmetric tensor $\epsilon_{ijk}$ has following properties:

Anti symmetricity
\begin{align} \epsilon_{\color{red}{\LARGE i}\color{blue}{\LARGE j}k}=-\epsilon_{\color{blue}{\LARGE j}\color{red}{\LARGE i}k},\qquad \epsilon_{i\color{red}{\LARGE j}\color{blue}{\LARGE k}}=-\epsilon_{i\color{blue}{\LARGE k}\color{red}{\LARGE j}},\qquad \epsilon_{\color{red}{\LARGE i}j\color{blue}{\LARGE k}}=-\epsilon_{\color{blue}{\LARGE k}j\color{red}{\LARGE i}} \end{align}
From this property, we notice that if two indices are the same, completely anti-symmetric tensor is equal to zero.
For example, $\epsilon_{iik}=-\epsilon_{iik}$ then $2\epsilon_{iik}=0$.
We conclude that
\begin{align} \epsilon_{\color{red}{\LARGE ii}k}=\epsilon_{k\color{red}{\LARGE ii}}=\epsilon_{\color{red}{\LARGE i}k \color{red}{\LARGE i}}=0 \end{align}

### Derivation (3 dimensions case)

##### Step. 1 Enumerate all the cases of $~\neq 0\Rightarrow$ Roughly determine the formula
Since $\epsilon_{ijk}$ is a completely anti-symmetric tensor, $\epsilon_{ijk}\epsilon^{ilm}\neq0$ only when
1. $j=l$, $k=m$,
2. $j=n$, $k=l$.
Then contraction formula can be represented as $$\epsilon_{ijk}\epsilon^{ilm}=a\cdot\delta_{jl}\delta_{km}+b\cdot\delta_{jm}\delta_{kl}$$ with two constants $a,b$.
##### Step. 2 Find unknown constants: Consider anti-symmetricity of indices
Since left-hand side is anti-symmetric with exchange of $j,k$ (or $l, m$), right-hand side must be the same*1: $$\epsilon_{ijk}\epsilon^{ilm}=a(\delta_{jl}\delta_{km}-\delta_{jm}\delta_{kl}).$$ Substituting $(j,k)=(l,m)=(2,3)$, we can get $a=1$. Now, we have got the following contraction formula!:
Contraction formula
$$\epsilon_{ijk}\epsilon^{ilm}=\delta_{jl}\delta_{km}-\delta_{jm}\delta_{kl}$$

### Exercise

Calculate following case (contractions of two pair indices): $$\epsilon_{\color{red}{\large i}\color{blue}{\large j}k}\epsilon^{\color{red}{\large i}\color{blue}{\large j}m}=?$$ Do not use above formula. Calculate in your head (you can calculate faster!).

### Higher dimensions case

The process is same as the three dimensions case.

*1:Or we can derive as follows: Since $\mathrm{LHS}=0$ when $j=k$ (or $l=m$), $a+b=0$.