# Proof of $\epsilon_{ijk}\epsilon_{klm}=\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl}$

I'm a student of physics. There is an identity in tensor calculus involving Kronecker deltas ans Levi-Civita pseudo tensors is given by $$\epsilon_{ijk}\epsilon_{klm}=\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl}$$ which is extensively used in physics in deriving various identities. I have neither found a proof of this in physics textbooks nor in Wikipedia. In particular, how does the above formula follow from the definition of $\epsilon_{ijk}$ tensor$$\epsilon_{ijk} = \begin{cases} +1 & \text{ for even permutations }, \\ -1 & \text{ for odd permutations } ,\\ \;\;\,0 & \text{ for repetition of indices }, \end{cases}$$ This is the only definition of I'm familiar with.

Here's an explicit proof that doesn't rely on knowning anything about determinants and just uses the definition of $$\epsilon_{ijk}$$ and $$\delta_{ij}$$.

Consider first

$$\epsilon_{ijk}\epsilon_{lmn}$$

This quantity is $$+1$$ if $$(lmn)$$ is an even permutation of $$(ijk)$$ and $$-1$$ if $$(lmn)$$ is an odd permutation of $$(ijk)$$. It is $$0$$ if $$(lmn)$$ is not a permutation of $$(ijk)$$. This means each of $$(ijk)$$ must have a pair in $$(lmn)$$. We can enumerate all possibilities.

\begin{align} \epsilon_{ijk}\epsilon_{lmn} = &\delta_{il}\delta_{jm}\delta_{kn} + \delta_{im}\delta_{jn}\delta_{kl} + \delta_{in}\delta_{jl}\delta_{km}\\ -&\delta_{in}\delta_{jm}\delta_{kl} - \delta_{il}\delta_{jn}\delta_{km} - \delta_{im}\delta_{jl}\delta_{kn} \end{align}

Let us now set $$l=i$$ and sum over $$i$$ (with Einstein summation notation implied)

\begin{align} \epsilon_{ijk}\epsilon_{imn} = &\delta_{ii}\delta_{jm}\delta_{kn} + \delta_{im}\delta_{jn}\delta_{ki} + \delta_{in}\delta_{ji}\delta_{km}\\ -&\delta_{in}\delta_{jm}\delta_{ki} - \delta_{ii}\delta_{jn}\delta_{km} - \delta_{im}\delta_{ji}\delta_{kn}\\ =& 3\delta_{jm}\delta_{kn} + \delta_{km}\delta_{jn} + \delta_{jn}\delta_{km}\\ -&\delta_{kn}\delta_{jm} - 3\delta_{jn}\delta_{km} - \delta_{jm}\delta_{kn}\\ =& \delta_{jm}\delta_{kn} - \delta_{jn}\delta_{km} \end{align}

By a change of index $$i\rightarrow k \rightarrow j\rightarrow i$$ and $$m\rightarrow l$$, $$n\rightarrow m$$ we get

\begin{align} \epsilon_{kij}\epsilon_{klm}=\epsilon_{ijk}\epsilon_{klm} = \delta_{il}\delta_{jm} - \delta_{im}\delta_{jl} \end{align}

We have that (the repeated index $$s$$ is summed and therefore it cannot enter into the result)

\begin{align} \epsilon_{sab}\epsilon_{sij}= A_{abij} \end{align}

where $$A_{abij}=-A_{baij}=-A_{abji}=A_{ijab}$$, which are obvious symmetries of the still unknown symbol $$A$$. Since $$\epsilon_{sab}$$ is non-zero only if $$s\neq a\neq b\neq s$$ (i.e. the indexes are all different), and similarly for $$\epsilon_{sij}$$, the only possibility to have a non-zero result is that $$a=i$$ and $$b=j$$ or that $$a=j$$ and $$b=i$$, so that

\begin{align} A_{abij} = \alpha \delta_{ai} \delta_{bj} + \beta \delta_{aj} \delta_{bi} \end{align}

To implement the expected symmetries of $$A$$ we must have $$\alpha = -\beta$$. The value $$\alpha = 1$$ is fixed by considering any non-zero particular case (e.g. $$a=i=1$$, $$b=j=2$$, gives $$\epsilon_{s12}\epsilon_{s12}= \epsilon_{312}\epsilon_{312}=1$$).