Kervaire invariant
Builds on Smooth manifoldQuadratic formSingular cohomologyCup product
Definition
Let and let be a closed smooth manifold of dimension with a framing of its stable normal bundle. Let be its middle singular cohomology with coefficients in . The cup product followed by evaluation on the fundamental class gives a nondegenerate bilinear form on , and for all . The framing determines a function such that Choose a basis of with and . The Kervaire invariant of is the Arf invariant of the quadratic form :
Remarks
The Arf invariant does not depend on the choice of the symplectic basis. It is the value that takes on more than half of the elements of .
Example: let with the framing given by its Lie group structure, so . Here with basis , dual to the two circle factors, and . For this framing , so .
Non-example: a closed framed manifold of dimension has no Kervaire invariant, since is not of the form . The construction needs an odd middle degree .
The Kervaire invariant is a framed cobordism invariant. By the Pontryagin–Thom construction it defines a homomorphism on the stable homotopy groups of spheres.
Reading path
To understand this page from the ground up, read these 5 pages in order. The foundations come first.
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- no evidence level
- Cup product concept
- Quadratic form concept
- Singular cohomology concept
- Smooth manifold concept
- Kervaire invariant definition
Further reading
About this page
| Tag | 7XTQKSHQEK |
|---|---|
| URI | m2://7XTQKSHQEK |
| Kind | definition |
| Writing | W0 stub |
| MSC | 55Q45, 57R15 |
| Defines | Kervaire invariant |
| Uses | Smooth manifold, Quadratic form, Singular cohomology, Cup product |
| Changed | 2 November 2026 |
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