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This is a generic question, a good answer to it may be a reference to a corresponding paper\textbook, but any useful comments would be okay too.

Let $\mathfrak{g}$ be a (simple) Lie algebra and $U_q(\mathfrak{g})$ be its q-deformation of its universal enveloping algebra. For example, for $\mathfrak{g} = \mathfrak{su}(2)$ with positive $\mathfrak{e}$, negative $\mathfrak{f}$ roots and Cartan element $\mathfrak{h}$ the commutation relation read

$[\mathfrak{e},\mathfrak{f}] =\mathfrak{h},\quad [\mathfrak{h},\mathfrak{e}] = 2\mathfrak{e},\quad [\mathfrak{h},\mathfrak{e}] = -2\mathfrak{f}.$

After q-deformation the first relationship above turns into

$[\mathfrak{e},\mathfrak{f}] =[\mathfrak{h}]_q :=\frac{q^\mathfrak{h}-q^{-\mathfrak{h}}}{q-q^{-1}}$,

where $q \in\mathbb{C}^\ast$.

Let us now consider Lie group $G$ such that $\mathfrak{g}$ is its Lie algebra. Thinking in terms of differential geometry, Lie group can be seen as a smooth (real) manifold. Let some coordinate patch be parametrized by local coordinates $x_1,\dots x_n$. Here we can implement a noncommutative deformation of the coordinates. There are two examples I'm aware of: $xy-yx=\hbar$ and $xy = t yx$ for some nonzero $t$.

My goal is to understand relationship (if any) between the above mentioned deformations: q-deformation of the Lie algebra and noncommutative deformation of the coordinates on the corresponding Lie group. Returning back to $\mathfrak{su}(2)$, we have $SU(2)=S^3$ as its Lie group. The question is, does the noncommutative $S^3$ have anything to do with $U_q(\mathfrak{su}(2))$?

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2 Answers 2

up vote 10 down vote accepted

There is certainly a way to quantize the algebra of functions on a Lie group in a way that is compatible with the $q$-deformation of the universal enveloping algebra of its Lie algebra. The standard way to do it is this: let $G$ be a simple, connected, simply connected complex Lie group with Lie algebra $\mathfrak{g}$. Construct $U_q(\mathfrak{g})$ as you have done in your example, using the Cartan data of $\mathfrak{g}$.

Then it is a theorem that, up to some characters acting on the Cartan part of $U_q(\mathfrak{g})$ (i.e. on the $K_i$'s) the finite-dimensional representation theory of $U_q(\mathfrak{g})$ is the same as that of $\mathfrak{g}$. That is, the representations have the same dimensions, the same weight structures, the same decompositions of tensor products, etc.

For any finite-dimensional representation $V$ of $U_q(\mathfrak{g})$, one can define its matrix coefficients: for $v \in V$ and $f \in V^*$, the matrix coefficient is the linear functional $$ c^V_{f,v} : U_q(\mathfrak{g}) \to \mathbb{C}$$ given by $$ c^V_{f,v}(a) = f (a\cdot v).$$

Denote by $\mathcal{O}_q(G)$ the span of all of the $c^V_{f,v}$ inside the dual of $U_q(\mathfrak{g})$. With the algebraic operations defined by duality from those in $U_q(\mathfrak{g})$, this is a Hopf subalgebra of the finite dual $U_q(\mathfrak{g})^{\circ}$ which separates points of $U_q(\mathfrak{g})$. This is the analogue of the algebra of polynomial functions on $G$. For some values of $q$, you can introduce a $*$-structure and a norm and take a completion to get the analogue of the algebra of continuous functions.

Basically this is just mimicking the Peter-Weyl decomposition of the algebra of functions on a compact group. You can do lots of stuff with this, for instance look at quantum homogeneous spaces, building analogues of the de Rham complex, etc.

A good reference is Klimyk and Schmudgen, Quantum Groups and Their Representations.

