I know classification of 2 manifolds and geometrization for 3 manifolds. Why for dimension great or equal to 4, this task become impossible?
edit: Or should I ask "why geometrization won't be possible for 4 or higher dimension?"
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1$\begingroup$ What is "Classificatoin"? $\endgroup$– user5810Commented Aug 27, 2011 at 3:24
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2$\begingroup$ If you are willing to accept "Morse" decompositions, then higher dimensional manifolds are actually easier to classify. $\endgroup$– MattCommented Aug 27, 2011 at 3:31
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24$\begingroup$ C'mon Ricky, you have power to edit rather than snark. $\endgroup$– Allen KnutsonCommented Aug 27, 2011 at 3:47
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3$\begingroup$ I was not aware that classification is synonymous with geometrization. Can the OP please clarify whether he or she is primarily interested in geometrization, or in some other notion of classification (cf. Matt's comment above) $\endgroup$– Yemon ChoiCommented Aug 27, 2011 at 5:15
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2$\begingroup$ @Ricky: Yeah, I noticed that too, but s/he spelled it correctly in the actual questoin. $\endgroup$– Allen KnutsonCommented Aug 28, 2011 at 2:31
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2 Answers
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I'm guessing that you heard this from someone whose reasoning goes "Every finite presentation of a group can be made to give the $\pi_1$ of a smooth 4-manifold. If we could put any 4-manifold into the Magic List of All, then we could recognize presentations of the trivial group. But no algorithm can do that."
Often people worry about classifications of simply connected manifolds, and don't have to deal with this. (Of course in three dimensions this becomes Perelman's theorem.)
answered Aug 27, 2011 at 3:51
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21$\begingroup$ Something that bugs me whenever this is discussed: Papakyriakopoulos solved the homeomorphism problem for 2-complexes, which have all finitely presented groups among their fundamental groups. So you can tell two presentation 2-complexes apart (up to homeomorphism), even though you can't tell if their fundamental groups are isomorphic. So Markov's theorem about there being no solution to the homeomorphism problem for 4-manifolds is a little more subtle than everyone lets on. $\endgroup$ Commented Aug 27, 2011 at 22:25
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2$\begingroup$ I had trouble finding an actual copy of the Papakyriakopoulos paper, but I did find this later paper [Whittlesey '58] with a proof of the same result, so I thought others might find the link useful: jstor.org/stable/2033313. I'd be grateful for a digital version of the Papakyriakopoulos paper. $\endgroup$ Commented Nov 14, 2017 at 22:56
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As pointed out in a comment by Autumn Kent to Allen Knutson's answer, the problem is a bit more subtle than it may appear. In order to prove that the homeomorphism problem for compact 4-manifolds, say in the topological category, is recursively unsolvable, it is not enough to know that (1) every finitely presented group can be realized as the fundamental group of some compact 4-manifold, and (2) the isomorphism problem for finitely presented groups is recursively unsolvable.
Instead, what you do is give a construction which to any finite presentation $< S | P >$ of a group associates a 4-manifold $M(S,P)$ in such a way that $\pi_1(M(S,P))$ is isomorphic to the group defined by the presentation $< S | P >$, and moreover two such manifolds are homeomorphic if and only if they have isomorphic fundamental groups.
Then you have constructed a class of 4-manifolds for which the homeomorphism problem is equivalent to the isomorphism problem for finitely presented groups, and therefore unsolvable.
About "geometrization for manifolds of dimension 4 or higher", well as far as I know there is no theorem which says it is impossible. It depends on what you mean by `geometric structure', and what you want those structures to do for you.
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$\begingroup$ So is such a construction possible (referring to the second paragraph)? $\endgroup$ Commented Dec 11, 2021 at 18:07
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$\begingroup$ @Colescu Yes, this was done by Markov in 1958. $\endgroup$ Commented Feb 23, 2022 at 3:17