3
$\begingroup$

The fundamental group of a knot $K$ (otherwise known as the knot group) is the fundamental group of the knot complement $S^{3} \backslash K $ in $S^{3} $.

In "Virtual Knots: The State of the Art" (http://books.google.com.au/books?id=WaCJ_-MpdBYC) on page 7, it says that the fundamental group of a knot recognizes prime knots. Does this mean that the knot group of a prime knot cannot be isomorphic to the amalgamated free product of two non-trivial knot groups- and in general can the knot group of a prime knot be an amalgamated free product?

Improve this question
$\endgroup$
1
  • $\begingroup$ To be somewhat pedantic, the fundamental group $\pi := \pi_1(S^3 \setminus K)$ itself doesn't determine the knot complement, since mirror knot complements have isomorphic fundamental groups. However, the data $(\pi, m, l)$ does determine the knot (where $m$ and $l$ are the meridian and longitude of the knot). See mathoverflow.net/questions/35680 $\endgroup$
    – Peter Samuelson
    Commented Jul 10, 2013 at 15:50

1 Answer 1

Reset to default
7
$\begingroup$

The statement on page 7 is referring to Corollary 2.1 of Gordon and Luecke's "Knots are determined by their complements."

Also a prime knot can not be an amalgamated free product of non-trivial knot groups by Corollay 2.5 of W. Thurston's "Three dimensional manifolds, Kleinian groups and hyperbolic geometry."

However, a knot group can be an amalgamated free product provided the subgroup you are amalgamating along is more complicated. For example, a knot complement with an essential, embedded quasi-Fuchsian surface has such a knot group. The basic idea of how to construct such knots goes as follows, take a genus $g$ handlebody and embed a knot in it such that the knot punctures any compressing disk in at least two points, then embedded the handlebody in $S^3$ such that the boundary of the handlebody is incompressible. This process is done in full rigor by Adams and Reid in "Quasi-Fuchsian surfaces in hyperbolic knot complements."

Improve this answer
$\endgroup$
2
  • $\begingroup$ The link for Adams and Reid seems not to work. $\endgroup$
    – Dietrich Burde
    Commented Jul 10, 2013 at 11:21
  • $\begingroup$ @DietrichBurde: Thank you. I fixed the broken link. $\endgroup$
    – Neil Hoffman
    Commented Jul 10, 2013 at 13:31

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .