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Consider the Kernel $K_n$ of the natural group homomorphism from the $n$-th 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 $n$ ?

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Jim's offered a fine answer, also the following question may also be relevant: – James Griffin Nov 11 '10 at 11:14
In fact, these form a group consisting of all pseudo-Anosov elements, as proved by Kim Whittlesey in her thesis: – Ian Agol Nov 11 '10 at 17:00
up vote 11 down vote accepted

Anything 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.

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.)

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Either I am misunderstanding something or for $n=3$ the element $[\sigma_1,\sigma_2]$ does not lie in the Kernel $K_n$. Its image is $[(1,2),(2,3)]=(1,2,3)\in \Sigma_3$. – HenrikRüping Nov 11 '10 at 13:11
Sorry, you should take pure braid generators to make this work! I will edit this to reflect that. – Jim Conant Nov 11 '10 at 13:42

Ted Stanford has a paper on this topic, giving a set of generators for the kernel. See

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