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Let $L$ denote the Borromean rings. Then we have a presentation: $\pi = \pi_1(S^3 - L) = <x,y,z | [x,[z,y^{-1}]], [y,[x,z^{-1}]], [z,[y,x^{-1}]]>$

Let $F$ be the free group on 3 letters. If $L$ where the unlink, then $\pi/\pi^{(n)} \cong F/F^{(n)}$ for all $n$, where $G^{(n)}$ denotes the $n^{th}$ term in the lower central series of $G$, but $L$ is not an unlink.

We have $\pi / \pi^{(2)} \cong \mathbb{Z}^3 \cong F / F^{(2)}$.

In this particular example, I would like to understand, ideally through a group-theoretic lens, two things:

(1) Why is $\pi / \pi^{(3)} \cong F / F^{(3)}$? There is presumably just a nice way to write down a presentation for $F / F^{n}$, and then in this case the isomorphism is just visible.

(2) Why is $\pi / \pi^{(4)} \ncong F/F^{(4)}$? I saw this mentioned in a book by Fenn where I am told to just look at the Magnus expansion of $[x,[z,y^{-1}]]$ and see that it is not in $F^{(4)}$, however to me that just tells me that the obvious map from $F$ doesn't yield an isomorphism, not that no isomorphism exists.

And as a general question:

(3) Are there links where $\pi / \pi^{(n)} \cong F/F^{(n)}$ for all $n$ but the links are not the unlink (where here $F$ is the free group on the appropriate number of letters).

user101010
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  • There's a chance computing the basic commutators as in https://math.stackexchange.com/a/480919/172988 might help for (1). There's a line in https://arxiv.org/abs/math/0609484v3 about a theorem of Stallings implying $\pi/\pi_n$ is an invariant of the concordance type of the link, though I don't know if it's a complete invariant. (Beware: usually $G^{(n)}$ denotes terms of the derived series.) – Kyle Miller Jul 22 '19 at 20:14

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