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I am working on Artin Groups, so called Dual Artin groups and the conjecture that they are isomorphic. Tuples of $n$ group elements can be acted on by the braid group $B_n$ in a particular way called the Hurwitz action. The following combinatorial problem has emerged and I wanted to know if it has a name, or if anyone can point me in any sort of direction.

  • Define an action $r$ on tuples where $r(x_1, \ldots, x_n) = (x_n, \ldots, x_2, x_1, x_2, \ldots x_n)$. For example $r(x_1) = (x_1)$, $r(x_1,x_2) = (x_2, x_1, x_2)$.

I will describe the problem for the specific case $n=3$.

Start with the tuple of single element tuples like so $h = ((1),(2),(3))$.

You act on adjacent pairs in $h$. For each pair we can act by swapping $h_1$ in to the second position, and $h_2$ concatenated by $r(h_1)$ in to the first position. Call this action $\sigma_1$. We can also do the opposite, move $h_2$ in to first position and $h_1$ concatenated by $r(h_2)$ in to second position. Call this $\sigma_1^{-1}$. Here is the result of these actions. I will omit the main surrounding brackets for readability. $$ \sigma_1 \cdot h = (2,1),(1),(3) \\ \sigma_1^{-1} \cdot h = (2),(1,2),(3) $$ We have equivalent actions on the other adjacent pair by $\sigma_2$ and $\sigma_2^{-1}$. $$ \sigma_2 \cdot h = (1),(3,2),(2) \\ \sigma_2^{-1} \cdot h = (1),(3),(2,3) $$ For clarity, let me give a couple more examples. $$ \sigma_1^2 \cdot h = (1,1,2,1),(2,1),(3) \\ \sigma_1\sigma_2 \cdot h = (3,2,1),(1),(2) $$

For any word in the letters $\{\sigma_1, \sigma_2, \sigma_1^{-1}, \sigma_2^{-1}\}$, we can compute the action of that word on $h$ by repeating the actions above. This generalizes to starting tuples $((1), \ldots , (n))$ in the obvious way. You would have $\sigma_1, \ldots \sigma_{n-1}$ and their inverses as generators of this action.

What I would love to know is: For some $n$, what kind of tuples can we produce from this action? E.g. for $n=3$, we can produce (1,1,2,1), because that is tuple in $\sigma_1^2 \cdot h$.

Side note: You may notice this is not a group action, but if we can cancel equal adjacent pairs, e.g. $(1,3,2,2)$ cancels to $(1,3)$, then this does become a braid group action.

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  • $\begingroup$ It seems that this does not define an action of the braid group; $\sigma_1 \sigma_2\sigma_1\cdot (1),(2),(3)=(3,1,2,1),(2,1),(1)$, but $\sigma_2\sigma_1\sigma_2 \cdot (1),(2),(3) = (3,2,1),(2,1),(1)$. $\endgroup$ Commented Nov 5 at 6:05
  • $\begingroup$ @TetsuyaIto, yep! I made a glaring error in explaining this. Apologies. This has now been fixed. Unfortunately, the likelihood of this being an already explored problem now seems very unlikely. $\endgroup$ Commented Nov 5 at 14:22

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