# How to improve this argument for a restriction of the universal $R$-matrix to $U_{q}^{+}(\mathfrak{g})\otimes U_{q}^{0}(\mathfrak{g})$?

The standard universal $R$-matrix for quantum group algebra $U_{q}\left(\mathfrak{g}\right)$, where $\mathfrak{g}$ is of finite type, is

$$R_{q}=exp\left(q\sum_{i,j}\left(B^{-1}\right)_{ij}H_{i}\otimes H_{j}\right)\prod_{\beta}exp_{q_{\beta}}\left[\left(1-q_{\beta}^{-2}\right)E_{\beta}\otimes F_{\beta}\right]$$

where the product is over all the positive roots of $\mathfrak{g}$, and the order of the terms is such that $\beta_{r}$ appears to the left of $\beta_{s}$ if $r\ge s$ [CP95, Theorem 8.3.9].

I conjecture that brading on the Borel subalgebra $U_{q}^{+}\left(\mathfrak{g}\right)\otimes U_{q}^{0}\left(\mathfrak{g}\right)$ is determined by the restriction of this universal $R$-matrix to that subalgebra, and that restriction is

$$R_{q}=exp\left(q\sum_{i,j}\left(B^{-1}\right)_{ij}H_{i}\otimes H_{j}\right)$$

Indeed, consider the braiding in $U_{q}^{+}\left(\mathfrak{g}\right)\otimes U_{q}^{0}\left(\mathfrak{g}\right)$. The expressions that define it can include $H_i$, $E_\beta$, but not $F_\beta$. The also have to be consistent with the universal $R$-matrix for $U_{q}^{0}\left(\mathfrak{g}\right)$.

Since the definition of $U_{q}^{+}\left(\mathfrak{g}\right)\otimes U_{q}^{0}\left(\mathfrak{g}\right)$ is the same as of $U_{q}^{0}\left(\mathfrak{g}\right)$ except that expressions with $F_\beta$ are omitted or $F_\beta$ is considered to be zero as appropriate, the restriction of the universal $R$-matrix to $U_{q}^{+}\left(\mathfrak{g}\right)\otimes U_{q}^{0}\left(\mathfrak{g}\right)$ is obtained by omitting the terms with $F_\beta$ / setting $F_\beta$ to zero, which makes the arguments of the $exp_{q\beta}$'s zero.

Question

How can this argument be improved? I would appreciate any suggestions or helpful references.

Reference

[CP95] Vyjayanthi Chari and Andrew Pressley, A guide to quantum groups, 1st paperback (with corrections) ed., Cambridge University Press, Cambridge, 1995.