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Let $B=S^3\setminus K$ for some (tame) knot $K$. Suppose we have a covering $E\to B$ with a finite fiber.

Question: is $E$ homeomorphic to a knot/link complement?

On this question I found only the paper Gonzales-Acuna, Whitten. Imbedding knot groups in knot groups (1987), which says that if the fiber is fixed then there are at most 2 such covering spaces.

It is very interesting how can one construct $E$ that is not a knot complement. Any links are welcome.

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    $\begingroup$ A result in the Gonzales-Acuna & Whitten paper is that "any covering between knot exteriors is cyclic", so this gives some restrictions for your question. For example, it shows that if $B$ is either the complement of (1) the two-bridge knot; or (2) a composite knot; or (3) a knot with a quadratic Alexander polynomial different from $t^2-t+1$ (i.e. that of the figure-eight knot); then in any of these cases there are no finite-fiber coverings $E$ as in the question. So in all those cases for $B$, the answer is (vacuously) "yes" to your question. $\endgroup$
    – Carl-Fredrik Nyberg Brodda
    Commented Mar 5, 2023 at 12:52
  • $\begingroup$ @Carl-FredrikNybergBrodda I claim that for any $n$ there is an $n$-covering over $S^3\setminus K$ -- just take for its fundamental group the kernal of the composition $\pi_1(S^3\setminus K)\to H_1(S^3\setminus K)=\mathbb{Z}\to\mathbb{Z}_n$ $\endgroup$
    – Andrey Ryabichev
    Commented Mar 5, 2023 at 13:54
  • $\begingroup$ Sorry, me saying that the answer is "vacuously yes" to your question is of course a typo and completely backwards, it should be: the answer is "always no". In the sense that for e.g. $B$ as in Gonzales-Acuna & Whitten none of the covering spaces $E$ you can construct will be knot complements (e.g. the obvious covering you mention induced by surjecting $\mathbb{Z}_n$). $\endgroup$
    – Carl-Fredrik Nyberg Brodda
    Commented Mar 5, 2023 at 14:12

2 Answers 2

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For this answer I will consider knots to be links (with one component).

In general the answer is "no". For example, consider $K = 4_1$, the figure-eight knot. Let $X = S^3 - K$. Neither the double nor triple cover of $X$ is a link complement. One of the four-fold covers of X is a link complement, but the other is not. One of its five-fold covers is a link complement, but the other three are not. None of its six-fold covers are link complements. [All of these computations are done in SnapPy - we either find a homeomorphism to a link complement or we find torsion in $H_1$.]

As we take larger and larger covers, we generically expect torsion to appear in $H_1$; such covers do not embed in $S^3$. On the other hand, the figure-eight does have infinitely many covers which are link complements. This is because the "good" four-fold cover is the complement of the link $10^2_{138}$ which has an unknotted component.

I believe that the story will be much the same for any non-trivial knot.

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Gonzalez-Acuña and Whitten answered this question for coverings by knot exteriors, as opposed to link exteriors more generally, in chapter 3 of "Imbeddings of three-manifold groups". They prove that a covering between two knot exteriors must be a regular cyclic covering (corollary 3.2) and then conclude (modulo geometrization):

Theorem 3.4(1). If you have an $n$-fold cover $E_J \to E_K$ ($n\geq 2$) between the exteriors of knots $J$ and $K$ in $S^3$, where $J$ is nontrivial and $K$ is not a torus knot, then either $n$-surgery or $-n$-surgery on $K$ is a lens space. (The torus knot case corresponds to lens space surgeries in a similar way, but the slope need not be integral. In any case torus knot exteriors are Seifert fibered, so any knot exteriors covering them must again be Seifert fibered and hence torus knot exteriors.)

The converse direction is easier to see: suppose that the $n$-surgery $S^3_n(K)$ is a lens space. The $n$-fold cyclic cover of $S^3_n(K)$ is $S^3$, and the core $\tilde{K} \subset S^3_n(K)$ of the surgery lifts in that cover to a knot $J$; then the exterior $E_J$ is the $n$-fold cyclic cover of the exterior of $\tilde{K}$, which is the same as $E_K$.

Anyway, this theorem gives strong restrictions on coverings of $E_K$ specifically by knot exteriors, because we know via Heegaard Floer homology that relatively few knots admit lens space surgeries: $K$ must be fibered, either $K$ or its mirror must be strongly quasipositive, it must have the same knot Floer homology as a Berge knot, and so on. It also restricts the value of $n$: for example we must have $n \geq 2g(K)-1$, and $n$ cannot be $2,3,4$ because no nontrivial knots admit lens space surgeries of these orders.

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  • $\begingroup$ Just a remark on the torus knot case: the monodromy of the fibration over the circle is infinite order, so it will self-cover infinitely many times. Moreover, different torus knots can cover each other (e.g. the right- and left-handed trefoils cover each other). $\endgroup$
    – Ian Agol
    Commented Mar 10, 2023 at 5:10

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