Irreducible decomposition of Lorentz tensors with Young tableaux 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? 
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. 
 A: Here's what I do (mostly with SO(3), so jumping to SO(4) consider AccidentalFourierTransform's comment):
Use Schur-Weyl Duality / Robinson-Schensted correspondence to get the symmetric, antisymmetric, and for rank greater then 2, the mixed symmetries of the tensors. You then use the hook length formula to compute the dimensions of the subspaces. For rank-2:
$$ 4 \otimes 4 = 10 \oplus 6 $$
where the ${\bf 10}$ corresponds to:
$$ S_{\mu\nu}=\frac 1 2 [T_{\mu\nu} + T_{\nu\mu}\,]$$
To proceed, you have subtract traces, since $S_{\mu\mu}$ transforms like a scalar:
$$ N_{\mu\nu} = S_{\mu\nu} - \frac 1 4 S_{\sigma\sigma}$$
(whether 1/4 is only valid for SO(4) and not for SO(3,1), I can't say).
${\bf N}$ is now a natural (i.e. trace free) rank-2 tensor.
Moving on to rank 3, the Young tableaux tell you:
$$ 4 \otimes 4 \otimes 4 = 20 \oplus 20 \oplus 20 \oplus 4 $$
where the 1st $\bf{20}$ is rank-3, and represents the fully symmetrize indices $S_{\mu\nu\sigma}$.
At this point you have to subtract traces. Note that:
$$ S_{\mu\mu\nu} = S_{\mu\nu\mu} = S_{\mu\mu\nu} $$
so there is only one independent trace, and it transforms like a 4-vector, so that:
$$ {\bf 20} \rightarrow {\bf 16} \oplus {\bf 4} $$
The fully anti-symmetrized part (${\bf 4}$) transforms like a 4 vector, so is irreducible.
For the mixed symmetry parts, you have to evaluate the Weyl modules defined by the standard Young tableaux (there are 2 standard filling of the the diagram, so you get 2 mixed $\bf{20}$'s), e.g:
$$M_{\mu\nu} = \frac 1 3 [T_{\mu\nu\sigma} +T_{\nu\mu\sigma} -T_{\sigma\nu\mu} -T_{\sigma\mu\nu}]$$
At this point you have 3 possible traces, that reduce to one independent value:
$$ M_{\mu\mu\nu} = - M_{\nu\mu\mu} $$
and
$$ M_{\mu\nu\mu} = 0$$
so that part transforms as a 4 vector that can be subtracted off.
With SO(3) you'd now have $8 = 5 +3$, which means the mixed part is a rank 2 tensor and a vector, and it works out beautifully.
What I have produced is another $20 = 16 + 4$, which I don't quite understand. Perhaps the $16$ can be decomposed into $16=10+6=9+6+1$, I'll have to think about more.
In summary: the Young tableaux give index permutations that are irreducible subspaces. To reduce them further, you have to start taking traces (which isn't index permutation, it's also summation).
Maybe ask Math?
A: Another alternative for irreducible decompositions is by the aid of Lorentz group projectors as explained in  https://www.mdpi.com/2218-1997/5/8/184 and sections 3 and 4 there.
