7
$\begingroup$

I need a reference (or a short proof) for the following statement:

Suppose a closed manifold $N$ is the result of a surgery (along an embedded sphere) on a closed manifold $M$. Then the difference $\sum dim H_i(N) - \sum dim H_i(M)$ (the homology is taken with coefficients in a field) is at most 2.

$\endgroup$

1 Answer 1

12
$\begingroup$

To say that a smooth, closed manifold $N$ is obtained by surgery along a (framed) sphere in $M$ is to say that there is a cobordism $P$ from $M$ to $N$ and a Morse function $f\colon P\to [0,1]$, with $f^{-1}(0)=M$, $f^{-1}(1)=N$, and exactly one critical point $c$. By Morse theory, $H_*(P,M)$ is then 1-dimensional, generated by the descending disc of $c$. Likewise, $H_*(P,N)$ is 1-dimensional, generated by the ascending disc of $c$.

By the homology exact sequence of the pair $(P,M)$, $\dim H_*(M)$ differs from $\dim H_*(P)$ by $1$. By the homology exact sequence of the pair $(P,N)$, $\dim H_*(N)$ also differs from $\dim H_*(P)$ by $1$. Hence $|\dim H_*(M) - \dim H_*(N)|$ is $0$ or $2$.

$\endgroup$
1
  • $\begingroup$ Thank you, Tim! It is exactly what I need. My proof is longer - I considered pairs like $(M,M\S^{\lambda})$ and it requires to study cases... $\endgroup$
    – Petya
    Feb 17, 2010 at 23:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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