I'm looking at motivating the standard deformation of $U(\mathfrak{sl}(2))$. As an algebra $U(\mathfrak{sl}(2))$ is generated by $X,Y$ and $H$ and subject to the relations $[X,Y] = H$, $[H,X] = 2X$ and $[H,Y] = -2Y$. From $q$-analysis I have that an integer $n$ is deformed according to $[n] = \frac{q^{n}-q^{-n}}{q-q^{-1}}$.


This seems to motivate, at least partially, the means of deforming $U(\mathfrak{sl}(2))$. That is, set $[X,Y] = [H] = \frac{q^{H}-q^{-H}}{q-q^{-1}}$ and then set $K = q^{H}$ and $K^{-1} = q^{-H}$. But what motivates the relations $KX = q^{2}XK$ and $KY = q^{-2}YK$?



Quantum group Uq(sl(2))
Hi Ryan,


You can prove that if $a,b$ are some elements in an algebra such that $[a,b]=\lambda b$ for $\lambda$ a scalar, then (in a context where this expression makes sense) $q^a b q^{-a}=q^{\lambda} b$: rewrite the relation as
$$ab=b(a+\lambda)$$
then
$$a^nb=b(a+\lambda)^n$$
therefore
$$q^ab=\sum \frac{\log(q)^na^n}{n!}b=b\sum \frac{\log(q)^n(a+\lambda)^n}{n!}= bq^{(a+\lambda)}=q^{\lambda}bq^a$$
since $\lambda$ commutes with everything.


Therefore it's in fact not a deformation but the same relation with $q^H$ instead of $H$.


In my opinion, what is more miraculous is the existence of a non-trivial deformation of the Hopf structure. Although it does not leads to explicit formulas, the conceptual explanation for that comes from the Kohno-Drinfeld theorem, or more generally from the Etingof-Kazhdan quantization functor.


