Classical limit and Drinfelds realization of quantum groups

Question

Let $$\hat{\mathfrak{g}}=\mathfrak{g}\otimes\mathbb{C}[t,t^{-1}]\oplus\mathbb{C}c\oplus\mathbb{C}d$$ be untwisted affine Lie algebra (as defined in V.G.Kac, Infinite-Dimensional Lie Algebras, 3d ed. Cambridge University Press, 1990). For $x\in\mathfrak{g}$ we can define series $$x(z):=\sum_{k\in\mathbb{Z}}x(k)z^k:=\sum_{k\in\mathbb{Z}}x\otimes t^{k}.$$

Lusztig proved that, roughly speaking, classsical limit $q\to 1$ of quantum affine algebra $U_{q}(\hat{\mathfrak{g}})$ is universal enveloping algebra $U(\hat{\mathfrak{g}})$ and that Chevalley generators of $\hat{\mathfrak{g}}$, $e_i, f_i$, $i=0,1,...,n$, correspond to Chevalley generators of $U_{q}(\hat{\mathfrak{g}})$, $e_i, f_i$, $i=0,1,...,n$. Precise formulation (and proof) can be seen for example in J.Hong, S.-J. Kang, Introduction to quantum groups and crystal bases, AMS, 2002.

Drinfeld found realization of quantum affine algebras $U_{q}(\hat{\mathfrak{g}})$ (here is one article about it http://arxiv.org/abs/q-alg/9610035) in terms of generators $x_{i}^{\pm}(k), a_{i}(l), K_{i}^{\pm 1}, \gamma^{\pm 1/2}, q^{\pm d}$, $i=1,2,...,n$, $k,l\in\mathbb{Z},l\neq 0$. Classical limit $q\to 1$ of Drinfeld generators $x_{i}^{\pm}(0)\in U_{q}(\hat{\mathfrak{g}})$ are Chevalley generators $e_{i}, f_{i}\in \hat{\mathfrak{g}}\subset U(\hat{\mathfrak{g}})$ for every $i=1,2,...,n$.

My question is following:

Is classical limit of Drinfeld generators $x_{i}^{\pm}(k)\in U_{q}(\hat{\mathfrak{g}})$, $k\in\mathbb{Z}$, $i=1,2,...,n$, equal to elements $e_{i}(k), f_{i}(k)\in \hat{\mathfrak{g}}\subset U(\hat{\mathfrak{g}})$? In other words, is subalgebra of $U(\hat{\mathfrak{g}})$ generated by elements $e_{i}(k)$ (or $f_{i}(k)$), $k\in\mathbb{Z}$, $i=1,2,...,n$, classical limit of subalgebra of $U_{q}(\hat{\mathfrak{g}})$ generated by elements $x_{i}^{+}(k)$ ($x_{i}^{-}(k)$), $k\in\mathbb{Z}$, $i=1,2,...,n$?

Edit Classical limit of Drinfelds relations for $U_{q}(\hat{\mathfrak{sl}}_{n})$ should give corresponding relations in Lie algebra $\hat{\mathfrak{sl}}_{n}$, but:

In quantum algebra $U_{q}(\hat{\mathfrak{sl}}_{n})$ we have relation (Drinfelds realization) $$(z_{1}-q^{2}z_{2})x_{i}^{+}(z_{1})x_{i}^{+}(z_{2})=(q^{2}z_{1}-z_{2})x_{i}^{+}(z_{2})x_{i}^{+}(z_{1}).$$ If classical limit of $x_{i}^{+}(z)$ is $e_{i}(z)$, then the classical limit of the above relation would give $$(z_{1}-z_{2})[e_{i}(z_{1}),e_{i}(z_{2})]=0.$$ But in algebra $U(\hat{\mathfrak{sl}}_{n})$ should hold $$[e_{i}(z_{1}),e_{i}(z_{2})]=0.$$

Answer 1

It is more like comment, but seems too long.

I would say yes. I cannot give precise reference, but by all the idealogy it is yes. Or you see some "underwater stones" - problems ?

Ideas are like this: If you start with RLL=LLR description, then you need to make "Gauss" or "tringular" decomposition of L = LowTriangular*D*UpperTrianular to extract Drinfeld's currents.

In the classical limit "everything" takes the form A = A_{cl} + O(h). Now if you take L = L_{cl} , Triangular = Triangular_{cla} + O(h)

The simple fact that should hint that the "yes" answer is the following. So the decomponsition L = LowTriangular*D*UpperTrianular in classical limit corresponds to L_{cl} = LowTriangular_{cl} + D_{cl} + UpperTrianular_{cl}

You see multiplication in the first order corresponds to addition. And this means that corresponding classical currents are just currents to appropriate upper-lower triangular parts - which corresponds to x(z).