Consider the Kernel $K\_n$ of the natural group homomorphism from the $n$-th [braid group](https://en.wikipedia.org/wiki/Braid_group) to the symmetric group. Then one can delete the $m$-th braid. This is a well defined homomorphism $d\_m:K\_n\rightarrow K\_{n-1}$. So is there for every $n\in \mathbb{N}$ a braid $1\neq b\in K\_n$ with $d\_m(b)=0$ for all $m$.


This is clearly true for $n=2$, as $K\_1$ is trivial and it is also true for $n=2$ (The "standard" braid does the job). What about higher $n$. Is there a nice construction, that works for every $n$ ?



Borromean braids
Certain elements in the $n-1$st term of the lower central series of the pure braid group should work. 
The pure braid group is generated by generators $\beta\_{i,j}$ where the $i$th strand pushes a finger over the intervening strands and hooks with the $j$th strand.
Then, when $n=3$, the commutator $[\beta\_{1,2},\beta\_{1,3}]$ is Brunnian in in your sense. For $n=4$ you can consider $[\beta\_{1,2},[\beta\_{1,3},\beta\_{1,4}]]$, etc. You need to make sure your commutator includes every $\beta\_{1,k}$.


Something like $[\sigma\_1^2,[\sigma\_2^2,\sigma\_3^2]]$ would also work.


The reason this works is that deleting a strand from the braid kills at least one generator involved in the iterated commutator, so that it collapses to $1$. That's why you need to include a generator $\beta\_{i,j}$ that involves each strand.


(This has been edited to remove inaccuracies of previous versions.)


