# Why is the Hopf link the only link with knot group $\mathbb{Z} \oplus \mathbb{Z}$?

We can use the Loop Theorem to show that if $\Sigma$ is a minimal-genus Seifert surface for a link $L$, then $\pi_1(\Sigma)$ injects into the knot group $\pi_1(S^3 \setminus L)$. An orientable connected surface with nonempty boundary and abelian fundamental group must be a disk or an annulus. Therefore $\pi_1(S^3 \setminus L)$ is abelian only if it is $\mathbb{Z}$ or $\mathbb{Z}\oplus \mathbb{Z}$, implying that $L$ is an unknot or a (nonsplit) two-component link with genus zero.

If $\,\pi_1(S^3 \setminus L) \cong \mathbb{Z} \oplus \mathbb{Z}$, how do we know that $L$ is the Hopf link?

I know that this was proven by Neuwirth in his paper "A note on torus knots and links determined by their groups" (1961). I can't access that paper, and I would be satisfied with an overview of his proof (or another reference). But it would be great to know if there have been any simple arguments discovered since then.

Edit: See my first answer below for a summary of Neuwirth's proof and my second answer for another argument and a proof that $\pi_1(\Sigma)$ injects into $\pi_1(S^3 \setminus L)$.

Here's another argument:

Lemma. If $\Sigma$ is a minimal-genus Seifert surface for a nonsplit link $L$, then there are inclusions $\pi_1(\Sigma) \hookrightarrow \pi_1(S^3 \setminus \Sigma) \hookrightarrow \pi_1(S^3 \setminus L)$.

Proof. By Gabai's Foliations and the Topology of 3-Manifolds, a Seifert surface for a nonsplit link is minimal if and only if the link complement has a $C^\infty$ foliation that contains $\Sigma$ as a compact leaf and has no Reeb components. Pair this with a result of Novikov: If $\mathscr{F}$ is a $C^2$ foliation on any manifold $M \neq S^2 \times S^1$ that contains no Reeb components, then $\pi_1(\operatorname{leaf})\to \pi_1(M)$ is injective for any leaf in $\mathscr{F}$. It follows that $\pi_1(\Sigma)\to \pi_1(S^3 \setminus L)$ is injective. Since this map factors through $\pi_1(S^3 \setminus \Sigma)$, the claim follows. $\square$

Claim. If $\pi_1(S^3 \setminus L)$ is abelian, then it is isomorphic to $\mathbb{Z}$ or $\mathbb{Z}\oplus \mathbb{Z}$ and $L$ is the unknot or the Hopf link, respectively.

Proof. If $L$ is a split link, then we can use van Kampen's theorem to see that $\pi_1(S^3 \setminus L)$ is a free product of two nontrivial groups and is nonabelian. Thus $L$ is nonsplit and the lemma implies that $\pi_1(\Sigma)$ injects into the knot group. It follows that $\pi_1(\Sigma)$ is abelian, so $\Sigma$ is either a disk or an annulus. If $\Sigma$ is a disk, then $L$ is the unknot and $\pi_1(S^3 \setminus L) \cong \mathbb{Z}$. Otherwise, $\Sigma$ is an annulus. The core circle $C$ of this annulus must be unknotted; if not, then $\pi_1(S^3 \setminus \Sigma)\cong\pi_1(S^3 \setminus C)$ would be nonabelian and therefore could not inject into $\pi_1(S^3 \setminus L)$. It follows that $\Sigma$ is a standardly-embedded annulus with $n$ twists. We can easily compute the knot group of the boundary of a standard $n$-twisted annulus and, as in Neuwirth's proof, it is seen to be abelian only when $n=1$. Thus $L$ is the Hopf link and $\pi_1(S^3 \setminus L)\cong \mathbb{Z}\oplus \mathbb{Z}$. $\square$

• Note that we can prove the "Lemma" for a knot without appealing to the results of Gabai (and Thurston) and Novikov: The first inclusion comes from the Loop theorem. Let $U=\mathring \Sigma \times (-1,1)$ be a normal neighborhood of $\mathring \Sigma$ in $S^3$, with $\Sigma^+$ and $\Sigma^-$ denoting the upper and lower "boundaries" of $U$. If the induced map $\pi_1(\Sigma^+) \to \pi_1(S^3 \setminus U)$ has nontrivial kernel, the Loop theorem implies that there is an embedded disk in $S^3 \setminus U$ with boundary lying in $\Sigma^+$. – Kyle Jun 4 '15 at 13:30
• [Continued] Cutting $\Sigma^+$ along this disk reduces the genus of $\Sigma^+$ (which is still bounded by a pushoff of $L$), which is a contradiction. To set up the the second inclusion, observe that $S^3 \setminus L$ is homeomorphic to the space obtained from $S^3 \setminus U$ by identifying $\Sigma^+$ and $\Sigma^-$. – Kyle Jun 4 '15 at 13:31
• [Continued] Then van Kampen's theorem (or, rather, the special case giving an HNN extension) says \begin{align*} \pi_1(S^3 \setminus L) &\cong \pi_1\left( \frac{S^3 \setminus U}{\Sigma^+ \sim \Sigma^-} \right) \\ &\cong \left\langle \pi_1(S^3 \setminus U),t \mid t \gamma t^{-1} = \phi(\gamma) \ \text{for all } \gamma \in \pi_1(\Sigma^+)\subset \pi_1(S^3 \setminus U) \right\rangle, \end{align*} where $\phi: \pi_1(\Sigma^+) \to \pi_1(\Sigma^-)$ is the obvious isomorphism. – Kyle Jun 4 '15 at 13:32
• [Continued] It follows that $\pi_1(S^3 \setminus L)$ contains a subgroup isomorphic to $\pi_1(S^3 \setminus U)$, which itself is isomorphic to $\pi_1(S^3 \setminus \Sigma)$. – Kyle Jun 4 '15 at 13:32

Thanks to Mike Miller, I got a chance to read Neuwirth's proof that $\pi_1(S^3 \setminus L) \cong \mathbb{Z} \oplus \mathbb{Z}$ implies that $L$ is the Hopf link. The argument begins by using Alexander duality and Dehn's lemma to show that $L$ is a two-component link whose components are unknotted. A generalization of Dehn's lemma (due to Shapiro-Whitehead) and some homology arguments (made possible by the fact that $\pi_1(S^3 \setminus L)$ is isomorphic to $H_1(S^3 \setminus L)$ because it is already abelian) show that $L$ bounds an annulus $A$. Moreover, $A$ is unknotted because its boundary components are unknotted. Therefore $A$ is isotopic to a standardly-embedded annulus with $n$ twists. We can't have $n=0$ because then $L$ is the 2-component unlink, which has nonabelian knot group. If $n\geq 1$, then $\pi_1(S^3 \setminus L) \cong \langle a,b : ab^n=b^n a\rangle$. This surects onto $\langle a,b: b^n=1\rangle \cong \mathbb{Z} * (\mathbb{Z}/n\mathbb{Z})$, which is nonabelian if $n\geq 2$. We conclude that $n=1$, hence $L$ is the Hopf link.