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Jun 18, 2017 at 6:00 comment added Ali Taghavi Thank you very much for your answer. I try to understand the detail.
Jun 16, 2017 at 23:36 comment added mme I guess you could just as well verify that $T^1 S^n$ never arises as an $(n-2)$-surgery on $S^{2n-1}$, which should be elementary calculation once you know there aren't very many knots.
Jun 16, 2017 at 22:56 history answered mme CC BY-SA 3.0