Even though a computer can deal with this computation, the quick explosion of the number of terms puts the question of an explicit formula for the expanding of the \({\mathbb {R}}_p\)’s.
First of all, what are the terms involved in such a decomposition? If we write \({\mathcal {S}}_n\) the symmetric group on a set of n elements, we get the formula:
$$\begin{aligned} {\mathbb {R}}_{p}&= \dfrac{1}{2^p}\delta _{\alpha _1 \beta _1 \ldots \alpha _p \beta _p}^{\gamma _1 \delta _1 \ldots \gamma _p \delta _p} \mathrm {R}_{\gamma _1 \delta _1}^{\alpha _1 \beta _1} \ldots \mathrm {R}_{\gamma _p \delta _p}^{\alpha _p \beta _p} \\&= \dfrac{1}{2^p} \sum _{\sigma \in {\mathcal {S}}_{2p}} \varepsilon (\sigma ) \mathrm {R}_{\mu _1 \mu _2}^{\mu _{\sigma (1)} \mu _{\sigma (2)}} \ldots \mathrm {R}_{\mu _{2p-1} \mu _{2p}}^{\mu _{\sigma (2p-1)} \mu _{\sigma (2p)}}. \end{aligned}$$
This sum is on (2p)! terms, which are obviously not linearly independent. The extraction of a basis among these terms is a problem which has been solved in [7], using tools of the group representation theory which we shall shortly present here.
For all \(k \ge 1\), every k-tensor \(T_{a_1 \ldots a_k}\) can be mapped to a representation of \({\mathcal {S}}_k\). Just as this representation can be decomposed into irreducible representations of \({\mathcal {S}}_k\) encoded by Young diagrams of size k, the tensor \(T_{a_1 \ldots a_k}\) can be decomposed as well on a basis \({\mathcal {T}}_k\) of k-tensors corresponding to Young diagrams of size k. More precisely, the tensors of \({\mathcal {T}}_k\) have peculiar symmetries which are encoded in standard Young tableaux of size k. We note with \(\longleftrightarrow \) this correspondence.
Let us explain how to proceed to such a decomposition: for each tableau of size k, take \(T_{a_1 \ldots a_k}\), symmetrise on the indices lying in each row, and then antisymmetrise on the indices lying in each column. For instance,
where the combinatorial factors \(\frac{1}{3}=\frac{2}{6}\) and \(\frac{1}{6}\) are the normalised numbers of standard Young tableaux of the same Young diagram. The normalisation constant is the sum on the diagrams of the squared number of standard tableaux of this diagram, which turns out to be k!. For example, for \(k=3\),
and we find \(1^2+2^2+1^2 = 3!\). Thus each 3-tensor can be decomposed into:
The product of two tensors \(U_{a_1 \ldots a_k} \in {\mathcal {T}}_k\) and \(U'_{b_1 \ldots b_l} \in {\mathcal {T}}_l\), whose symmetries correspond to the diagrams \({\mathfrak {U}}\) and \({\mathfrak {U}}'\), corresponds itself to the outer product of \({\mathfrak {U}}\) and \(\mathfrak {U'}\), namely \({\mathfrak {U}}\cdot {\mathfrak {U}}'\). This product can in turn be decomposed into irreducible representations of \({\mathcal {S}}_{k+l}\), following the Littlewood–Richardson rule. For example, if \(U_{abc}\) and \(U'_{abc}\) are 3-tensors represented by
then
, and
An interesting theorem proved in [7] states that among these irreducible representations of \({\mathcal {S}}_{k+l}\), those of which all rows have an even length are invariant under the orthogonal group \(O(k+l)\). Moreover, they form a basis of these invariant representations. In terms of tensors on a manifold, the invariant representations under the action of \(O(k+l)\) correspond to the scalars built from the contractions of \(U_{a_1 \ldots a_k}\) and \(U'_{b_1 \ldots b_l}\). Hence, the space of the scalars built from the contractions of \(U_{a_1 \ldots a_k}\) and \(U'_{b_1 \ldots b_l}\) has the same dimension as the space of the irreducible representations of \(S_{k+l}\) which are invariant under \(O(k+l)\) and which appear in the decomposition of \({\mathfrak {U}}\cdot {\mathfrak {U}}'\). So, a basis of independent scalars will have the same cardinal as the number of diagrams with even row length in the decomposition of \({\mathfrak {U}}\cdot {\mathfrak {U}}'\).
Intuitively, contracting a pair of indices of the tensors can be understood as crossing a pair of cells off the Young diagram. If the two cells lie in different rows, the result vanishes because of the antisymmetrisation between the rows. If the two cells lie in the same row, the contraction is nontrivial. Hence, all rows must have even lengths so that the resulting empty diagram correspond to a nontrivial scalar.
However, the equality of the dimensions does not imply a simple bijective correspondence between the even row length diagrams and the independent scalars: most of the time, such a correspondence does not exist. Keeping our example, we have two even row length representations:
and two independent scalars built from the contractions of \(U_{a_1 \ldots a_k}\) and \(U'_{b_1 \ldots b_l}\):
$$\begin{aligned} U_{ab}^{a} {U'}_c^{cb} \qquad \text {and} \qquad U_{abc}{U'}^{cab}, \end{aligned}$$
but no canonical bijection between them.
In case \(U_{a_1 \ldots a_k} = U'_{b_1 \ldots b_l}\), the list of irreducible representations is restricted by the symmetries under the exchange of the two tensors: this is not an outer product anymore, but a new operation called a plethysm, \(\otimes \). This is the case we are interested in.