Math-1479 · mathematics · Topology

Math-1479

Kirby Problem 4.118

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Statement

Definitions and mathematical statement

An index-$k$ handle in dimension four is $D^k\times D^{4-k}$, attached to the boundary of the manifold already constructed along $\partial D^k\times D^{4-k}$, for $0\le k\le4$. A smooth handle decomposition constructs a compact smooth manifold by finitely many such attachments and rounds the resulting corners. For a closed manifold the construction ends with empty boundary.

Let $X$ be a closed connected simply connected smooth $4$-manifold. Does $X$ admit a handle decomposition with no index-one handles? More strongly, does it admit a handle decomposition with neither index-one nor index-three handles? Such a decomposition may be taken to have one index-zero handle and one index-four handle, with the remaining handles of index two in the stronger version.

Simple connectivity is an algebraic property of the whole manifold: its fundamental group is trivial. Absence of one-handles asks for a particular geometric presentation, and is commonly called geometric simple connectivity. The question is whether every $X$ with the former property admits a presentation with the latter.

The desired handle decomposition need not be obtained from a given one by a prescribed bounded sequence of moves. One is free to choose any smooth decomposition of $X$. No stabilization by $S^2\times S^2$, connected sum with another manifold, or change of smooth structure is allowed. Closedness is essential to the stated domain; contractible manifolds with boundary are not included.

Short English statement

Does every closed simply connected smooth four-manifold have a handle decomposition without one-handles, and even without both one- and three-handles?

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