# Properties of the so(n) lie algebra

1. Nov 6, 2015

### spaghetti3451

1. The problem statement, all variables and given/known data

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 $n>2$.

2. Relevant equations

3. 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?

Last edited: Nov 6, 2015
2. Nov 11, 2015