I am familiar with the notion of irreps. My question refers simply to tensor representations (not tensor products of representations) and how can we decompose them into irreducible parts? For example, a rank 2 tensor is decomposed into an antisymmetric part, a traceless symmetric and its trace. What is the generalization of that for higher rank tensors? Could someone provide an example for, say rank 3 or 4? Thank you
[Physics] Irreducible decomposition of higher order tensors
group-representationstensor-calculus
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You can decompose a rank two tensor $X_{ab}$ into three parts:
$$X_{ab} = X_{[ab]} + (1/n)\delta_{ab}\delta^{cd}X_{cd} + (X_{(ab)}-1/n \delta_{ab}\delta^{cd}X_{cd})$$
The first term is the antisymmetric part (the square brackets denote antisymmetrization). The second term is the trace, and the last term is the trace free symmetric part (the round brackets denote symmetrization). n is the dimension of the vector space.
Now under, say, a rotation $X_{ab}$ is mapped to $\hat{X}_{ab}=R_{a}^{c}R_{b}^{d}X_{cd}$ where $R$ is the rotation matrix. The important thing is that, acting on a generic $X_{ab}$, this rotation will, for example, take symmetric trace free tensors to symmetric trace free tensors etc. So the rotations aren't "mixing" up the whole space of rank 2 tensors, they're keeping certain subspaces intact.
It is in this sense that rotations acting on rank 2 tensors are reducible. It's almost like separate group actions are taking place, the antisymmetric tensors are moving around between themselves, the traceless symmetrics are doing the same. But none of these guys are getting rotated into members "of the other team".
If, however, you look at what the rotations are doing to just, say the symmetric trace free tensors, they're churning them around amongst themselves, but they're not leaving any subspace of them intact. So in this sense, the action of the rotations on the symmetric traceless rank 2 tensors is "irreducible". Ditto for the other subspaces.
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
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.