Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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),pp109) 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 correspendence) 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),pp146--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 $K_q[x,y]$?

share|improve this question
3  
Aren't $U(\mathfrak g)$ and $U_q(\mathfrak g)$ isomorphic as algebras, only different as coalgebras? –  Allen Knutson Apr 11 '12 at 16:55
    
@Allen: I'm confused. How would you construct an algebra isomorphism here? @tzhang: What meaning do you attach to the symbol q here? It's used in more than one sense in the literature, sometimes as an indeterminate or arbitrary complex number or root of unity. Does it matter for your question? –  Jim Humphreys Apr 11 '12 at 20:10
1  
@Allen, I think the isomorphism occurs when your coefficient ring is a power series ring in the indeterminate $q$. If your coefficient ring is only $\mathbb{C}[q]$, or if $q$ is specialized to a complex value, then there is no reason to expect an algebra isomorphism. –  Christopher Drupieski Apr 11 '12 at 20:33
    
U(g) and Uq(g) are different as algebras and coalgebras, the relationship between “different” quantum deformations are given in mathoverflow.net/questions/55647/… –  tzhang Apr 12 '12 at 5:33
2  
I also heard that U_q and U are isomorphic q=e^h and working over C[h], as I remember it mentioned in the Cartier's Bourbaki talks. As far as I understand reason is simple H^2(g) are trivial - so we cannot deform Lie algebra structure and this also implies we cannot deform U(g) is a reasonably non-trivial way. Another to look on it - let us look on the category of finite-dim representations of U(g) - it is semi-simple - so there is no deformation of the category structure, the only thing we can deform tensor product, this indeed can be done, and tensor category of U_q(g) modules is ... –  Alexander Chervov Apr 12 '12 at 6:32
show 1 more comment

3 Answers 3

Two seemingly easy to read references are given in the papers 'An introduction to Quantum Lie Algebras' by Delius which is available here: http://arxiv.org/pdf/q-alg/9605026v1.pdf and 'Quantum Lie Algebras associated to $U_q(\mathfrak{gl}_2)$ and $U_q(\mathfrak{sl}_2)$' available here: http://arxiv.org/pdf/q-alg/9508013v1.pdf

They explicitly deal with quantized $\mathfrak{sl}_2$; the first link in terms of $U_h(\mathfrak{sl}_2)$ and the second in terms of $U_q(\mathfrak{sl}_2)$ which it would appear is the case you are interested in.

These are from the nineties so I am sure more modern references are available.

share|improve this answer
    
Thank you for the references you have given, but it seems that the axioms of "Quantum Lie Algebras" there are not very algebraic explictly. I have found that V. Lyubashenko, A.Sudbery "Quantum Lie algebras of type A_n" arxiv.org/abs/q-alg/9510004 and "Quantum deformations of simple Lie algebras" cms.math.ca/10.4153/CMB-1997-017-6 more usefull to the problems. Thank you all the same! –  tzhang Apr 12 '12 at 5:30
add comment

You might be interested in the notion of a braided Lie algebra due to Majid. Roughly speaking this is a coalgebra ${L}$ in a braided category (ie $L$ is an object in a braided category category, with morphisms $\Delta:L \otimes L \to L$, and $\epsilon:L \to C$ satisfying the natural generalization of the axioms of a coalgebra), and in addition a morphism $$ [ , ]:L \otimes L \to L, $$ satisfying a "braided version" of the axioms of a Lie algebra.

The notion of the universal enveloping algebra of a Lie algebra generalizes to this context, and, quoting from Majid's paper http://arxiv.org/pdf/hep-th/9303148v1.pdf,

... the standard quantum deformations $U_q({\frak g})$ are understood as the enveloping algebras of such underlying braided Lie algebras ...

The best place to starting learning about these structures is probably Majid's Quantum Groups Primer book.

The paper arxiv.org/abs/q-alg/9510004 mentioned in Jake's answer contains some discussion of these structures.

share|improve this answer
add comment

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:

  1. They are non trivial deformation of the universal enveloping algebra as a Hopf algebra.
  2. 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)

share|improve this answer
    
@Nicola Ciccoli nice answer. Concerning the group-like elements let me point the paper "Free-Field Representation of Group Element for Simple Quantum Group" Alexei Morozov, Luc Vinet arxiv.org/abs/hep-th/9409093, their claim is that in quantum case group-like elements live in in U_q(g)\otimes Fun_q(G), and the formula exp(g) is substituted by roughly speaking by exp_q(g*t). It seems to me that this paper is not so well-known. –  Alexander Chervov Apr 17 '12 at 12:54
    
@Alexander Thanks. I am not sure I caught the point of Morozov-Vinet's paper. But let me add here that the sudden switch form $u_h(\mathfrak g)$ to $U_q(\mathfrak g)$ in my answer is due to the fact that group-likes are elements of the form $K=q^H$ where $H$ is in the Cartan of the original Lie algebra, and therefore do require power series in the formal setting but can be considered as generators on the global one. –  Nicola Ciccoli Apr 17 '12 at 16:50
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.