96
$\begingroup$

This question is only motivated by curiosity; 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.

Improve this question
$\endgroup$
5
  • 3
    $\begingroup$ Excellent question; I'd like to hear the answer to this too. $\endgroup$
    – Todd Trimble
    Commented Oct 28, 2010 at 21:55
  • 3
    $\begingroup$ 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 $\endgroup$
    – Paul
    Commented Oct 29, 2010 at 2:17
  • 8
    $\begingroup$ @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. $\endgroup$
    – Andy Putman
    Commented Oct 29, 2010 at 2:40
  • 1
    $\begingroup$ @Paul the link in your comment is broken, here's a replacement: arxiv.org/abs/math/0610573 $\endgroup$
    – David Roberts ♦
    Commented Mar 29, 2022 at 7:30
  • $\begingroup$ Sorry but I am rather confused by all these answers and comments. What is the final conclusion then? $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Jul 7, 2023 at 22:40

4 Answers 4

Reset to default
47
$\begingroup$

Galewski-Stern proved

https://mathscinet.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.

Improve this answer
$\endgroup$
8
  • $\begingroup$ Your answer is about combinatorial triangulations. The question is about arbitrary ones. $\endgroup$
    – Igor Belegradek
    Commented Oct 29, 2010 at 0:59
  • $\begingroup$ @Igor : Galewski-Stern's theorem is definitely about noncombinatorial triangulations. $\endgroup$
    – Andy Putman
    Commented Oct 29, 2010 at 1:06
  • 3
    $\begingroup$ 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. $\endgroup$
    – Igor Belegradek
    Commented Oct 29, 2010 at 1:31
  • 1
    $\begingroup$ I've found the paper. Question settled: a "PL acyclic manifold" is the same as an acyclic PL manifold. $\endgroup$
    – algori
    Commented Apr 29, 2011 at 20:29
  • 22
    $\begingroup$ 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 $\endgroup$
    – Jeffrey Giansiracusa
    Commented Mar 12, 2013 at 10:19
33
$\begingroup$

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: https://www.maths.ed.ac.uk/~v1ranick/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.

Improve this answer
$\endgroup$
5
  • $\begingroup$ Thanks! Those slides say that the problem is solved in dimension 4 (all manifolds are triangulable), attributed to Casson. $\endgroup$
    – Charles Rezk
    Commented Oct 28, 2010 at 22:38
  • 4
    $\begingroup$ The slides say (on page 5) that NOT every 4-manifold is triangulable. $\endgroup$
    – Igor Belegradek
    Commented Oct 28, 2010 at 22:49
  • 1
    $\begingroup$ 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. $\endgroup$
    – Igor Belegradek
    Commented Oct 28, 2010 at 23:19
  • 1
    $\begingroup$ @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. $\endgroup$
    – Paul
    Commented Oct 29, 2010 at 1:03
  • $\begingroup$ Ah, I misread it. $\endgroup$
    – Charles Rezk
    Commented Oct 29, 2010 at 1:10
10
$\begingroup$

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 Fragments of geometric topology from the sixties by Sandro Buoncristiano. Also see the presentation From Triangulations to 4-Manifolds: In Honor of Takao Matumoto’s 60th Birthday by Ron Stern.

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 Ron Stern's 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, the book Lectures on the Topology of 3-manifolds: An Introduction to the Casson Invariant (p.168, Theorem 18.4) by Nikolai Saveliev seems to contain a result that strengthens the one mentioned in Paul's answer.

Added: The paper Piecewise linear structures on topological manifolds (22.5. Example) by Yuli B. Rudyak explicitly gives an example of "A topological manifold which is homeomorphic to a polyhedron but does not admit any PL structure".

Improve this answer
$\endgroup$
7
$\begingroup$

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

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .