# Ways to prove that $n$-component Brunnian link is nontrivial

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.

• 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
– Dora
Dec 29, 2022 at 4:55
• 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. Dec 29, 2022 at 5:28
• . . . and technically, any non-trivial knot is a non-trivial Brunnian 1-link. Dec 29, 2022 at 5:43
• 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. Dec 29, 2022 at 5:58
• 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.
– HJRW
Dec 29, 2022 at 14:10

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).