11
$\begingroup$

In the first answer for this question is writen, about the braided category of representation of the enveloping algebra $U(\frak{g})$, for $\frak{g}$ a semisimple Lie algebra:

The space of deformations of the braided tensor structure on these particular categories is one dimensional (I believe this is a result of Drinfeld), so we get a universal family of braided deformations of U(g)-mod by varying q in U_q(g).

What is the precise reference for this result? I guess here by tensor, the answerer means equivalently monoidal?

What happend if we take out "braided"? Does there exist deformations of the monoidal category of $U(\frak{g})$-modules which are not equivalent to the cateogry of modules of $U_q(\frak{g})$, for some $q$?

Improve this question
$\endgroup$

3 Answers 3

Reset to default
13
$\begingroup$

That the only monoidal deformations of the category of representations of $U(\mathfrak{g})$ is the category of representations of $U_q(\mathfrak{g})$ is known in Type A from Kazhdan-Wenzl (Adv. Soviet Math, 1993), for the category of vector representations in Type BCD by combining results of Wenzl-Tuba with Liu, and for $G_2$ by Kuperberg. It is open for the other exceptional Lie algebras. I think it's open in type BD if you include the spin representations, though Wenzl has some partial results.

Improve this answer
$\endgroup$
4
  • 1
    $\begingroup$ I care more about the $q$-deformations, so thanks you for your answer. As Adrien has pointed out in his answer, the question seems to be settled for $h$-adic case, so why is it not possible to "transfer" the proof to the $q$-case? $\endgroup$
    – Bas Winkelman
    Commented Nov 3, 2018 at 20:52
  • $\begingroup$ That's a good question. I'd think that the tangent space to the space of q-deformations has an injection to the space of h-deformations, which would say that the former is bounded if you know the latter... The results that I quoted are actually stronger than just saying you know the q-deformations. So there are two options: 1) I'm wrong and just to classify deformations is easier, or 2) there's something subtle going on like maybe the h-adic results assume something more (e.g. that the fiber functor also deforms). $\endgroup$
    – Noah Snyder
    Commented Nov 4, 2018 at 0:45
  • 1
    $\begingroup$ @BasWinkelman Compare with the statement that every formal deformation of $U(\mathfrak g)$ is isomorphic to $U(\mathfrak g)[[\hbar]]$ as an algebra (for $\mathfrak g$ simple). There isn't an analogous statement for the $q$ version. Now any deformation over $\mathbb C[q,q^{-1}]$ embeds as an algebra into $U[[\hbar]][\hbar^{-1}]$ but you definitely cannot conclude that any two $q$-deformations are equivalent in any sense. So the $q$-version is definitely more subtle. In particular you really need to work at the categorical level, and to restrict to (locally) finite-dimensional ones. $\endgroup$
    – Adrien
    Commented Nov 4, 2018 at 9:31
  • 1
    $\begingroup$ So @NoahSnyder I'd say this is 1), classifying formal deformations is definitely easier, inequivalent $q$-deformations could potentially becomes equivalent as formal ones. $\endgroup$
    – Adrien
    Commented Nov 4, 2018 at 9:37
10
$\begingroup$

$\newcommand{\g}{\mathfrak g}$ I think the statement Scott Carnahan was refeering to in his answer concerns in fact formal deformations of representations of $\g$, i.e. deformations over the ring $\mathbb C[[\hbar]]$. In that case the reference is Drinfeld's "On quasitriangular quasi-Hopf algebras and on a group that is closely connected with $Gal(\bar{\mathbb Q}/\mathbb Q)$". If I remember correctly, it says more generally that for any finite-dimensional Lie algebra there is a bijection between equivalence classes of such deformations as braided monoidal categories and $S^2(\g)^\g[[\hbar]]$, and if $\g$ is simple this is one dimensional. I think his results also implies that in the simple case those are the only deformations as monoidal categories as well (this is definitevely true, I'm just unsure if it can be extracted from his paper directly).

