Math-1625 · mathematics · Topology

Math-1625

Kirby Problem 4.83

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Statement

Definitions and mathematical statement

Let $X$ be a topological four-manifold that admits a smooth structure $\mathcal S$. A self-homeomorphism $h:X\to X$ is smoothable relative to $\mathcal S$ if it is topologically isotopic to a diffeomorphism of $(X,\mathcal S)$. An isotopy is a continuous path through self-homeomorphisms beginning at $h$. For manifolds with boundary use boundary-fixed maps and isotopies relative to the boundary, as in the source's relative convention.

Classify the topological four-manifolds $X$ for which there exists a smoothing $\mathcal S$ and a self-homeomorphism $h$ that is not smoothable in this sense. Address closed, compact-with-boundary, and open manifolds, distinguishing their boundary and support conventions. No compact-support condition is imposed on an open-manifold homeomorphism unless a more restrictive version is stated.

For fixed $\mathcal S$, the condition can equivalently be expressed as failure of surjectivity of \[ \pi_0\operatorname{Diff}(X,\mathcal S) \longrightarrow\pi_0\operatorname{Homeo}(X), \] with relative groups in the boundary case. Components are understood using the usual smooth and compact-open topologies.

The question concerns an entire topological isotopy class missing the smooth image. A map that is not itself differentiable may still be isotopic to a diffeomorphism and thus would not qualify. Conversely, the existence of two inequivalent smooth structures does not by definition supply such a self-map. The smoothing and homeomorphism are both existentially quantified for each proposed underlying manifold.

Short English statement

Which four-manifolds have a smoothing with a self-homeomorphism that cannot be isotoped to any diffeomorphism?

Sources

Catalogue entry

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