The knot group of a knot $K$ is the fundamental group of $\mathbb R^3 \smallsetminus K$; that is, the set of possibly self-crossing closed paths (starting and ending at any single point in space) which you can take through real-space which are not equivalent to each other if you forbid passing through the locus of the knot; and where you can compose the paths in the natural way. For instance, the knot group of the un-knot is just $\mathbb Z$, corresponding to the number of times that a path winds about a cycle in space.

We can consider similar questions for links ("knots" having more than one closed loop) as for knots of a single component: given a link, we can consider the group of closed paths in space which avoids crossing either of the two components.

I have a friend who has a tattoo of a knot (more precisely, a link of two components) in the shape of a maple leaf; we were curious what properties it had. A presentation of the link is given below. How would one obtain a presentation of the link group? Does the link group of this particular link correspond to any relatively nice group?

Maple leaf link

  • $\begingroup$ Have you considered asking SnapPea? $\endgroup$
    – user641
    Apr 10, 2012 at 20:00
  • $\begingroup$ I asked it - the group has a nasty presentation, with three relators of length 200. $\endgroup$
    – user641
    Apr 10, 2012 at 20:24
  • $\begingroup$ @SteveD: Should I take this to mean that it is quite likely that there's no reasonable approach to determine it by hand? That is, is SnapPea usually very good at yielding concise answers when they exist? $\endgroup$ Apr 10, 2012 at 22:48
  • 1
    $\begingroup$ You could compute a presentation by hand using the Wirtinger presentation for a knot group. However, the presentation would involve 16 generators (one for each strand in the picture) and 16 relations (one for each crossing), and there is no reason to think it could be simplified considerably. Indeed, Steve's SnapPea results seem to indicate that the knot group is unlikely to be anything particularly simple. $\endgroup$
    – Jim Belk
    Apr 10, 2012 at 23:06
  • $\begingroup$ @NieldeBeaudrap: Yes, SnapPea (and I also checked with GAP) does a very good job of simplifying relations via Tietze transformations. I don't think it always gets the simplest presentation, but I would take the results I got as an almost certain indication this group is complicated. $\endgroup$
    – user641
    Apr 10, 2012 at 23:46

2 Answers 2


I plugged this into SnapPy and got the following presentation:

Generators: a,b,c,d

Relators: abCBACADadcDAdabCBADadCDAdadcDAdabcBADadCDAdacabcBADadCBCbcDAdabCBACdcDadCDcabcBADadCBcbcDAd, abCBCbDAdacACADadBcbcDAd, abCBACdcDAdCDcDadCDADadcDAdabCBADadCDAdadcDAdCdcDadCDcabcBADadCBcbcDAd

(That is, four generators, three relators.) The link complement has volume approximately 33.331392948. Since the link is alternating and has no flypes, the diagram shown is minimal - thus the crossing number is 16. One more thing - the diagram has an involution about the central vertical line. You could mod out by this involution to get a smaller orbifold - there might be something known about the quotient...

  • $\begingroup$ Looks like a $C^{\prime}(1/6)$ presentation! (I can see no piece of length greater than $4$). If I am bored, I may verify this...but I'd have to be pretty bored... $\endgroup$
    – user1729
    May 27, 2013 at 11:51
  • $\begingroup$ @user1729 - If it was $C'(1/6)$ then I believe that the group would be word hyperbolic. However link groups contain copies of $\mathbb{Z}^2$, so are never word hyperbolic. $\endgroup$
    – Sam Nead
    May 28, 2013 at 8:34
  • $\begingroup$ Oh. I didn't know that about link groups. Interesting. $\endgroup$
    – user1729
    May 28, 2013 at 8:57
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    $\begingroup$ The $\mathbb{Z}^2$ comes from the so-called "peripheral torus". Take a small neighborhood of a component of the link (to get a solid torus) and take the boundary of that (to get a two-torus). The fundamental group of that two-torus injects into the fundamental group of the link complement. $\endgroup$
    – Sam Nead
    Jun 2, 2013 at 11:10

Using appcontour I get the following Alexander polynomial:


Defined up to a change of base of $Z^2$

The second Alexander ideal is interesting, it is generated by $<1133, v^2 - 531 v + 1, v - 10 u + 481 uv>$ (obtained using a combination of appcontour and Macaulay2)

The third Alexander ideal is the whole ring.


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