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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.

edited Oct 29 2010 at 3:29

Romeo
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asked Oct 28 2010 at 21:04

Charles Rezk
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Excellent question; I'd like to hear the answer to this too. – Todd Trimble Oct 28 2010 at 21:55

The fact that the double suspension of a homology sphere is homeo to a sphere is due to Bob Edwards. It finally appeared (after 30 years!) on the ArXiv in 06 front.math.ucdavis.edu/0610.5573 – Paul Oct 29 2010 at 2:17

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@Paul : That's not quite true. The double suspension theorem is due to Jim Cannon and was published a 1979 paper entitled "Shrinking cell-like decompositions of manifolds. Codimension three". Edwards proved a "triple suspension theorem" and also proved the double suspension theorem for many examples, including the Poincare homology 3-sphere. – Andy Putman Oct 29 2010 at 2:40

Paul · Accepted Answer · 2010-10-29 00:53:26Z UTC

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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.

answered Oct 29 2010 at 0:53

Paul
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Your answer is about combinatorial triangulations. The question is about arbitrary ones. – Igor Belegradek Oct 29 2010 at 0:59

@Igor : Galewski-Stern's theorem is definitely about noncombinatorial triangulations. – Andy Putman Oct 29 2010 at 1:06

No. "combinatorial triangulation" is synomymous with "PL structure". See the MR. You are probably mistaking it with the Kirby Siebenmann invariant, which is the obstruction to finding a PL structure on a Topological manifold. – Paul Oct 29 2010 at 1:06

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I take it back, Andy and Paul are right. By the way, the introduction to "The Hauptvermutung Book" explains this all in some detail: See maths.ed.ac.uk/~aar/books/haupt.pdf. – Igor Belegradek Oct 29 2010 at 1:31

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Ciprian Manolescu has just posted a paper in which he claims to prove that no such homology 3-sphere exists. arXiv:1303.2354 Pin(2)-equivariant Seiberg-Witten Floer homology and the Triangulation Conjecture – Jeffrey Giansiracusa Mar 12 at 10:19

show 3 more comments

Junyan Xu · Answer 2 · 2012-01-26 16:07:17Z UTC

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".

Andrew Ranicki · Answer 3 · 2011-03-21 08:28:48Z UTC

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

answered Mar 21 2011 at 8:28

Andrew Ranicki
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Andy Putman · Answer 4 · 2013-03-12 01:43:27Z UTC

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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.

edited Mar 12 at 1:43

answered Oct 28 2010 at 22:03

Andy Putman
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Thanks! Those slides say that the problem is solved in dimension 4 (all manifolds are triangulable), attributed to Casson. – Charles Rezk Oct 28 2010 at 22:38

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The slides say (on page 5) that NOT every 4-manifold is triangulable. – Igor Belegradek Oct 28 2010 at 22:49

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It seems that the fact that E8 is not triangulable should follow from basic properties of Casson invariant (of which I know next to nothing). I am curious to see how the argument goes. – Igor Belegradek Oct 28 2010 at 23:19

@IB the relationship is explained in Akbulut-McCarthy's book in detail but I've forgotten the argument. There's an outline here math.niu.edu/~rusin/known-math/96/Triangulations which asserts that if E8 were triangulable it could be smoothed in the complement of its vertices. Removing a nbd of each vertex yields a smooth manifold with boundary a union of homotopy (?) 3-spheres whose total Rohlin invariant is 1 (since $\sigma(E8)=1$). But Casson's invariant is zero on homotopy spheres and is a lift of Rohlin. – Paul Oct 29 2010 at 1:03

Ah, I misread it. – Charles Rezk Oct 29 2010 at 1:10

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Which manifolds are homeomorphic to simplicial complexes?

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