Not every orientable 3-manifold is a double cover of $S^3$ branched over a link. For example, the 3-torus isn't. However, in 1975 Montesinos conjectured (Surjery on links and double branched covers of $S^3$, in: "Knots, groups and 3-manifolds", papers dedicated to the memory of R. Fox) that every orientable 3-manifold is a double branched cover of a sphere with handles i.e. the connected sum of a certain number of copies of $S^1\times S^2$ (this number can be zero, in which case we get $S^3$). Notice that this time $T^3$ does not provide a counter-example since if we take the quotient of $T^3$ by the involution $(x,y,z)\mapsto (x^{-1}, y^{-1},z)$ we get $S^2\times S^1$.

I was wondering what the status of this conjecture is.


1 Answer 1


It is false. For example, there are closed, orientable, aspherical 3-manifolds that admit no nontrivial action of a finite group whatsoever. The first examples were due to F. Raymond and J. Tollefson in the 1970s, I believe.

  • $\begingroup$ I don't understand the answer. Why a manifold without nontrivial action cannot a branch cover of $S^1\times S^2$? Compare it with that every 3-manifold is a 3-fold branch cover of $S^3$. $\endgroup$
    – Wolffo
    May 11, 2010 at 14:53
  • 1
    $\begingroup$ It's because the question asks for a 2-fold cover, which would have to be regular. $\endgroup$ May 11, 2010 at 14:59
  • $\begingroup$ Then is the conjecture true when restricting the condition on manifold with $\mathbb{Z}$ action? $\endgroup$
    – Wolffo
    May 11, 2010 at 15:41

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