# Rational homology spheres and knots

It is known that the boundary of a branched double cover of the four ball branched over a surface bounded by a knot in is a rational homology $3$-sphere. Two questions come to my mind simply out of curiosity:

1. Which rational homology $3$-spheres arise this way? By this I mean, is this set large or small (in the most vague terms, nothing formal) in the set of R.H.3S.'s? Is there an invariant capable of detecting when a RH3S arises this way?
2. If we now substitute usual knots for embeddings $\mathbb{S}^2 \hookrightarrow \mathbb{S}^4$ (knotted spheres), and look at the branched double covers of the $5$ ball branched over a $3$-manifold bounded by the knotted sphere, is their boundary a RH4S?

Many thanks!

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Your question is phrase in an odd way: you only need to refer to the branched cover of $S^3$ over a knot, the branched cover over the 4-ball is irrelevant. Similar for part 2. – Ian Agol May 21 '12 at 6:09
Hi Agol, I phrased the question that way because I was trying to connect this somehow to the "differential" part of the topology. From this point of view the DBCs on the four ball are relevant as from them arises a bound to the slice genus by the knot signature. I am not sure if this is good enough reason for you, if not I can always edit the question. – Juan OS May 21 '12 at 10:07

Montesinos, José M. Surgery on links and double branched covers of $S^3$. Knots, groups, and 3-manifolds (Papers dedicated to the memory of R. H. Fox), pp. 227–259. Ann. of Math. Studies, No. 84, Princeton Univ. Press, Princeton, N.J., (1975; MR0380802).
A $\mathbb{Q}$HS arises as a double branched cover of a knot in $S^3$ (equivalent to your construction) if and only if it is obtained by (rational) surgery on a strongly invertible link in $S^3$.
This is something you can detect, either by Casson-Walker invariants (see papers by Chbili- you will recall that these have surgery presentations and therefore take a rather special form if the surgery link is strongly invertible), or more directly, using result of Meeks-Yau and of Thurston to reduce the problem to one of identifying isometries on geometric manifolds, and then checking whether the whole JSJ structure admits a $\mathbb{Z}/2\mathbb{Z}$ action. This is a very strong condition of course, so the set of $\mathbb{Q}HS$s which arise via this construction is small.