Deriving the commutation relations of the so(n) Lie algebra The generators $(A_{ab})_{st}$ of the $so(n)$ Lie algebra are given by:
$$(A_{ab})_{st} = -i(\delta_{as}\delta_{bt}-\delta_{at}\delta_{bs}) = -i\delta_{s[a}\delta_{b]t}$$,
where $a,b$ label the number of the generator, and $s,t$ label the matrix element.
Now, I need to prove the following commutation relation using the definition above:
$$([A_{ij},A_{mn}])_{st} = -i(A_{j[m}\delta_{n]i}-A_{i[m}\delta_{n]j})_{st}$$.
Here's my attempt.
$([A_{ij},A_{mn}])_{st}$
$ = (A_{ij})_{sp}(A_{mn})_{pt}-(ij \iff mn)$
$= -\delta_{s[i}\delta_{j]p}\delta_{p[m}\delta_{n]t}-(ij \iff mn)$
$= -\delta_{s[i}\delta_{j][m}\delta_{n]t}+\delta_{s[m}\delta_{n][i}\delta_{j]t}$
Could you please suggest the next couple of steps? Should I expand all the antisymmetrised Kronecker delta's, or is there some sneaky shortcut to get to the answer?
 A: Notice that $-i(A_{j[m}\delta_{n]i}-A_{i[m}\delta_{n]j})_{st}$ = $-i(A_{jm}\delta_{ni}-A_{jn}\delta_{mi}-A_{im}\delta_{nj}+A_{in}\delta_{mj})_{st}$.
Now, 
$-\delta_{s[i}\delta_{j][m}\delta_{n]t}+\delta_{s[m}\delta_{n][i}\delta_{j]t}$
$=-\delta_{si}\delta_{j[m}\delta_{n]t}+\delta_{sj}\delta_{i[m}\delta_{n]t}+\delta_{sm}\delta_{n[i}\delta_{j]t}-\delta_{sn}\delta_{m[i}\delta_{j]t}$
$=-\delta_{si}\delta_{jm}\delta_{nt}+\delta_{si}\delta_{jn}\delta_{mt}+\delta_{sj}\delta_{im}\delta_{nt}-\delta_{sj}\delta_{in}\delta_{mt}+\delta_{sm}\delta_{ni}\delta_{jt}-\delta_{sm}\delta_{nj}\delta_{it}-\delta_{sn}\delta_{mi}\delta_{jt}+\delta_{sn}\delta_{mj}\delta_{it}$
$=(-i)(-i)(\delta_{si}\delta_{jm}\delta_{nt}-\delta_{si}\delta_{jn}\delta_{mt}-\delta_{sj}\delta_{im}\delta_{nt}+\delta_{sj}\delta_{in}\delta_{mt}-\delta_{sm}\delta_{ni}\delta_{jt}+\delta_{sm}\delta_{nj}\delta_{it}+\delta_{sn}\delta_{mi}\delta_{jt}-\delta_{sn}\delta_{mj}\delta_{it})$
$=(-i)(-i)(\delta_{sj}\delta_{mt}\delta_{ni}-\delta_{sm}\delta_{jt}\delta_{ni}-\delta_{sj}\delta_{nt}\delta_{mi}+\delta_{sn}\delta_{jt}\delta_{mi}-\delta_{si}\delta_{mt}\delta_{nj}+\delta_{sm}\delta_{it}\delta_{nj}+\delta_{si}\delta_{nt}\delta_{mj}-\delta_{sn}\delta_{it}\delta_{mj})$
$=(-i)(-i)(\delta_{s[j}\delta_{m]t}\delta_{ni}-\delta_{s[j}\delta_{n]t}\delta_{mi}-\delta_{s[i}\delta_{m]t}\delta_{nj}+\delta_{s[i}\delta_{n]t}\delta_{mj})$
$=-i((A_{jm})_{st}\delta_{ni}-(A_{jn})_{st}\delta_{mi}-(A_{im})_{st}\delta_{nj}+(A_{in})_{st}\delta_{mj})$
Thus, the proof is complete.
