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As is well-known, the quantized enveloping algebra $U_q(\frak{sl}_2)$ is not well-defined when $q=1$ because of the relation $$ [E,F] = \frac{K-K^{-1}}{q-q^{-1}}. $$ To address this problem, one has an alternative formulation of $U_q(\frak{sl}_2)$: One removes the offensive relation, introduces an extra generator $G$, along with extra relations, $$ [G,E] = E(qK+q^{-1} K^{-1}), ~~~~~~~~~ [G,F] = -(qK + q^{-1} K^{-1})F, $$ and $$ [E,F] = G, ~~~~~~~~ (q-q^{-1})G = K - K^{-1}. $$ and canonical extensions of the Hopf algebra maps. This new algebra can be shown isomorphic to $U_q(\frak{sl}_2)$, yet is well-defined at $q=1$. It is not equal to $U(\frak{sl}_2)$, but to a double cover of $U(\frak{sl}_2)$.

What happens in the general $U_q(\frak{g})$ case? Again $U_q(\frak{g})$ is not well-defined since we the relation $$ [E_i,F_j] = \delta_{ij}\frac{K_i-K^{-1}}{q-q^{-1}}. $$ Judging by the representation theory, there should exist a reformulation of $U_q(\frak{g})$, where the offensive relations are removed, $n$ new generators $G_i$ (where $n$ is the rank of $\frak{g}$) are introduced, along with an extra set of new relations, which I would guess as $$ [G_i,E_j] = \delta_{ij}E_i(qK_i+q^{-1} K_i^{-1}), ~~~ [G_i,F_j] = -\delta_{ij}(qK_i + q^{-1} K_i^{-1})F_i, $$ and $$ [E_i,F_j] = \delta_{ij}G_i, ~~~~~~~ (q-q^{-1})G_i = K_i - K_i^{-1}. $$ This should then be equal to $U_q(\frak{g})$, and is clearly well-defined when $q=1$. I would then guess that at $q=1$ we have an $n$-fold cover of $U(\frak{g})$, which one should consider as justification for the $2^n$ representations types of $U_q(\frak{g})$. Am I OK in reasoning here?

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The key point here is to understand how the parameter $q$ is fixed to a complex number. Usually $U_q(\mathfrak g)$ is introduced as a $\mathbb C(q)$-Hopf algebra with generators and relations, and relations are singular at $q=1$. This means that to obtain a $\mathbb C$-Hopf algebra from $U_q(\mathfrak g)$ one has first to fix a so-called integer form, i.e. a $\mathbb C[q,q^{-1}]$--subalgebra $\cal U$ such that ${\cal U}\otimes\mathbb C(q)=U_q(\mathfrak g)$.

What you are writing is one possible choice of this integer form. However the choice of integer form is, in general, not unique. Basically, if I remember correctly, for the semisimple case any lattice in between the weight lattice and the root lattice should provide you a different integer form, this should be explained in older papers by De Concini (back in the 90ies) on Qg's at roots of unity (i.e., the chapter De Concini and Procesi - Quantum groups in D-modules, representation theory, and quantum groups).

Your choice should correspond to what is called the simply connected integer form. The key (not so easy) part is to show that indeed the subalegbra generated by the elements you show is a $\mathbb C[q,q^{-1}]$-integer form. Details on such fact should be given in the cited reference.

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  • $\begingroup$ I prefer to work with the definition of $Uq_(𝔤)$ as the noncommutative algebra over $C$, where q is some fixed value of q. How this relates to the case of $C(q)$ definition is not clear to me. $\endgroup$ Apr 5, 2013 at 13:13
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Even if strictly speaking it's not true, it's always usefull to remember that morally $$K_i=q_i^{H_i}:=\exp(H_i\log(q_i))$$ where $q_i=q^{d_i}$ for some integer $d_i$ attached to the Cartan matrix of $\mathfrak g$, and where $H_i$ is the $i$th Cartan generator of $\mathfrak g$. This explain why the classical limit of $K_i$ should be 1.

Now set $$G_i=\frac{K_i-K_i^{-1}}{q_i-q_i^{-1}}$$

Then using again this heuristic it's easliy seen that you have $$\lim_{q\rightarrow 1} G_i= \lim_{q\rightarrow 1}\frac{q_i^{H_i}-q_i^{- H_i}}{q_i-q_i^{-1}}= H_i$$

It's now rather clear how to prove that it indeed work, since $G_i$ is precisely the R.H.S. of the relation you want to modify, so just replace it by $$[E_i,F_j]=\delta_{i,j} G_i$$ and add the relaiton $$(q-q^{-1})G_i=K_i-K_i^{-1}$$ which is nothing but the definition of $G_i$. That you get the same algebra is obvious because you did not really change anything. The former relation does not depend on $q$ anymore so gives you the appropriate relation at the classical limit, and the latter relation just become trivial.

Then following again the heuristic that $K_i$ should goes to 1 at the limit, take the quotient by the ideal generated by $K_i-1$, and you get the classical envelopping algebra.

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