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I need to prove a commutator relation, but I'm getting stuck at the definition of the matrices.

$(L_{ab})_{cd} = \delta_{ac} \delta_{bd} - \delta_{ad} \delta_{bc}$ with $a<b$ and $a, b \in 1,2,3,4$. What does this definition of the matrix mean? Can someone explain this to me?

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Let me illustrate in the case $2\times2$.

The matrix $L_{ab}$ would be $$ \left[\begin{array}{ccc} \delta_{a1}\delta_{b1}-\delta_{a1}\delta_{b1} &\delta_{a1}\delta_{b2}-\delta_{a2}\delta_{b1}\\ \delta_{a2}\delta_{b1}-\delta_{a1}\delta_{b2} &\delta_{a2}\delta_{b2}-\delta_{a2}\delta_{b2} \end{array}\right] .$$

Then for $L_{11}$ we have $$ \left[\begin{array}{ccc} \delta_{11}\delta_{11}-\delta_{11}\delta_{11} &\delta_{11}\delta_{12}-\delta_{12}\delta_{11}\\ \delta_{12}\delta_{11}-\delta_{11}\delta_{12} &\delta_{12}\delta_{12}-\delta_{12}\delta_{12} \end{array}\right] = \left[\begin{array}{ccc} 0&0\\ 0&0 \end{array}\right] $$

But for $L_{12}$ $$ \left[\begin{array}{ccc} \delta_{11}\delta_{21}-\delta_{11}\delta_{21} &\delta_{11}\delta_{22}-\delta_{12}\delta_{21}\\ \delta_{12}\delta_{21}-\delta_{11}\delta_{22} &\delta_{12}\delta_{22}-\delta_{12}\delta_{22} \end{array}\right] = \left[\begin{array}{ccc} 0&1\\ -1&0 \end{array}\right] $$

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  • $\begingroup$ Thank you! But what do the second indices 'cd' mean. So, I'd understand it if it was only $L_{ab}$, but I'm very confused about the $(L_{ab})_{cd}$ notation. $\endgroup$ – Darius Nov 15 '15 at 17:47
  • $\begingroup$ the $cd$ indexes determines the entry in the $c$-row and $d$-column of the matrix $L_{ab}$ $\endgroup$ – janmarqz Nov 15 '15 at 17:51
  • $\begingroup$ you oughtta use that $\delta_{ij}=1$ when $i=j$ and $\delta_{ij}=0$ when $i\neq j$... this is the famous delta Kronecker $\endgroup$ – janmarqz Nov 15 '15 at 17:55
  • $\begingroup$ Thanks, now I got it! So basically I could build this matrix in 4x4 by following your idea. Now I have to prove this commutator relation: $\left[L_{ab}, L_{cd} \right] = \delta_{ad}L_{bc} + \delta_{bc}L_{ab} - \delta_{ac} L_{bd} - \delta_{bd} L_{ac}$. DO you have any hint on proving this? I guess writing out the matrices and doing the multiplication is pretty straight forward and not so nice. $\endgroup$ – Darius Nov 15 '15 at 22:30

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