Deriving the structure constants of the SO(n) group

[spaghetti3451]

The commutation relations for the $\mathfrak{so(n)}$ Lie algebra is:$([A_{ij},A_{mn}])_{st} = -i(A_{j[m}\delta_{n]i}-A_{i[m}\delta_{n]j})_{st}$.where the generators $(A_{ab})_{st}$ of the $\mathfrak{so(n)}$ 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.I would like to show that the structure constants $f_{ij,mn}^{ks}$ of the $\mathfrak{so(n)}$ Lie algebra such that$[A_{ij},A_{mn}] = if_{ij,mn}^{ks}A_{ks}$are given by$f_{ij,mn}^{ks} = \delta_{k[j}\delta_{i][m}\delta_{n]s}$.Can someone help me out with this?

 

[fzero]

$([A_{ij},A_{mn}])_{st} = -i(A_{j[m}\delta_{n]i}-A_{i[m}\delta_{n]j})_{st}$.
...
I would like to show that the structure constants $f_{ij,mn}^{ks}$ of the $\mathfrak{so(n)}$ Lie algebra such that$[A_{ij},A_{mn}] = if_{ij,mn}^{ks}A_{ks}$

From these equations, you have
$$f_{ij,mn}^{ks}A_{ks}=-(A_{j[m}\delta_{n]i}-A_{i[m}\delta_{n]j}) .$$
You can expand the RHS using identities like $A_{jm} = \delta_{jk}\delta_{ms} A_{ks}$ to derive the form of the structure constants.

 

[spaghetti3451]

Thanks! I got it!

 

[spaghetti3451]

It is easy to show that the generators $(A_{ab})_{st}$ of the $\mathfrak{so(n)}$ Lie algebra given by

$(A_{ab})_{st} = -i(\delta_{as}\delta_{bt}-\delta_{at}\delta_{bs}) = -i\delta_{s[a}\delta_{b]t}$

reduce to the three generators $A_{23},A_{31}, A_{12}$ of the $so(3)$ Lie algebra.

The generators $A_{ii}$ for $i=1,2,3$ are redundant as they all equal to $0$.
The generators $A_{32},A_{13},A_{21}$ are redundant as $A_{32}=-A_{23},A_{13}=-A_{31},A_{21}=-A_{12}$.
The redundancy comes about because the matrices $A_{ab}$ themselves form the matrix elements of an antisymmetric matrix element.

Am I correct?

 

[fzero]

It is easy to show that the generators $(A_{ab})_{st}$ of the $\mathfrak{so(n)}$ Lie algebra given by

$(A_{ab})_{st} = -i(\delta_{as}\delta_{bt}-\delta_{at}\delta_{bs}) = -i\delta_{s[a}\delta_{b]t}$

reduce to the three generators $A_{23},A_{31}, A_{12}$ of the $so(3)$ Lie algebra.

The generators $A_{ii}$ for $i=1,2,3$ are redundant as they all equal to $0$.
The generators $A_{32},A_{13},A_{21}$ are redundant as $A_{32}=-A_{23},A_{13}=-A_{31},A_{21}=-A_{12}$.
The redundancy comes about because the matrices $A_{ab}$ themselves form the matrix elements of an antisymmetric matrix element.

Am I correct?

Yes, the generators are antisymmetric in both pairs of indices:
$$(A_{ab})_{st} = - (A_{ba})_{st} = -(A_{ab})_{ts} .$$

 

[spaghetti3451]

I would also like to show that the structure constants $f_{ij,mn}^{ks}$ of the $\mathfrak{so(n)}$ Lie algebra reduce to the structure constants $\epsilon_{ij}^{k}$ of the $so(3)$ Lie algebra defined as follows:

$[X_{p},X_{q}]=i\epsilon_{pqr}X_{r}$

where $X_{1}=A_{23},X_{2}=A_{31},X_{3}=A_{12}$.

Now, however must I try, I cannot show that $f_{ij,mn}^{ks} = \delta_{k[j}\delta_{i][m}\delta_{n]s}$ reduces to $\epsilon_{pqr}$ under the above identification $p=ij,q=mn,r=ks$.

To illustrate,

$f_{ij,mn}^{ks}$
$= \delta_{k[j}\delta_{i][m}\delta_{n]s}$
$= \delta_{kj}\delta_{i[m}\delta_{n]s}-\delta_{ki}\delta_{j[m}\delta_{n]s}$
$= \delta_{kj}\delta_{im}\delta_{ns}-\delta_{kj}\delta_{in}\delta_{ms}-\delta_{ki}\delta_{jm}\delta_{ns}+\delta_{ki}\delta_{jn}\delta_{ms}$

Now, $\epsilon_{123} = 1$, but $f_{23,31}^{12} \neq 1$.

