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I want to understand the irreducible decomposition of Lorentz tensors by using Young tableaux. Let me start with a trivial example. Suppose we work in $n=4$ dimensions, and that we have a rank 2 homogeneous tensor $T_{ab}$. Doing the Young tableau, we find that $4\otimes 4=10\oplus 6$, where the tensor splits into a symmetric and an antisymmetric part with subspace dimensions 10 and 6, respectively. If the antisymmetric part is real, then it is irreducible. Assume that the tensor is not traceless. The symmetric part can be further reduced to a symmetric traceless part and the trace. How is this further reduction visible when using Young operators? Shouldn't the Young tableau directly give the irreducible parts of this tensor, i.e. $4\otimes 4=1\oplus 6\oplus 9$? I know this is probably not the case, because doing Young tableaux we only consider (anti-)symmetrisations. I ask then, what is the logic behind this further reduction of the symmetric part? What are the correct conditions for applying the Young tableau method.

Furthermore, increasing the rank, there exists for example the usual irreducible decomposition of the torsion tensor into an axial, a vectorial and a traceless tensorial part. I do not see how this split can be obtained with Young diagrams. Again, while the physical motivation may be clear, I cannot find a general mathematical prescription for this type of decomposition, nor have I found a formula for tensors of arbitrary rank. I would definitely appreciate some good references, since I have some gaps to fill in this topic.

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