Talk:The Multivariable Alexander Polynomial: Difference between revisions
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:Dror doesn't understand the multivariable Alexander polynomial well enough to give simple topological reasons for the vanishing of the said polynomial for these knots. |
:Dror doesn't understand the multivariable Alexander polynomial well enough to give simple topological reasons for the vanishing of the said polynomial for these knots. |
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The multivariable Alexander polynomial is zero precisely when H1 of the universal abelian cover has non-zero rank (as a module over the group-ring of covering transformations). Equivalently, if H2 of the universal abelian cover is non-trivial. In this case, H2 is free on one generator, which is represented by a map of a genus 2 surface into the link complement. So far I haven't found a very appealing description of this surface, but it's there... |
The multivariable Alexander polynomial is zero precisely when H1 of the universal abelian cover has non-zero rank (as a module over the group-ring of covering transformations). Equivalently, if H2 of the universal abelian cover is non-trivial. In this case, H2 is free on one generator, which is represented by a map of a genus 2 surface into the link complement. So far I haven't found a very appealing description of this surface, but it's there... -ryan budney |
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Revision as of 14:12, 13 April 2007
Regarding this comment:
- Dror doesn't understand the multivariable Alexander polynomial well enough to give simple topological reasons for the vanishing of the said polynomial for these knots.
The multivariable Alexander polynomial is zero precisely when H1 of the universal abelian cover has non-zero rank (as a module over the group-ring of covering transformations). Equivalently, if H2 of the universal abelian cover is non-trivial. In this case, H2 is free on one generator, which is represented by a map of a genus 2 surface into the link complement. So far I haven't found a very appealing description of this surface, but it's there... -ryan budney
