most recent 30 from http://mathoverflow.net 2013-05-20T22:26:25Z http://mathoverflow.net/feeds/question/97506 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/97506/rational-homology-spheres-and-knots Juan OS 2012-05-20T20:45:50Z 2012-05-21T02:15:50Z

<p>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:</p> <ol> <li>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?</li> <li>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? </li> </ol> <p>Many thanks! </p>

http://mathoverflow.net/questions/97506/rational-homology-spheres-and-knots/97519#97519 Daniel Moskovich 2012-05-21T02:15:50Z 2012-05-21T02:15:50Z

<p>For Question 1, I believe that the answer follows from:</p> <p>Montesinos, José M. <i>Surgery on links and double branched covers of $S^3$.</i> 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).</p> <p>Namely:</p> <blockquote> 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 $S^3$. </blockquote> <p>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 <em>small</em>.</p>