# Problem calculating commutator with Casimir

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 $0$.

• Try checking your index structure -- halfway you switch from having good indices to having three copies of some indices. – knzhou Oct 23 '17 at 17:31

It's the same through line 5. Line 6 starts being different: $$\left[N^2,N_j\right] {=N^2N_j-N_jN^2\nonumber\\ =(N_iN_i)N_j-N_j(N_iN_i)\nonumber\\ =N_iN_jN_i+N_i[N_i,N_j]-N_jN_iN_i\nonumber\\ =N_jN_iN_i+[N_i,N_j]N_i+N_ii\epsilon^{ijk}N_k-N_jN_iN_i\nonumber\\ =i\epsilon^{ijk}N_kN_i+N_ii\epsilon^{ijk}N_k\nonumber\\ =i\epsilon^{ijk}N_kN_i+i\epsilon^{ijk}N_iN_k\nonumber\\ =i\epsilon^{ijk}N_kN_i-i\epsilon^{kji}N_iN_k\nonumber\\ =i\epsilon^{ijk}N_kN_i-i\epsilon^{ijk}N_kN_i\nonumber \quad relabeled \quad indices\\ =0}$$