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.

At the end of this article, you will be able to calculate contraction formula in your head!

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

### 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
- $j=l$, $k=m$,
- $j=n$, $k=l$.

__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 $l=m$), $a+b=0$.