# Computing the ribbon element of $\operatorname{U_q}(\mathfrak{sl}_2)$

I'm trying to understand the computation of the ribbon element of $\operatorname{U_q}(\mathfrak{sl}_2)$. I've laid out an argument by Andre Henriques below. I don't understand how the conclusion (bold sentence in last paragraph) is arrived at. The result $q^{\frac{1}{2}i^2+i}$ comes from what I call equation '***' directly above that sentence in bold. However, it seems to only be true for the case $n=1$. My question is, what allows us to forget about the other values of $n$? Here is the argument:

According to a computation by Andre Henriques, the $R$-matrix for quantum group $\operatorname{U_q}(\mathfrak{sl}_2)$ is $$\sum_{n,m}\frac{1}{n!}(\frac{1}{2}\operatorname{log}(q))^nq^{m \choose 2}\frac{1}{[m]_{q}!}H^nE^m\otimes H^nF^m$$

With the $R$-matrix in hand we proceed with computing the ribbon element. The ribbon element is defined as $\nu=m(1\otimes \mu^{-1})R=m(1\otimes \mu)\bar{R}$. Here, Turaev tells us that $\mu=\prod K^{2n_i}_i$ where $n_i\in \mathbb{Z}$ come from a particular linear combination of simple roots $\alpha_i$ satisfying $2\rho=\sum n_i \alpha_i$. Here, $\rho$ is the element a weight space that satisfies $\langle 2\rho, \alpha_i\rangle=\langle \alpha_i, \alpha_i \rangle$. For the case $\operatorname{U_q}(\mathfrak{sl}_2)$ this just means that $\mu = K$ since the root system for $\operatorname{U_q}(\mathfrak{sl}_2)$ is one dimensional.

Look again at the definition of the ribbon element $\nu\in \operatorname{U_q}(\mathfrak{sl}_2)$. The definition is telling us that $\nu$ is an element that behaves the same for different compositions of $R$, $R^{-1}$, $\mu$, $\mu^{-1}$ and $m$. So let's do just that, compose both ways and see what falls out when we let $\nu$ act on a representation. The $R$ matrix for $\operatorname{U_q}(\mathfrak{sl}_2)$ is $$\sum_{n,m}\frac{1}{n!}(\frac{1}{2}\operatorname{log}(q))^nq^{m \choose 2}\frac{1}{[m]_{q}!}H^nE^m\otimes H^nF^m$$ so $$\nu=m(1\otimes \mu^{-1})R=\sum_{n,m}\frac{1}{n!}(\frac{1}{2}\operatorname{log}(q))^nq^{m \choose 2}\frac{1}{[m]_{q}!}H^nE^m K^{-1}H^nF^m\quad\quad \text{*}$$ and $$\nu=m(1\otimes \mu)\bar{R}=\sum_{n,m}\frac{1}{n!}(\frac{1}{2}\operatorname{log}(q))^nq^{m \choose 2}\frac{1}{[m]_{q}!}H^nF^m KH^nE^m\quad\quad \text{**}$$ Now we are interested in how $\nu$ acts on a category of representations. In particular, we want to check that $*$ and $**$ above act in the same way on each representation. Note that for an irreducible representation a central element should act by a scalar. Fix a representation $W_i$ and take the lowest weight vector $w_{-i}$ of that representation. For $\nu=m(1\otimes \mu^{-1})R$ we get \begin{align*} \nu(w_{-i})&=\big (\sum_{n,m}\frac{1}{n!}(\frac{1}{2}\operatorname{log}(q))^nq^{m \choose 2}\frac{1}{[m]_{q}!}H^nE^m K^{-1}H^nF^m \big ) w_{-i}\\ &=\big (\sum_{n}\frac{1}{n!}(\frac{1}{2}\operatorname{log}(q))^nH^nK^{-1}H^n\big ) w_{-i}\quad\quad \text{since $F$ on the lowest weight vector is 0}\\ &= \big (\sum_{n}\frac{1}{n!}(\frac{1}{2}\operatorname{log}(q))^nH^{2n}K^{-1}\big ) w_{-i}\quad\quad \text{***} \end{align*} In other words, $\nu$ acts on each representation of $\operatorname{U_q}(\mathfrak{sl}_2)$ by multiplication of the scalar $q^{\frac{1}{2}i^2 +i}$. Taking the highest weight vector $w_i\in W_i$ and acting on it by $\nu=m(1\otimes \mu)\bar{R}$ gives the same scalar action $q^{\frac{1}{2}i^2+i}$ on each representation as we were hoping.