Action of universal R-matrix of U_q(sl_2)

Question

My question is really simple but requires a few definitions. No special knowledge of quantum groups should be needed, it is more about tensor algebra. Let $q \in \mathbb{C}$ with $q \neq 0, \pm 1$. Denote by $U_q(\mathfrak{sl}_2)$ the complex associative $\mathbb{C}$-algebra generated by the elements $E,F,K,K^{-1}$ subject to the relations $KK^{-1}=K^{-1}K=1$, $KEK^{-1}=q^2E$, $KFK^{-1}=q^{-2}F$, and $[E,F] = \frac{K-K^{-1}}{q-q^{-1}}$. We create another element $H$ by setting $K = q^H$ with $q=e^h$. This doesn't actually make perfect sense, but this is a subtlety I think we can ignore here.

Consider a universal R-matrix for $U_q(\mathfrak{sl}_2)$, a special invertible element of $U_q(\mathfrak{sl}_2) \otimes U_q(\mathfrak{sl}_2)$:

$R = \left( \sum_{n=0}^\infty q^{\frac{n(n+1)}{2}} \frac{(1-q^2)^n}{[n]_q!} E^n \otimes F^n \right) q^{-\frac{1}{2}(H \otimes H)}$

where $[n]_q! = [n]_q [n-1]_q \cdots [1]_q$, and $[n]_q = \frac{q^n-q^{-n}}{q-q^{-1}}$. Let $\lambda \neq 0$ and let $V_\lambda = \langle v_0, v_1, v_2, \ldots \rangle$ be the $\mathbb{C}$-vector space generated by distinct elements $\{v_0,v_1,\ldots\}$. We define an action of $U_q(\mathfrak{sl}_2)$ by

$Kv_l = q^{\lambda - 2l}v_l$,

$Fv_l = v_{l+1}$,

$Ev_l = [l]_q[\lambda+1-l]_q v_{l-1}$

Now we define a function $c: V_\lambda \otimes V_\lambda$ by $c(v_i \otimes v_j) = \tau(R(v_i \otimes v_j))$, where $\tau: V_\lambda \otimes V_\lambda \to V_\lambda \otimes V_\lambda$ is the flip operation.

We see that $c(v_0 \otimes v_0) = q^{-\frac{1}{2}\lambda^2} v_0 \otimes v_0$. Only $q^{-\frac{1}{2}H \otimes H}$ acts on $v_0 \otimes v_0$, since $E^n \otimes F^n$ annihilates $v_0 \otimes v_0$ for $n \geq 1$.

The source I am reading says that $c(v_0 \otimes v_1) = q^{-\frac{1}{2}\lambda(\lambda-2)}v_1 \otimes v_0$. Why is this? I think that $(E \otimes F)(v_0 \otimes v_1) = (E v_0 \otimes F v_1) = 0$. I seem to be wrong but I don't see why.

Answer 1

I found the answer on page 178 of Kassel's Quantum Groups. Given a braided bialgebra $(H,\mu,\eta,\Delta,\epsilon,R)$, with $V,W$ left $H$-modules, and a universal R-matrix $R = \sum_i s_i \otimes t_i \in H \otimes H$, we have that

$c_{V,W}^R(v \otimes w) = \tau_{V,W}(R(v \otimes w)) = \sum_i t_i w \otimes s_i v$

for all $v \in V$, $w \in W$. That gets the answer I wanted in my question. Though to me it still looks like the flip is applied before $R$ rather than after, so I am not totally satisfied.

Edit: I realized now that the flip is actually being applied to $R$ before $R$ gets applied to $v \otimes w$. The notation just doesn't look right to me. Can anybody explain why its $\tau(R(v \otimes w))$ instead of $R(\tau(v \otimes w))$?

Answer 2

$c$ means "act the element $R$, then do $\tau$". So claiming $c(v_0 \otimes v_1) = q^{-\lambda (\lambda-2)/2} v_1\otimes v_0$ is equivalent to $R (v_0\otimes v_1) = q^{-\lambda(\lambda-2)/2} v_0 \otimes v_1$. This follows from your description of $R$: you're right that $E^n \otimes F^n$ acts as zero on $v_0 \otimes v_1$ for $n \geq 1$, so only the action of $q^{-H\otimes H /2}$ matters, and since $K=q^H$ acts as $q^{\lambda -2}$ on $v_1$ the scalar is correct.