Math-1442 · mathematics · Topology

Math-1442

Kirby Problem 4.6

Status Open · unsolvedmath.comLLM status Open unverified

Top Open Problems Not reported cataloguedOpenOpen

Statement

Definitions and mathematical statement

Fix a closed connected orientable smooth $3$-manifold $Y$. A smooth filling is a compact connected smooth $4$-manifold $W$ with a boundary identification $\partial W\cong Y$. If orientations are chosen, use the induced boundary orientation in this identification. Two fillings $W_0,W_1$ form an absolutely exotic pair if they are homeomorphic as manifolds with boundary but no diffeomorphism between them exists, even when arbitrary maps on their boundaries are allowed.

Does every $Y$ admit such a pair with both $W_0$ and $W_1$ orientable? In concrete terms, find for each $Y$ two compact orientable smooth $4$-manifolds having boundary diffeomorphic to $Y$, a homeomorphism $W_0\to W_1$, and an obstruction to every diffeomorphism $W_0\to W_1$.

The requested smooth distinction is absolute. Showing that no diffeomorphism realizes one preferred identification of the boundary would prove only a relative distinction, because a different boundary identification might extend. The homeomorphism is not required to be the identity under the selected boundary markings.

There is no requirement that a filling be contractible, a homology ball, simply connected, or symplectic. Imposing any of those would strengthen the problem. The universal quantifier is on $Y$, and the topological filling and its two smoothings may depend on $Y$. Both fillings must be orientable; examples with nonorientable fillings do not settle the specified question.

Short English statement

Does every closed orientable three-manifold bound two homeomorphic orientable smooth four-manifolds admitting no diffeomorphism between them?

Sources

Catalogue entry

Statement licence: CC-BY-4.0 · Source file

Contributions 0

No attempts yet. Significant partial results and substantial reductions are welcome. Contribute an attempt.