Irreducible Representations of SO(n) tensors

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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Related question by OP: physics.stackexchange.com/q/71905/2451 –  Qmechanic Jul 23 at 16:20
You duplicate questions, there is one below with essentially the same question –  John Jul 23 at 18:58

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?

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So if I'm getting this right, the way to decompose a rank n tensor would be to work with the method you suggest for the prodocut $1\otimes 1 \otimes ... \otimes 1$ with n copies of 1?? –  Cala Jul 23 at 17:03
You do not need Clebsch-Gordan, this is too detailed. What you need is just Littlewood-Richardson and its variation for $so(n)$. –  John Jul 23 at 18:58
@Cala - you are quite right! –  Edward Hughes Jul 23 at 23:31
@John - thanks for correcting me. I hadn't come across Littlewood-Richardson before, but a quick Google makes your point abundantly clear! I shall certainly refer to it in future. –  Edward Hughes Jul 23 at 23:32
Computing (decomposing) tensor product is not something that can be done for general representations in a closed form, but there is an algorithm that allows to do this for any given representations. Сomprehensive discussion can be found in Hing, Kang, "Quantum groups and Crystal bases" (you can just look in the middle). arxiv.org/pdf/hep-th/9912205v3.pdf page 100 for SU(N), which is simpler and in fact SO(n) rules can be reduced to a combination of SU(n)-rules. There are examples in pages.uoregon.edu/jcomes/TensorAppendix.pdf –  John Jul 24 at 20:01
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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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