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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 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 $$[e_{i}(z_{1}),e_{i}(z_{2})]=0.$$

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Not sure, but .......... please take into acount that e(z) contain only positive powers of z ! , it might be that it will imply what you want.... sorry at the moment I did not do all these for quite long time.. – Alexander Chervov Feb 11 '13 at 7:30
Yep, it seems it work: C[[z1,z2]] does not have zero divisors that it why we can cancel (z1-z2). It is important that we work with "positive currents" here - i.e. x(z), e(z) contain only z^k for k>=0 – Alexander Chervov Feb 11 '13 at 16:15
x(z) and e(z) contain all integer powers of z: $$x_{i}^{+}(z)\in U_{q}(\hat{\mathfrak{g}})[[z^{\pm 1}]]$$ and $$e_{i}(z)\in \hat{\mathfrak{g}}[[z^{\pm 1}]].$$ – Neph Feb 11 '13 at 23:27
Neph, I am not sure, but as far as I remember to work with RLL=LLR and to make this limit it seems it is necessary to work with positive currents and negative currents separately. And to see that x^{+-} is limit of corresponding e^{+-}. For example this work by my coursemate treats the case of the Yangian (only positive currents vs Yangian double - Z-currents). See formulas (*) page 6. – Alexander Chervov Feb 12 '13 at 10:05
Thank you. I'll take a look at the article. – Neph Feb 12 '13 at 12:53

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).

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I edited my question by adding a comment in which is described what bothers me. Probably I'm missing something but I don't know what. Thank you for responce. – Neph Feb 11 '13 at 0:27

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