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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
  • $\begingroup$ Another tool in the arsenal is to go one more step and drill out the Hopf-linking circle for your link in the n=1 case. If you can argue its exterior is hyperbolic, you could potentially argue your family is also hyperbolic, via covering spaces and filling. $\endgroup$
    – Ryan Budney
    Dec 29, 2022 at 5:58
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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 are hyperbolic (thus non-trivial).

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