Ramified cover of 3-ball - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T19:45:05Z http://mathoverflow.net/feeds/question/53809 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/53809/ramified-cover-of-3-ball Ramified cover of 3-ball Anton Petrunin 2011-01-30T18:35:08Z 2011-02-07T02:27:53Z <p>I it true that any connected oriented 3-manifold with nonempty boundary can be obtained as a ramified cover of 3-ball with a ramification at a <em>link</em>? </p> <ul> <li>link = a 1-dimensional submanifold with possibly nonempty boundary.</li> </ul> <p>If answer is "YES", can we choose in addition the restriction of the covering at the boundary?</p> http://mathoverflow.net/questions/53809/ramified-cover-of-3-ball/54057#54057 Answer by Agol for Ramified cover of 3-ball Agol 2011-02-02T03:26:47Z 2011-02-02T03:26:47Z <p><a href="http://www.ams.org/journals/tran/1979-247-00/S0002-9947-1979-0517687-9/home.html" rel="nofollow">Berstein and Edmonds</a> prove in Cor. 6.3 that for an orientable 3-manifold $W$ with connected boundary, with a branched cover $\varphi: \partial W\to S^2$ of degree $n>3$, then there is a branched cover $\Phi: W\to D^3$ such that $\Phi_{|\partial W}=\varphi$. In <a href="http://www.ams.org/mathscinet-getitem?mr=553345" rel="nofollow">another paper</a>, Edmonds claims in Theorem 2.1 that Cor. 6.3 extends to maps $f: W\to D^3$ such that the boundary map is a branched cover of the same degree as $f$ (<em>allowable</em>). One can easily construct an allowable map $f:W\to D^3$ by mapping $\partial W$ to $S^2$ by a branched cover so that each component of $\partial W$ has positive degree $>2$ (with respect to the orientation induced by $W$), and extend to all of $W$ by coning off. Theorem 2.1 implies that this map is homotopic to a branched cover.</p> <p>The hypothesis of degree $>2$ is necessary, since for example if one has a knot $K\subset S^3$ which is not (strongly) <a href="http://en.wikipedia.org/wiki/Invertible_knot#Non-invertible_knots" rel="nofollow">invertible</a>, then $M=S^3-\mathcal{N}(K)$ is a manifold with torus boundary such that there is a degree 2 map $T^2=\partial M\to S^2$ which is the quotient of the elliptic involution, but which doesn't extend over $M$. </p>