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  "body": "\n## Definition\n\nLet $k \\ge 0$ and let $M$ be a closed [[smooth manifold|F45VQ47GPV]] of dimension $4k + 2$ with a framing of its stable normal bundle.\nLet $V = H^{2k+1}(M; \\F_2)$ be its middle [[singular cohomology|6N7BC70CX0]] with coefficients in $\\F_2$.\nThe [[cup product|JX47MS1R1H]] followed by evaluation on the fundamental class gives a nondegenerate bilinear form $x \\cdot y$ on $V$, and $x \\cdot x = 0$ for all $x$.\nThe framing determines a function $q \\colon V \\to \\F_2$ such that\n$$q(x + y) = q(x) + q(y) + x \\cdot y \\quad \\text{for all } x, y \\in V.$$\nChoose a basis $a_1, \\dots, a_m, b_1, \\dots, b_m$ of $V$ with $a_i \\cdot b_j = \\delta_{ij}$ and $a_i \\cdot a_j = b_i \\cdot b_j = 0$.\nThe *Kervaire invariant* of $M$ is the Arf invariant of the [[quadratic form|0RBQS1JJCR]] $q$:\n$$\\Phi(M) = \\sum_{i=1}^{m} q(a_i)\\, q(b_i) \\in \\F_2.$$\n\n## Remarks\n\nThe Arf invariant does not depend on the choice of the symplectic basis.\nIt is the value that $q$ takes on more than half of the elements of $V$.\n\nExample: let $M = S^1 \\times S^1$ with the framing given by its Lie group structure, so $k = 0$.\nHere $V = \\F_2^2$ with basis $a$, $b$ dual to the two circle factors, and $a \\cdot b = 1$.\nFor this framing $q(a) = q(b) = 1$, so $\\Phi(M) = 1$.\n\nNon-example: a closed framed manifold of dimension $4$ has no Kervaire invariant, since $4$ is not of the form $4k + 2$.\nThe construction needs an odd middle degree $2k + 1$.\n\nThe Kervaire invariant is a framed cobordism invariant.\nBy the Pontryagin–Thom construction it defines a homomorphism $\\pi^s_{4k+2} \\to \\Z/2$ on the stable homotopy groups of spheres.\n",
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