Heegaard splitting of covering hyperbolic manifold. - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T16:34:48Z http://mathoverflow.net/feeds/question/86800 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/86800/heegaard-splitting-of-covering-hyperbolic-manifold Heegaard splitting of covering hyperbolic manifold. yanqing 2012-01-27T07:27:40Z 2012-04-04T02:19:32Z <p>I am curious about how the Heegaard genus changes after a finite covering. </p> <p>Is there anyone constructing an closed hyperbolic 3-manifold \$N\$ such that </p> <p>the Heegaard genus of a finite covering of \$N\$ is less than the Heegaard genus of \$N\$? </p> <p>Thank you!</p> <p>Note: Heegaard genus of a 3-manifold means the minimal genus of all Heegaard splittings.</p> http://mathoverflow.net/questions/86800/heegaard-splitting-of-covering-hyperbolic-manifold/86831#86831 Answer by Michael Siler for Heegaard splitting of covering hyperbolic manifold. Michael Siler 2012-01-27T15:05:32Z 2012-01-27T15:05:32Z <p>There are examples like this. Check out section 4.5 of Shalen's paper "Hyperbolic volume, Heegaard genus and ranks of groups." It's here: <a href="http://arxiv.org/abs/0904.0191" rel="nofollow">http://arxiv.org/abs/0904.0191</a></p> <p>He gives a reference for a genus 3 example by Alan Reid and a sketch of a technique for producing examples by Hyam Rubinstein. Shalen also conjectures that the genus can drop by at most 1 in a finite cover of a closed hyperbolic 3-manifold.</p> http://mathoverflow.net/questions/86800/heegaard-splitting-of-covering-hyperbolic-manifold/93062#93062 Answer by Yo'av Rieck for Heegaard splitting of covering hyperbolic manifold. Yo'av Rieck 2012-04-04T02:19:32Z 2012-04-04T02:19:32Z <p>Hyam Rubinstein and me have results about the behavior of the Heegaard genus under double covers for non-Haken manifolds, see <a href="http://arxiv.org/abs/math/0607145" rel="nofollow">http://arxiv.org/abs/math/0607145</a>. Essentially, we show that the Heegaard genera of the two manifolds bound each other linearly. (The statement is a little more complicated for branched covers.)</p>