Math-1484 · mathematics · Topology

Math-1484

Kirby Problem 5.8

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Statement

Definitions and mathematical statement

A closed topological $5$-manifold $M$ is compact, connected and boundaryless, Hausdorff and second countable, and locally homeomorphic to $\mathbb R^5$. It is aspherical if its universal cover is contractible. Equivalently, since a manifold has the homotopy type of a CW complex, all higher homotopy groups $\pi_i(M)$ vanish for $i\ge2$.

The manifold is triangulable if it is homeomorphic to the geometric realization of a simplicial complex. This is a homeomorphic triangulation of the actual topological space, not merely a simplicial complex with the same homotopy type. No requirement that the triangulation give a particular smooth or piecewise-linear structure is built into this definition.

Does there exist a closed aspherical $5$-manifold which is not triangulable? Thus a witness must combine a contractible universal cover with an obstruction to every possible simplicial triangulation of its underlying topology.

Orientability is not imposed in the question. Neither a finite fundamental group nor any specific presentation of it is assumed. The dimension must be exactly five: examples in dimension four or in dimensions at least six would not resolve this case. An obstruction only to smoothing, or only to a chosen PL structure, would be insufficient unless it also ruled out homeomorphism to every simplicial complex. The assertion asks for one example, without a classification of all such manifolds.

Short English statement

Is there a closed aspherical five-manifold whose topological space admits no simplicial triangulation?

Sources

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