In Drinfeld's paper "Quasi-Hopf Algebras" he illuminates a process by which you can replace the $R \in A \otimes A$ associated to a quasi-Hopf QUE-algebra $(A, \Delta, \varepsilon, \Phi)$ over $k[[h]]$ with an $\overline{R}$ which satisfies the unitary condition. I describe the process here:


$\overline{R} = R\*(R^{21}\*R)^{-\frac{1}{2}}$


Using Sweedler notation if $R = \Sigma (r\_1 \otimes r\_2)$ then $R^{21} := \Sigma(r\_2 \otimes r\_1)$. The unitary condition states $RR^{21} = 1$. His argument for why $\overline{R}$ satisfies this condition is as follows:


$\overline{R}^{21}\overline{R} = R^{21}\*(R\*R^{21})^{-\frac{1}{2}}\*R\*(R^{21}\*R)^{-\frac{1}{2}} = R^{21}\*(R\*R^{21})^{-1}\*R = 1$


My question is about this second equality here. How does he arrive at this? Keep in mind that $A$ is not a commutative algebra. If there are any clarification questions please ask! Thank you so much!



Why does Drinfeld Unitarization work?
The short answer to your question is that if $x,y$ are elements in an algebra in topologically free $k[[\hbar]]$-modules whose constant term is 1, then they have a unique square root whose constant term is also one, and if $x,y$ commute then say the square root of $x$ also commutes with $y$. Indeed if $a$ is the square root of $x$, then
$$(yay^{-1})^2=ya^2y^{-1}=x$$
and because $yay^{-1}$ also has constant term equal to 1, we get $yay^{-1}=a$. This shows at once that $(RR^{2,1})^{\frac12}$ commutes with $R$.


One way to think about it is as follow (this is also explained in Joel's paper). Any finitely generated group $G$ has a so-called pro-unipotent aka malcev aka rational completion $G(\mathbb{Q})$. One of its definition is that it is the univrsal uniquely divisible group having a morphism from $G$. In other words, it is the universal group in which images of elements of $G$ have a unique $n$th root for any $n$. So roughly elements of this groups are the $x^{\lambda}$ where $x \in G$ and $\lambda \in \mathbb{Q}$. Now the same argument as above shows if $x,y$ commute, then so do any possibly rational power of their image in $G(\mathbb{Q})$ (uniqueness is again key here).


This has a relative version, where in the case at hand you roughly speaking apply this construction to the pure braid group $P\_n$ inside of the braid group $B\_n$: you get a certain group $B\_n(\mathbb{Q})^{rel}$ fitting into an exact sequence
$$1 \rightarrow P\_n(\mathbb{Q}) \rightarrow B\_n(\mathbb{Q})^{rel} \rightarrow S\_n\rightarrow 1. $$


Long story short you get this way a morphism from the so-called cactus group $\Gamma\_n$ (the group of which coboundary categories give representations) into $B\_n(\mathbb{Q})^{rel}$ by taking square roots of the generators of $P\_n$ inside there. Now for any quantized quasi-Hopf algebra, of more generally in any braided tensor category over $k[[\hbar]]$ in which the braiding satisfies
$$\beta\_{U,V}\beta\_{V,U} =id\_{U\otimes V} +O(\hbar)$$


the representations of $B\_n$ you get factor through $B\_n(\mathbb{Q})^{rel}$, hence restrict to representations of $\Gamma\_n$.


