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The attached image shows a way to construct an $n$-component Brunnian link for any $n\geq 3$. That is, this link is not trivial, but deleting any of its components makes the new link trivial. The latter property is obvious from the picture, however I would like to have a strict proof of nontriviality. What bothers me is that nowhere in the literature there is an example of proving nontriviality of any Brunnian $n$-link.

Let us denote this link by $L=K_1\cup K_2 \cup \dotsb \cup K_{n-1}\cup K_n$. Also, it might be useful to examine $L'=L\setminus K_n=K_1\cup K_2 \cup \dotsb \cup K_{n-1}$. I tried to look at the link group $\pi_1(L').$ Since $L'$ is equivalent to $n-1$ disjoint circles, I conclude that $\pi_1(L')=F_{n-1}$. Then I ask what $[K_n]\in \pi_1(L')$ would be in the case if $L$ were trivial. In that case I would be able to move $K_n$ a little bit and get a homotopic loop that is a clean circle, thus $[K_n]\in \pi_1(L')$ is the identity. Now, for contradiction I would like to calculate explicit form of $[K_n]\in \pi_1(L')$ (or to somehow just prove that it is not the identity). I tried doing it with Wirtinger presentation, but it was messy and, even if I did find explicit form for $[K_n]\in \pi_1(L')$ in terms of group’s generators correctly, determining whether $[K_n]$ is equal to the identity in this finitely generated group is a well-known undecidable problem.

Is there a way to improve my approach in order to prove nontriviality? If not, do you see another way to prove that? Actually, just an example of proof of nontriviality for any $n$-component Brunnian link would be enough for me, chances are the idea would work for many Brunnian links if not for all.

enter image description here

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    $\begingroup$ You can prove this using Milnor’s invariants, which are generalizations of linking numbers. See his paper “Link groups”. By the way, the classical decision problems for fundamental groups of 3-manifolds like link exteriors are all decidable. See here for a summary of the literature on them: arxiv.org/abs/1205.0202 $\endgroup$
    – Dora
    Dec 29, 2022 at 4:55
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    $\begingroup$ Your statement that "NOWHERE in the literature is there an example of proving nontriviality of any Brunnian n-link", technically this is false. You might want to qualify your statement. Are you talking about an infinite family? Milnor invariants work, similarly finite-type invariants work, like Koschorke invariants. I think perhaps you are not quite using English correctly, as the Whitehead links and Borromean rings are known to be non-trivial. $\endgroup$
    – Ryan Budney
    Dec 29, 2022 at 5:28
  • $\begingroup$ . . . and technically, any non-trivial knot is a non-trivial Brunnian 1-link. $\endgroup$
    – Ryan Budney
    Dec 29, 2022 at 5:43
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    $\begingroup$ It's maybe worth noting that the fundamental argument you suggest does not work for the link in the picture. Each component is freely homotopic (not isotopic!) in the complement of the other (n-1) components to the unlink. $\endgroup$
    – mme
    Dec 29, 2022 at 6:40
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    $\begingroup$ There seems to be some confusion about decision problems. The word problem is indeed undecidable in an arbitrary finitely presented group, but is easily decidable in a free group, which is your situation. You do this by writing your conjugacy class as a cyclically reduced word. $\endgroup$
    – HJRW
    Dec 29, 2022 at 14:10

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In a comment to this question I give some of the details of a hyperbolic geometry proof that the link shown is non-trivial. That proof works only when $n$ is large; in fact all links in the family (except for the first) are hyperbolic and thus non-trivial. The first link in the family is actually a knot -- it is the connect sum of a pair of trefoils and so is non-trivial.

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