Prove a Levi Cevita Epsilon identity

Question

I'm trying to show that $$\frac{1}{4} \epsilon_{ijp}\epsilon_{kmq}T_{ijkm}=-T_{rqrp}$$ where $T_{ijkm}$ is a rank 4 tensor anti symmetric in $(i,j)$ and in $(k,m)$ which satisfies $T_{ijij}=0$.

No idea where to start really, I know there's a formula for expanding out the double epsilon terms as Kroncker Deltas but we've only been lectured on the case where they share a summed over term (i.e the $\epsilon_{ijk}\epsilon_{pqk}$ one) so I don't think that's the intended way to go about this.

Answer 1

We make repeated use of the following facts:

• If $M_{ij}$ is antisymmetric, then $\exists W_k$ such that $$M_{ij} = \varepsilon_{ijk} W_k$$
• We have the following $\varepsilon_{ijk}$ identity: $$\varepsilon_{kij} \varepsilon_{kpq} = \delta_{ip} \delta_{jq} - \delta_{iq} \delta_{jp} \tag{$\star$}$$

I'll assume $(\star)$ since you state that you're familiar with it. The former is straightforward to show:

Let $W_k = \frac{1}{2}\varepsilon_{kpq} M_{pq}$. Then $$\varepsilon_{ijk} W_k = \frac{1}{2}\varepsilon_{kij} \varepsilon_{kpq} M_{pq} = \frac{1}{2}(\delta_{ip} \delta_{jq} - \delta_{iq} \delta_{jp}) M_{pq} = \frac{1}{2} \left(M_{ij} - M_{ji}\right) = M_{ij}$$

Before showing your result, there's a few small notes to make.

• I'll let $n$ denote the dimension of your space. In this case, since we're working with $\varepsilon_{ijk}$, we're taking $n = 3$. Note also that $\delta_{ii} = n$.
• It's also useful to contract a further two indices in the identity $(\star)$. Contracting $i$ and $p$, we have $$\varepsilon_{kij}\varepsilon_{kiq} = \delta_{ii} \delta_{jq} - \delta_{iq} \delta_{ji} = 3 \delta_{jq} - \delta_{jq} = 2 \delta_{jq}$$

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Now we're ready to show the result. We'll exploit the antisymmetry of $T$ in its various indices to deduce its form. We first use the antisymmetry in $(i, j)$ (the indices $k$ and $m$ just "go along for the ride" here - if you like, we can think of doing this for each possible choice $k$ and $m$) to write $$T_{ijkm} = \varepsilon_{ija} Y_{kma}$$ Simlarly, antisymmetry of $Y_{kma}$ in $k$ and $m$ (this time, the $a$ just goes along) gives $$Y_{kma} = \varepsilon_{kmb} X_{ab}$$ hence $$T_{ijkm} = \varepsilon_{ija} \varepsilon_{kmb} X_{ab}$$ Since we're told $T_{ijij} = 0$, we're probably going to need this, and now is likely to be a good time to see what this tells us. In particular, $$T_{ijij} = \varepsilon_{ija}\varepsilon_{ijb} X_{ab} = (n-1) \delta_{ab} X_{ab} = (n-1)X_{aa} = 0$$ i.e. $X$ is traceless.

Now we just substitute our expression for $T_{ijkm}$ into the equation and see what pops out. We have \begin{align*} \varepsilon_{ijp} \varepsilon_{kmq} T_{ijkm} &= \varepsilon_{ijp} \varepsilon_{kmq} \varepsilon_{ija} \varepsilon_{kmb} X_{ab} \\ &= (\varepsilon_{ijp}\varepsilon_{ija}) \cdot (\varepsilon_{kmq} \varepsilon_{kmb}) X_{ab} \\ &= 4 \delta_{pa} \delta_{qb} X_{ab} \\ &= 4 X_{pq} \tag{$\star\star$} \end{align*} where we made a sensible grouping of the epsilons based on our $(n-1) \delta_{jp}$ result. Now note that \begin{align*} -T_{rqrp} &= -\varepsilon_{rqa} \varepsilon_{rpb} X_{ab} \\ &=-(\delta_{qp}\delta_{ab} - \delta_{qb}\delta_{ap}) X_{ab} \\ &= - \delta_{qp} X_{aa} + X_{pq} \\ &= X_{pq} \end{align*} where we've used the fact that $X_{aa} = 0$ that we deduced earlier.

Substituting $X_{pq} = -T_{rqrp}$ into $(\star\star)$ gives $$\varepsilon_{ijp} \varepsilon_{kmq} T_{ijkm} = -4T_{rqrp}$$ which gives the result you need.