Properties of the so(n) lie algebra

[spaghetti3451]

Homework Statement

The generators of the ##SO(n)## group are pure imaginary antisymmetric ##n \times n## matrices. Therefore, the dimension of the ##SO(n)## group is ##\frac{n(n-1)}{2}##. Therefore, the basis for the so(n) Lie algebra is given by the ##\frac{n(n-1)}{2}## basis vectors as follows: ##(A_{ab})_{st} = -i(\delta_{s[a}\delta_{b]t})##.

In the above, ##ab##, where ##a < b##, labels the generator, and ##st## labels the matrix element.

1. Prove that the commutator in the defining representation is given by: ##([A_{ij},A_{mn}])_{st} = -i(A_{j[m}\delta_{n]i}-A_{i[m}\delta_{n]j})_{st}##.

Now, define the so(n) algebra using ##([A_{ij},A_{mn}]) = -i(A_{j[m}\delta_{n]i}-A_{i[m}\delta_{n]j})##.

2. Show that ##[A_{ij},A_{mn}] = i\delta_{k[j}\delta_{i][m}\delta_{n]s}A_{ks}##.

Now, define ##[A_{ij},A_{mn}] = if_{ij,mn}^{ks}A_{ks}##. where ##f_{ij,mn}^{ks}=\delta_{k[j}\delta_{i][m}\delta_{n]s}##.

3. Show that the Cartan metric tensor ##g_{ij,ps} = 2(n-2)\delta_{ij,ps}##. Hence, show that the group ##SO(n)## is semisimple and compact for ##n>2##.

Homework Equations

The Attempt at a Solution

1. Here goes nothing.

##([A_{ij},A_{mn}])_{st}##
##=(A_{ij}A_{mn})_{st}-(A_{mn}A_{ij})_{st}##
##=(A_{ij})_{su}(A_{mn})_{ut}-(A_{mn})_{su}(A_{ij})_{ut}##
##=(A_{ij})_{su}[-i(\delta_{u[m}\delta_{n]t})]-(A_{mn})_{su}[-i(\delta_{u[i}\delta_{j]t})]##
##=-i[(A_{ij})_{su}(\delta_{um}\delta_{nt}-\delta_{un}\delta_{mt})-(A_{mn})_{su}(\delta_{ui}\delta_{jt}-\delta_{uj}\delta_{it})]##
##=-i[(A_{ij})_{sm}\delta_{nt}-(A_{ij})_{sn}\delta_{mt}-(A_{mn})_{si}\delta_{jt}+(A_{mn})_{sj}\delta_{it}]##
##=-[\delta_{si}\delta_{jm}\delta_{nt}-\delta_{sj}\delta_{im}\delta_{nt}-\delta_{si}\delta_{jn}\delta_{mt}+\delta_{sj}\delta_{in}\delta_{mt}-\delta_{sm}\delta_{ni}\delta_{jt}+\delta_{sn}\delta_{mi}\delta_{jt}+\delta_{sm}\delta_{nj}\delta_{it}-\delta_{sn}\delta_{mj}\delta_{it}]##
##=-i(-i\delta_{sj}\delta_{mt}\delta_{ni}+i\delta_{sm}\delta_{jt}\delta_{ni}+i\delta_{sj}\delta_{nt}\delta_{mi}-i\delta_{sn}\delta_{jt}\delta_{mi}+i\delta_{si}\delta_{mt}\delta_{nj}-i\delta_{sm}\delta_{it}\delta_{nj}-i\delta_{si}\delta_{nt}\delta_{mj}+i\delta_{sn}\delta_{it}\delta_{mj})##
##=-i((A_{jm})_{st}\delta_{ni}-(A_{jn})_{st}\delta_{mi}-(A_{im})_{st}\delta_{nj}+(A_{in})_{st}\delta_{mj})##
##=-i(A_{jm}\delta_{ni}-A_{jn}\delta_{mi}-A_{im}\delta_{nj}+A_{in}\delta_{mj})_{st}##
##-i(A_{j[m}\delta_{n]i}-A_{i[m}\delta_{n]j})_{st}##

Is this the best way to prove part 1?

 

[mighty2000]

I feel like there must be a simpler way, but I can't think of one.

2. I'm not sure how to proceed with this one. I'm not sure what the notation ##[A_{ij},A_{mn}] = i\delta_{k[j}\delta_{i][m}\delta_{n]s}A_{ks}## means. Does it mean that ##[A_{ij},A_{mn}] = i\delta_{k[j}\delta_{i][m}\delta_{n]s}A_{ks}## is defined for all ##i,j,m,n,k,s##?

3. Again, not sure how to proceed. I'm also not sure what the Cartan metric tensor is or how it relates to the rest of the problem. Any hints or guidance would be greatly appreciated.