Math-1463 · mathematics · Topology

Math-1463

Kirby Problem 4.59

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Statement

Definitions and mathematical statement

For integers $p\ge1$ and $q$ coprime to $p$, let $L(p,q)$ be the lens space obtained from $S^3\subset\mathbb C^2$ by the free cyclic action generated by $(z_1,z_2)\mapsto(e^{2\pi i/p}z_1,e^{2\pi iq/p}z_2)$. Use the quotient orientation. It has first integral homology $\mathbb Z/p\mathbb Z$.

Two oriented closed $3$-manifolds $Y_0,Y_1$ are topologically integral-homology cobordant if there is a compact oriented topological $4$-manifold $W$ with $\partial W=(-Y_0)\sqcup Y_1$ such that each boundary inclusion induces an isomorphism \[ H_i(Y_j;\mathbb Z)\longrightarrow H_i(W;\mathbb Z) \quad\text{for every }i,\ j=0,1. \] No smooth structure on $W$ is required.

Are two lens spaces homology cobordant in this sense if and only if they are orientation-preservingly homeomorphic? The homeomorphism implies the cobordism direction by taking a product. The difficult direction asks whether an arbitrary topological homology cobordism forces that homeomorphism.

The coefficients are integers, not rationals: rational homology cobordism is a weaker relation. The hypotheses immediately force the two lens spaces to have the same $p$, since their first homology groups are isomorphic, but the gluing parameter $q$ still matters. Orientations are compared consistently; reversing the chosen orientation of one lens space changes the oriented instance of the question.

Short English statement

Are lens spaces topologically integral-homology cobordant exactly when they are homeomorphic, with the corresponding orientation convention?

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