Which manifolds are homeomorphic to simplicial complexes? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T10:54:15Z http://mathoverflow.net/feeds/question/44021 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/44021/which-manifolds-are-homeomorphic-to-simplicial-complexes Which manifolds are homeomorphic to simplicial complexes? Charles Rezk 2010-10-28T21:04:15Z 2013-03-12T01:43:27Z <p>This question is only motivated by curiousity; I don't know a lot about manifold topology.</p> <p>Suppose $M$ is a compact topological manifold of dimension $n$. I'll assume $n$ is large, say $n\geq 4$. The question is: <em>Does there exist a simplicial complex which is homeomorphic to $M$?</em></p> <p>What I think I know is:</p> <ul> <li><p>If $M$ has a piecewise linear (PL) structure, then it is triangulable, i.e., homeomorphic to a simplicial complex.</p></li> <li><p>There is a well-developed technology ("<a href="http://en.wikipedia.org/wiki/Kirby-Siebenmann_invariant" rel="nofollow">Kirby-Siebenmann invariant</a>") which tells you whether or not a topological manifold admits a PL-structure.</p></li> <li><p>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 <a href="http://en.wikipedia.org/wiki/Homology_sphere" rel="nofollow">homology sphere</a> $S$ (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 $M$. </p></li> </ul> <p>This leaves open the possibility that there are topological manifolds which do not admit <em>any</em> PL-structure but are still homeomorphic to some simplicial complex. Is this possible?</p> <p>In other words, what's the difference (if any) between "triangulable" and "admits a PL structure"?</p> <p><a href="http://en.wikipedia.org/wiki/4-manifold" rel="nofollow">This wikipedia page on 4-manifolds</a> 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.</p> http://mathoverflow.net/questions/44021/which-manifolds-are-homeomorphic-to-simplicial-complexes/44025#44025 Answer by Andy Putman for Which manifolds are homeomorphic to simplicial complexes? Andy Putman 2010-10-28T22:03:55Z 2013-03-12T01:43:27Z <p>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: <a href="http://www.maths.ed.ac.uk/~aar/slides/orsay.pdf" rel="nofollow">http://www.maths.ed.ac.uk/~aar/slides/orsay.pdf</a></p> <p>EDIT : Hot off the press is a <a href="http://arxiv.org/abs/1303.2354" rel="nofollow">paper</a> 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.</p> http://mathoverflow.net/questions/44021/which-manifolds-are-homeomorphic-to-simplicial-complexes/44039#44039 Answer by Paul for Which manifolds are homeomorphic to simplicial complexes? Paul 2010-10-29T00:53:26Z 2010-10-29T00:53:26Z <p>Galewski-Stern proved</p> <p><a href="http://www.ams.org/mathscinet-getitem?mr=420637" rel="nofollow">http://www.ams.org/mathscinet-getitem?mr=420637</a></p> <p>" 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."</p> <p>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.</p> http://mathoverflow.net/questions/44021/which-manifolds-are-homeomorphic-to-simplicial-complexes/59045#59045 Answer by Andrew Ranicki for Which manifolds are homeomorphic to simplicial complexes? Andrew Ranicki 2011-03-21T08:28:48Z 2011-03-21T08:28:48Z <p>For a discussion of the 4-dimensional case see <a href="http://www.map.him.uni-bonn.de/Questions_about_surgery_theory" rel="nofollow">http://www.map.him.uni-bonn.de/Questions_about_surgery_theory</a></p> http://mathoverflow.net/questions/44021/which-manifolds-are-homeomorphic-to-simplicial-complexes/62535#62535 Answer by Junyan Xu for Which manifolds are homeomorphic to simplicial complexes? Junyan Xu 2011-04-21T13:29:49Z 2012-01-26T16:07:17Z <p>Regarding Charles Rezk's second question:</p> <p><em>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?</em></p> <p>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 &lt;8). See Problem 3 of <a href="http://www.maths.ed.ac.uk/~aar/haupt/sandro.pdf" rel="nofollow">http://www.maths.ed.ac.uk/~aar/haupt/sandro.pdf</a>. Also see <a href="http://math.uci.edu/~rstern/Hiroshima2_18_06.pdf" rel="nofollow">this presentation</a>.</p> <p>For dimension >4, <a href="http://eom.springer.de/t/t093230.htm" rel="nofollow">Springer Online Reference Works</a> 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 <a href="http://math.uci.edu/~rstern/Hiroshima2_18_06.pdf" rel="nofollow">this presentation</a> 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.</p> <p>In addition, <a href="http://books.google.com/books?id=gz2CbgR5RbwC&amp;printsec=frontcover&amp;source=gbs_atb#v=onepage&amp;q&amp;f=false" rel="nofollow">This book</a> (p.168, Theorem 18.4) seems to contain a result that strengthens the one mentioned in Paul's answer.</p> <p>Added: <a href="http://arxiv.org/abs/math.AT/0105047" rel="nofollow">This paper</a> (22.5. Example) explicitly gives an example of "A topological manifold which is homeomorphic to a polyhedron but does not admit any PL structure".</p>