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I know that rank 2 tensors can be decomposed as such. But I would like to know if this is possible for any rank tensors?

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A (higher) $n$-rank tensor $T^{\mu_1\ldots \mu_n}$ with $n\geq 3$ cannot always be decomposed into just a totally symmetric and a totally antisymmetric piece. In general, there will also be components of mixed symmetry.

The symmetric group $S_n$ acts on the indices $$(\mu_1,\ldots ,\mu_n)\quad \longrightarrow\quad (\mu_{\pi(1)},\ldots ,\mu_{\pi(n)})$$ via permutations $\pi\in S_n$. One can decompose the tensor $T^{\mu_1\ldots \mu_n}$ according to irreps (irreducible representations) of the symmetric group.

Each irrep corresponds to a Young tableau of $n$ boxes. For instance, a single horizontal row of $n$ boxes corresponds to a totally symmetric tensor, while a single vertical column of $n$ boxes corresponds to a totally antisymmetric tensor. But there are also other Young tableaux with a (kind of) mixed symmetry.

Here is a Google search for further reading.

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No.

Definitions:

  • A rank-n tensor is a linear map from n vectors to a scalar.
  • A symmetric tensor is one which in which the order of the arguments doesn't matter.
  • An antisymmetric tensor in which transposing two arguments multiplies the result by -1.

Suppose we have some rank-3 tensor $T$ with symmetric part $S$ and anti-symmetric part $A$ so

$$T(a,b,c) = S(a,b,c) + A(a,b,c)$$

where $a,b,c\,$ are arbitrary vectors. Transposing $c$ and $a$ on the right hand side, then transposing $a$ and $b$, we have

$$T(a,b,c) = S(c,a,b) + A(c,a,b)$$

so we conclude

$$T(a,b,c) = T(c,a,b)$$

it is trivial to construct a counterexample, so not all rank-three tensors can be decomposed into symmetric and anti-symmetric parts.

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Is it possible to find a more general decomposition into tensors with certain symmetry properties under permutation of the input arguments? –  Lagerbaer Nov 28 '12 at 22:35
    
@Lagerbaer see physics.stackexchange.com/q/18228 –  Mark Eichenlaub Nov 28 '12 at 22:53
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