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Thanks! I know Klimyk and Kassel's books, however I was not able to get this information form them. So the parameter in the Moyal $\ast$ product is exactly the $q$ from the Lie algebra, right? Why not all values of $q$ are allowed? Did you mean those which are roots of unity? –  Peter Feb 16 '11 at 23:06
    
The quantized function algebras are definitely in Klimyk-Schmudgen, I think in Chapter 9. The $q$ in $U_q(\mathfrak{g})$ and in $\mathcal{O}_q(G)$ are the same. The stuff on $*$-structures comes in right after they define $U_q(\mathfrak{g})$, in Chapter 6. Look at Theorem 20 in that chapter. I don't know about Moyal $*$-products or their relation to quantum groups, though. –  MTS Feb 16 '11 at 23:19
    
Oh yeah, I forgot to mention that the $*$-structure on $\mathcal{O}_q(G)$ also comes in by duality from the one on $U_q(\mathfrak{G})$. –  MTS Feb 16 '11 at 23:54

This is not at all an innocent question as there are really many notions of "quantizing stuff" around. A systematic comparison is probably not available (yet) for several reasons. Let me just try to illustrate this:

The $q$-deformations usually take the deformation parameter $q$ to be a complex number. The deformation process works by first presenting the classical (hopf, bi-, associative, whatever-) algebra in terms of generators and relations and then, as a second step, pluging in some $q$'s in the relations. In this approach it is typically easy to see that one stays with the same kind of algebraic structure (hopf, bi, asso, ...). This is guaranteed usually by design. The crucial point why this deserves the name deformation (at least for me) is that in these examples the new $q$-dependent algebra is still "as big as" the classical one. More precisely, one has a PBW like theorem also in the deformed setting.

Other approaches to deformation (usually with a parameter denoted by $\hbar$) stay with the same underlying vector space and try to deform the classical product structure (Lie, coasso, asso, whatever) in a $\hbar$-dependent way. Thus the "as big as" property is trivially fulfilled but the non-trivial question is whether one can keep the same kind of algebraic features (Lie, coasso, asso, ...) as one has classically. Moreover, it is usually rather difficult (at least in deformation quantization) to have a "good" dependence on $\hbar$. That is why one usually is forced to take formal series in $\hbar$ instead of nice convergent expressions, unless one has a special situation (on vector spaces, nice class of functions...).

So how can one relate the two: well, by the original (partly physical) motivation, the deformation parameters are related by $q = \exp(\hbar)$. So what one needs to do is to write out the relations in the $q$-deformed picture in terms of $\hbar$, interpret everything as a formal series if necessary (it usually will be). And then, this is the crucial step, one has to use the PBW property to pull-back the quantum algebraic structure to the classical underlying vector space. This allows to compare the classical and the quantum algebraic structures directly.

More concretely, I had a diploma student once working on this kind of comparison. He identified certain $q$-deformed spaces with certain star products for particular quadratic Poisson structures. I should remember this better (shame shame) but he discussed several versions of quantum $\mathbb{R}^4$ and $\mathbb{S}^3$.

In conclusion: there are definitely ways to compare things. But usually only on a very algebraic level of formal power series in the deformation parameter. This is sort of disapointing as in particular for the $q$-deformed world one knows good analytic properties like $C^*$-algebraic features for the Connes-Landi spheres etc.

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Thanks, I knew that $q=e^\hbar$ from physical considerations. If you'll find the references to those papers by your student please let me know. –  Peter Feb 16 '11 at 18:38
    
By the way, Peter, you ought to vote the answer up! –  Sammy Black Feb 16 '11 at 19:31
1  
@Peter: his name is Klaus Zimmermann, he is still in Freiburg Physics Institute quantum.uni-freiburg.de/… (sorry, MO corrupts this long link somehow..) It was in his master-like thesis, so it's not on the web I guess. But he will send you a copy, I guess. –  Stefan Waldmann Feb 17 '11 at 8:15
    
Thanks a lot, Stefan! –  Peter Feb 17 '11 at 19:02

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