Does the coefficient of the meridian determine the coefficient of the longitude?(on Dehn surgery) - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T02:06:03Z http://mathoverflow.net/feeds/question/93849 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/93849/does-the-coefficient-of-the-meridian-determine-the-coefficient-of-the-longitude Does the coefficient of the meridian determine the coefficient of the longitude?(on Dehn surgery) HJ 2012-04-12T10:44:28Z 2012-04-12T17:32:35Z <p>I'm studying Dehn surgery, and it says that the coefficient $(p,q)$ which says how the meridian curve on solid torus is attached will determine the entire resulting manifold. I'm wondering whether the coefficient $(p,q)$ of the meridian also determine the coefficient for longitude. If not, then can any $(r,s)$ with $ps-qr=\pm 1$ be the coefficint for the longitude?</p> http://mathoverflow.net/questions/93849/does-the-coefficient-of-the-meridian-determine-the-coefficient-of-the-longitude/93857#93857 Answer by Kevin Walker for Does the coefficient of the meridian determine the coefficient of the longitude?(on Dehn surgery) Kevin Walker 2012-04-12T11:35:53Z 2012-04-12T17:32:35Z <p>Yes, the longitude of the Dehn surgery solid torus can go to any $(r, s)$ such that the 2x2 determinant $ps-qr = \pm 1$. This is because the various choices for the longitude all differ by homeomorphisms of the torus which extend to homeomorphisms of the solid torus. </p> <p>More generally, consider gluing together two manifolds $M$ and $N$ along a homeomorphism $f:\partial M \to \partial N$. Let $h:\partial N\to \partial N$ be a homeomorphism which extends to a homeomorphism $h':N\to N$. Then gluing via $f$ and gluing via $h\circ f$ yield homeomorphic manifolds.</p>