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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.

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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}

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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 $\epsilon_{s12}\epsilon_{s12}= \epsilon_{312}\epsilon_{312}=1$).

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