Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
add comment

2 Answers 2

up vote 8 down vote accepted

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.

share|improve this answer
    
Thanks, Allan! I've found their papers. –  algori Dec 21 '09 at 2:29
add comment

I don't understand the answer. Why a manfold without nontrivial action cannot a branch cover of $S^1\timesS^2$? Compare it with that every 3-manfold is a 3-fold branch cover of $S^3$.

share|improve this answer
    
It's because the question asks for a 2-fold cover, which would have to be regular. –  Richard Kent May 11 '10 at 14:59
    
I see. Thank you. –  Wolffo May 11 '10 at 15:09
    
Then is the conjecture true when restricting the condition on manifold with $\mathbbZ$ action? –  Wolffo May 11 '10 at 15:41
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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