Integer homology sphere as subsequent surgery on integer homology spheres

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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.