What exactly is the problem?

 

[fzero]

For $SO(3)$ we can use the invariant $\epsilon_{ijk}$ to project a pair of indices onto a single index. So the expression that you should compute is $\epsilon_{aks}{\epsilon_b}^{ij}{\epsilon_c}^{mn} f^{ks}_{ij,mn}$.

 

[spaghetti3451]

All right. Let me first perform the computation.

$\epsilon_{aks}{\epsilon_b}^{ij}{\epsilon_c}^{mn} f^{ks}_{ij,mn}$
$= \epsilon_{aks}{\epsilon_b}^{ij}{\epsilon_c}^{mn} (\delta_{kj}\delta_{im}\delta_{ns}-\delta_{kj}\delta_{in}\delta_{ms}-\delta_{ki}\delta_{jm}\delta_{ns}+\delta_{ki}\delta_{jn}\delta_{ms})$
$= \epsilon_{aks}({\epsilon_b}^{mk}{\epsilon_c}^{ms} - {\epsilon_b}^{nk}{\epsilon_c}^{sn} -{\epsilon_b}^{km}{\epsilon_c}^{ms}+{\epsilon_b}^{kn}{\epsilon_c}^{sn})$
$= \epsilon_{aks}({\epsilon_b}^{mk}{\epsilon_c}^{ms} + {\epsilon_b}^{nk}{\epsilon_c}^{ns} +{\epsilon_b}^{mk}{\epsilon_c}^{ms}+{\epsilon_b}^{nk}{\epsilon_c}^{ns})$
$= 4 \epsilon_{aks}{\epsilon_b}^{mk}{\epsilon_c}^{ms}$
$= 4 \epsilon_{aks}{\epsilon_m}^{kb}{\epsilon_m}^{sc}$
$= 4 \epsilon_{aks}(\delta^{ks}\delta^{bc}-\delta^{kc}\delta^{bs})$
$= -4 \epsilon_{acb}$
$= 4 \epsilon_{abc}$

Is there some way to get rid of the factor of 4 in front of $\epsilon_{abc}$?

 

[fzero]

All right. Let me first perform the computation.

$\epsilon_{aks}{\epsilon_b}^{ij}{\epsilon_c}^{mn} f^{ks}_{ij,mn}$
$= \epsilon_{aks}{\epsilon_b}^{ij}{\epsilon_c}^{mn} (\delta_{kj}\delta_{im}\delta_{ns}-\delta_{kj}\delta_{in}\delta_{ms}-\delta_{ki}\delta_{jm}\delta_{ns}+\delta_{ki}\delta_{jn}\delta_{ms})$
$= \epsilon_{aks}({\epsilon_b}^{mk}{\epsilon_c}^{ms} - {\epsilon_b}^{nk}{\epsilon_c}^{sn} -{\epsilon_b}^{km}{\epsilon_c}^{ms}+{\epsilon_b}^{kn}{\epsilon_c}^{sn})$
$= \epsilon_{aks}({\epsilon_b}^{mk}{\epsilon_c}^{ms} + {\epsilon_b}^{nk}{\epsilon_c}^{ns} +{\epsilon_b}^{mk}{\epsilon_c}^{ms}+{\epsilon_b}^{nk}{\epsilon_c}^{ns})$
$= 4 \epsilon_{aks}{\epsilon_b}^{mk}{\epsilon_c}^{ms}$
$= 4 \epsilon_{aks}{\epsilon_m}^{kb}{\epsilon_m}^{sc}$
$= 4 \epsilon_{aks}(\delta^{ks}\delta^{bc}-\delta^{kc}\delta^{bs})$
$= -4 \epsilon_{acb}$
$= 4 \epsilon_{abc}$

Is there some way to get rid of the factor of 4 in front of $\epsilon_{abc}$?

I get the same factor of 4. It turned out that I forgot some factors of 2 in the mapping I suggested. Let us define $A_{ij} =\gamma \epsilon_{ijp} X_p$ for some constant $\gamma$. Then we can multiply this again by $\epsilon$ to show that $X_p = (1/(2\gamma)) \epsilon_{pij} A_{ij}$. Now consider
$$ [X_p, X_q] = \frac{1}{4\gamma^2} \epsilon_{pij} \epsilon_{qmn} [A_{ij},A_{mn}]
= \frac{1}{4\gamma^2} \epsilon_{pij} \epsilon_{qmn} f^{ks}_{ij,mn} \gamma \epsilon_{ksr} X_r.$$
We find precisely the factor of 4 that we needed and can therefore set $\gamma=1$.

 

[spaghetti3451]

Thank you so much! I get it now!