Foulkes' conjecture for sixth powers and quadratic stabilization

Foulkes' conjecture is claimed for sixth powers, with a Lean-proved bound past which a natural map is onto.

2 papers September 25, 2026 Explained by AI

Foulkes' conjecture compares two ways of nesting polynomial spaces and says one always fits inside the other. One paper proves that a natural map between them is onto once b is at least a(a − 1), in any dimension. For a = 6 that settles every b from 30 up. The other paper claims the remaining range, so the sixth case holds for every b ≥ 6.

First step

1950

H. O. Foulkes, the plethysm question

Significance

Solid

It settles the next open case of a 75-year-old conjecture and gives a uniform bound for every case.

Trust

Formal proof listed

Field

Algebra

2 papers

The explanation

Take three variables. The polynomials of degree 3 in them form a space of dimension 10. Now take polynomials of degree 2 whose variables are those 10 basic cubics. That space has dimension 55. Swap the roles. Degree-2 polynomials in three variables form a space of dimension 6. Degree-3 polynomials built from those 6 have dimension 56. The two nested constructions are called plethysms, and here 55 is at most 56.

Foulkes' conjecture says this is no accident. Write Symᵏ for the space of degree-k polynomials. For a ≤ b, the space Symᵃ(Symᵇ V) should fit inside Symᵇ(Symᵃ V)Source 1. Fitting inside means more than a dimension count. It must respect the symmetries of V, so the fit is called equivariant. Equivalently, every irreducible building block appears at least as often on the right as on the leftSource 2. With two variables the two sides are always equal, a fact called Hermite reciprocitySource 3. In the example with a = 6 and b = 7, two variables give 1,716 on each side. Three variables give 4,496,388 on the left and 5,379,616 on the right.

Foulkes raised the question in 1950Source 4. The cases a = 1 and a = 2 follow from classical formulasSource 5. Dent and Siemons proved a = 3Source 6. McKay showed that one good starting case spreads to all larger bSource 7. Computer work then gave a = 4 and a = 5Source 8. Evseev, Paget and Wildon checked every case where a and b add up to at most 19Source 9. Brion proved the conjecture for each fixed a once b is large, but with no uniform bound on how largeSource 10.

These papers address the sixth case and a bound for every a. One paper proves that a natural map from Symᵇ(Symᵃ V) to Symᵃ(Symᵇ V) is onto whenever b is at least a(a − 1)Source 11. The bound does not depend on the number of variablesSource 12. A map that is onto in that direction yields the embedding Foulkes asked forSource 13. For a = 6 this covers every b from 30 upSource 13. The other paper handles the remaining range, b from 6 to 29, and claims the full sixth caseSource 14.

The single obstacle is that the natural map can fail. At (5, 5) and (6, 6), the canonical map has a nonzero kernel in large enough dimension, even though both sides matchSource 15. So one cannot simply prove the obvious map is one-to-one. The sixth-case paper instead compares the counts of building blocks directly. It bounds whole families of counts and leaves only a small band for exact computationSource 16.

How sure is this? A computer proof checker called Lean has a formal proof of the quadratic bound. It shows the canonical map is onto whenever b is at least a(a − 1), in every dimension. That proof also covers the sixth case for b at least 30. The range from 6 to 29 rests on the paper's exact computations, outside the Lean check. No specialist review of either paper has been published.

What it doesn't show. This does not settle Foulkes' conjecture for a = 7 or larger when b is below a(a − 1).

Where it came from

This result comes from an unreleased OpenAI model, published in the openai/math release on 6 October 2026.

Hype check

Solid. The quadratic bound has a Lean proof, which covers the sixth case for b ≥ 30; b from 6 to 29 rests on exact computation. No specialist review is published.

