restrictions on the fundamental group of a knot complement?

knot-theory

1,011

Solution 1

(Maybe it shouldn't be an answer, as I won't say anything no one said before.)

A knot group is indeed a f.p. (obvious) group whose H1 is Z and H2 is 0 (by Alexander's duality). Moreover, as the meridian normally generates the group (that's a clear consequence of the Wirtinger's presentation), its weight is one. (The weight is the minimal number of "normal" generators. As far as I know, it's still a very mysterious group invariant.)

One of the early successes of surgery theory was to prove that these properties characterize knot groups of high dimension. (Michel Kervaire, Les nœuds de dimension supérieure (1965), available here.)

But, as always, the dimension three is more restrictive. For example, a knot exterior is a Haken manifold so Waldhausen's work imply a lot of restrictions. For a start, the word problem for this knot has to be solvable.

I'm sure that all the things we know about geometric decomposition and 3-manifold groups give various restrictions (for example, Stallings's fibration theorem imply that if you have an epimorphism from a knot group to a f.g. group whose kernel if f.g., then it comes from a fibration on the circle; in particular, the kernel is a surface group and the image is Z).

But I think that no definitive answer is known. Globally, 3-manifold groups are still mysterious (we still don't know if they are linear!) and they are just very particular quotients of link groups so I think our knowledge of them is still limited.

Solution 2

Famously, knot groups are torsion free. One proof of this uses the fact that $S^3 - K$ is a $K(\pi, 1)$.

Another nice fact - knots have Seifert surfaces and so knot groups contain free groups.

Another one - knot complements have Heegaard splittings. It follows that any knot group has a presentation with one more generator than relation.

I think that combining Waldhausen's work with Gordon-Luecke should prove that the knot group (possibly equipped with a $Z^2$ "peripheral" subgroup) determines the knot in $S^3$.

Neuwirth wrote a book entitled "Knot groups", published in 1965. There is lots more to know since then!

Solution 3

If $\pi$ is a knot group, in addition to $H_1(\pi)=\mathbb Z$, you also have $H_2(\pi)=0$. This is because $H_2(S^3\setminus K)=0$, and $K(\pi,1)=(S^3\setminus K)$ union cells of dimension $\geq 3$. So $H_2(S^3\setminus K)$ surjects onto $H_2(K(\pi,1))=H_2(\pi)$.

1,011

yoyo

Updated on August 01, 2022

Comments

yoyo 10 months

Are there any restrictions on $\pi_1(S^3\backslash K)$ for a (tame) knot $K$ besides having $\pi_1^{\text{ab}}(S^3\backslash K)=\mathbb{Z}$?
So we have 1) finite presentation, 2) $H_1=\mathbb{Z}$, 3) $H_2=0$, and I saw somewhere 4) the knot group must be the normal closure of a single element.

Apparently this characterizes knot groups in dimensions $n\geq5$ (i.e. $\pi_1(S^n\backslash S^{n-2})$), but fails to do so in lower dimensions. Basically I just wanted to know if there is simple characterization of knot groups (not "it has a presentation that looks like a wirtinger presentation")
- Pete L. Clark about 12 years
  
  Well, it has to be finitely presented, for instance.
- Qiaochu Yuan about 12 years
  
  Aren't there restrictions that come just from it being the fundamental group of a 3-manifold?
Admin over 11 years

In fact, groups with weight one are precisely quotients of knot groups.