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Deriving the structure constants of the SO(n) group

  1. Nov 19, 2015 #1
    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?
     
  2. jcsd
  3. Nov 19, 2015 #2

    fzero

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    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.
     
  4. Nov 21, 2015 #3
    Thanks! I got it!
     
  5. Nov 21, 2015 #4
    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?
     
  6. Nov 21, 2015 #5

    fzero

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    Yes, the generators are antisymmetric in both pairs of indices:
    $$(A_{ab})_{st} = - (A_{ba})_{st} = -(A_{ab})_{ts} .$$
     
  7. Nov 21, 2015 #6
    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?
     
    Last edited: Nov 21, 2015
  8. Nov 21, 2015 #7

    fzero

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    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}##.
     
  9. Nov 21, 2015 #8
    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}##?
     
    Last edited: Nov 21, 2015
  10. Nov 21, 2015 #9

    fzero

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    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##.
     
  11. Nov 21, 2015 #10
    Thank you so much! I get it now!
     
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