4
$\begingroup$

If one has a smooth simply connected manifold $M^n$ which we know to bound an $n+1$ dimensional manifold $N$, what can be said about a handle decomposition for one in terms of a handle decomposition for another?

I am particularly interested in the 4 dimensional case. For instance, if we have a closed simply connected smooth 4-manifold which bounds a 5-manifold, can we say anything about a handle decomposition for this 5-manifold?

Thank you

Rob

Improve this question
$\endgroup$
5
  • 6
    $\begingroup$ Handle decomposition of N will be more complicated than the one of M. In order to formulate a better question, think about the case when M is the sphere and N is an arbitrary close n+1 manifold with a disk removed. $\endgroup$
    – Misha
    Commented Mar 21, 2013 at 15:18
  • $\begingroup$ Maybe the intended question was not about every N but about some N. $\endgroup$
    – Tom Goodwillie
    Commented Mar 22, 2013 at 2:38
  • 2
    $\begingroup$ Tom: You might be right, but then the only meaningful question I see is to get an upper bound on the number of handles needed to construct some N if some handle info about M is given. In any case, it is OP's responsibility to state the precise problem he/she has in mind. $\endgroup$
    – Misha
    Commented Mar 22, 2013 at 3:39
  • $\begingroup$ Apologies for lack of clarity, I am particularly interested in the 4 dimensional case as stated but was wondering if anything was known in general. Mainly I was wondering if we could bound the degree of the handles for instance if I have a 4 manifold which has 0, 1, 2 and 4 handles which bounds a 5 manifolds can we say that the 5 manifold only has 1 and 2 handles or are there any examples where the 5 manifold requires 3 handles. For instance the case with $S^4$ where there are no 1, 2 or 3 handles we know that the 5 manifolds is in fact the 5 ball and so also has no 1,2 or 3 handles. $\endgroup$
    – Robert
    Commented Mar 22, 2013 at 10:42
  • 2
    $\begingroup$ Robert: The 5-manifold can have any handles you like: Every manifold has handle decomposition with handles of all possible indices. Moreover, you can take your 5-manifold and form its connected sum with a closed 5-manifold of your choice. Did you read Milnor's book on Morse theory? $\endgroup$
    – Misha
    Commented Mar 22, 2013 at 13:53

1 Answer 1

Reset to default
4
$\begingroup$

This question was answered by Misha in a comment.

Nothing useful can be said about the handle decomposition of $N$, because, among other things, if an $n$-manifold $N$ has boundary $M$, then the connect-sum of $N$ with any closed $n$-manifold still has boundary $M$. For example, given an arbitrary closed $5$-manifold $N$, we may remove a small $4$-ball to obtain a manifold with boundary $S^4$.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .