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?
[Physics] Can any rank tensor be decomposed into symmetric and anti-symmetric parts
representation-theorytensor-calculus
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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.
You need the Clebsch-Gordan decomposition, at least in the case $n = 3$. The reason that we decompose a rank $2$ tensor in the way you describe is that
$$\mathbf{1} \otimes \mathbf{1} = \mathbf{2}\oplus \mathbf{1} \oplus \mathbf{0} $$
where the bold numbers denote spin representations.
Here's a bit more detail. In quantum physics we are really interested in representations of the Lie algebra of $SO(n)$ namely $\mathfrak{so}(n)$. The most useful case for physical purposes is $n = 3$, where there is an isomorphism
$$\mathfrak{so}(3) = \mathfrak{su}(2)$$
The Clebsch-Gordon result comes from the structure of representations for $\mathfrak{su}(2)$. In brief, $\mathfrak{su}(2)$ has irreps $\mathbf{n}$ for each half-integer $n$. Each irrep has $2n+1$ characteristic labels called weights, evenly spaces between $-n$ and $n$. Physically one interprets these as the component $j_3$ of spin.
When you take a tensor product of irreps the weights add up, to give you weights for the tensor product representation. A theorem says that this decomposes into the direct sum of irreps in the only way that uses up all these weights.
In case that all sounds absurd, let's do a concrete example. The tensors you mention are elements of tensor products of the vector representation of $\mathfrak{su}(2)$ typically denoted $\mathbf{1}$. We want to prove the result above that
$$\mathbf{1} \otimes \mathbf{1} = \mathbf{2}\oplus \mathbf{1} \oplus \mathbf{0} $$
Well $\mathbf{1}$ has weights $+1,0,-1$ so the tensor product will have weights
$$-2,-1,-1,0,0,0,+1,+1,+2$$
which are all possible ways of adding the weights for $\mathbf{1}$. Now rewrite this list suggestively
$$-2,-1,0,+1,+2,\ \ \ \ \ \ -1,0,+1,\ \ \ \ \ \ 0$$
These are just the weights for a $\mathbf{2}$ plus the weights for a $\mathbf{1}$ plus the weights for a $\mathbf{0}$.
Now it's not hard to identify $\mathbf{2}$ with the traceless symmetric matrices, $\mathbf{1}$ with the antisymmetric ones and $\mathbf{0}$ with the trace, checking that these all transform correctly under the relevant representations.
As an exercise you now have all the tools to prove that
$$\mathbf{1} \otimes \mathbf{1} \otimes \mathbf{1} = \mathbf{3}\oplus \mathbf{2} \oplus \mathbf{1} \oplus \mathbf{0}$$
Can you identify what these are, in terms of decomposing the rank $3$ tensor? Hint: there exist totally symmetric tracefree tensors, totally antisymmetric tensors, a trace term, and tensors of mixed symmetry.
Here's a good reference for the Lie algebra stuff. Let me know if you need any further details!
P.S. I don't know what one can do for general $n\neq 3$. The Clebsch-Gordan niceness is a specific property of $\mathfrak{su}(2)$ so I expect it becomes quite messy. Perhaps somebody else has some expertise here?
Best Answer
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.