Decomposition of tensors into symmetry classes according to Schur functors

Question

I am mainly looking for references on this subject, as I was unable to find any, at least any that answers what I am looking for to a satisfactory degree.

As it is well-known and extremely easy to show if $T\in T^2(V)$ is a tensor of degree 2 over a vector space $V$ (we may assume $V$ to be finite dimensional and over a characteristic 0 field), it may be uniquely split as $T=S+A$, where $S$ is symmetric and $A$ is alternating.

If instead $T\in T^k(V)$ is a tensor of degree $k$, the corresponding decomposition is $$ T^k(V)=\bigoplus_\lambda S_\lambda(V)^{\oplus n_\lambda}, $$ where the $S_\lambda$ is the Schur functor corresponding to the Young tableau $\lambda$ and $n_\lambda$ is a multiplicity. I deliberately left this formula somewhat incomplete because I don't remember the details. I know most of this from Fulton/Harris and I have always found that book to be quite confusing tbh, but it does have the details.

The problem with this formula is that 1) the spaces $S_\lambda(V)$ are unique up to isomorphism, but their particular realizations as subspaces of $T^k(V)$ depends on the particular conventions one have for Young tableaux, 2) there are multiplicities. Hence this direct sum decomposition is an outer, rather than inner direct sum.

I would like an inner direct sum decomposition which allows one to write explicitly given a tensor $T$ that $$ T=\sum_{i\in I}T^{(i)}, $$ where each term in this sum is a unique and $\mathrm{GL}(V)$-invariant irreducible term.

From the non-uniqueness of the Schur functors(' images) as well as the existence of multiplicities, I assume such a decomposition is not unique. But I also assume that if one chooses a particular numbering scheme or conventions for the Young tableaux then it is possible to get an irreducible decomposition which is unique with respect to that scheme, including the ways the various multiplicities appear as isomorphic but nontheless different subspaces of $T^k(V)$.

I find it hard to imagine that there are no explicitly specifiable such schemes with explicit projectors, seeing that in problems involving tensors in algebra and physics, such decompositions can be rather important. Yet I was unable to find any references that considers this "embedding problem". Pretty much all references I have found are interested in this decomposition for the purposes of representation theory, where the outer direct sum is sufficient.

So, I am asking for references, any kind of paper or textbook that treats the problem of decomposing a tensor into $\mathrm{GL}(V)$-irreducible terms as an inner direct sum with hopefully rather explicit descriptions of the resulting terms.

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Edit: Upon further reflection, my question can be narrowed down and simplified as follows.

I am not sure if my terminology is standard so let's say that a Young diagram $\lambda=(\lambda_1,\dots,\lambda_q)$ with $k$ boxes is canonical if it is numbered consecutively as eg.$$ \begin{matrix}\boxed{1} & \boxed{2} & \boxed{3} \\ \boxed{4} & \boxed{5} \\ \boxed{6} & \boxed{7} \\ \boxed{8}\end{matrix}. $$ A Young tableau is standard if all rows and columns are strictly increasing.

Then - as I looked up - in the decomposition $$ T^k(V)=\bigoplus_\lambda S_\lambda(V)^{\oplus n_\lambda} $$ the sum is over all shapes which can be identified with canonical tableaux and each multiplicity $n_\lambda$ is the number of standard tableaux with shape $\lambda$.

So we can define the projectors $\pi_\lambda:T^k(V)\rightarrow T^k(V)$, where $\lambda$ is any standard tableau with $k$ boxes, and $\pi_\lambda$ acts by first symmetrizing over all rows of the tableau and then alternating over all columns, with an appropriate normalizing factor so that $\pi_\lambda$ is an idempotent.

So what is obvious is that $$ \pi_\lambda(T^k(V))\cong S_{\text{shape of }\lambda}(V). $$

I suspect/hope that it is true that $\pi_{\lambda}\pi_{\lambda^\prime}=0$ when $\lambda\neq\lambda^\prime$ (here each $\lambda$ is a numbered standard tableau, not just a shape!) and $\sum_{\lambda}\pi_\lambda=\mathrm{id}_{T^k(V)}$, which would realize the required inner direct sum decomposition.

But I don't know - and this is what I am unable to find - how the $\pi_\lambda$ and the $\pi_{\lambda^\prime}$ are related when $\lambda$ and $\lambda^\prime$ are different standard tableaux with the same shape.

Is it true that the corresponding subspaces are independent in $T^k(V)$?

Do we actually have the decomposition $\mathrm{id}_{T^k(V)}=\sum_{\lambda\in\text{standard tableaux with k boxes}}\pi_{\lambda}$?

If so, is each term here unique, assuming we fixed a convention for the Young symmetrizers (eg. first symmetrize then alternate)?

