Does the fundamental group detect the unlink?

It is known that the unknot is the only knot whose complement has fundamental group $\mathbb{Z}$.

Does this fact generalize to links? That is, suppose that $\ell= \ell_1 \cup \dots \cup \ell_n$ is an $n$-component link in $\mathbb{R}^3$ such that $\pi_1(\mathbb{R}^3-\ell)$ is a free group on $n$-generators. Is $\ell$ the unlink on $n$-components?

Solution 1:

A useful result here is

Theorem. (Kneser's conjecture) If $M$ is a compact orientable $3$-manifold with incompressible boundary, and if $\pi_1(M)\approx G_1*G_2$, then there is a connect sum decomposition $M=M_1\mathbin{\#}M_2$ with $\pi_1(M_i)\approx G_i$ for $i=1,2$.

This paper suggests Epstein, "Free products with amalgamation and 3-manifolds", 1961, and Hempel, section 7 for details. (Calegari's notes, Theorem 3.9, has the case of $M$ closed.)

If $L$ is a link with $M=S^3-\nu(L)$ having a compressible boundary, then one of the components is an unknot split from the rest of the link. This is because the loop theorem gives an embedded disk whose boundary is a link component, and the boundary of a regular neighborhood of this disk is a splitting sphere.

Otherwise, if $S^3-\nu(L)$ has incompressible boundary and $\pi_1(S^3-\nu(L))=G_1*G_2$ with each $G_i$ nontrivial, then Kneser's conjecture gives an embedded sphere that splits $L$ into two sublinks.

Recall that $H_1(S^3-L)= \mathbb{Z}^n$, where $n$ is the number of components of $n$. If $\pi_1(S^3-L)$ is a free group of $n$ generators, then this paired with the above implies that $L$ is a completely split link, where each component $L_i$ has $\pi_1(S^3-L_i)=\mathbb{Z}$. The only knot whose knot group is $\mathbb{Z}$ is the unknot, hence each $L_i$ is an unknot.

So, yes, if $\pi_1(S^3-L)$ is a free group on $n$ generators, then $L$ is an $n$-component unlink.