---
tag: Z95Q7B022R
kind: problem
name: Kervaire invariant problem
msc: [55Q45, 57R15]
uses: [P4YXE08ZXF, HGZAYPQBXW]
about: [P4YXE08ZXF]
sources:
  - url: https://en.wikipedia.org/wiki/Kervaire_invariant
history:
  posed_by: "Michel Kervaire"
  first_stated: "1960"
---

## Problem

For which integers $n = 4k + 2$ is there a closed framed smooth manifold of dimension $n$ whose [[Kervaire invariant|P4YXE08ZXF]] equals $1$?
Equivalently, in which degrees $n$ do the [[stable homotopy groups of spheres|HGZAYPQBXW]] $\pi^s_n$ contain an element of Kervaire invariant $1$?

## Remarks

Kervaire showed in 1960 that the invariant vanishes in dimension $10$.
Browder showed in 1969 that the invariant can be nonzero only in dimensions $n = 2^j - 2$.
Framed manifolds of Kervaire invariant $1$ exist in dimensions $2$, $6$, $14$, $30$ and $62$.
Hill, Hopkins and Ravenel showed that none exist in dimensions $2^j - 2$ with $j \ge 8$, that is, in dimension $254$ and above.
This left dimension $126$ as the only open case; in 2024 Weinan Lin, Guozhen Wang and Zhouli Xu announced that a manifold of Kervaire invariant $1$ exists in dimension $126$.

## Explanations

### L1 human

The Kervaire invariant is a yes-or-no number attached to certain framed manifolds of dimension $4k + 2$.
The problem asks in which dimensions it can be "yes".
The answer turned out to be a short and strange list: $2$, $6$, $14$, $30$, $62$, and, by the 2024 announcement, $126$.
The question was a guide for decades of work on the homotopy groups of spheres.