TITLE: Does there exist any "quantum Lie algebra" embeded into the quantum enveloping algebra U_q(g)?
QUESTION [11 upvotes]: We have known that any finite dim Lie algebra can be embeded into it's enveloping algebra $U(\mathfrak{g})$, my question is: is there any "quantum Lie algebra" embeded into the quantum enveloping algebra $U_q(\mathfrak{g})$?
The related question is, take $sl(2)$ generated by $\{X,Y,H|[XY]=H, [HX]=2X, [HY]=-2Y\}$ for example, consider the representation on polynomial $K[x,y]$, $K[x,y]$ is in fact a module-algebra over $U(sl(2))$, the  elment of $sl(2)$ can be represented by $X=x\frac{\partial}{\partial y}, Y=y\frac{\partial}{\partial x}, H=x\frac{\partial_q}{\partial x}-y\frac{\partial_q}{\partial y}$ (see  Kassel "Quantum groups" (GTM155), pp. 109). In fact, $\{x\frac{\partial}{\partial y}, y\frac{\partial}{\partial x}, x\frac{\partial_q}{\partial x}-y\frac{\partial_q}{\partial y}\}$ generated a three dim Lie subalgbebra (isomorphic to $sl(2)$ under the above correspondence) of derivation algebra of $K[x,y]$.
Similariy, Is there quantum Lie algebra contained in $U_q(sl(2))$? In fact, by Kassel "Quantum groups" (GTM155), pp. 146-149, there is an action of $U_q(sl(2))$ on quantum plane $K_q[x,y],  E=x\frac{\partial_q}{\partial y}, E=y\frac{\partial_q}{\partial x}, K=\sigma_x\sigma_y^{-1}, K^{-1}=\sigma_y\sigma_x^{-1}$ , so is there any finite dim quantum Lie algebra generated by $E,F,K,K^{-1}$, or does the operators  $x\frac{\partial_q}{\partial y}, y\frac{\partial_q}{\partial x}, \sigma_x, \sigma_y^{-1}, \sigma_y, \sigma_x^{-1}$  generate a Lie subalgebra of  of derivation algebra of $K_q[x,y]$?

REPLY [4 votes]: I hope no one gets offended if I summarize a couple of comments adding few details to it. Truly $U_h(\mathfrak g)$ as an associative algebra is not different from $U(\mathfrak g)[[h]]$ (here $\mathfrak g$ is semisimple, everything is char=0). 
This results is just a rigidity result on the associative algebra $U(\mathfrak g)$ (which reflects rigidity of the Lie algebra $\mathfrak g$). What I wrote inside the bracket seems innocent but it is not: one may say it depends on the fact that the Hochschild cohomology of $U(\mathfrak g)$ is isomorphic to the Chevalley-Eilenberg cohomology of $\mathfrak g$. I find this is nicely explained in http://people.mpim-bonn.mpg.de/crossi/LectETHbook.pdf .
When I first saw this result (which is already in the by now classical Chari-Pressley's book) my first impression was "so what's all the fuzz about quantum groups?". The point is that: 

They are non trivial deformation of the universal enveloping algebra as a Hopf algebra.
The isomorphism as associative algebras is neither explicit not canonical. We know it exists from purely cohomological arguments...

How does this connect to the embedding $\mathfrak g\hookrightarrow U(\mathfrak g)$? The canonical way to reconstruct $\mathfrak g$ inside $U(\mathfrak g)$ is to identify it with the set of primitive elements (primitive means $\Delta X=X\otimes 1+1\otimes X$). Therefore this embedding depends on the whole Hopf algebra structure (coproduct to determine primitive elements and product to show that they form a Lie algebra and generate a PBW basis). 
The set of primitive elements in $U_h(\mathfrak g)$ is trivial and certainly does not allow to reconstruct a PBW basis.
One may look for twisted primitive elements with respect to some group-like element different from 1. This is an interesting object, it contains the analogue of simple root vectors, but its algebraic properties are rather weak. Still: starting from twisted primtive elements and performing $q$-commutators, in the context of global quantization $U_q(\mathfrak g)$ it is possible to reconstruct all "root vectors" giving a PBW basis. But there is no obvious algebraic structure even on this set of $q$-root vectors.
Of course one may try to understand some kind of embedding $\mathfrak g_h\hookrightarrow U_h(\mathfrak g)$ as a deformation of $\mathfrak g\hookrightarrow U(\mathfrak g)$ as was done in some of the mentioned reference but everything is non canonical and, in my opinion, in the long run it just turns out to be a way of building up an explicit algebra isomorphism; which is known to be technically very complicated.
(this comment does not touch on the "braided" side of the story; that, I do not understand)