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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!

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    $\begingroup$ 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). $\endgroup$ – Noah Snyder Apr 16 at 17:14
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    $\begingroup$ @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. $\endgroup$ – Olivia Borghi Apr 16 at 17:36
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    $\begingroup$ 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. $\endgroup$ – Noah Snyder Apr 16 at 17:53
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    $\begingroup$ 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. $\endgroup$ – Noah Snyder Apr 16 at 17:55
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    $\begingroup$ 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. $\endgroup$ – Noah Snyder Apr 16 at 18:08
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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 $\Gamma_n$.

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  • $\begingroup$ 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! $\endgroup$ – Olivia Borghi Apr 19 at 0:42
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    $\begingroup$ 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 $\endgroup$ – Adrien Apr 19 at 7:54
  • $\begingroup$ 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. $\endgroup$ – Olivia Borghi Apr 19 at 15:52
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    $\begingroup$ 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 ! $\endgroup$ – Adrien Apr 19 at 18:27
  • $\begingroup$ I did work through that but sanity checks are always nice. Thanks again for everything! $\endgroup$ – Olivia Borghi Apr 19 at 21:48
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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!

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

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