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 link?

  • link = a 1-dimensional submanifold with possibly nonempty boundary.

If answer is "YES", can we choose in addition the restriction of the covering at the boundary?

  • 5
    $\begingroup$ For the second part, the answer will be no in general. If one has a boundary which is a torus, and you require that the covering map be the elliptic involution, then the map won't extend in general over the 3-manifold, since this would extend to an involution on the manifold, which in general doesn't exist. $\endgroup$ – Ian Agol Jan 30 '11 at 19:09
  • 4
    $\begingroup$ See Cor. 6.3 of ams.org/journals/tran/1979-247-00/S0002-9947-1979-0517687-9/… $\endgroup$ – Ian Agol Jan 30 '11 at 19:31
  • $\begingroup$ @Agol, Thank you so much, your ref gives more than I wanted :) $\endgroup$ – Anton Petrunin Jan 30 '11 at 20:14

Berstein and Edmonds 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 another paper, 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 $f$ (allowable). 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.

The hypothesis of degree $>2$ is necessary, since for example if one has a knot $K\subset S^3$ which is not (strongly) invertible, 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 $M$.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.