In an solution to a problem I was attempting it uses the fact that,

$$\epsilon_{ijk}\delta_{ij} = 0$$ The explanation I am given says: "the levi-civita is antisymmetric under swaps of i and j whilst the kronecker delta is symmetric under swaps of i and j".

I understand what that means but I am struggling to use that fact to prove the statement above.

  • $\begingroup$ Do we follow the Einstein convention i.e are we talking about $$\epsilon_{ijk}\delta_{ij}=\sum_{i,j}\epsilon_{ijk}\delta_{ij}$$ $\endgroup$
    – marwalix
    Commented Mar 31, 2021 at 13:29

1 Answer 1


Oh nevermind I've figured it out. $$\delta_{ij}\epsilon_{ijk} = -\delta_{ij}\epsilon_{jik}=-\delta_{ji}\epsilon_{jik}=-\delta_{ij}\epsilon_{ijk} \\ \therefore \delta_{ij}\epsilon_{ijk} = -\delta_{ij}\epsilon_{ijk} = 0$$

  • $\begingroup$ Indeed. Though it’s perhaps simpler (and somewhat more illuminating if you work a lot with $\delta$) to just note that if $\delta{ij}\neq 0$, then $i=j$ and thus $\epsilon_{ijk}=\epsilon_{iik}=\epsilon_{jik}=-\epsilon{ijk}$ so that $\epsilon_{ijk}=0$. $\endgroup$
    – Aphelli
    Commented Mar 31, 2021 at 13:42
  • $\begingroup$ Ah yes that's a good way to look at it $\delta_{ij} = 1 \rightarrow i = j \rightarrow \epsilon_{ijk} = 0$ $\endgroup$ Commented Mar 31, 2021 at 14:09

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