I've been trying to solve Problem 2.4 in Srednicki's Quantum Field Theory textbook. This involves proving the identity $$[J_i,J_j]=i\hbar\epsilon_{ijk}J_k$$ where $J$ is the angular momentum operator, given the relations $$ [M^{\mu\nu},M^{\rho\sigma}]=i\hbar (g^{\mu\rho}M^{\nu\sigma}-g^{\nu\rho}M^{\mu\sigma}-g^{\mu\sigma}M^{\nu\rho}+g^{\nu\sigma}M^{\mu\rho}) $$ $$ J_i=\frac{1}{2}\epsilon_{ijk}M^{jk} $$ I know that one way of approaching the problem is to take $i=1,j=2$ and show $[J_1,J_2]=i\hbar J_3$, then cyclically permute the indices to cover all distinct combinations of $i$ and $j$, but this seems somewhat inelegant.

On the other hand, trying to simplify the general expression $[J_i,J_j]$ yields immediately $$\frac{i\hbar\epsilon_{ilm}\epsilon_{juv}}{4}(g^{lu}M^{mv}-g^{mu}M^{mv}-g^{lv}M^{mu}+g^{mv}M^{lu})$$ which doesn't seem to simplify.

Is it possible to solve this problem using index notation without cyclic permutations?

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    $\begingroup$ See Wikipedia for how to reduce the product of two Levi-Civita symbols with no indices contracted. Look under Properties > Three dimensions > Product. $\endgroup$ – G. Smith Oct 4 at 17:13
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    $\begingroup$ Stating a problem in simpler terms (in this case using indices 1, 2, 3 and then permute indices) is awesomely elegant. $\endgroup$ – Marc Plana Caballero Oct 4 at 18:40
  • $\begingroup$ It may be simpler to write the problem in scalar indices rather than variable indices, but it would be nice if the notation itself captured that elegance. $\endgroup$ – Alekxos Oct 4 at 18:42
  • $\begingroup$ @G.Smith Thanks, I was aware of the contracted epsilon identity but not the simplification with no indices contracted. $\endgroup$ – Alekxos Oct 4 at 18:44

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