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.

  • 3
    $\begingroup$ Excellent question; I'd like to hear the answer to this too. $\endgroup$
    – Todd Trimble
    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
    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. $\endgroup$ 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
    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? $\endgroup$ Jul 7, 2023 at 22:40

4 Answers 4


Galewski-Stern proved


" 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.

  • $\begingroup$ Your answer is about combinatorial triangulations. The question is about arbitrary ones. $\endgroup$ Oct 29, 2010 at 0:59
  • $\begingroup$ @Igor : Galewski-Stern's theorem is definitely about noncombinatorial triangulations. $\endgroup$ 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. $\endgroup$ 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
    Apr 29, 2011 at 20:29
  • 21
    $\begingroup$ 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 $\endgroup$ Mar 12, 2013 at 10:19

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.

  • $\begingroup$ Thanks! Those slides say that the problem is solved in dimension 4 (all manifolds are triangulable), attributed to Casson. $\endgroup$ Oct 28, 2010 at 22:38
  • 4
    $\begingroup$ The slides say (on page 5) that NOT every 4-manifold is triangulable. $\endgroup$ 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. $\endgroup$ 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
    Oct 29, 2010 at 1:03
  • $\begingroup$ Ah, I misread it. $\endgroup$ Oct 29, 2010 at 1:10

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".


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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.