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By Lickorish and Wallace , any closed,connected, orientable 3-manifold can be gotten as a surgery on a link in $S^3$. Let say our manifold, M, is an integer homology sphere and L = $ L_1 \cup L_2 \cup \dots L_n $ be one such link in $ S^3$ for $M$, with specified surgery coefficient . Now if we do subsequent surgeries one by one on each of the link component (starting with $L_1$), we get a new manifold $M_k$ such that $M$ can be gotten from $M_k$ and link $ L^k = L_k^ \prime \cup \dots L_n ^ \prime $ in $M_k$ with appropriately changed surgery coefficient. My question is : Can we make sure that each of this $M_k$'s are integer homology spheres, when we know that $M$ is such?

My Guess is : Yes, by applying Kirby Calculus appropriately. Is it true?

Subhankar D.
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Yes, each integral homology sphere $M$ has a surgery description $L=L_1\cup\cdots\cup L_n$ such that the intermediate manifolds $M_k$ are all integral homology spheres.

In Refined Kirby calculus for integral homology spheres, Habiro remarks "It is well known that every integral homology sphere can be expressed as the result from $S^3$ of surgery along a framed link of diagonal linking matrix with diagonal entries $\pm 1$" (page 2 of the linked paper). Such a surgery description yields intermediate manifolds $M_k$ that are integral homology spheres (as noted on the Wikipedia page for homology spheres).

So I've given you an affirmative answer, but I haven't provided you the proof (because I don't know it). Hopefully this points you in the right direction.

Adam Lowrance
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Here is a proof using Mapping class group.

You can costruct homology sphere from elements of Torelli groups. Basically consider a Heegard splitting of $S^3$ and then reglue those solid handle bodies with precomposing by an element of Torelli group. And since elelemt of Torelli group doesn't affect the homology, it will give you a homology sphere.

It is well known that all homology spheres arise in this way. Morita proved that all homology sphere can be generated from elelemnts of Torelli group generated by Dehn-twist along separating curves.

Now follow the strategy of proof of Lickorish and Wallace the way they proved that every closed oriented three manifold can be constructed from $S^3$ by doing Dehn surgeries along links ...Here the set of link will be the generating set of separating simple closed curves and then at each step you will get homology sphere since you are doing dehn surgery along separating curves so it won't affect the homology. And in the end you will get the desired homology sphere.

Anubhav Mukherjee
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  • Can you please state the statement of Morita's theorem? – Subhankar D. Mar 03 '18 at 21:51
  • @SubhankarD. In that paper he mainly proved the statement which I mentioned in the 2 nd paragraph. I think the actual statement is a bit more complicated in terms of Johnson's homomorphism. If you want I can send you the paper. Are you convinced with the proof? – Anubhav Mukherjee Mar 04 '18 at 01:52
  • Not really! In your first paragraph itself, I think you'll get back S^3 itself ,since action of an element of torelli group gets you back the same curve in the homology of a heegaard surface ! So, you don't actually get any new homology sphere , but S^3 itself! Isn't it? – Subhankar D. Mar 04 '18 at 04:01
  • @SubhankarD. Torelli group action doesn't fix any curve. – Anubhav Mukherjee Mar 04 '18 at 04:08
  • It's a trivial action on the homology. That doesn't imply it fixed curves. – Anubhav Mukherjee Mar 04 '18 at 04:09
  • Yes, I understand.But see, if you just look at the simplest Heegaard decomposition of S^3 i.e the one with genus one heegaard surface and longitude and meridian being the attaching curves , then does acting on an element of the torelli group gives back anything new? I found a paper that said that dehn twist on separating curves and bounding pair maps generate torelli group and both of these doesn't change the attaching curves,isn't it? I've very small knowledge in Mapping class group, so let me know where I'm thinking wrong.Thanks. – Subhankar D. Mar 04 '18 at 04:20
  • @SubhankarD. Here I am not restricting genus... genus one case is the trivial... the theorem says that you can constructed homology sphere from torelli element when you allowed to vary your genus. – Anubhav Mukherjee Mar 04 '18 at 04:22
  • That is, given somne homology sphere there exists some heegard g- splitting of S^3 and some torelli elemet from that g-genus surface such that , regluing that will give you the resultant manifold. – Anubhav Mukherjee Mar 04 '18 at 04:24
  • Torelli group generated by bounding pair of maps, but what Morita proved that, for constructing homology sphere all you need is torelli element genberated by sperating curves. – Anubhav Mukherjee Mar 04 '18 at 04:25
  • A link of the paper or the results you said will be appreciated. Thanks – Subhankar D. Mar 04 '18 at 04:51
  • @subhankar I'm sorry..I don't know any link. But I've read that paper sometimes back. And as I said the paper used more technical terms like Johnson homomorphism and all. – Anubhav Mukherjee Mar 04 '18 at 04:54
  • But you can ask me questions, I can clear your doubts. – Anubhav Mukherjee Mar 04 '18 at 04:59
  • Okay; Please share the name of the paper (or its arxiv link, may be) if you find it sometimes. I'd like to understand that paper. Thanks – Subhankar D. Mar 04 '18 at 05:01
  • I may email you... Prof Dan Margalit send me that paper once. I don't have my laptop with me now. But if you believe that result, is my proof make sense to you now? – Anubhav Mukherjee Mar 04 '18 at 05:06
  • Let us continue this discussion in chat. – Subhankar D. Mar 04 '18 at 05:09
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This follows rather immediately from the classification of indefinite unimodular quadratic forms. Recall such a form is classified by its signature and parity. In particular any odd indefinite form is equivalent to $I^n \oplus -I^m$ (where $I^n$ is the form corresponding to the identity matrix of rank $n$). For $X$ a 4-manifold which is just 2-handles attached to the ball a handle slide of the component $\alpha$ over the component $\beta$ has the effect on $H_2$ of replacing $[\alpha]$ with $[\alpha] \pm [\beta]$ (depending on the orientations of the band move). Since $SL(n,\Bbb Z)$ is generated by such elementary matrices (ones which are a diagonal except a $\pm 1$ at one coordinate), from any 2-handlebody with indefinite and odd intersection form we can get a handlebody structure with the intersection form presented as $I^n \oplus -I^m$.

So all you need to is blowup once to force your intersection form to be odd and indefinite, and then realize the transformation diagonalizing your form as a sequence of handleslides.

PVAL-inactive
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  • I think I do understand your argument. So, after getting a link-surgery description of an integer homology sphere, you look at the intersection matrix (i.e linking numbers in the off diagonals and framings in the diagonals) and it has to be of +1 or -1 determinant , hence it represents an unimodular quadratic form, as you mentioned and then one follows your argument. Isn't that right!? – Subhankar D. Mar 04 '18 at 04:13
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    @SubhankarD. yes that's correct. – PVAL-inactive Mar 04 '18 at 04:18