# Why “Classification” of 4 manifolds is NOT possible?

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?"

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

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.)

-
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