# Homework Help: 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

### Staff: Admin

Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?

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