I am trying to understand how the Knot Group sits inside of knot theory.

I am giving a talk on a short paper I read in the subject, and I would like to motivate it a little. Not being well read in this matter, my naive assumptions are these:

  1. Knot Theory (mostly) cares about the classification of knots - being able to tell two of them apart.
  2. The Knot Group is an invariant, hence different knot groups imply different knots.

But from these basic assumptions I'm not sure how to answer the question "Why do people study homomorphisms going out of knot groups?"

I could give an answer like "because its interesting" but that feels unsatisfying. I would appreciate any insight on this.


Even a person interested solely in classifying knots should be interested in studying homomorphisms going out of knot groups, for the following reasons.

Suppose you have a knot $K$ that you suspect is not equivalent to the trivial knot. How would you prove it?

Since the fundamental group of the complement of a trivial knot is an infinite cyclic group, one method would be to prove that the group $\pi_1(S^3-K)$ is not infinite cyclic. How would you do that?

As an example, let $K \subset S^3$ be the figure eight knot. One reason $K$ is not a trivial knot is because there is a surjective homomorphism from the group $\pi_1(S^3-K)$ onto the dihedral group of order $10$. This is an exercises about the knot group that might be found in knot theory books, and I will leave it to you to look this up, or to figure it out using some presentation of $\pi_1(S^3-K)$ such as the Wirtinger presentation. This suffices to prove that the group $\pi_1(S^3-K)$ is not infinite cyclic, because the only groups onto which an infinite cyclic group can surject are other cyclic groups; but the dihedral group of order $10$ is not cyclic, it is not even abelian.

  • $\begingroup$ I thought the Trefoil has a homomorphism onto the Dihedral group of order 6, while the figure eight knot has a homomorphism onto the Dihedral group of order 10. $\endgroup$ – Josh B. Sep 6 '18 at 4:02
  • $\begingroup$ There is a mistake in the claim that there is a surjection $\pi_1 (S^3 - K) \to D_6$, as @N.Owad pointed out here math.stackexchange.com/questions/3388030/… . $\endgroup$ – Minkowski Oct 11 '19 at 8:24
  • $\begingroup$ Oops, thanks for that correction folks. And @JoshB., sorry that I never even noticed your correction for the last year. $\endgroup$ – Lee Mosher Oct 11 '19 at 12:02

Knots themselves are real-life physical objects (unlike for instance topological spaces) and their study seems to be a rather fundamental question in physics. I think this makes motivating knot theory rather easy. The way people use invariants in practice is not just to tell two things apart. There are rather interesting results in knot theory which use the fundamental group but are interesting independent of the fundamental group.

The easiest interesting examples of studying homomorphisms from a knot group $\pi_1(S^3-K) \to G$ are Fox colorings (https://en.wikipedia.org/wiki/Fox_n-coloring). In this case $G=D_{2n}$ and there is a completely visual way of picturing these morphisms, and very easy proofs of things like the non-triviality of the trefoil fall out as well.


The Gordon-Luecke theorem implies that if two knots have isomorphic knot groups, then they are equivalent (as a connect sum representation, up to mirror images of the prime components). That is, the knot group contains almost everything there is to know about the knot.

One way to study groups is to study their representations, which is another word for "homomorphisms out of the group." If two groups have different representations, then they must not be isomorphic groups. If two groups have exactly the same representations (for all codomains), then Gordon-Luecke applies as it did before.

Some common representations are actions on a finite set or actions on a vector space (i.e., homomorphisms $G\to S_n$ or homomorphisms $G\to \mathrm{GL}(V)$). Sometimes the group is restricted to some finite group (like $D_{2n}$) or to a particular kind of linear map (like $\mathrm{SU}(2)$). Since knot groups are infinite, the linear representation theory tends to be rather complicated, however.

One application of representations of a knot group is to twist the chain complex associated with the universal cover of the knot complement. "Torsions" calculate a summary of the representation that can be used to distinguish different groups. The Alexander polynomial is one torsion which is associated with the abelianization, which is a particularly simple representation.

Another application is demonstrating that a knot is nontrivial, which a representation with nonabelian image certifies.

  • $\begingroup$ Is the trefoil a ribbon knot? Because there is a ribbon knot which has the same knot group as the trefoil. To get the knot, I just use the method in ams.org/journals/proc/1980-078-01/S0002-9939-1980-0548101-9/… . This seems to go against what you say about the knot group. How am I mistaken? $\endgroup$ – Prototank Jul 4 '17 at 22:43
  • $\begingroup$ @Prototank No, the trefoil knot is not a ribbon knot because it's not a slice knot. I'm not familiar with what calculation you did, but if there were a ribbon knot with the same knot group as the trefoil, then the Alexander polynomial (which is a knot group invariant) of the trefoil would factor as $f(t)f(t^{-1})$ for some Laurent polynomial $f$, but the Alexander polynomial of the trefoil is irreducible. Without any further information, I would guess what you found is a knot whose knot group surjects onto the trefoil knot's knot group, but it can't be an isomorphism. $\endgroup$ – Kyle Miller Jul 5 '17 at 3:04
  • $\begingroup$ You are misstating the G-L theorem. It only claims that if knot complements are homeomorphic then knots are isotopic. For prime knots, indeed, isomorphism of $\pi_1$ implies isotopy. I think, this is due to W.Whitten. (Otherwise, there are easy counter examples, just take the connected sum of a chiral knot $K$ and its mirror as $L$ and the sum of $K$ with itself as $L'$. Then $\pi_1(L)=\pi_1(L')$ but they are not isotopic.) $\endgroup$ – Moishe Kohan Mar 2 '18 at 19:50
  • $\begingroup$ @MoisheCohen Thanks. That was something I was confused about back when I wrote this. I believe I corrected the statement now, that one can (theoretically) read off the prime decomposition of a knot up to mirror images of the components. For isomorphisms of $\pi_1$ for prime knots implying isotopy, isn't the trefoil a counterexample? $\endgroup$ – Kyle Miller Mar 2 '18 at 20:11

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