There is a nice interpretation of this and of similar results in the framework of shifted Poisson geometry, I highly recommend having a look at Pavel Safronov's very cool paper "Poisson-Lie structures as shifted Poisson structures" (arXiv:1706.02623). In particular this basically settle the question for monoidal deformations and arbitrary finite-dimensional $\g$.

Edit: My first answer was claiming something wrong in the general case, I corrected.

Improve this answer
$\endgroup$
5
  • $\begingroup$ Is this really deformations of the tensor category, or just of the Hopf algebra? $\endgroup$
    – Noah Snyder
    Commented Nov 3, 2018 at 22:18
  • 1
    $\begingroup$ If $\mathfrak{g}$ is a semisimple Lie algebra with $k$ simple components then the space of invariant symmetric bilinear forms is $k$-dimensional, and the same is true for invariant symmetric 2-tensors. (One can independently rescale the Killing form on each component by its own scalar.) $\endgroup$
    – Victor Protsak
    Commented Nov 4, 2018 at 5:08
  • $\begingroup$ @VictorProtsak RIght, of course, I edited. $\endgroup$
    – Adrien
    Commented Nov 4, 2018 at 8:55
  • $\begingroup$ @NoahSnyder Drinfeld's paper deals with twist-equivalence classes of deformations of $U(\mathfrak g)$ as a (quasi-triangular-)quasi-Hopf algebra, so you're right it's not entirely obvious that it also classifies deformations as (braided) tensor categories, but I think at least for semi-simple $\mathfrak g$ this is the case. $\endgroup$
    – Adrien
    Commented Nov 4, 2018 at 9:12
  • $\begingroup$ Note that while "being modules for an Hopf algebra" is an extra structure while "being modules for a quasi-Hopf algebra" is a property, which I think is preserved under formal deformations under reasonable assumptions... $\endgroup$
    – Adrien
    Commented Nov 4, 2018 at 9:57
5
$\begingroup$

The following is not really an answer but a rather too-long comment, with respect to your second question:

Does there exist deformations of the monoidal category of $U(\frak{g})$-modules which are not equivalent to the cateogry of modules of $U_q(\frak{g})$, for some $q$?

2-parameter deformations of the universal enveloping algebras $U_{pq}(\frak{g})$ have been studied such as for example: $U_{pq}[sl(2)]$, $U_{pq}[sl(3,C)]$, $U_{pq}[u(2)]$, $U_{pq}[u(1,1)]$, etc. Most of these are based on the two-parameter deformation function $$ [x]_{pq}=\frac{q^{x}-p^{-x}}{q-p^{-1}} $$ I am not really sure as to whether the representation categories of such deformed algebras are braided or even monoidal or if they may be described as deformations of the symmetric monoidal categories of the representations of $U(\frak g)$.
However, in the first of the above references, the authors claim that:

the two-parametric quantum group denoted as $U_{pq}[sl(2)]$ admits a class of infinite-dimensional representations which have no classical (non-deformed) and one-parametric deformation analogues, even at generic deformation parameters

which may be considered -in my understanding- as an indication that the representation categories of $U_q[sl(2)]$ and $U_{pq}[sl(2)]$ cannot be equivalent.
Thus, if i am correct in this and given the excerpt from the OP, the categories of $U_{pq}[sl(2)]$-modules cannot be braided.
Also, given Noah Snyder's answer, such categories of representations of two-parameter deformed algebras cannot even be monoidal.

On the other hand, in the last of the above cited articles, a $pq$-deformed coproduct $\Delta_{pq}$ is given for $U_{pq}[u(2)]$ and an element $R_{pq}$ such that: $$ \Delta_{qp}=R_{qp}\Delta_{pq}R_{qp}^{-1} $$ For $p=q^{-1}$, $R_{qp}$ reduces for the universal $R$-matrix of Drinfeld. However, i do not generally know if categorical implications (such as an associated braiding or deformed braiding) have been explicitly studied for $R_{qp}$ and more generally for deformations of this kind.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .