Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Question 1. Let $X_i$ be an infinite family of closed, orientable, smooth 4-manifolds with the following properties:

a) $\pi_1(X_i) = \mathbb{Z}\times \mathbb{Z_{2}}$ for any $i = 1, 2, \cdots $

b) all the homology groups of $X_i$ and $X_j$ with integer coefficients are same

Is it true in this family there are infinitely many homeomorphic 4-manifolds?

Does this follow from Freedman's classification theorem since $\mathbb{Z}\times \mathbb{Z_{2}}$ is a "good" group?

Question 2. What if $X_i$ has a boundary? How much is known in non simply-connected case?

share|improve this question
    
In the boundary case you certainly get infinitely many non-homeomorphic manifolds distinguished by their 3-dimensional boundary. It suffices to pick one manifold and modify it via a boundary-connected sum with a contractible 4-manifold (there are infinitely many distinct contractible 4-manifolds, easily distinguished by their homology 3-sphere boundaries: they can be easily constructed as "Mazur manifolds"): you get a new manifold homotopic to the original one but with a different boundary (the previous boundary connected sum with a homology 3-sphere) –  Bruno Martelli May 19 '12 at 17:52
    
@Bruno I should have been more clear with my question. I assume the boundaries of all these $X_i$ are same, and there are only finitely many possible homotopy types for $X_i$. –  user23802 May 19 '12 at 18:43
add comment

1 Answer

up vote 5 down vote accepted

First, you also want to fix not just $H_2$ but $H_2(M, {\mathbb Z}[\pi_1(M)])$ together with the intersection form on this group. With this in mind, if $M$ is a closed 4-manifold whose fundamental group is infinite cyclic, then Freedman-style classification is indeed available for $M$, but requires extra work which was done by Stong and Wang in "Self-homeomorphisms of 4-manifolds with fundamental group ${\mathbb Z}$", where they corrected some errors in the book of Freedman and Quinn. (Wang may have done this earlier in his unpublished thesis.) In particular, in this setting, you get only finitely many topological types of the manifolds $M$. Stong and Wang also prove that a self-homeomorphism of such $M$'s are "almost" determined (up to pseudoisotopy) by its action on $H_2(M, {\mathbb Z}[\pi_1(M)])$.This result might take care of manifolds whose fundamental group is ${\mathbb Z}\times {\mathbb Z}_2$ (by considering free actions of topological involutions on manifolds $M$ above). However, you would have to check if two pseudo-isotopic involutions are topologically conjugate.

share|improve this answer
    
I guess you mean the homology of the universal cover, or the homology group with local coefficients $H_{2}(X_{i},{\mathbb{Z}}[{\pi_{1}(X_{i})}]$, otherwise there′s no action of $\pi_1(X_i)$. –  Fernando Muro May 19 '12 at 7:39
    
Dear Misha: Thank you very much for your answer. According to the paper by I. Hambleton "Intersection forms, fundamental groups, and 4-manifolds" gokovagt.org/proceedings/2008/ggt08-hambleton.pdf (page 142), Freedman's surgery theory applies for topological 4-manifolds whose fundamental groups are poly-cyclic. Does this seems to indicate that the classification holds for 4-manifolds with such fundamental groups? Does anyone knows what happens in the case when $X_i$ has a boundary? If I am not mistaken, S. Boyer has the classification result for simply connected ones. –  user23802 May 19 '12 at 12:43
    
@alex-lin: The statement that "4-d surgery works for pi_1=G" is not the same as "there exists a classification theorem for 4-d manifolds with pi_1=G". To get from the 2nd to the 1st, you need to solve a homotopy problem. when $\pi_1=1$, the homotopy type is determined by the intersection form, so that for any choice of $H_2$ there are finitely many homotopy types. To answer your question you'll want to know, given $\pi_1=ZxZ/2$ and rank$H_2$, how many homotopy types are possible. Then surgery will apply to tell you (with some work) how many homeomorphism types. Same for Boyer's result. –  Paul May 19 '12 at 16:10
    
typo: I mean from the 1st to the 2nd. –  Paul May 19 '12 at 16:12
    
Thanks Paul for your comment. In this specific case, I do have only finitely many homotopy types. –  user23802 May 19 '12 at 16:51
show 3 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.