Words used

Symmetric power Symᵏ V: The space of degree-k polynomials whose variables form a basis of V.
Plethysm: Nesting one symmetric power inside another, such as Symᵃ(Symᵇ V).
Equivariant: Compatible with every invertible linear change of the variables.
Irreducible building block: A piece of a symmetric space that cannot be split further; multiplicities count how often each appears.
Canonical Foulkes–Howe map: A natural map between the two plethysms built by multiplying polynomials.
Surjective: Onto: every element of the target is hit.

The original papers (2)

Foulkes' conjecture for the sixth symmetric power (PDF, opens in a new tab)
September 25, 2026
Quadratic stabilization of the canonical Foulkes--Howe map (PDF, opens in a new tab)
September 25, 2026 Formal proof listed

Lean notes

All quoted lines (16)

For a finite-dimensional complex vector space $V$ and integers $1\le a\le b$ , it asks for a $\GL(V)$ -equivariant injection $\Sym^a(\Sym^b V)\hookrightarrow\Sym^b(\Sym^a V).$
Foulkes' conjecture for the sixth symmetric power Back to text
Over $\CC$ , complete reducibility makes this equivalent to an inequality between the multiplicities of every irreducible polynomial representation.
Foulkes' conjecture for the sixth symmetric power Back to text
For binary forms, classical Hermite reciprocity gives an isomorphism between the two symmetric powers for all $a,b$ ; see.
Foulkes' conjecture for the sixth symmetric power Back to text
The conjecture originates in Foulkes' study of concomitants.
Foulkes' conjecture for the sixth symmetric power Back to text
Thrall's decomposition formulas imply $a=2$ ; see also.
Foulkes' conjecture for the sixth symmetric power Back to text
Dent and Siemons proved $a=3$ .
Foulkes' conjecture for the sixth symmetric power Back to text
McKay proved that injectivity for every $V$ at a seed $(a,b_0)$ , with $b_0\ge a$ , propagates to every $(a,b)$ with $b\ge b_0$ .
Foulkes' conjecture for the sixth symmetric power Back to text
The computation at $(4,4)$ by M\"uller and Neunh\"offer therefore yields $a=4$ ; Cheung, Ikenmeyer, and Mkrtchyan's injectivity computation at $(5,6)$ yields $a=5$ , with the diagonal comparison again immediate.
Foulkes' conjecture for the sixth symmetric power Back to text
Evseev, Paget, and Wildon verified Foulkes' multiplicity inequalities when the two parameters sum to at most $19$ , including the sixth case through $b=13$ .
Foulkes' conjecture for the sixth symmetric power Back to text
Brion proved eventual truth for fixed $a$ and later gave a dimension-dependent bound for surjectivity of the canonical map in the opposite direction.
Foulkes' conjecture for the sixth symmetric power Back to text
For every finite-dimensional complex vector space $V$ , we prove that the canonical Foulkes–Howe map $\Sym^b(\Sym^a V)\longrightarrow\Sym^a(\Sym^b V)$ is surjective whenever $a\ge2$ and $b\ge a(a-1)$ .
Quadratic stabilization of the canonical Foulkes--Howe map Back to text
This gives a quadratic stabilization bound independent of $\dim V$ .
Quadratic stabilization of the canonical Foulkes--Howe map Back to text
At $a=6$ , an invariant complement to its kernel supplies an embedding for $b\ge30$ .
Foulkes' conjecture for the sixth symmetric power Back to text
Theorem … resolves this case of Foulkes' conjecture positively, with no restriction on $\dim V$ .
Foulkes' conjecture for the sixth symmetric power Back to text
The canonical maps at $(5,5)$ and $(6,6)$ have nonzero kernels in suitable dimensions, although their source and target representations coincide.
Foulkes' conjecture for the sixth symmetric power Back to text
The present certificates combine such calculations with bounds on entire families of multiplicities, leaving only a small band for exact comparison.
Foulkes' conjecture for the sixth symmetric power Back to text

Other sources

openai/math release notes (2026): model provenance and review status manuscripts and supporting proof artifacts produced by an internal OpenAI model

Browse Algebra Named conjectures Open since the 1950s Lean-checked results Solid results