# Irreducible representation of tensor field

1. Jan 17, 2013

### su-ki

In Mark Srednicki's book "Quantum Field Theory"
He says that a tensor field $B^{αβ}$ with no particular symmetry can be written as :-

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

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

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.

Last edited: Jan 17, 2013
2. Jan 17, 2013

### Bill_K

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. Jan 17, 2013

### samalkhaiat

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

See posts #24 and 25 in