My primary reference for this question is the very good book *Quantum Groups and Knot Invariants* by C. Kassel, M. Rosso, and V. Turaev. I'm also drawing from P. Etingof and O. Schiffmann, *Lectures on quantum groups*, another very good book. If you want to see pictures of Lie bialgebras and quasitriangular structures, I will shamelessly self-promote my notes on Lie bialgebras, which I put together while studying for my quals last year.

Pick your favorite field of characteristic $0$, and let $\mathfrak g$ be a finite-dimensional Lie algebra. Abusing the language a bit, let me define a **(quadratic) Casimir** to be any symmetric $\mathfrak g$-invariant element $t\in \mathfrak g^{\otimes 2}$. (Equivalently, $t$ is symmetric and its image in the universal enveloping algebra $\mathcal U\mathfrak g$ is central.) I'd like to compare two constructions of "quantum" categories that start with this Lie theoretic data.

## Quantization of infinitesimally-braided categories

So pick a quadratic Casimir $t$. Let $\mathcal S$ be the category of finite-dimensional $\mathfrak g$-modules. For any two modules $(\pi\_U,U)$ and $(\pi\_V,V) \in \mathcal S$, we have an element $t\_{U,V} \in {\rm End}\_{\mathfrak g}(U\otimes V)$ given by $t_{U,V} = (\pi_U \otimes \pi_V)(t)$; it is a $\mathfrak g$-morphism because $t$ is $\mathfrak g$-invariant. Moreover, under the canonical "flip" map $\sigma_{U,V}: U\otimes V \to V\otimes U$, $t_{U,V}$ maps to $\sigma_{U,V}t_{U,V}\sigma_{V,U} = t_{V,U}$, because $t$ is a symmetric element of $\mathfrak g^{\otimes 2}$. Along with the rule for how $\mathfrak g$ acts on tensor products (determining the action of $t_{U,V\otimes W}$ on $U\otimes V\otimes W$), it follows that $t$ defines on $\mathcal S$ the structure of a **infinitesimally braided category**.

Then recall the following very general construction. Let $\Phi$ be a **Drinfeld associator**: i.e., $\Phi(A,B)$ is a formal power series in non-commuting variables of the form $\exp($a Lie series$)$, satisfying certain nonlinear conditions (a "pentagon" and two "hexagon"s) that make the following construction work. (Only one Drinfeld associator is explicitly known, given by the solution to the Knizhnik-Zamolodchikov differential equation, and its coefficients are real but transcendental. But the equations defining $\Phi$ at each order are an over-determined linear system in rational coefficients, so if any solution exists, a rational one does. And by a theorem of Le and Murakami, up to equivalence of braided monoidal categories, the output of this construction does not depend on the choice of associator.)

Then define a new category $\mathcal S[[\hbar]]$. The objects are the same as those of $\mathcal S$, and $\hom_{\mathcal S[[\hbar]]}(V,W) = \hom_{\mathcal S}(V,W)[[\hbar]]$ (formal power series of morphisms) with composition given simply by the composition in $\mathcal S$ extended $\hbar$-linearly (and adicly), i.e. following the rule for multiplication of formal power series.

Moreover, give $\mathcal S[[\hbar]]$ the tensor product inherited from $\mathcal S$. However, give it nontrivial associativity and braiding constraints. Namely, define the associator by $a_{123} = \Phi(\hbar t_{12}, \hbar t_{23})$ and $c = \sigma \exp(\hbar t / 2) = \exp(\hbar t / 2)\sigma$. The axioms for the Drinfeld associator $\Phi$ imply that $\mathcal S[[\hbar]]$ is a (weak) braided monoidal category.

## Quantization of quasitriangular Lie bialgebras

Again let's start with $t$ a quadratic Casimir. But let's suppose a bit more: let's suppose that $t$ is the symmetrization of some element $r\in \mathfrak g^{\otimes 2}$ satisfying the (very over-determined) **classical Yang-Baxter equation** (CYBE). Namely, let $\beta: \mathfrak g^{\otimes 2} \to \mathfrak g$ be the Lie bracket, and use the obvious index notation. Then the CYBE says:
$$ \bigl[(\beta\_{13} \otimes {\rm id}\_2 \otimes {\rm id}\_4) + ({\rm id}\_1 \otimes \beta\_{23} \otimes {\rm id}_4) + ({\rm id}\_1 \otimes {\rm id}\_3 \otimes \beta\_{24}) \bigr] (r\_{12}\otimes r\_{34}) = 0$$
A choice of such an $r$ is a **quasitriangular structure** on $\mathfrak g$. Consider the map $\delta: \mathfrak g \to \mathfrak g^{\wedge 2}$ given by antisymmetrizing the output of $x \mapsto {\rm ad}_x(r)$, where ${\rm ad}_x$ is the action of $x\in \mathfrak g$ on $\mathfrak g^{\otimes 2}$. It follows from the CYBE and the fact that $t$ is $\mathfrak g$-invariant that $\delta$ satisfies the co-Jacobi identity.

Then by a general theorem of Etingof and Kazhdan, the data $(\mathfrak g,r)$ as above along with a choice of Drinfeld associator $\Phi$ determines a (noncommutative, noncocommutative) Hopf algebra $\mathcal U_\hbar \mathfrak g$ (over power series in $\hbar$) and a nontrivial braiding on the (strongly-associative monoidal) category of (finitely-generated topologically free) $\mathcal U_\hbar\mathfrak g$-modules. As an algebra, $\mathcal U_\hbar\mathfrak g \cong \mathcal U\mathfrak g[[\hbar]]$, I think, so as a category (although maybe not as a braided monoidal category?), $\mathcal S[[\hbar]] \cong (\mathcal U_\hbar\mathfrak g)$-mod.

## Questions

The two constructions above are similar: both start with a Lie algebra and a Casimir, and both end up with braided monoidal categories. But they're not the same. The first construction had to deform the associator to make everything work out, whereas in the second the associator is what I called **strongly associative**: the category embeds naturally in the category of vector spaces, and the associator is the same as the essentially-trivial associator there that we all know and love. (So it's almost "strictly associative"; "strong" seems like a good antonym for "weak".) On the other hand, the second construction required more data: it required choosing a classical $r$-matrix.

So: how related are these two constructions? For example, suppose I run the first construction, but happen to know that $t$ comes from an $r$-matrix. Does this tell me anything about more about the structure of $\mathcal S[[\hbar]]$? Conversely, if I take the first construction, and then do some Mac Lane-style strictifying, how close can I get to the second construction?