We know that for n $\ge$ 5, a manifold admits a piecewise linear structure if and only if its Kirby-Siebenmann class vanish and Galewski and Stern showed the existence of a similar invariant to test for the existence of a triangulation whenever n $\ge$ 5 . So is there a similar invariant to detect if a manifold admits a CW structure. Likewise, what could be the obstructions that can arise that stops a CW complex being a manifold ( I know of Poincare duality being one for finite CW complexes).
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1$\begingroup$ I believe the question of whether or not every finitely presented Poincaré Duality group is the fundamental group of a closed aspherical manifold is open in all dimensions >2. So the "obstructions that can arise that stop a CW complex being a manifold" are not fully understood (though, as you say, Poincaré duality is certainly a necessary condition). $\endgroup$– HJRWCommented Dec 2 at 9:27
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$\begingroup$ Can you provide precise definition of a manifold you're using? For example, if you do not require metrizability, then you have an example by Calabi-Rosenlicht (which is separable and Hausdorff), which can be informally described as a sphere with continuum-many punctures along the big circle. It is weakly equivalent to a CW complex, but not homotopy equivalent. So, metrisability is necessary (and also sufficient, by Milnor). $\endgroup$– Denis TCommented Dec 2 at 14:54
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2$\begingroup$ Also, "admitting CW structure" can be understood at least in two different ways: being homeomorphic to a CW, or being homotopy equivalent. (It is not clear whether you're interested in the setting of general topology or algebraic topology.) $\endgroup$– Denis TCommented Dec 2 at 15:13
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$\begingroup$ @DenisT By manifold, I meant the ones which are metrisable and by admitting CW structure, I meant homeomorphism. $\endgroup$– TyrannosaurusCommented Dec 2 at 15:32
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1$\begingroup$ See also my answer here. $\endgroup$– Moishe KohanCommented Dec 3 at 11:39
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1 Answer
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Every compact topological manifold $M$ has the homotopy type of CW-complex. So, in the compact case, there is no such invariant of the type you are looking for (or, at least, such an invariant cannot also be a homotopical invariant).
Moreover, if $M$ is closed and $\dim(M) \geq 6$, then $M$ is homeomorphic to a CW-complex.
References.
Kirby, R. C.; Siebenmann, L. C., On the triangulation of manifolds and the Hauptvermutung, Bull. Am. Math. Soc. 75, 742-749 (1969). ZBL0189.54701." Bull AMS 75 (1969).
More details can be found in the answers to MO36838.
answered Dec 2 at 10:06
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4$\begingroup$ Neither the paper nor the MO post you link seem to mention the $\dim\geq 6$ result you mention. Would you happen to have a reference for it? $\endgroup$– WojowuCommented Dec 2 at 11:18
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3$\begingroup$ I don't see how the result of the first paragraph precludes existence of a (homotopy-invariant) invariant for a CW structure. It may be that the obstruction is only applicable for spaces which are closed manifolds. So unless we have that any closed manifold has homotopy type of a CW complex which is also a closed manifold, I fail to see the conclusion. $\endgroup$– WojowuCommented Dec 2 at 11:21
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3$\begingroup$ References can be found in the proofs of Theorems 3.13 and 3.16 of this very helpful survey: arxiv.org/pdf/1910.07372 $\endgroup$ Commented Dec 2 at 14:32
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2$\begingroup$ @Wojowu A comment by Igor Belegradek below the linked MO answer says that the dim $\geq 6$ result is given on p. 107 of Kirby-Siebenmann. $\endgroup$ Commented Dec 2 at 14:37
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