# Why does Drinfeld Unitarization work?

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!

• My recollection is that $RR^{21}$ is central and you're taking a central square root of it (i.e. by plugging it into the power series for square root). – Noah Snyder Apr 16 at 17:14
• @NoahSnyder do you happen to know where this is elaborated on? This explanation is not included in the paper as far as I can see. – Olivia Borghi Apr 16 at 17:36
• I haven't read this Drinfeld paper carefully, so I can't address whether it's talked about there. That square root means using the power series for square root is mentioned in Kamnitzer-Henriques which is where I learned about this stuff. Morally the central-ness is that the square of the braiding is a natural transformation of the identity functor, and such natural transformations correspond to central elements in the Hopf algebra. – Noah Snyder Apr 16 at 17:53
• The reason this is related to the square of the braiding is that $R^{21}R = \tau R \tau R = B^2$ where $\tau$ is the swap map and $B$ is the braiding. – Noah Snyder Apr 16 at 17:55
• It's possible that I'm getting confused. You can get by with less, you only need that it commutes with $R$ and $R^{21}$, and it's certainly central in the 2-strand braid group. – Noah Snyder Apr 16 at 18:08

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 $$\Gamma_n$$.

• Thank you so much! I appreciate you taking the time to help me and this helped with a different question I had about Cactus group to malcev complete braid group homomorphism in that paper! – Olivia Borghi Apr 19 at 0:42
• You're welcome. I should have mentioned that Khoroshkhin--Willwacher have an interesting operadic take on this trick which you might find useful: arxiv.org/abs/1905.04499 – Adrien Apr 19 at 7:54
• This paper is actually what inspired the question in the first place! Specifically they say $\overline{R}$ exists in the Malcev completion of the second braid group. Feels like $R$ should correspond to $b_1 \in B_2$ but its hard to see how $b_1(b_1b_1)^{-\frac{1}{2}}$ isn't the identity. – Olivia Borghi Apr 19 at 15:52
• That's the thing, $x=b_1^2$ is an element of $P_2$, and you take the square root of its image in the completion $P_2(\mathbb{Q})$ which is again an element of this group, in particular not equal to $b_1$. In other words has two square roots in the group I denoted by $B_2(\mathbb{Q})^{rel}$, one is $b_1$, the other is the unique one which is in $P_2(\mathbb{Q})$, and you're taking their ratio which is thus of order two. Apologies if you already figured that out and were mentioning this just for context ! – Adrien Apr 19 at 18:27
• I did work through that but sanity checks are always nice. Thanks again for everything! – Olivia Borghi Apr 19 at 21:48

Indeed those papers of Drinfeld are very terse. [My most sucessful use of MathOverflow] was also an explanation of an offhand remark in such a paper.

I think that the best reference for this construction is the paper by Berenstein-Zwicknagl https://arxiv.org/abs/math/0504155, see especially the later half of section 1.

In this paper, they describe $$\bar R$$ in the case where $$A$$ is a quantum group (I mean the quantized universal envelopping algebra of a semisimple Lie algebra). Pick two representations $$V, W$$ that you want to braid. Consider the action of $$R$$ on $$V \otimes W$$. All of the eigenvalues of $$R$$ will be of the form $$\pm q^k$$ for some $$k \in \mathbb Z$$. Then this factor $$(R^{21} R)^{-1/2}$$ is a diagonal matrix which gets rid of these powers of $$q$$.

Hope that helps!

• Thank you so much, Joel! I really appreciate you taking the time to help me out. – Olivia Borghi Apr 19 at 0:42