# 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.

• $\mathrm{SO}(4)\cong(\mathrm{SU}(2)\times\mathrm{SU}(2))/\mathbb Z_2$ is not simple, so you can't really use Young tableaux. $\mathrm{SO}(n)$ is simple for $n=3$ and $n\ge5$. For example, for $n=4$ (and unlike for $n=3$ or $n\ge5$), the anti-symmetric representation is reducible (into the so-called self-dual and anti-self-dual parts). In any case, this is really a pure-math question, so I think it's better for you to ask on math.SE. – AccidentalFourierTransform Sep 21 '18 at 1:11
• What if you consider the identity component isomorphism with $sl(2,C)$, where the latter is simple. You're right however, I'm posting this on math. – kospall Sep 21 '18 at 1:25

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).

• The arithmetic factor for the trace is the same for the Lorentz group indeed. From your answer, I understand that it is always the case that we try to subtract the trace. Is that right? Now, for the mixed symmetry you have to take anti-symmetrisation and symmetrisation of column and row indices respectively. From the possible anti-symmetrisations you get the $M_{\mu\nu\sigma}$ part, but where is the $1/3$ factor coming from, since for a rank 3 tensor the Young tableau looks like a $\Gamma$, a box with a box on the right and a box below. – kospall Sep 21 '18 at 11:15
• @kospall I'm looking at my code and it says the normalization divisor is the "hook product", which is the product of all the hook-length of each of the 3 cells in the diagram. The hook length is 1 plus the arm and leg length. The arm (leg) length is the number of cells left-of (below) a cell, so its [3, 1], [1]. Product is 3, Hence, $1/3$. Why that works, I can't explain. That's why it's called "the remarkable hook-length formula". – JEB Sep 23 '18 at 0:44