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Irreducible representation of tensor field

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su-ki
#1
Jan17-13, 10:43 AM
P: 2
In Mark Srednicki's book "Quantum Field Theory"
He says that a tensor field [itex] B^{αβ} [/itex] with no particular symmetry can be written as :-

[itex] B^{αβ} = A^{αβ} + S^{αβ} + (1/4) g^{αβ} T(x) [/itex] Equn. 33.6

where A - Antisymmetric, S = symmetric and T(x) = trace of [itex] B^{αβ} [/itex] .

Is there any reason for explicit addition of trace term ?
Coz generally we split things into symmetric and antisymmetric parts and trace is included in symmetric part.
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Bill_K
#2
Jan17-13, 12:05 PM
Sci Advisor
Thanks
Bill_K's Avatar
P: 4,160
If you're talking about the general linear group GL(n), the irreducible representations are the tensors whose indices have been symmetrized in a particular way. When you go to the orthogonal group, there are fewer transformations in the group, and some of these representations are no longer irreducible. The operation of contraction (forming a trace) commutes with the orthogonal transformations.
samalkhaiat
#3
Jan17-13, 02:59 PM
Sci Advisor
P: 892
Quote Quote by su-ki View Post
In Mark Srednicki's book "Quantum Field Theory"
He says that a tensor field [itex] B^{αβ} [/itex] with no particular symmetry can be written as :-

[itex] B^{αβ} = A^{αβ} + S^{αβ} + (1/4) g^{αβ} T(x) [/itex] Equn. 33.6

where A - Antisymmetric, S = symmetric and T(x) = trace of [itex] B^{αβ} [/itex] .

Is there any reason for explicit addition of trace term ?
Coz generally we split things into symmetric and antisymmetric parts and trace is included in symmetric part.

This is true only if the symmetric part is traceless. In general, for any rank-2 tensor, we write
[tex]B^{ ab } = B^{ (ab) } + B^{ [ab] } .[/tex]
Then we take the symmetric part and decompose it as
[tex]
B^{ (ab) } = \left( B^{ (ab) } - \frac{ 1 }{ 4 } g^{ ab } B \right) + \frac{ 1 }{ 4 } g^{ ab } B .
[/tex]
The tensor (call it [itex]S^{ ab }[/itex]) in the bracket on the left-hand side is symmetric and traceless, because
[tex]B = \mbox{ Tr } ( B^{ (ab) } ) = g_{ ab } B^{ (ab) }[/tex]
So, our original tensor can now be written as
[tex]
B^{ ab } = A^{ ab } + S^{ ab } + \frac{ 1 }{ 4 } g^{ ab } B ,
[/tex]
where [itex]A^{ ab } = - A^{ ba } \equiv B^{ [ab] }[/itex]

See posts #24 and 25 in

http://www.physicsforums.com/showthr...=192572&page=2

Sam

su-ki
#4
Jan17-13, 03:03 PM
P: 2
Irreducible representation of tensor field

yea, i got just it, thank u :)


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