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Irreducible representation of tensor field 
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#1
Jan1713, 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. 


#2
Jan1713, 12:05 PM

Sci Advisor
Thanks
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.



#3
Jan1713, 02:59 PM

Sci Advisor
P: 910

This is true only if the symmetric part is traceless. In general, for any rank2 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 lefthand 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 


#4
Jan1713, 03:03 PM

P: 2

Irreducible representation of tensor field
yea, i got just it, thank u :)



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