Given the universal enveloping algebra, $U(\mathfrak{sl}(2))$ the coalgebra structure is defined such that the generators $X,Y$ and $H$ are primitive elements. From this, is there a "nice" way to motivate the coproduct for $U\_{q}(\mathfrak{sl}(2))$? Of course, this question can be generalized to: given $U(\mathfrak{g})$ how does one discover what the coproduct for $U\_{q}(\mathfrak{g})$ should be based on what the coproduct for $U(\mathfrak{g})$ is? Most texts simply plunk down the structure without really motivating where it comes from or how it was arrived at (or the various ways it could be arrived at).


Given that the generators for $U\_{q}(\mathfrak{sl}(2))$ are $E,F,K,K^{-1}$, I surmised that since $KK^{-1} = K^{-1}K = 1$ and $\Delta$ is to be an algebra morphism, then 
$$\Delta(KK^{-1}) = \Delta(K)\Delta(K^{-1})=1\otimes 1 = \Delta(K^{-1})\Delta(K) = \Delta(K^{-1}K)$$
seems to "naturally" suggest the assignment $\Delta(K) = K\otimes K$ and $\Delta(K^{-1}) = K^{-1}\otimes K^{-1}$. With this I then looked at
$$\Delta(EF-FE) = \Delta\left(\frac{K-K^{-1}}{q-q^{-1}}\right)$$
and was able to use some algebraic manipulation to, sort of, "coax out" that 
$$\Delta(E) = E\otimes K+1\otimes E \qquad \Delta(F) = F\otimes 1+K^{-1}\otimes F$$
This all seems well and good, but nevertheless pedestrian (and a tad artificial). Is there or are there some more elegant ways (revealing deeper connections) to motivate the coalgebra structure (or the entire Hopf structure) of $U\_{q}(\mathfrak{sl}(2))$ or more generally $U\_{q}(\mathfrak{g})$? Thanks in advance.



Hopf structure of Uq(sl(2))
Let me add a few motivations to the nice ones already provided.


The first goes through the dual Hopf algebra $O\_q(SL\_2)$; it is more roundabout, but each step is more naturally motivated. (this approach is explained in, e.g. Kassel's book on Quantum Groups. First, let me admit the "quantum plane", whose algebra of functions is defined as


$\mathbb{C}\_q[x,y]:=\mathbb{C}\langle x,y\rangle / (yx-qxy).$


If we choose to regard this as an algebra of functions on a quantum $\mathbb{C}^2$, then we should expect that whatever our definition of quantum matrices on $\mathbb{C}^2$, its algebra $O\_q(Mat\_2)$ should have a co-action, i.e. a map $\Delta:C\_q[x,y]\to C\_q[x,y]\otimes O\_q(Mat\_2)$. Let us define elements $a,b,c,d$ by the equations


$\Delta(x)=x\otimes a + y\otimes b$, $\Delta(y)=x\otimes c+y\otimes d$.


Note that these are the same equations as for the usual coaction of $O(Mat\_2)$ on $\mathbb{C}[x,y]$, and are just the formula for matrix multiplication written in a funny way.


Now ask that $O\_q(Mat\_2)$ be generated by those elements, subject to the requirement that $\Delta$ is a map of algebras. You find relations on $a,b,c,d$ as follows:


$\Delta(xy)=\Delta(x)\Delta(y)=x^2\otimes ac + xy\otimes ad + yx\otimes cb + y^2\otimes bd$.


Setting this equal to $1/q\Delta(yx)$ gives you relations like
$ca=qac$, $bc=cb$, $ad-da = (q-q^{1})bc$,


and so on, precisely the defining relations of $O\_q(Mat\_2)$. Direct computation tells you that $det\_q=ad-q^{-1}bc$ (if i remember correctly) is the unique central element in degree 2, and so quotienting by $det\_q-1$ defines $O\_q(SL\_2)$. Now, define $U\_q(sl\_2)$ as the dual Hopf algebra w.r.t to $O\_q(SL\_2)$, and using this, recover the relations for $U\_q(sl\_2)$ uniquely.




---


Now let me give another motivation for the relations in $U\_q(sl\_2)$ coming not from the quantum geometry point of view, but from braided tensor categories. First, for any number $q$, note that there is a braided tensor category structure on the category of $\mathbb{Z}$-graded vector spaces, where $\sigma(v\_k\otimes v\_l)=q^{kl} v\_l\otimes v\_k,$ for $v\_k, v\_l$ in degrees $k$ and $l$. Ranging over $q$ this essentially exhausts the possible braidings on $Z$-graded vector spaces. Let's call this category $C\_q$. Let's denote by $V\_k$ the one dimensional vector space concentrated in degree $k$.


Now, consider the tensor algebra of $V\_1$, $T(V\_1)$ in this category. As any tensor algebra, it admits the free coproduct $\Delta: V\_1\to V\_1\otimes V\_0 + V\_0\otimes V\_1$, making $T(V\_1)$ into a Hopf algebra in $C\_q$. Now if we have modules $M,N$ in $C\_q$, we can act on their tensor product $M\otimes N$ by the co-product, but we have to be careful!


$v . (m \otimes n) = \Delta(v) m \otimes n$, but now $T(V\_1)\otimes T(V\_1)$ acts on $m\otimes n$, by first braiding the second factor of $T(V\_1)$ past $m$, then acting in the obvious way:


$T(V\_1)\otimes T(V\_1)\otimes m \otimes n \xrightarrow{\sigma} T(V\_1)\otimes m \otimes T(V\_1)\otimes n$


In particular if $v\in V\_1$, then the braiding adds a factor of $q^{|m|}$ to one of the summands of $\Delta(v)$ when you braid past.


Now in this story, $T(V\_1)$ is the free algebra on $E$, which is $U\_q(n\_+)$ in this case. A $\mathbb{Z}$-graded vector space is the same as an integral $\mathbb{C}[K,K^{-1}]$-module, since the grading is determined by eigenspaces of $K$, and a $U\_q(n\_+)$-module in the category $C\_q$ is the same as an integral $U\_q(b)$-module. So in this latter way of phrasing things the extra $q^{|m|}$ we found has to be put in by hand, by putting a $K$ in the second term of the co-product: $\Delta(E)=E\otimes 1 + K\otimes E$.


On the one hand, this second explanation probably seems rather artificial; on the other hand it is saying that there is nothing at all mysterious about the appearance of $K$'s in the formula for the co-product, it's just that you were working in the symmetric category of vector spaces, rather than the category of graded vector spaces, where you should have worked. The penalty you pay is putting $K$'s in places to keep track of what the braiding was keeping track of for you.


