Let $K$ be a knot, $G=\pi\_1(S^3 - K)$ and $G^{(n)}$ is the $n$-th term
of the derived series. Then $G/G^{(n)}$ is poly-torsion-free abelian since it is known that each $G^{(i)}/G^{(i+1)}$ is torsion-free. Suppose each $G^{(i)}/G^{(i+1)}$ is infinite for $i\geq 0$. Consider a tower of iterated universal abelian covers of $G$


$\cdots \rightarrow M\_{i+1} \rightarrow M\_{i}\rightarrow \cdots \rightarrow M\_2 \rightarrow M\_1 \rightarrow M\_0 = M$, where $\pi\_1(M\_i)=G^{(i)}$. Does there exist a limit $M\_\omega$ of the above tower? If so, is there an ordinal $\omega$ such that $\pi\_1(M\_\omega) = G^{(\omega)}=1$?



Does there exist a limit of (infinite) iterated universal abelian covers of a knot group?
If you take a non-trivial cable of a knot with trivial Alexander polynomial, then I think $G^{(i)}/G^{(i+1)}$ will be infinite for $i\geq 0$, yet one will never have $G^{(\alpha)}=1$ for any ordinal $\alpha$. 


A knot with Alexander polynomial $=1$ (such as a Whitehead double) has infinite cyclic cover acyclic, and hence will have $G^{(2)}=G^{(1)}$. Therefore, when one takes a cable of this knot, its derived series will agree with that of a torus knot. For a torus knot, $G^{(1)}$ is a free group (the fundamental group of the fiber), and so $G^{(\omega)}=1$, where $\omega$ is the first infinite ordinal. Hence when we take a cable of a knot with trivial Alexander polynomial, $G^{(\omega)}$ will be perfect (and equal to the commutator subgroup of the pattern knot). 


