The difference between $q$-deformations and $h$-deformations

What is the difference between $q$-deformations and $h$-deformations of universal enveloping algebras?

In chapter XVI of Quantum groups by Kassel, a very precise definition of a quantum enveloping algebra is given. Such an algebra is called an $h$-deformation. In chapter XVII the Drinfeld-Jimbo algebras are introduced and it is proved that they are $h$-deformations. Chapter XVIII gives some uniqueness results of these $h$-deformations (the rigidity theorems).

Along the way, $q$-deformations are also introduced (although there is not really a proper definition). One result states that you can view a $q$-deformation as a Hopf-subalgebra of an $h$-deformation. One major difference is the quasi-triangular structure. By definition $h$-deformations have a quasi-triangular structure, but $q$-deformations don't. By Kassel's definition, this implies that $q$-deformations are not quantum enveloping algebras.

An obvious question arises: Are there still rigidity results for $q$-deformations?

Further I'd like to point out that other good books such as 'Quantum groups and their represenations' of Klimyk and Schmüdgen introduce Drinfeld-Jimbo algebras using the parameter $q$. However, Drinfeld introduced $h$-deformations first. Whenever people talk about $q$-deformations, they seem to be interested only in the representation theory and seem to forget what a deformation should be.

Other than the above question, I would very much appreciate it if anyone could shed some light on these issues. In particular I am confused why some many people accept the $q$-deformations without rigidity results (or I can't find these). For sake of completeness I would like to state that I also asked this question today, which is related but focuses on something different. I am not trying to spam the site with a bunch of questions.

EDIT: I would like to mention that the Hall algebra approach to quantum groups seems to recover $q$-deformations, but it looks very unlikely to find $h$-deformations in this way (how on earth would you get $e^h$ from counting extensions?)

1 Answer

While $\hbar$-quantum groups are really to be understood as deformations, $q$-quantum groups are somewhat different.

A quantum group in the $\hbar$-setting is a $\mathbb C[[\hbar]]$-Hopf algebra $\mathcal A_\hbar$. As such, the deformation parameter can be given a specific value only at $\hbar=0$, by letting $\mathcal A_0=\mathcal A_\hbar/(\hbar\mathcal A_\hbar)$. On the contrary a $q$--quantum group is a $\mathbb C(q)$-Hopf algebra $A_q$. After fixing a $\mathbb C[q,q^{-1}]$-rational form (and depending upon this choice) the parameter $q$ can be given any value in $\mathbb C\setminus 0$. In a sense the $\mathcal A_\hbar$ is a local deformation, while $A_q$ is a global deformation.

The difference is indeed made clear by rigidity results. Rigidity results for $\mathcal A_\hbar$ means that in an infinitesimal neigbourhood of $0$ any deformation of $\mathcal A_0$ is isomorphic to $\mathcal A_0$ itself (usually what one proves is rigidity of the associative algebra structure). On the other hand it is not true that in a neigbourhood of $q=1$ in the complex plane any specialized $q$-algebra is isomorphic to $A_1$. In fact, any neigbourhood of $q=1$ contains some roots of unity and therefore contains some special algebras which are not isomorphic as associative algebras to $A_1$. It is true, however, that if you restrict $q$ to be real inside $]0,1[$ you can prove some rigidity results.

As for the relation between the two, I haven't clear the embedding result that you mention. It is however possible that having fixed $q\in\mathbb R$ one can restrict $\mathcal A_\hbar$ to those power series that are convergent when $e^\hbar=q$.

ADDED

To answer your additional question:

$q$-deformations, as I said, are those that allow you to put the deformation parameter to a complex number therefore:

• They allow phenomena like the roots of unity case, which is the part of the theory that connects to braid invariants and was, historically, one of the main sources of interest in qg-theory.
• They allow to put a $C^*$--algebra structure on the quantum group thus connecting to operator algebraic approaches. You do not have anything like a $C^*$-algebra on $\mathbb C[[\hbar]]$-deformations.
• $C^*$-algebraic approach is essential to try to connect quantum group theory to NC geometry à la Connes.
• Through representation theory, $q$--deformations give rise to $q$-special functions and their properties like summation formulas, thus generalinzing the representation theoretic approach to classical special function theory.

This are but few (personal, idiosincratic) of the reasons of interests in $q$- deformations. Of course reconciling the two approaches is very interesting. For example: - on the $\mathbb C[[\hbar]]$-side much is known about NC index theorems. Understanding completely how this relates to the $q$-side of the story is what would bring in an understanding of NC index theory in terms of global geometry of the underlying Poisson group.

As for $q$ rigidity results I think you should read this: Reference request quantum SU(3)

• Thank you for the clarification. So why are people so interested in $q$-quantum groups other than their appearance in mathematical physics? What is the mathematical meaning (or even definition) of these $q$-quantum groups? Also do you know any reference on specific rigidity results in the $q$-case? Dec 23 '17 at 16:36
• It answer my questions, however, these motivations for looking at $q$-deformations come from a $C^*$-algebraic approach. There are also many algebraists working on $q$-deformations (even recovering some of these using Hall algebras). I don't really understand a purely algebraic reason for studying $q$-deformations other than we can actually do it. Feb 2 '18 at 12:21