Splitting of homomorphism from cactus group to permutation group

We all learned in kindergarten that the category of finite-dimensional (type I, say) $U_q(\mathfrak{g})$-modules is braided monoidal for $\mathfrak{g}$ a complex semisimple Lie algebra. This gives an action of the braid group $B_n$ on $V^{\otimes n}$ for any object $V$ in the category.

In order to construct certain maps on $V^{\otimes n}$, e.g. $q$-analogues of symmetrizers, antisymmetrizers, there is a problem: classically, the symmetrizer is defined as a sum over permutations, whereas in this case we have only an action of the braid group, not the symmetric group. To resolve this, one constructs a nice splitting of the quotient map $B_n \to S_n$ as follows: write a permutation as a minimal product of adjacent transpositions, and then replace each transposition with the braiding. Along the way you have to prove that any two such minimal decompositions of a permutation can be transformed into one another using only the braid relations; this proves that the splitting map is well-defined. Note that this map isn't a homomorphism, but it has some nice properties.

For various reasons, it is sometimes nice to consider a modification of the braiding. Drinfeld calls this a unitarized R-matrix, while some others (e.g. Kamnitzer, Henriques, Tingley) call this the cactus commutor. This guy squares to the identity but does not satisfy the braid relations, although there is something else called the cactus relation or the coboundary relation. This gives an action of the cactus group $J_n$ on $V^{\otimes n}$.

Question: is there an analogous splitting of the quotient map $J_n \to S_n$?

-
I think you went to a very different kindergarten than I did... – Andy Putman May 18 '11 at 4:51
Interesting question! My best guess was to try writing elements of S_n as a product of the generators s_{p,q} from Kamnitzer-Henriques where they generators are "nested" with the largest generators written first. However, a quick count shows that although this works for n=3, for n=4 there are only 22 products of that form so this doesn't work. – Noah Snyder May 18 '11 at 4:51
Also, if you're interested in quantum symmetrizers and anti-symmetrizers and their relation to the cactus group, then you might want to look at Zwicknagl's papers. – Noah Snyder May 18 '11 at 4:53
@Noah: I was led to this stuff by looking at Zwicknagl's papers! – MTS May 18 '11 at 5:53
A curious question, my intuition says that the answer is no, but I think that the question is too open for a negative answer. If you were to look at the group which drops the squaring to the identity, call it BJ_n, then I suspect that the map BJ_n &rarr; J_n would 'split', but that's not the answer to your question. Also does this question deserve an operad tag? – James Griffin May 18 '11 at 11:50