# Powers of meridians in knot groups

Given a (tame) knot $$K \subset S^3$$, let $$t \in G = \pi_1(S^3 - K)$$ be any meridian. The Wirtinger presentation shows that $$\langle \langle t \rangle \rangle = G$$, where the notation indicates the normal closure of $$t$$ in $$G$$.

Let $$I$$ be an infinite subset of the natural numbers. A linking number argument shows that $$\bigcap_{i \in I} \langle \langle t^i \rangle \rangle \subset [G,G].$$ My question is: when is this intersection trivial? When does $$\bigcap_{i \in I} \langle \langle t^i \rangle \rangle = \{1\}?$$ In particular, does this hold if $$I$$ is the set of all powers of $$2$$? Is there a known example where the above does not hold?

• This is true for hyperbolic knots via the theory of hyperbolic Dehn filling. I’ll try to dig up a reference when I get a chance. I’m not sure about torus knots - they self-cover themselves, and the intersection might be the commutator subgroup? Mar 30 at 0:35
• Thanks! I'll try to work this out, but would definitely appreciate a reference if this is true for any infinite set of natural numbers. Mar 30 at 14:11
• My comment about torus knots wasn’t correct (as follows from Ito’s answer), the point being that in subgroups isomorphic to the torus knot group, the meridian of the subgroup is not a power of the meridian in the ambient group (I hadn’t thought about this carefully when I made my original comment). Mar 30 at 16:34

For hyperbolic knots, this holds for any infinite set $$I\subset \mathbb{N}$$ via hyperbolic Dehn filling. One may use Thurston’s original theorem, or the geometric group theory version of Groves-Manning and Osin. I will describe a version using the Gromov-Thurston $$2\pi$$ theorem since this is how I think about it intuitively. To understand this argument you need to know a bit about Riemannian geometry and hyperbolic orbifolds as well as consult some of the linked references.

Consider a tame knot $$K\subset S^3$$ with hyperbolic metric $$\rho$$ on $$M_K=S^3-K$$. Let $$x\in \pi_1(M_K)-\{1\}$$. There is a peripheral torus $$T=\partial (\mathcal{N}(K))\subset M_K$$ with $$\pi_1$$-injective fundamental group, such that $$\pi_1(T)\subset \pi_1(M_K)$$ is generated by elements $$\mu$$ the meridian, realized by a little loop in $$T$$ linking $$K$$, and $$\lambda$$ the longitude which is trivial in $$\pi_1(T)\to H_1(T)\to H_1(M_K)$$. If $$x$$ is not peripheral, then it is realized by a closed geodesic $$\gamma:S^1\to M_K$$ immersed in $$M_K$$, meaning that $$x=\gamma_\#(s)$$, where $$s\in \pi_1(S^1)$$ is a generator (I am suppressing basepoints in the notation).

In the metric $$\rho$$ on $$H_K$$, there is a tubular neighborhood $$C\subset M_K$$ called a horocusp which is the image of a horoball in the universal cover, such that $$\partial C$$ is isotopic to $$T$$ in $$M_K$$. We may shrink $$C$$ down to a smaller horocusp $$C_x$$ such that $$C_x \cap \gamma(S^1)= \emptyset$$. Choose a number $$n\in I$$ such that $$length([n\mu]) > 2\pi$$, where $$[n\mu]$$ is a geodesic curve in $$\partial C_x$$ realizing the homology class $$n\mu\in H_1(C_x)=H_1(T)$$. Then the Gromov-Thurston $$2\pi$$ theorem implies that one may fill in an orbifold negatively curved cone metric with cone angle $$2\pi/n$$ along the meridian (this manifold has the notation $$M_K(n,0)$$ in SnapPea/SnapPy). The orbifold fundamental group is $$\pi_1(M_K)/<<\mu^n>>$$, the group obtained by killing the $$n$$th power of the meridian. Then the element $$x\in \pi_1(M_K)$$ will be non-trivial in $$\pi_1(M_K(n,0))$$, since it is realized by a closed geodesic. The case that $$x$$ is peripheral I will leave as an exercise (hint: the subgroup $$<\lambda, \mu>/<>$$ will inject into $$\pi_1(M_K(n,0))$$).

I think this proof should generalize to arbitrary knots, but I haven’t thought it through carefully.

Such an intersection is trivial if $$I$$ is whole $$\mathbb{N}$$, mainly because the knot groups $$G$$ are residually finite.

Assume that $$1 \neq x \in \bigcap_{i \in I} \langle \! \langle t^i \rangle \! \rangle$$. Since $$G$$ is residually finite, there exists a homomorphism $$\phi:G \rightarrow F$$ to a finite group $$F$$ such that $$\phi(x) \neq 1$$. On the other hand, since $$F$$ is finite $$\phi(t^n) = 1$$ for some $$n$$, so $$\phi(\langle \! \langle t^n \rangle \! \rangle) = 1$$. Since $$x \in \bigcap_{i \in I} \langle \! \langle t^i \rangle \! \rangle \subset \langle \! \langle t^n \rangle \! \rangle$$, this contradicts $$\phi(x) \neq 1$$.

(I think that instead of residually finiteness, using the property that the knot group is virtually residually $$p$$ all but finitely many $$p$$, we can say the trivialy of intersection for more general $$I$$. However, I do not know whether the same conclusion holds or not when $$I$$ is the set of all powers of $$2$$.)