# Prove commutator relation

I need to prove a commutator relation, but I'm getting stuck.

$(L_{ab})_{cd} = \delta_{ac} \delta_{bd} - \delta_{ad} \delta_{bc}$ with $a<b$ and $a, b \in 1,2,3,4$.

Now I need to show that:

$\left[L_{ab}, L_{cd} \right] = \delta_{ad} L_{bc} + \delta_{bc} L_{ab} - \delta_{ac} L_{bd} - \delta_{bd} L_{ac}$.

Can someone give me a hint on how to show that?

• You need to explain more what you're working with. What are the $L_{ab}$, for example? – Paul Nov 16 '15 at 16:15
• Well, the matrix $L_{ab}$ is defined by the delta function as described above. As said in the exercise it is a generator of a rotation matrix of so(4). – Darius Nov 16 '15 at 16:16
• Did you try writing it out? – A.P. Nov 16 '15 at 16:29
• I am not sure if this would be the right way as i would multiplicity two 4x4 matrices with large entries.. – Darius Nov 16 '15 at 16:34

Your commutator relation has an error in its second summand. The correct equation is $$[L_{ab}, L_{cd}]=\delta_{ad}L_{bc}+\delta_{bc}L_{ad}−\delta_{ac}L_{bd}−\delta_{bd}L_{ac}.$$
The remaining part is straightforward if you know how to multiply matrices. You have to calculate (sum rule) $$([L_{ab}, L_{cd}])_{nm} = (L_{ab})_{nk}(L_{cd})_{km} - (L_{cd})_{nk}(L_{ab})_{km}.$$
• @Darius $([L_{ab},L_{cd}])_{nm}$ is just the element on row $n$ and column $m$ of $[L_{ab},L_{cd}]$. You used the same notation in your question! As for multiplying two matrices by using just their entries... try to check what is the $ij$-th entry of the product of two generic matrices... – A.P. Nov 16 '15 at 19:19
• Thank you so much!I did it and I think I got the right solution. But I still have questions: I don't really get the second line of your first answer, how do you get this line, especially with the indices k,n,m? And why is, for example $\delta_{an} \delta_{cm} = L_{ac}$? – Darius Nov 16 '15 at 20:08