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Timeline for Alexander invariant of torus knot

Current License: CC BY-SA 4.0

17 events
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Jul 30, 2022 at 2:43 comment added Gerry Myerson Some time ago, an edit was made that eliminated the (1), (2), (3), (4) tags but left in a reference to them.
May 9, 2022 at 14:20 history edited YCor CC BY-SA 4.0
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May 9, 2022 at 13:27 answer added Giacomo Bascapè timeline score: 5
May 6, 2022 at 22:33 comment added Giacomo Bascapè I still don't see solution: I can't compute the Smith Normal form with the natural algorithm because Z[t,t^-1] is not a PID
May 18, 2014 at 16:53 review Close votes
May 19, 2014 at 18:01
May 18, 2014 at 16:38 comment added Ryan Budney I like Ian's suggestion. One way to do it efficiently would be to apply Reidemeister-Schrier to your presentation of the fundamental group.
May 11, 2014 at 1:49 history edited Jacob.Z.Lee CC BY-SA 3.0
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May 11, 2014 at 1:26 comment added Jacob.Z.Lee yeah. We can get the result by other methods ,like free calculus .
May 11, 2014 at 1:19 comment added Jacob.Z.Lee I want to get the relation $\Delta(t)\alpha=0$ by eliminate the gennerator $\beta$.So we get the $\Lambda-$module $H_1(\tilde{X})\cong(\alpha\mid \Delta(t)\alpha=0)\cong \Lambda/(\Delta(t))$
May 10, 2014 at 18:19 comment added Dietrich Burde See mathoverflow.net/questions/129717/….
May 10, 2014 at 17:36 comment added Ian Agol Just a remark: torus knot complements are mapping tori of finite-order automorphisms, from which one may deduce the Alexander polynomial by taking the characteristic polynomial.
May 10, 2014 at 13:15 history edited Jacob.Z.Lee CC BY-SA 3.0
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May 10, 2014 at 5:56 comment added Alex Degtyarev Convert the matrix to its Smith normal form. In this particular case, this is doable over the integers.
S May 10, 2014 at 3:47 history suggested gaoxinge CC BY-SA 3.0
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May 10, 2014 at 3:18 review Suggested edits
S May 10, 2014 at 3:47
May 10, 2014 at 1:36 history edited Jacob.Z.Lee CC BY-SA 3.0
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May 10, 2014 at 1:30 history asked Jacob.Z.Lee CC BY-SA 3.0