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

It is known that a manifold $B$ bounds a compact contractible topological manifold if and only if $B$ is a homology sphere. The "only if" direction follows by excising a small ball in the interior of the contractible manifold, and noting that its boundary sphere has the same homology as $B$ by Poincare duality. The "if" direction is due to Freedman in dimension $3$ and to Kervaire in dimensions $>3$.

Question. Is there a characterization of boundaries of noncompact contractible manifolds?

Note that if $B$ is a boundary component of a contractible $n$-manifold $W$, then the following holds.

  1. If $B$ is compact, then $W$ is compact and $\partial W=B$.
    (If $V$ denotes the union of $B$ and the interior of $W$, then the long exact sequence of the pair gives $H_{n-1}(B)\cong H_{n}(V,B)$ and Poincare duality gives implies that $H_{n-1}(B)$ is nontrivial, and hence so is $H_{n}(V,B)\cong H^0_c(V)$ which is only possible if $V$ is compact.)

  2. $B$ is stably parallelizable (because $W$ is parallelizable).

  3. If $U$ is a proper open subset of $B$, then $U$ bounds a noncompact contractible manifold, namely, $U\cup\mathrm{Int}(W)$.

  4. The product of $B$ with any open contractible manifold bounds a noncompact contractible manifold.

  5. If $B'$ bounds a contractible manifold $W'$, then the connected sum $B\# B'$ bounds a contractible manifold, namely, the boundary connected sum of $W$, $W'$.

The above question may be too hard, so I would be happy with any addition to 1--5.

share|improve this question
2  
6. Intersection products in $B$ are trivial. –  Tom Goodwillie Feb 27 '12 at 1:28
2  
By 3, $B$ can have the homotopy type of any finite complex, by taking a thickening of some embedding in Euclidian space. So for example any $k$ dim'l complex has the homotopy type of a $n=2k+1$ manifold which is the boundary of a contractible $n+1$ manifold. –  Paul Feb 27 '12 at 3:17
1  
Every cycle in $B$ is the boundary of a chain in $W$. So basically given a $p$-cycle and an $(n-p-1)$-cycle in $B$ their intersection is the boundary of the intersection of a $(p+1)$-chain and an $(n-p)$-chain in $W$, so the boundary of a $1$-chain, so a homologically trivial $0$-chain. To make this rigorous I guess you would need to say something about Poincare duality and compactly supported cohomology. –  Tom Goodwillie Feb 27 '12 at 5:14
1  
Maybe the answer is any open subset of a homology sphere? Linking numbers vanish also. I can't see how to find a compact $n$-manifold with boundary $M$ whose interior bounds a contractible manifold but which doesn't embed in a homology $n$-sphere. For $n=3$, it seems like you can add 2-handles to $M$ to get a homology ball, and perhaps in higher dimensions you could add handles to get a homology ball. –  Paul Feb 27 '12 at 13:17
1  
@Igor: I meant the $Q/Z$ valued linking form on the torsion homology classes. I was thinking a warm up problem is to determine which $B=$interior$(M)$, $M$ compact, are boundaries, and suggesting maybe all such embed as codim 0 submanifolds of a homology sphere. –  Paul Feb 27 '12 at 13:54
show 10 more comments

Know someone who can answer? Share a link to this question via email, Google+, Twitter, or Facebook.

Your Answer

 
discard

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

Browse other questions tagged or ask your own question.