My interest is purely in $\text{SO}(n)$ tensors and how one works out their irrep decomposition. For instance, for rank 2 tensors we simply split into an antisymmetric part, a traceless symmetric part and the trace. Is there a more general, recursive procedure for higher rank tensors? Even if not, what is the usual method when trying to do it for say rank 3 or 4? Any pointers to the literature would be more than helpful.
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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?
Edward Hughes
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There is a on-line web page where you can have some results about the product of $2$ representations: (in the first page, choose tensor product decomposition and choose $Bx =SO(2x+1)$ or $Dx = SO(2x)$ , then, in the second page, choose the Dynkin indices of the 2 representations, and you get the Dynkin indices of the representations which enter in the decomposition).
(If you want the product of 3 representations, use 2 steps.)
From an other reference (Ref : Pierre Ramond, Group Theory, A physicist's Survey, Cambridge), it seems that there is a pattern which concerns (at least for $n \ge7$) the multiplication of a fundamental representation $n$ by the (2-totally antisymmetric) adjoint representation : $\frac{n(n-1)}{2}$ . We have : $$n \otimes \frac{n(n-1)}{2} = n \oplus \frac{n(n-1)(n-2)}{6} \oplus\frac{n(n^2-4)}{3}$$
The second term is the 3-totally antisymmetric representation. For instance, we have :
$$7_{(100)}\otimes 21_{(010)}=7_{(100)} \oplus35_{(002)} + \oplus105_{(110)}$$ $$8_{(1000)}\otimes 28_{(0100)}=8_{(1000)} \oplus56_{(0011)} + \oplus160_{(1100)}$$ $$9_{(1000)}\otimes 36_{(0100)}=9_{(1000)} \oplus84_{(0010)} + \oplus231_{(1100)}$$ $$10_{(10000)}\otimes 45_{(01000)}=10_{(10000)} \oplus120_{(00100)} + \oplus320_{(11000)}$$ where the Dynkin indices are written for the representations.
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