Powers of meridians in knot groups

Question

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?

Answer 1

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 $<\lambda, \mu>/<<n\mu>>$ 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.

Answer 2

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 $2$.)