How to Decompose finite dimensional irreducible representation of $SO(d+1,1)$ into irreps of the subgroup...
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I want to know how can one decompose a non-trivial finite dimensional irreducible representation (obviously not unitary) of generalized Lorentz Group like $SO(d+1,1)$ into irreducible representations of its subgroup $SO(d)times SO(1,1)$. For example, consider the spin-2 tensor representation of $SO(d+1,1)$ ,that expressed by a Young Tableaux having a single row with two boxes, gets decomposed into the corresponding Young Tableaux representations of $SO(d)$ with associated $SO(1,1)$ charges. How can one determine this spectral decomposition?
representation-theory
asked Nov 15 at 11:32
parthiv haldar
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I want to know how can one decompose a non-trivial finite dimensional irreducible representation (obviously not unitary) of generalized Lorentz Group like $SO(d+1,1)$ into irreducible representations of its subgroup $SO(d)times SO(1,1)$. For example, consider the spin-2 tensor representation of $SO(d+1,1)$ ,that expressed by a Young Tableaux having a single row with two boxes, gets decomposed into the corresponding Young Tableaux representations of $SO(d)$ with associated $SO(1,1)$ charges. How can one determine this spectral decomposition?
representation-theory
asked Nov 15 at 11:32
parthiv haldar
62
New contributor
parthiv haldar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I want to know how can one decompose a non-trivial finite dimensional irreducible representation (obviously not unitary) of generalized Lorentz Group like $SO(d+1,1)$ into irreducible representations of its subgroup $SO(d)times SO(1,1)$. For example, consider the spin-2 tensor representation of $SO(d+1,1)$ ,that expressed by a Young Tableaux having a single row with two boxes, gets decomposed into the corresponding Young Tableaux representations of $SO(d)$ with associated $SO(1,1)$ charges. How can one determine this spectral decomposition?
representation-theory
asked Nov 15 at 11:32
parthiv haldar
62
New contributor
parthiv haldar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I want to know how can one decompose a non-trivial finite dimensional irreducible representation (obviously not unitary) of generalized Lorentz Group like $SO(d+1,1)$ into irreducible representations of its subgroup $SO(d)times SO(1,1)$. For example, consider the spin-2 tensor representation of $SO(d+1,1)$ ,that expressed by a Young Tableaux having a single row with two boxes, gets decomposed into the corresponding Young Tableaux representations of $SO(d)$ with associated $SO(1,1)$ charges. How can one determine this spectral decomposition?
representation-theory
representation-theory
asked Nov 15 at 11:32
parthiv haldar
62
New contributor
parthiv haldar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Nov 15 at 11:32
parthiv haldar
62
New contributor
parthiv haldar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Nov 15 at 11:32
parthiv haldar
62
New contributor
parthiv haldar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Nov 15 at 11:32
parthiv haldar
62
asked Nov 15 at 11:32
parthiv haldar
62
62
New contributor
parthiv haldar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
parthiv haldar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
parthiv haldar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
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parthiv haldar is a new contributor. Be nice, and check out our Code of Conduct.
parthiv haldar is a new contributor. Be nice, and check out our Code of Conduct.
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