Answer 1

This is in fact rather complicated stuff, and not that well known, even though it was discovered by Alfred Young in his sixth paper on quantitative substitutional analysis from 1932.

I also thought the identity $\pi_{\lambda}\pi_{\lambda'}=0$ for $\lambda\neq\lambda'$ was true, from reading the article "Diagrammatic Young Projection Operators for $U(n)$" by Elvang et al. (see top of page 5). However, this is false. See this previous MO question:

A basic question about Young symmetrizers

The construction of orthogonal idempotents is a bit complicated and called Young's seminormal representation.

A good and short presentation for a physics audience is the article "Hermitian Young Operators" by Keppeler and Sjödahl. A longer pedagogical introduction in the same spirit is the online textbook "The Special Unitary Group, Birdtracks, and Applications in QCD" by Alcock-Zeilinger.

For a math audience, the best introduction I know are two sets of lectures notes by Adriano Garsia. They have now appeared as the first two chapters of the book "Lectures in Algebraic Combinatorics".

Finally, note that this question is related to what I called CG2 or the explicit Clebsch-Gordan problem in this other MO post

Clebsch–Gordan decomposition formula for algebraic groups

Much of it, as far as I know is still open.

Answer 2

By Schur-Weyl duality, this amounts to a question about the group algebra of the symmetric group.

Imagine we have a formula for $1 \in \mathbb{Q}S_k$

$$ 1 = \pi_1 + \pi_2 + \cdots + \pi_d $$

where each $\pi_i \in \mathbb{Q}S_k$ is idempotent, and for $i \neq j$ we have $\pi_i \cdot \pi_j = 0 = \pi_j \cdot \pi_i$. Then, for every right $S_k$-module $V$ and vector $v \in V$, we have a canonical decomposition $$ v = v \cdot \pi_1 + v \cdot \pi_2 + \cdots + v \cdot \pi_d $$ giving $V$ as an (internal) direct sum $$ V = \sum_{i=1}^d V \cdot \pi_i. $$

By semisimplicity, the group algebra is isomorphic to a product of matrix algebras, one for each irreducible representation. The map sends a permutation to the square matrices that give its action in each irrep. In the product-of-matrix-algebras basis, the best idempotents are clear: place a single one somewhere on the diagonal. Then, we get our $\pi_i$'s by applying the inverse of the isomorphism. Here is one way to do it in sage:

def matrix_coefficients_basis(k):
    entries = [
        [
            e
            for partition in Partitions(k)
            for row in SymmetricGroupRepresentation(partition, 'specht').representation_matrix(permutation).rows()
            for e in row
        ]
        for permutation in Permutations(k)
    ]
    return matrix(entries)


def coordinate_lookup(k):
    coordinates = [
        (partition, i, j)
        for partition in Partitions(k)
        for i in range(partition.dimension())
        for j in range(partition.dimension())
    ]
    return {triple: position for position, triple in enumerate(coordinates)}


def internal_decomposition(partition):
    p = Partition(partition)
    k = sum(partition)
    group_algebra = SymmetricGroupAlgebra(QQ, k)
    permutations = [group_algebra(permutation) for permutation in Permutations(k)]
    lookup = coordinate_lookup(k)
    interesting_rows = [lookup[p, i, i] for i in range(p.dimension())]
    decomposition = matrix_coefficients_basis(k).inverse()
    return [
        sum(coefficient * permutation for coefficient, permutation in zip(decomposition.row(i), permutations))
        for i in interesting_rows
    ]

Here are some example formulas:

sage: internal_decomposition([2])                                                                                                                      
[1/2*[1, 2] + 1/2*[2, 1]]
sage: internal_decomposition([1,1])                                                                                                                    
[1/2*[1, 2] - 1/2*[2, 1]]
sage: internal_decomposition([3])                                                                                                                      
[1/6*[1, 2, 3] + 1/6*[1, 3, 2] + 1/6*[2, 1, 3] + 1/6*[2, 3, 1] + 1/6*[3, 1, 2] + 1/6*[3, 2, 1]]
sage: internal_decomposition([2,1])                                                                                                                    
[1/3*[1, 2, 3] + 1/3*[1, 3, 2] - 1/3*[3, 1, 2] - 1/3*[3, 2, 1],
 1/3*[1, 2, 3] - 1/3*[1, 3, 2] - 1/3*[2, 3, 1] + 1/3*[3, 2, 1]]
sage: internal_decomposition([1,1,1])                                                                                                                  
[1/6*[1, 2, 3] - 1/6*[1, 3, 2] - 1/6*[2, 1, 3] + 1/6*[2, 3, 1] + 1/6*[3, 1, 2] - 1/6*[3, 2, 1]]

Of course, these results depend entirely on the chosen bases for irreps, and the resulting formulas are not combinatorial or efficient to compute. Nevertheless, I hope they are of some value!