Obviously, I tumbled over
[Classification of (compact) Lie groups](https://mathoverflow.net/questions/6079/classification-of-compact-lie-groups)
- are the quantum Lie groups (or make that: algebras) easier to classify?
Or does the whole q-deformation thingie make it even more complicated?


(The classification scheme from the linked post is already to technical
for me, although I understand the basic idea. If an exhaustive list of
quantum Lie groups exists, I would already be happy with that for now.)



Classification of quantum Lie groups
What Scott's comment is getting at is that you need to have an abstract definition of "quantum Lie group" if you want to have a classification result. As the theory of quantized enveloping algebras and quantized coordinate algebras is currently formulated, this is not really how it works.


Rather, you start with a (finite-dimensional, semisimple, complex) Lie algebra $\mathfrak{g}$, and from its Cartan data you write down the Hopf algebra $U\_q(\mathfrak{g})$. I assume here that $q$ is not a root of unity. This is functorial in the somewhat limited sense that if the Dynkin diagram for $\mathfrak{g}\_1$ includes into the Dynkin diagram for $\mathfrak{g}\_2$ then there is a corresponding map of Hopf algebras $U\_q(\mathfrak{g}\_1) \to U\_q(\mathfrak{g}\_2)$ (so diagram automorphisms give rise to Hopf algebra automorphisms). However, this is different from defining a class of Hopf algebras satisfying some properties and then showing that they must all be of the form $U\_q(\mathfrak{g})$ for some $\mathfrak{g}$.


If you take $U\_q(\mathfrak{g})$ as given, however, then you can construct quantized coordinate algebras corresponding to all of the connected groups $G$ which have Lie algebra $\mathfrak{g}$.


First, let $G$ be the connected, simply connected Lie group with Lie algebra $\mathfrak{g}$. Now $U\_q(\mathfrak{g})$ has (morally) the same finite-dimensional representation theory as $U(\mathfrak{g})$ does. If we restrict to Type 1 representations, i.e. those in which the Cartan generators $K\_i$ act as powers of $q$ on weight vectors (as opposed to acting as negative powers of $q$, which is well-defined since $q$ isn't a root of unity), then the finite-dimensional representations are in 1-1 correspondence.


For each of these Type 1 representations, we can define matrix coefficients. Say $V$ is a finite-dimensional representation of $U\_q(\mathfrak{g})$, and let $v \in V$ and $\phi \in V^\*$. Then the associated matrix coefficient is $ c^V\_{\phi,v} : U\_q(\mathfrak{g}) \to \mathbb{C}, $ defined by
$$ c^V\_{\phi,v}(X) = \phi(X\cdot v). $$
These matrix coefficients actually live in the finite dual $U\_q(\mathfrak{g})^\circ$, and you define $\mathcal{O}\_q(G)$ to be the (Hopf) subalgebra of $U\_q(\mathfrak{g})^\circ$ generated by all of the matrix coefficients of all of the finite-dimensional representations. (Of course it suffices just to take irreducibles.) This gives you the quantized analogue of the Peter-Weyl decomposition of the algebra of functions on $G$.


Now there are other connected, but not simply connected, groups which have the same Lie algebra. Let's pick one and call it $G'$. Since $G'$ is a quotient of $G$, then (classically) the coordinate algebra $\mathcal{O}(G')$ is a subalgebra of $\mathcal{O}(G)$. In terms of the Peter-Weyl decomposition, $\mathcal{O}(G')$ is generated by matrix coefficients of all of the finite-dimensional representations of $\mathfrak{g}$ which integrate to give a representation of $G'$. Then you can define $\mathcal{O}\_q(G')$ to be the subalgebra of $\mathcal{O}\_q(G)$ generated by only those matrix coefficients.


Anyway, I don't know if that's what you were looking for exactly. Hope it's helpful. If you think of a more precise question to ask then you may find more answers forthcoming.


