Degree 8 multilinear operations on Jordan algebras

Question

I am interested in the dimension, or, even better, in the $S_8$-module structure of the space of degree 8 multilinear operations on Jordan algebras.

Recall that a Jordan algebra is a commutative but not generally associative algebra in which the product satisfies $$ (xx)(yx)=((xx)y)x, $$ in other words, multiplications by $x$ and by $x^2$ commute.

Elementary formulation: Let $J(x_1,\ldots,x_8)$ be the free Jordan algebra on eight generators, and let $V$ be the subspace of $J$ consisting of elements that have degree exactly one in each of the eight generators. What is the dimension of $V$, and how does $V$ decompose into irreducibles under the obvious $S_8$-action?

Operadic formulation: consider the operad $\mathcal{J}$ generated by one symmetric binary operation subject to the $S_4$-module of relation generated by the multilinearization of the Jordan identity. Concretely, that identity is $$ ((a_1a_2)a_3)a_4+((a_1a_4)a_3)a_2+((a_2a_4)a_3)a_1=(a_1a_2)(a_3a_4)+(a_1a_3)(a_2a_4)+(a_1a_4)(a_2a_3) . $$ Compute $\dim\mathcal{J}(8)$ and decompose that space as $S_8$-module.

Why 8? This question obviously depends on a positive integer parameter $n$. Up to $n=7$ the situation is well understood; the corresponding dimensions were computed by Glennie in late 1960s, are equal, respectively, to 1, 1, 3, 11, 55, 330, 2345, and are documented in the paper

C. M. Glennie, Identities in Jordan algebras, Computational Problems in Abstract Algebra. Pergamon, Oxford, 1970, pp. 307-313.

A scanned copy of that paper is stored on the OEIS website, where those dimensions are collected in the sequence A001776. The symmetric group decompositions in that range are also known, though it is harder to give a precise reference.

Of course, the case $n=8$ is the first "interesting" one, since this is where the story of "special" Jordan algebras begin. (A Jordan algebra is called special if it is a subalgebra of an associative algebra equipped with the operation $a\circ b=ab+ba$; there is, in addition to Jordan identities, an identity of degree $8$, also found by Glennie, that is satisfied only in special Jordan algebras.) However, one would expect that in the 50+ years that elapsed since Glennie's work, at least the step from $7$ to $8$ could have been made, and I wonder if anyone knows if that is indeed the case.

What I tried? I used two computational approaches, the Albert nonassociative algebra system and the operadic Gröbner basis calculator based on my own work, and alas my computer cannot go beyond $n=7$ in less than a day in either of them (and I did not leave it on for longer so far).

Many thanks for any possible leads!

Answer 1

I managed to run Albert on a very powerful computer at work, and the computation of the desired dimension converged: it seems equal to 19089. I would very much like to confirm that it is correct (I am using this number to test various conjectures, and its independent verification is very desirable). Information about $n=9$ would also be very important.

Answer 2

I know a way to do it, but I'm not sure how it compares to the ways you've already tried. It is very possible that it is worse than these methods.

There are two steps. First, we build a presentation of the desired left $S_8$-module, and second, we tensor it with every right irrep of $S_8$. This will build multiplicity spaces for various Schur functors, and the final answer can be evaluated using plethysm in the ring of symmetric functions.

The required presentation is given by a matrix with entries in $\mathbb{Q}S_8$. There will be a row for every possible binary tree on eight leaves. These will be the basis vectors for our free module of generators. There will be a column for every possible way of plugging in trees for the four variables $a_1, a_2, a_3, a_4$ in your multilinearized identity. These will be the basis vectors for our free module of relations. Each incarnation of the identity can then be written as a $\mathbb{Q}S_8$-combination of the generators; indeed, each term is a tree with some coefficient and some with its leaves permuted somehow.

Once you have this presentation matrix, you choose an irrep of $S_8$. Because we need a right module, we transpose the action matrices. Then, we apply the representation to the entries of the presentation matrix to get a big block matrix of rational numbers. The corank of this matrix then gives the multiplicity of the Schur functor corresponding to our irrep. If we run through every irrep, we get every multiplicity.

I think there are 429 binary trees with eight leaves, and the largest rep of $S_8$ has dim 90, so the number of rows for the hardest multiplicity space is 38610. (If your Grobner theory can pick standard forms that limit the number of tree shapes we need, then you could get this number much lower.) Then, use the usual algorithms to estimate the rank of this matrix.