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


1 Answer 1


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.

  • $\begingroup$ You are totally right.. I had so(n) symmetry in mind. My main question is whether an antisymmetric tensor is always irreducible with respect to rotations regardless of rank. I take from your answer that it is not always the case? $\endgroup$
    – Cala
    Jul 22, 2013 at 22:49
  • $\begingroup$ 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 $\endgroup$
    – John
    Jul 23, 2013 at 18:50
  • $\begingroup$ @Cala It seems John's answer is what you need, aside from an explicit example, which you can now build. The only reason I make this comment is that it is annoying not to be aknowledged once one has done the work of writing an answer up. You're probably excited to have the answer and have forgotten to accept or acknowledge (I've been guilty of this myself). $\endgroup$ Oct 21, 2013 at 0:02

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