The Kervaire invariant problem at the prime three

Claims to settle which odd-primary Kervaire classes survive at the prime three: stems 10, 106 and 322.

1 paper September 24, 2026 Explained by AI

Stable homotopy groups of spheres record ways to map one high-dimensional sphere onto another. A spectral sequence gives candidate classes, and some of them do not survive to real maps. At odd primes, one family of candidates is called the Kervaire classes. At primes 5 and up, all but the first fail. This paper claims that at the prime three exactly three survive, in stems 10, 106 and 322, each with a representative of order three.

First step

1978

Douglas Ravenel, odd-primary Kervaire classes ruled out for primes 5 and above

Significance

Major

It claims to finish the odd-primary Kervaire invariant problem at the one prime left open by Ravenel's 1978 work.

Trust

Claimed, not checked

Field

Topology

1 paper

The explanation

Stable homotopy groups of spheres list the ways to map one high-dimensional sphere onto another, up to deformation. The stem is the difference in dimension. The Adams spectral sequence is a step-by-step calculation that offers candidate classes in each stem. Some candidates survive to real maps. Others are killed by a differential, a map between pages of the calculation. At the prime three, the Kervaire candidate b_j sits in stem 4 × 3^(j+1) − 2. For j = 0, 1, 2, 3 and 4 this gives stems 10, 34, 106, 322 and 970.

The paper asks two questions. Does b_j survive? And if so, does it come from a map of order three, one that gives zero when added to itself three times? Survival fixes a class only up to error terms, so the order is a separate questionSource 1. Toda showed that b₁ supports a nonzero differential at every odd primeSource 2. Ravenel ruled out every b_j with j at least 1 for primes 5 and upSource 3. At three this conclusion failsSource 4. The survivor in stem 106 comes from a corrected combination of two classes. One of those classes itself does not surviveSource 5. That is the obstacle. A class's fate in the related Adams–Novikov calculation need not decide its fate in the Adams calculation.

The paper claims to settle the prime three completely. Exactly b₀, b₂ and b₃ survive, and each has a representative of order threeSource 6. So there are classes in stems 10, 106 and 322. The class b₁ and every b_j with j at least 4 do not surviveSource 7. In particular, the paper claims a new order-three Kervaire element in stem 322Source 8. In Amelotte's notation this new class is θ₄Source 9.

The result also has consequences for spaces before stabilizing. One example concerns loop spaces of fibres built from spheres. A loop space is the space of loops in a space. The paper claims these split as a product of two pieces only for n equal to 3, 27 or 81, among n above 1Source 10. For n = 81 the spaces involved have dimensions 163 and 487. The prime-two Kervaire problem is a different question with a different familySource 11.

How sure is this? There is no Lean check. A computer proof checker has not looked at any part of it. The proof combines long spectral sequence calculations with careful tracking of correction terms. No specialist review of the paper has been published.

What it doesn't show. This does not address the classical prime-two Kervaire invariant one problem, which has a different family and dimension convention.

Where it came from

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

Hype check

Major if the proof holds. It covers only the prime three and only the standard family. There is no Lean check and no published specialist review.

Words used

Stable homotopy group of spheres: The set of ways to map one high-dimensional sphere onto another, up to deformation, once dimensions are large.
Stem: The difference between the dimensions of the two spheres in such a map.
Adams spectral sequence: A step-by-step calculation that offers candidate classes in each stem.
Differential: A map between pages of the calculation; a class that it moves does not survive.
Kervaire class: One of a standard family of candidate classes b_j, named by analogy with Kervaire's prime-two problem.
Order three: A map that gives zero when added to itself three times.
Loop space: The space of all loops in a given space, starting and ending at one point.

The original paper

The Kervaire invariant problem at the prime three (PDF, opens in a new tab)
September 24, 2026

All quoted lines (11)

These are different questions: permanence determines an associated-graded class, while the order of a representative also depends on an extension in actual homotopy.
The Kervaire invariant problem at the prime three Back to text
Toda proved the first exclusion in the odd-primary family: $b_1$ supports a nonzero differential at every odd prime.
The Kervaire invariant problem at the prime three Back to text
For every odd prime $p\geq5$ , it excludes all standard Kervaire classes $b_j$ with $j\geq1$ .
The Kervaire invariant problem at the prime three Back to text
At three, the same conclusion fails.
The Kervaire invariant problem at the prime three Back to text
The classical survivor in stem $106$ is detected by a corrected combination of $\beta_{9/9}$ and $\beta_7$ , although $\beta_{9/9}$ itself supports a differential.
The Kervaire invariant problem at the prime three Back to text
These classes survive precisely at indices $0$ , $2$ , and $3$ , and each surviving detection coset contains an element of additive order three.
The Kervaire invariant problem at the prime three Back to text
The class $b_1$ and every $b_j$ with $j\geq4$ fail to survive.
The Kervaire invariant problem at the prime three Back to text
In particular, there is an order-three Kervaire element in stem $322$ .
The Kervaire invariant problem at the prime three Back to text
His class $\theta_r$ is detected by $b_{r-1}$ ; thus $b_2$ here corresponds to $\theta_3$ , and the new $b_3$ class corresponds to $\theta_4$ .
The Kervaire invariant problem at the prime three Back to text
For every integer $n>1$ , the underlying space $\Omega S^{2n+1}\{3\}$ has a nontrivial product decomposition if and only if $n\in\{3,27,81\}$ .
The Kervaire invariant problem at the prime three Back to text
The problem here is the standard odd-primary Adams-family problem; the classical prime-two Kervaire invariant one problem has a different family and dimension convention.
The Kervaire invariant problem at the prime three 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

308Finite Smith–Toda complexes at every heightTopology
310Quillen's conjecture in rational homologyTopology
315The four-dimensional Singer conjectureTopology
316Curtis’s conjectureTopology
307Failure of rational injectivity for maximal coarse assemblyTopology

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