What is the relationship between universal enveloping algebra $U(g)$ and the QUE algebras $U_{q}(g)$

Question

I just know $U_{q}(sl_2)$ is universal enveloping algebra $U(sl_2)$ when $q$ tends to $1$. Due to Qiaochu Yuan's comment, this is true in general. Conversely, Suppose $sl_2$ has three basises $e, h, f$, then we let $E=e, F=f, K=q^{h}, K^{-1} = q^{-h}$, we can get $U_{q}(sl_2)$. What is the general case?

Answer 1

Let me assume $\mathfrak{g}$ semisimple for simplicity (I think that also works in general).

The answer to your question is yes, we always have $\lim\limits_{q \to 1} U_q = U$.

This is what Qiaochu said in the comments section. The structure of $U_q$ is richer mainly because of the interesting coproduct, leading to the notion of $R$-matrix.

To check it, $U_q$ has generators $E_i, F_i, K_i^{\pm1}$ and different scalars $q_i$ so that each $(E_i,F_i,K_i^{\pm 1},q_i)$ is isomorphic to $U_q(\mathfrak{sl}_2)$. You also want the $K_i$ to commute and same with the $q_i$. Finally there is a last type of relations, the quantum Serre relation and this is basically the only thing you need to check.

For example, in type $A_2$ you need to check that $$ \lim\limits_{q \to 1} (E_1^2E_2 - (q+q^{-1})E_1E_2E_1 + E_2E_1^2) = e_1^2e_2 - 2e_1e_2e_1 + e_2e_1^2 $$ which is clear.

Here I took the Lusztig presentation indexed by a set of simple roots but of course it also holds with different generators.

Just a remark about representation theory, if $q$ is not a root of unity $U_q$ and $U$ basically have the same representation theory. When $q$ is a root of unity, this is more complicated and look like representation of $U$ but in positive characteristic. This is well explained in Jantzen's book on quantum groups.