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I am a little confused about what a cyclic covering of a knot is. Rather, I have a definition I don't understand which I am comparing to some intuition which I'm not sure is correct.

On one hand, we have the definition from the representation of the knot $G$ (my version comes from Fox (1962). The abelianizing homomorphism maps this group to the infinite cyclic group $\mathbb{Z}$. This is the simpleist possible thing, send every element $g\in G$ to $t$ which generates $\mathbb{Z}$. For any positive integer $a$ there is a unique homomorphism $\mathbb{Z}\to \mathbb{Z}_a$ ("The modulo map" is how I usual think), so there is also a unique homomorphism from $G$ to $\mathbb{Z}_a$. The infinite cyclic branched coverings of the base space $\mathbb{S}^3$ over the knot are those that belong to the kernel of $G\to \mathbb{Z}$, and those belonging to the kernel of $G\to \mathbb{Z}_a$ are the finite cyclic ones.

So I think of the covering of knots as being given by representations of the symmetric group onto the knot group. For instance, the knot group

$$G=(x,a:a^2x=xa)$$ has the representation $x\to (0 ~1)$ and $a\to (0~1~2)$ ("over the element $x$ in the cover the first two covers trade places while the third is unchanged"). My understanding of a finite cyclic covering would be a covering in which each element is a cycle of order equal to the order of the covering. An infinite cyclic covering would be a covering map of infinite order with the permutation label $(0~1~...)$.

But if that intuition is correct (Edit below suggests it is), cyclic $n$-coverings are pretty special - every element of a group presentation would have to have be of order $n$, which must also satisfy the group relators. So I'm guessing I don't have the right idea about what a cyclic covering is in this context. And further, I'm not sure how the kernel of the map $G\to \mathbb{Z}$ is related to this permutation of infinite order.

Can anyone help?

EDIT: I have been able to pin down where my "intuition" came from, and that is the book "Knots, Links, Braids, and 3-Manifolds" by Prasolov and Sossinsky. They pretty clearly say that starting with a line $l\subset \mathbb{R}^3$ one can get an $n$-fold cyclic branched covering $p:\mathbb{S}^3\to \mathbb{S}^3$ by adding a point, so the branch set is $l\cup p$ and colored $(1~2~...n)$.

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I've been working on this, so I guess I'll answer my own question. This is all basically from Fox (1962).

The key is how a subgroup of the fundamental group of a knot uniquely defines a covering space. Given a subgroup $\Gamma \subset G$, this subgroup defines a set of cosets $\Gamma=\Gamma_0, \Gamma_1,...\Gamma_{g-1}$. The action of each element $a\in G$ on these cosets permutes the cosets, which gives a representation of $G$ onto $\mathbb{S}_{g-1}$,

$$\rho(a)=\left(\begin{array}[cccc] \Gamma_0&\Gamma_1&...&\Gamma_{g-1}\\ \Gamma_0a&\Gamma_1a&...&\Gamma_{g-1}a\end{array}\right)$$

So a cyclic covering is when the subgroup chosen is the kernel of the Abelianizing homomorphism, and the representation of the symmetric group can be determined by chasing down how each element $a\in G$ acts on each coset.

So let's do a concrete example to see why my intuition was actually correct. A particular knot group is given by

$$G=(a,x:a^2x=xa)$$

Let's find the infinite cyclic covers of this knot group. Under the Abelianizing homomorphism $A:G\to \mathbb{Z}$ we have to enforce $ax=xa$, which implies $A(a)=1$ and $A(x)=t$, so $Ker(A)=<a>=\Gamma$. The cosets of $\Gamma$ are indexed by the order of $x$ in them:

$$\Gamma_0=\Gamma a=(a^n|n\in \mathbb{Z})$$ $$\Gamma_1=\Gamma x=(a^nx=a^{n-2}xa=a^{n-4}xa^2=...|n\in\mathbb{Z})$$ $$\Gamma_2=\Gamma x^2=(a^nx^2=a^{n-2}xax=a^{n-4}xa^2x=...|n\in\mathbb{Z})$$

And so forth. Since the action of $x$ on each coset moves $\Gamma_i$ to $\Gamma_{i+1}$, this give us a representation of $G$ onto $S_{\infty}$ of

$$\rho(a)=1,\qquad \rho(x)=(0~1~2~3~...)$$

as expected by my intuition.

We can find a cyclic covering of order 3 as well, by finding the kernel of the map $f\circ A$, where $f:\mathbb{Z}\to \mathbb{Z}_3$ by $f(t^3)=1$. The kernel of this map is $\Gamma=(a,x^{3i}:i\in\mathbb{Z})$, and there are three cosets, $$\Gamma_0=\Gamma,~\Gamma_1=\Gamma x^{3i+1},~\Gamma_2=\Gamma x^{3i+2}$$ And the representation on $S_3$ is given by $$\rho(a)=1,\qquad \rho(x)=(0 1 2)$$

So the essential connection between the representation of $S_n$ on the knot and the cyclic covers is through how the kernel of the Abelianization permutes the cosets. I would have liked the answer to be a little simplier - for instance, is there an easy way to determine if a particular representation is cyclic or not? Based on this, I know that the representation

$$\rho(a)=(0~1~2),\qquad \rho(x)=(0~1)$$

is not cyclic, but only because I know the forms for all the cyclic covers already. A more general approach would be interesting, I think.

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