# Irreducible decomposition of higher order tensors

I am familiar with the notion of irreps. My question refers simply to tensor representations (not tensor products of representations) and how can we decompose them into irreducible parts? For example, a rank 2 tensor is decomposed into an antisymmetric part, a traceless symmetric and its trace. What is the generalization of that for higher rank tensors? Could someone provide an example for, say rank 3 or 4? Thank you

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Essentially a duplicate of physics.stackexchange.com/q/45368/2451 – Qmechanic Jul 22 '13 at 19:02

When you say tensor there is also a need to specify what is the group/algebra it is a tensor of. That you said that rank-two decomposes into symmetric, antisymemtric and trace, I think you have in mind either $so(d)$ or $sp(2m)$. For $gl(d)$ there is no trace. In any case, suprisingly, decomposing a rank-$k$ tensor that have a-priori no symmetries is equivalent to computing $\otimes^k V$, where $V$ is a vector representation.
For example, take $T^{ab|c}$ of $so(d)$ and assume that it is, say, symmetric and traceless in $ab$ (we know how to decompose rank-two tensors). Then one finds $T^{ab|c}=S^{abc}+H^{ab,c}+\left(\eta^{ac}V^b+\eta^{bc}V^a-\frac2d \eta^{ab}V^c\right)$ where $S^{abc}$ is totally-symmetric and traceless. $V^a$ parameterizes the trace $T^{ab|c}\eta_{bc}$ and $H^{ab,c}$ is traceless and obeys $H^{ab,c}+H^{bc,a}+H^{ca,b}\equiv0$. $H$ is neither totally symmetric nor antisymmetric, it has a mixed symmetry.
No, totally antisymmetric tensor is almost always irreducible. The only exception is if you have $so(2n)$ and the tensor is of rank-$n$, then one can impose (anti)-selfduality condition – John Jul 23 '13 at 18:50