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  "body": "\n## Statement\n\nLet $M$ be a closed, oriented, [[smooth|F45VQ47GPV]] [[4-manifold|8H58Q6YJCH]] that admits a [[spin structure|RSPZ9TCFQ8]].\nThen the [[signature|07EZA80NEH]] $\\sigma(M)$ of its [[intersection form|WK8GKRKAVD]] is divisible by $16$.\n\n## Remarks\n\nVladimir Rokhlin proved the theorem in 1952.\nOne modern proof uses the Atiyah–Singer index theorem: the index of the Dirac operator of $M$ equals $-\\sigma(M)/8$.\nIn dimension $4$ the spinor bundles are quaternionic, so the kernel and cokernel of the Dirac operator are quaternionic vector spaces and the index is even.\nHence $\\sigma(M)/8$ is even.\nThe K3 surface is a spin $4$-manifold of signature $-16$, so $16$ cannot be replaced by a larger number.\nWithout smoothness the theorem fails: the [[$E_8$ manifold|BNDRN0G9BR]] is a closed topological $4$-manifold with even form and signature $8$.\nWithout the spin condition it fails: the complex projective plane has signature $1$.\nAlgebra alone gives less: an even unimodular form always has signature divisible by $8$.\nFor a simply connected $M$, being spin is the same as having an even intersection form.\n\n## Explanations\n\n### L1 human\n\nThe signature is a single integer that measures the crossings of surfaces in a four-dimensional space.\nRokhlin's theorem says that for smooth spaces of a special kind, called spin, this integer is always a multiple of $16$.\nAlgebra only predicts a multiple of $8$, and the gap of a factor $2$ is a purely smooth phenomenon.\n\n### L2 human\n\nThe theorem is the first sign that smooth $4$-manifolds are more rigid than topological ones.\nIt shows that the $E_8$ manifold has no smooth structure.\nIt defines the Rokhlin invariant of an integral homology $3$-sphere: the signature of a smooth spin $4$-manifold it bounds, divided by $8$, modulo $2$.\nThe [[11/8 conjecture|90RA6TAT7H]] and Furuta's $10/8$ theorem refine it for spin $4$-manifolds.\n",
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