Not counting equivalent braids, are there finite or infinite numbers of 3-braids whose closures are trivial knot or links? If the answer is infinite, are there some patterns in those infinite numbers of braids, e.g. there exists some repeated parts?



Are there infinite number of 3-braids with trivial closure?
Suppose that $\beta$ is a three-braid whose closure is trivial. Suppose that $\gamma$ is any three-braid. Then $\gamma \beta \gamma^{-1}$ is again a three-braid with trivial closure. However, if $\beta$ is non-trivial, then for "generic" $\gamma$, the braids $\beta$ and $\gamma \beta \gamma^{-1}$ will not be equivalent.


To explain the word generic, note that $\beta$ commutes with its powers (and its roots, if any). Thus taking $\gamma$ to be a power (or a root, or a power of a root) of $\beta$ will result in $\gamma \beta \gamma^{-1}$ being equivalent to $\beta$. More generally, $\gamma$ should not lie in the centraliser of $\beta$. However, the centraliser is typically a very small subset of the braid group. (One notable exception to this is when $\beta$ lies in the centre of the braid group.)




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Here is a simple version of HJRW's suggestion: Let $\beta$ be the braid $\sigma\_1 \sigma\_2^{-1}$. This is a three-strand pseudo-Anosov braid. Thus its centraliser (in the braid group) is virtually $\mathbb{Z}^2$ (as it contains $\beta$ and the central element $\Delta$).


A bit surprisingly, this is also the *only* pseudo-Anosov three-braid (conjugacy class) closing to the unknot, due to the work of Birman-Menasco (cited in the comments).


