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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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What is "Classificatoin"? – Ricky Demer Aug 27 '11 at 3:24
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If you are willing to accept "Morse" decompositions, then higher dimensional manifolds are actually easier to classify. – Matt Aug 27 '11 at 3:31
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C'mon Ricky, you have power to edit rather than snark. – Allen Knutson Aug 27 '11 at 3:47
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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) – Yemon Choi Aug 27 '11 at 5:15
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@Ricky: Yeah, I noticed that too, but s/he spelled it correctly in the actual questoin. – Allen Knutson Aug 28 '11 at 2:31
up vote 24 down vote accepted

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.)

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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. – Richard Kent Aug 27 '11 at 22:25

As pointed out in a comment by Richard 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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