Decomposition of $SO(10)$ into $SU(4)$











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I would like to apply branching rule to decompose a representation of $SO(10)$ into that of $SU(4)$. So I did the following, and I'm not sure if it's correct. I use the following chain of decomposition:



$$SO(10)hspace{1mm}rightarrowhspace{1mm}SU(2)times SU(2) times SU(4) hspace{1mm} rightarrow hspace{1mm}SU(4).$$



In particular, I'm interested in the adjoint $mathbf{45}$ of $SO(10)$:
begin{eqnarray}
mathbf{45} &rightarrow &mathbf{(3,1,1) + (1,3,1) + (1,1,15) + (2,2,6)} nonumber\
%
&rightarrow & mathbf{(3times 1) + (3times 1) + 15 + (4times 6)}. nonumber\
%
end{eqnarray}

My concern is in the second line of the above equation. Since I'm only interested in the $SU(4)$ part and not the $SU(2)times SU(2)$ part, can I just simply convert the $SU(2)times SU(2)$ representation into the coefficients counting the number of representation of $SU(4)$?










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    up vote
    1
    down vote

    favorite












    I would like to apply branching rule to decompose a representation of $SO(10)$ into that of $SU(4)$. So I did the following, and I'm not sure if it's correct. I use the following chain of decomposition:



    $$SO(10)hspace{1mm}rightarrowhspace{1mm}SU(2)times SU(2) times SU(4) hspace{1mm} rightarrow hspace{1mm}SU(4).$$



    In particular, I'm interested in the adjoint $mathbf{45}$ of $SO(10)$:
    begin{eqnarray}
    mathbf{45} &rightarrow &mathbf{(3,1,1) + (1,3,1) + (1,1,15) + (2,2,6)} nonumber\
    %
    &rightarrow & mathbf{(3times 1) + (3times 1) + 15 + (4times 6)}. nonumber\
    %
    end{eqnarray}

    My concern is in the second line of the above equation. Since I'm only interested in the $SU(4)$ part and not the $SU(2)times SU(2)$ part, can I just simply convert the $SU(2)times SU(2)$ representation into the coefficients counting the number of representation of $SU(4)$?










    share|cite|improve this question


























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      I would like to apply branching rule to decompose a representation of $SO(10)$ into that of $SU(4)$. So I did the following, and I'm not sure if it's correct. I use the following chain of decomposition:



      $$SO(10)hspace{1mm}rightarrowhspace{1mm}SU(2)times SU(2) times SU(4) hspace{1mm} rightarrow hspace{1mm}SU(4).$$



      In particular, I'm interested in the adjoint $mathbf{45}$ of $SO(10)$:
      begin{eqnarray}
      mathbf{45} &rightarrow &mathbf{(3,1,1) + (1,3,1) + (1,1,15) + (2,2,6)} nonumber\
      %
      &rightarrow & mathbf{(3times 1) + (3times 1) + 15 + (4times 6)}. nonumber\
      %
      end{eqnarray}

      My concern is in the second line of the above equation. Since I'm only interested in the $SU(4)$ part and not the $SU(2)times SU(2)$ part, can I just simply convert the $SU(2)times SU(2)$ representation into the coefficients counting the number of representation of $SU(4)$?










      share|cite|improve this question















      I would like to apply branching rule to decompose a representation of $SO(10)$ into that of $SU(4)$. So I did the following, and I'm not sure if it's correct. I use the following chain of decomposition:



      $$SO(10)hspace{1mm}rightarrowhspace{1mm}SU(2)times SU(2) times SU(4) hspace{1mm} rightarrow hspace{1mm}SU(4).$$



      In particular, I'm interested in the adjoint $mathbf{45}$ of $SO(10)$:
      begin{eqnarray}
      mathbf{45} &rightarrow &mathbf{(3,1,1) + (1,3,1) + (1,1,15) + (2,2,6)} nonumber\
      %
      &rightarrow & mathbf{(3times 1) + (3times 1) + 15 + (4times 6)}. nonumber\
      %
      end{eqnarray}

      My concern is in the second line of the above equation. Since I'm only interested in the $SU(4)$ part and not the $SU(2)times SU(2)$ part, can I just simply convert the $SU(2)times SU(2)$ representation into the coefficients counting the number of representation of $SU(4)$?







      group-theory representation-theory lie-groups lie-algebras branching-rules






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      edited Nov 22 at 14:19









      Qmechanic

      4,75811852




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      asked Nov 21 at 8:08









      user195583

      211




      211






















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          Note that the branching rule $$so(10)quad supseteq quad so(4) oplus so(6)quad cong quad su(2)oplus su(2) oplus su(4)$$ only holds at the level of Lie algebras, not Lie groups. But apart from that then Yes, the number of $su(4)$-representation in OP's example is given by the product of dimensions of the two $su(2)$-representations.






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            Note that the branching rule $$so(10)quad supseteq quad so(4) oplus so(6)quad cong quad su(2)oplus su(2) oplus su(4)$$ only holds at the level of Lie algebras, not Lie groups. But apart from that then Yes, the number of $su(4)$-representation in OP's example is given by the product of dimensions of the two $su(2)$-representations.






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              up vote
              1
              down vote













              Note that the branching rule $$so(10)quad supseteq quad so(4) oplus so(6)quad cong quad su(2)oplus su(2) oplus su(4)$$ only holds at the level of Lie algebras, not Lie groups. But apart from that then Yes, the number of $su(4)$-representation in OP's example is given by the product of dimensions of the two $su(2)$-representations.






              share|cite|improve this answer























                up vote
                1
                down vote










                up vote
                1
                down vote









                Note that the branching rule $$so(10)quad supseteq quad so(4) oplus so(6)quad cong quad su(2)oplus su(2) oplus su(4)$$ only holds at the level of Lie algebras, not Lie groups. But apart from that then Yes, the number of $su(4)$-representation in OP's example is given by the product of dimensions of the two $su(2)$-representations.






                share|cite|improve this answer












                Note that the branching rule $$so(10)quad supseteq quad so(4) oplus so(6)quad cong quad su(2)oplus su(2) oplus su(4)$$ only holds at the level of Lie algebras, not Lie groups. But apart from that then Yes, the number of $su(4)$-representation in OP's example is given by the product of dimensions of the two $su(2)$-representations.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 22 at 14:18









                Qmechanic

                4,75811852




                4,75811852






























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