Math-1465 · mathematics · Topology

Math-1465

Kirby Problem 4.69

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Statement

Definitions and mathematical statement

Let $M$ be a closed smooth $3$-manifold and $f:M\to M$ a self-diffeomorphism. A topological pseudo-isotopy from the identity to $f$ is a homeomorphism \[ F:M\times[0,1]\longrightarrow M\times[0,1] \] with $F(x,0)=(x,0)$ and $F(x,1)=(f(x),1)$. A smooth pseudo-isotopy has the same boundary restrictions and is a diffeomorphism of the product smooth manifolds. Boundary collars can be included in the standard relative convention.

Does there exist a pair $(M,f)$ for which a topological pseudo-isotopy exists but no smooth pseudo-isotopy exists? The manifold $M$ and the diffeomorphism are both to be chosen; the source imposes no orientability or geometric-structure hypothesis.

A pseudo-isotopy need not preserve the interval coordinate. In particular its restrictions to intermediate slices need not be self-maps of those slices. This distinguishes pseudo-isotopy from an ordinary isotopy $f_t:M\to M$, which would give a level-preserving product map $(x,t)\mapsto(f_t(x),t)$.

The two endpoint maps are already smooth. The proposed obstruction concerns extending their prescribed behavior over the $4$-dimensional cylinder by a diffeomorphism, even though a homeomorphic extension exists. Showing only that $f$ is not smoothly isotopic to the identity does not suffice, because a smooth pseudo-isotopy could still fail to preserve levels. Nor is ordinary homotopy to the identity a substitute for the required topological extension.

Short English statement

Can a diffeomorphism of a closed three-manifold be topologically but not smoothly pseudo-isotopic to the identity?

Sources

Catalogue entry

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