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In his ICM Adress at Nice (Proceedings of the International Congress of Mathematicians Nice, September, 1970, Gauthier-Villars, editeur, Paris 6 e ,1971, Volume 2, pp. 133-163.),

Robion Kirby adresses the problem, whether the fourth topological Spin bordism group is $\mathbb{Z}$ or $\mathbb{Z}\oplus \mathbb{Z}/2$.

Depending on whether the triangulation obstruction $$\Delta:\Omega_{4}^{\rm Spin Top}\to \mathbb{Z}/2 $$ is zero or not. This being equivalent to the fact that Rokhlin's divisibility property of the signature still holds modulo 2, in the sense that the signature is eight times the triangulation obstruction modulo two.

This should be an example of a topological spin manifold which is not bordant to a smooth spin manifold.

¿Is this problem solved? Are there newer references for this problem I am not aware of?

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1 Answer 1

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The map $\Omega_4^{\text{Spin}} \to \Omega_4^{\text{SpinTop}}$ is taken isomorphically by the signature to the inclusion $16\Bbb Z \hookrightarrow 8\Bbb Z$, so that the groups are abstractly isomorphic but that the natural map is not an isomorphism. This is obtained as Theorem 13.1 on page 325 of Kirby and Siebenmann's (1977) foundational essays.

A representative for the nonzero element of $\Omega_4^{\text{SpinTop}}/\Omega_4^\text{Spin} = \Bbb Z/2$ is given by Freedman's E8 manifold: take the E8 plumbing, a smooth manifold with intersection form E8 and boundary the Poincare homology sphere $\Sigma(2,3,5)$, and cap the boundary off with a contractible topological manifold. The result is topologically spinnable because the manifold is simply connected and the intersection form is even.

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