Let $\Delta_n$ stand for the Garside element of the braid group $B_n$. It turns out that the family of all Garside elements have the following ``operadic'' property:$$\Delta_n\left[ \Delta_{k_1},\ldots, \Delta_{k_n} \right] = \Delta_{k_1 + \ldots + k_n}$$where for braids $\beta, \beta_1,\ldots,\beta_n$, the symbol $\beta \left[ \beta_1,\ldots, \beta_n \right]$ stands for cabling: replace the $i$ strand of $\beta$ with $\beta_i$.
I need this property for a paper and I have a straightforward (and a bit tedious) proof of this, but I would prefer to not have to write it. So my first question is:
- Has anybody else come across this? Is this written somewhere? If so a reference would be greatly appreciated.
- And my second question: I call this property ``operadic'' because braids form an operad under cabling, but I'm not using operads really in my work. Is there an other more suitable name for this concept, a family of operations in an operad that is closed under compositions?
Update and answer after five years. I ended up writing up a rather tedious inductive proof but my gut feeling about it was right: there is the following easy proof: Clearly any such cabling is positive, non-repeating, and induces the flip permutation, properties also enjoyed by the Garside element. Since the map from positive non-repeating braids to the symmetric group is a bijection if follows that any such cabling is indeed equal to the Garside element.