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?
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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.)
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).
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$\begingroup$ So... Is the answer yes or no? Even if we consider $\gamma$'s that don't lie in the centralizer of $\beta$, then it's not clear that an infinite number of $\gamma$'s will give pairwise distinct non-trivial braids. $\endgroup$ Commented Jan 12, 2023 at 15:16
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2$\begingroup$ @NajibIdrissi: The centraliser of a typical braid is small. For instance, if the braid is pseudo-Anosov then the centraliser is virtually cyclic. So there are many examples where the centraliser is of infinite index in the braid group, and in any of these cases Sam Nead's argument will give you an example. $\endgroup$– HJRWCommented Jan 12, 2023 at 16:06
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$\begingroup$ Thank you, that answers my questions. May I ask a bit further, if I exclude those conjugate braids, i.e., only considering braid not in the form of $\gamma\beta\gamma^{-1}$, is the answer still infinite? $\endgroup$ Commented Jan 13, 2023 at 1:32
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3$\begingroup$ Interesting... and true! A search finds the paper "Studying links via closed braids. III. Classifying links which are closed 3-braids" by Birman and Menasco. Their classification theorem says that there are only finitely many (at most three) conjugacy classes in $B_3$ closing to any fixed knot type. Apparently this result is older (due to Magnus and Peluso) for three-braids closing to the unknot. There one will need the relationship between the Burau representation and the Alexander polynomial. en.wikipedia.org/wiki/… $\endgroup$– Sam NeadCommented Jan 13, 2023 at 7:31