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I am trying to verify that the Casimir of the Lie group $SO(3)$ is actually $N^2=N_iN_i$, but I have problems, with indices surely, and I was wondering if someone could help me figuring out how to properly verify these commutation relation. Here is what I come up with:

$$ \left[N^2,N_j\right] {=N^2N_j-N_jN^2\nonumber\\ =(N_iN_i)N_j-N_j(N_iN_i)\nonumber\\ =N_iN_jN_i+N_i[N_i,N_j]-N_jN_iN_i\nonumber\\ =N_jN_iN_i+[N_i,N_j]N_i+N_ii\epsilon^{ijk}N_k-N_jN_iN_i\nonumber\\ =i\epsilon^{ijk}N_kN_i+N_ii\epsilon^{ijk}N_k\nonumber\\ =i\epsilon^{ijk}N_iN_k(2+[N_k,N_i])\nonumber\\ =i\epsilon^{ijk}N_iN_k(2+i\epsilon^{kij}N_j)\nonumber} $$

I don't see how this is supposed to give $0$.

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    $\begingroup$ Try checking your index structure -- halfway you switch from having good indices to having three copies of some indices. $\endgroup$ – knzhou Oct 23 '17 at 17:31
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It's the same through line 5. Line 6 starts being different: $$ \left[N^2,N_j\right] {=N^2N_j-N_jN^2\nonumber\\ =(N_iN_i)N_j-N_j(N_iN_i)\nonumber\\ =N_iN_jN_i+N_i[N_i,N_j]-N_jN_iN_i\nonumber\\ =N_jN_iN_i+[N_i,N_j]N_i+N_ii\epsilon^{ijk}N_k-N_jN_iN_i\nonumber\\ =i\epsilon^{ijk}N_kN_i+N_ii\epsilon^{ijk}N_k\nonumber\\ =i\epsilon^{ijk}N_kN_i+i\epsilon^{ijk}N_iN_k\nonumber\\ =i\epsilon^{ijk}N_kN_i-i\epsilon^{kji}N_iN_k\nonumber\\ =i\epsilon^{ijk}N_kN_i-i\epsilon^{ijk}N_kN_i\nonumber \quad relabeled \quad indices\\ =0} $$

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  • $\begingroup$ Oh wow, thanks a lot! I did not think about this particular relabeling trick. Must add it to my useful tools! Thanks again! $\endgroup$ – Soap312 Oct 23 '17 at 20:40

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