Mixing
Decomposition of $SO(10)$ into $SU(4)$
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
edited Nov 22 '18 at 14:19
Qmechanic
4,87711854
asked Nov 21 '18 at 8:08
user195583
211
add a comment |
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
edited Nov 22 '18 at 14:19
Qmechanic
4,87711854
asked Nov 21 '18 at 8:08
user195583
211
add a comment |
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
edited Nov 22 '18 at 14:19
Qmechanic
4,87711854
asked Nov 21 '18 at 8:08
user195583
211
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
group-theory representation-theory lie-groups lie-algebras branching-rules
edited Nov 22 '18 at 14:19
Qmechanic
4,87711854
asked Nov 21 '18 at 8:08
user195583
211
edited Nov 22 '18 at 14:19
Qmechanic
4,87711854
asked Nov 21 '18 at 8:08
user195583
211
edited Nov 22 '18 at 14:19
Qmechanic
4,87711854
edited Nov 22 '18 at 14:19
Qmechanic
4,87711854
edited Nov 22 '18 at 14:19
Qmechanic
4,87711854
4,87711854
asked Nov 21 '18 at 8:08
user195583
211
asked Nov 21 '18 at 8:08
user195583
211
asked Nov 21 '18 at 8:08
user195583
211
211
add a comment |
add a comment |
1 Answer
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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.
answered Nov 22 '18 at 14:18
Qmechanic
4,87711854
add a comment |
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1 Answer
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1 Answer
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active
oldest
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votes
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.
answered Nov 22 '18 at 14:18
Qmechanic
4,87711854
add a comment |
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.
answered Nov 22 '18 at 14:18
Qmechanic
4,87711854
add a comment |
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.
answered Nov 22 '18 at 14:18
Qmechanic
4,87711854
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.
answered Nov 22 '18 at 14:18
Qmechanic
4,87711854
answered Nov 22 '18 at 14:18
Qmechanic
4,87711854
answered Nov 22 '18 at 14:18
Qmechanic
4,87711854
answered Nov 22 '18 at 14:18
Qmechanic
4,87711854
4,87711854
add a comment |
add a comment |
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