Skip to main content
9 events
when toggle format what by license comment
Feb 11, 2021 at 21:09 comment added Daniel Moskovich @Ian Agol It was 10 years ago and I don't remember what I meant. 2021 Daniel wishes 2011 Daniel had made good on his word and updated the answer!
Feb 11, 2021 at 18:24 comment added Ian Agol @DanielMoskovich: is the point here that the the T-L signatures and Minkowski units are concordance invariant, and hence one can reconstruct the Blanchfield pairing, and hence S-equivalence class, from the Alexander module and concordance class invariants? At least, they seem to be concordance invariant according to this webpage: encyclopediaofmath.org/wiki/Knots_and_links,_quadratic_forms_of
Apr 29, 2011 at 15:40 comment added Daniel Moskovich @Andy I answered when I learnt more... actually, this answer should be updated, because I know a lot more now- maybe I will, when I get the energy. The T-L signatures and Minkowski units indeed turn out to be enough.
Apr 29, 2011 at 15:35 comment added Andy Putman I just noticed via the moderation tools that someone had flagged this as spam. Since we just had a disagreement about voting to close, I want to emphasize that I had nothing to do with this (it should go without saying, but I want to make sure there is no confusion). As to why someone (not me) downvoted it, my only guess is that it is considered a bit gauche to answer your own questions.
Apr 29, 2011 at 12:27 comment added Daniel Moskovich Why was this downvoted?
Jan 1, 2010 at 22:05 comment added Ryan Budney Trotters's "On S-equivalence of Seifert Matrices" (1973) shows S-equivalence is the same as isometry of the Blanchfield pairing. In the same paper he composes the Blanchfield pairing with what's now called the "Trotter trace" to show that rationally Blanchfield pairing classification reduces to the classification of a skew-symmetric rational-valued bilinear form on the Alexander module. From here the reference is Milnor's "On Isometries of inner product spaces". Scanning through the paper I think you might need real coefficients for the classification.
Dec 31, 2009 at 0:46 comment added Daniel Moskovich Can you give a reference for that? Integrally, I expect some sort of Minkowski units would be needed, but I don't know whether even that is enough.
Dec 30, 2009 at 22:53 comment added Ryan Budney I think it was Trotter that showed the T-L signatures (equivalently the Milnor signatures) determine the blanchfield pairing rationally. But integrally I doubt the answer is that clean.
Dec 9, 2009 at 8:21 history answered Daniel Moskovich CC BY-SA 2.5