Math-3259 · mathematics · Topology

Math-3259

Kirby Problem 4.110

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Statement

Definitions and mathematical statement

Shared notation.
$\mathbb N=\{1,2,\ldots\}$, and $\mathbb Z,\mathbb Q,\mathbb R,\mathbb C$ denote the integers, rationals, reals and complex numbers. Any other conventions are specified locally.

A compact smooth four-dimensional two-handlebody $W$ is a manifold admitting a handle decomposition using only handles $D^j\times D^{4-j}$ with $j\in\{0,1,2\}$. An exact symplectic structure with convex boundary is a one-form $\lambda$ such that $\omega=d\lambda$ is nondegenerate and the Liouville vector field $V$, defined by $\iota_V\omega=\lambda$, is transverse outward along $\partial W$. The resulting contact structure is $\xi=\ker(\lambda|_{\partial W})$; contact means that $\lambda|_{\partial W}\wedge d(\lambda|_{\partial W})$ never vanishes.

A Weinstein structure on $W$ is a Liouville structure $\lambda'$ together with a Morse function $\phi:W\to\mathbb R$ constant and maximal on the boundary for which the Liouville vector field is gradient-like. Two boundary contact manifolds are contactomorphic if a diffeomorphism identifies their cooriented contact plane fields.

Statement recorded in the supplied source.
Does there exist a smooth two-handlebody $W$ carrying an exact symplectic form with convex contact boundary $(\partial W,\xi)$, but no Weinstein structure on $W$ that fills the same contact boundary? More precisely, no choice of Weinstein structure on this $W$ should induce a contact structure contactomorphic to $\xi$. [S2]

Short English statement

Can a four-dimensional two-handlebody have an exact symplectic structure with convex boundary while admitting no Weinstein structure that fills that same boundary contact structure?

Sources

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