Every smooth manifold $M$ has a PL structure, and therefore a triangulation. Given a submanifold $N$ of $M$, does anyone know some nice conditions for $N$ to be the subcomplex of some triangulation of $M$, or isotopic to one?
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$\begingroup$ Perhaps I am missing an important adjective, but I think your first sentence is false: arxiv.org/abs/1303.2354 $\endgroup$– Adam SaltzCommented May 10, 2015 at 20:57
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2$\begingroup$ @Adam--you missed the important adjective "smooth"; with that adjective it's a famous theorem of Whitehead. Manolescu's theorem says that there is a (high-dimensional) topological manifold without a simplicial triangulation. It was already known by Kirby-Siebenmann that there are topological manifolds that are not PL (more restrictive than just triangulable) and by Casson-Taubes-Freedman that there are non-triangulable topological 4-manifolds. $\endgroup$– Danny RubermanCommented May 11, 2015 at 0:27
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It follows from Verona's solution to Thom's triangulation conjecture that the inclusion $N\hookrightarrow M$ is triangulable whenever it is proper and topologically stable, and $M$ and $N$ are without boundary.
Verona, Andrei, Stratified mappings - structure and triangulability, Lecture Notes in Mathematics. 1102. Subseries: Mathematisches Institut der Universität und Max-Planck-Institut für Mathematik, Bonn, Vol. 4. Berlin etc.: Springer-Verlag. IX, 160 p. DM 26.50 (1984) ZBL0543.57002.
answered May 10, 2015 at 15:19
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2$\begingroup$ Is topological stability checkable? $\endgroup$ Commented May 11, 2015 at 14:49
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$\begingroup$ @IgorRivin: That's a good question, I confess to not being an expert. Here's what I know: $\operatorname{Emb}(M,N)$ is open in $C^\infty(M,N)$, and dense if $2m\le n$. Also, the space of topologically stable maps $M\to N$ is open and dense in $C^\infty(M,N)$ (the Thom-Mather theorem). I believe it follows that every embedding is isotopic to a topologically stable, hence triangulable, embedding. $\endgroup$ Commented May 12, 2015 at 1:13
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$\begingroup$ I realize this is now a somewhat old question, but doesn't this follow even more directly from Theorem 7.8 in Verona's book by treating the pair (M,N) as giving an abstract stratification. The bundle condition is satisfied due to the tubular neighborhood theorem. Is there a reason that's not right? $\endgroup$ Commented Oct 29, 2015 at 23:12
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$\begingroup$ @GregFriedman: You may be right. I don't have access to the book right now, what is the precise statement of 7.8? $\endgroup$ Commented Oct 30, 2015 at 7:20
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1$\begingroup$ Theorem 7.8 says "Let A be an a.s [abstract stratification] of finite depth. Then there exists a smooth triangulation $(K,\phi)$ of $A$ [then there's a bit more about how you can choose it to so that a certain map to a manifold is simplicial if you also want that]." Of course it takes some unwinding of the definitions to know what a smooth triangulation is and what an abstract stratification is. But it's basically a Thom-Mather space, and if $N$ is a proper smooth submanifold of $M$, I think the pair should satisfy the Thom-Mather conditions. $\endgroup$ Commented Oct 31, 2015 at 0:41