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Jul 26, 2012 at 10:07 comment added HJRW Mark - yes, I'm aware of that. If you look at the UPDATE in my answer, you'll see that I assert that 'there may well be examples of such maps with no simple loops in the kernel'. I also suggest somewhere to look for examples.
Jul 24, 2012 at 15:00 comment added user6976 @Henry: The question in fact has nothing to do with 3-manifolds. What one needs is to kill long and complicated self-intersecting loops and at the same time kill the first homology by imposing relations of the form $x=$ long commutator where $x$ is a standard generator of the surface group. The goal is to produce a quotient with no simple loops in the kernel. I think it is quite doable and the example should exist. I am not sure now about the number of generators ($\le 3$) of the quotient, but that also should be possible to achieve with some effort.
Aug 23, 2011 at 15:02 comment added Thom Thanks Henry! I am convinced that the answer to my question is unknown.
Aug 23, 2011 at 0:54 vote accept Thom
Jul 20, 2012 at 23:43
Aug 22, 2011 at 19:03 comment added HJRW A quick search of MathSciNet shows that the Simple Loop Conjecture holds for Seifert-fibred 3-manifolds and, more generally, graph manifolds. As there are very complicated hyperbolic 3-manifolds with trivial H_1 and two generators, I would be surprised if those assumptions help much.
Aug 22, 2011 at 16:06 comment added Sam Nead @Thom - The simple loop conjecture holds for three-manifolds of the form surface cross interval, by work of Gabai.
Aug 22, 2011 at 14:36 comment added Thom Thanks! Are there cases of $3$-manifolds (with infinite $\pi_1)$) for which we know Simple Loop Conjecture holds? Would it be helpful if we assume that the group $G$ has three (or less) generators and abelianization of rank 0.
Aug 22, 2011 at 14:26 vote accept Thom
Aug 22, 2011 at 18:12
Aug 22, 2011 at 12:37 history edited HJRW CC BY-SA 3.0
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Aug 22, 2011 at 12:16 history answered HJRW CC BY-SA 3.0