Decomposing an arbitrary rank tensor into components with symmetries

Question

It is well known that an arbitrary rank 2 tensor (excuse me for my sloppy language) decomposes into a symmetric and anti-symmetric component: $$T_{ab}=T_{(ab)}+T_{[ab]}.$$ Is there a similar result for a rank 3 tensor? How about a tensor of arbitrary rank? I can see that, for example, $T_{abc}=T_{a(bc)}+T_{a[bc]}$. But I am looking for a decomposition which mixes all indices. It may look something like this $$T_{abc}=T_{(abc)}+T_{[abc]}+T_{[a|b|c]}+T_{(a|b|c)}+T_{a(bc)}+T_{a[bc]}\cdots .$$ I am more interested in the general case, but the rank 3 case provides a nice playground.

Answer 1

Yes, but it's complicated. A rank $3$ tensor has a symmetric part, an antisymmetric part, and a third part which is harder to explain (but which you can compute by subtracting off the symmetric and antisymmetric parts).

In general a rank $n$ tensor decomposes according to the irreducible representations of the symmetric group $S_n$, by Schur-Weyl duality. There are $p(n)$ of these where $p(n)$ is the partition function, so a rank $4$ tensor decomposes into $5$ parts, a rank $5$ tensor decomposes into $7$ parts, etc.