Publications Meetings The Profession Membership Programs Math Samplings Policy & Advocacy In the News About the AMS

Remote Access
Green Open Access
Proceedings of the American Mathematical Society
Proceedings of the American Mathematical Society
ISSN 1088-6826(online) ISSN 0002-9939(print)


How many knots have the same group?

Author: Jonathan Simon
Journal: Proc. Amer. Math. Soc. 80 (1980), 162-166
MSC: Primary 57M25
MathSciNet review: 574528
Full-text PDF Free Access

Abstract | References | Similar Articles | Additional Information

Abstract: Let K be a knot in $ {S^3},G = {\pi _1}({S^3} - K),n = $ number of prime factors of $ K,\nu (G) = $ number of topologically different knot-complements with group G and $ \kappa (G) = $ number of distinct knot types with group G.

Theorem. If K is prime, then $ \nu (G) \leqslant 2$. If $ n \geqslant 2$, then $ \nu (G) = \kappa (G) \leqslant {2^{n - 1}}$. For each $ n \geqslant 2$, the bound $ {2^{n - 1}}$ is the best possible. For K prime, we still have the conjecture $ \nu (G) = \kappa (G) = 1$. If K is a cable-knot, then $ \kappa (G) \leqslant 2$.

References [Enhancements On Off] (What's this?)

Similar Articles

Retrieve articles in Proceedings of the American Mathematical Society with MSC: 57M25

Retrieve articles in all journals with MSC: 57M25

Additional Information

PII: S 0002-9939(1980)0574528-5
Article copyright: © Copyright 1980 American Mathematical Society

Comments: Email Webmaster

© Copyright , American Mathematical Society
Contact Us · Sitemap · Privacy Statement

Connect with us Facebook Twitter Google+ LinkedIn Instagram RSS feeds Blogs YouTube Podcasts Wikipedia