Math-2182 · mathematics · Geometry

Math-2182

Boundaries of Groups and Kleinian Groups — Problem 39

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Statement

Definitions and mathematical statement

Let $M$ be a closed simply connected topological four-manifold which is not smoothable: there is no smooth manifold structure compatible with its topology. Closed means compact without boundary. Seek a metric $d$ inducing its given topology and a Borel measure $\mu$ for which there is $C\geq1$ satisfying \[ C^{-1}r^4\leq\mu(B_d(x,r))\leq Cr^4 \quad(x\in M,\ 0<r\leq\operatorname{diam}_dM). \] This is Ahlfors four-regularity. Also require linear local contractibility: some $\lambda\geq1$ has the property that every ball $B_d(x,r)$, for $0<r\leq\operatorname{diam}_dM/\lambda$, contracts inside $B_d(x,\lambda r)$.

Does every such nonsmoothable $M$ admit a metric with both properties? Even constructing such a metric on a standard example, such as the simply connected topological four-manifold with $E_8$ intersection form, is a significant subquestion in the source.

The contraction is a homotopy of the inclusion of the smaller ball to a constant map in the larger ball; it need not stay inside the smaller ball. Neither geodesicity of the metric nor a smooth tensor field is assumed. These are quantitative metric and topological requirements, weaker in formulation than supplying a quasiconformal atlas. A topological manifold already has locally contractible neighborhoods, but that fact gives no uniform linear control relative to an Ahlfors-regular metric.

Short English statement

Can a nonsmoothable simply connected four-manifold carry an Ahlfors four-regular, linearly locally contractible metric?

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