# Finding commutators without cyclic permutations

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?

• See Wikipedia for how to reduce the product of two Levi-Civita symbols with no indices contracted. Look under Properties > Three dimensions > Product. – G. Smith Oct 4 at 17:13
• Stating a problem in simpler terms (in this case using indices 1, 2, 3 and then permute indices) is awesomely elegant. – Marc Plana Caballero Oct 4 at 18:40
• 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. – Alekxos Oct 4 at 18:42
• @G.Smith Thanks, I was aware of the contracted epsilon identity but not the simplification with no indices contracted. – Alekxos Oct 4 at 18:44