Which manifolds are homeomorphic to simplicial complexes? - MathOverflow

Which manifolds are homeomorphic to simplicial complexes?

2010-10-28T21:04:15Z

This question is only motivated by curiousity; I don't know a lot about manifold topology.

Suppose $M$ is a compact topological manifold of dimension $n$. I'll assume $n$ is large, say $n\geq 4$. The question is: Does there exist a simplicial complex which is homeomorphic to $M$?

What I think I know is:

If $M$ has a piecewise linear (PL) structure, then it is triangulable, i.e., homeomorphic to a simplicial complex.
There is a well-developed technology ("Kirby-Siebenmann invariant") which tells you whether or not a topological manifold admits a PL-structure.
There are exotic triangulations of manifolds which don't come from a PL structure. I think the usual example of this is to take a homology sphere $S$ (a manifold with the homology of a sphere, but not maybe not homeomorphic to a sphere), triangulate it, then suspend it a bunch of times. The resulting space $M$ is supposed to be homeomorphic to a sphere (so is a manifold). It visibly comes equipped with a triangulation coming from that of $S$, but has simplices whose link is not homemorphic to a sphere; so this triangulation can't come from a PL structure on $M$.

This leaves open the possibility that there are topological manifolds which do not admit any PL-structure but are still homeomorphic to some simplicial complex. Is this possible?

In other words, what's the difference (if any) between "triangulable" and "admits a PL structure"?

This wikipedia page on 4-manifolds claims that the E8-manifold is a topological manifold which is not homeomorphic to any simplicial complex; but the only evidence given is the fact that its Kirby-Siebenmann invariant is non trivial, i.e., it doesn't admit a PL structure.

Answer by Andy Putman for Which manifolds are homeomorphic to simplicial complexes?

2010-10-28T22:03:55Z

I don't know about dimension 4, but for high dimensions this is a well-known open problem. I don't think much progress has been made on it for a while. I recommend Ranicki's lecture notes from Siebenmann's retirement conference for a good summary about what is known about this and related problems: http://www.maths.ed.ac.uk/~aar/slides/orsay.pdf

EDIT : Hot off the press is a paper of Manolescu claiming to disprove the conjecture of Galewski-Stern and construct manifolds in all dimensions $\geq 5$ which are not homeomorphic to simplicial complexes.

Answer by Paul for Which manifolds are homeomorphic to simplicial complexes?

2010-10-29T00:53:26Z

Galewski-Stern proved

http://www.ams.org/mathscinet-getitem?mr=420637

" It follows that every topological m-manifold, m≥7 (or m≥6 if ∂M=∅), can be triangulated if and only if there exists a PL homology 3-sphere H3 with Rohlin invariant one such that H3#H3 bounds a PL acyclic 4-manifold."

The Rohlin invariant is a Z/2 valued homomorphsim on the 3-dimensional homology cobordism group, $\Theta_3\to Z/2$, so if it splits there exist non-triangulable manifodls in high dimensions.

Answer by Andrew Ranicki for Which manifolds are homeomorphic to simplicial complexes?

2011-03-21T08:28:48Z

For a discussion of the 4-dimensional case see http://www.map.him.uni-bonn.de/Questions_about_surgery_theory

Answer by Junyan Xu for Which manifolds are homeomorphic to simplicial complexes?

2011-04-21T13:29:49Z

Regarding Charles Rezk's second question:

This leaves open the possibility that there are topological manifolds which do not admit any PL-structure but are still homeomorphic to some simplicial complex. Is this possible?

For dimension 4, it follows from the Poincare conjecture that a 4-manifold is triangulable iff smoothable (which is also equivalent to having PL structure for dimension <8). See Problem 3 of http://www.maths.ed.ac.uk/~aar/haupt/sandro.pdf. Also see this presentation.

For dimension >4, Springer Online Reference Works claims that "the imbedding $PL \subset TRI$ is also irreversible in the same strong sense (there exist polyhedral manifolds of dimension $\geq 5$ that are homotopy inequivalent to any PL-manifold)", but gives no examples. In this presentation it is stated that "All oriented closed 5-manifolds triangulable", so I think among them there may be some with nontrivial KS invariant and hence cannot bear PL structure.

In addition, This book (p.168, Theorem 18.4) seems to contain a result that strengthens the one mentioned in Paul's answer.

Added: This paper (22.5. Example) explicitly gives an example of "A topological manifold which is homeomorphic to a polyhedron but does not admit any PL structure".