The standard-module KZ connection has the reflection residue

For $f\in P$, the relation with a generator $x\in {\mathfrak{h}}^{\ast }$ gives $[y,x]=x(y)-{\sum }_s{c}_s{\alpha }_s(y)\frac{x-sx}{{\alpha }_s}s$. The same formula for products follows from $[y,fg]=[y,f]g+f[y,g]$ and $\frac{fg-s(fg)}{{\alpha }_s}=\frac{f-sf}{{\alpha }_s}s(g)+f\frac{g-sg}{{\alpha }_s}$. Induction on polynomial degree gives $[y,f]={\mathrm{\partial }}_{y}f-{\sum }_s{c}_s{\alpha }_s(y)\frac{f-sf}{{\alpha }_s}s$. Each quotient is polynomial because $f-sf$ vanishes on the reflecting hyperplane of $s$.

Since $y(1\otimes v)=0$ in the induced module, step 1.1 gives $y(f\otimes v)={\mathrm{\partial }}_y(f)\otimes v-{\sum }_s{c}_s{\alpha }_s(y)\frac{f-sf}{{\alpha }_s}\otimes sv$. The final $s$ acts on the $E$ factor because the group element in the commutator acts on the inducing vector.

Add the correction term in [F4]. On $f\otimes v$, $(1-s)(f\otimes v)=f\otimes v-sf\otimes sv$. Combining this with step 2.1 cancels the terms containing $sf\otimes sv$ and yields ${\mathrm{\partial }}_y(f\otimes v)={\mathrm{\partial }}_y(f)\otimes v+{\sum }_s{c}_s{\alpha }_s(y)\frac{f}{{\alpha }_s}\otimes (1-s)v$. This is the connection $d+{\sum }_s{c}_s(d{\alpha }_s\mathrm{/}{\alpha }_s)(1-s_E)$ in the stated trivialization.

At a point of $H$ outside the other reflecting hyperplanes, every ${\alpha }_s$ with fixed hyperplane different from $H$ is nonzero, so its connection term is holomorphic there and has zero logarithmic residue. The reflections with fixed hyperplane $H$ are exactly $s_H^{\mathrm{\ell }}$, $1\le \mathrm{\ell }<e_H$; for each, ${\alpha }_{s_H^{\mathrm{\ell }}}$ is a nonzero scalar multiple of ${\alpha }_H$, so $d{\alpha }_{s_H^{\mathrm{\ell }}}\mathrm{/}{\alpha }_{s_H^{\mathrm{\ell }}}=d{\alpha }_H\mathrm{/}{\alpha }_H$. The coefficient of this logarithmic form is therefore $R_H={\sum }_{\mathrm{\ell }=1}^{e_H-1}{c}_{H,\mathrm{\ell }}(1-s_H^{\mathrm{\ell }})$.

Since $W_H$ is cyclic over $\mathbb{C}$, its representation on $E$ is a direct sum of eigenspaces. On the eigenspace where $s_H$ acts by ${\zeta }_H^j$, each $s_H^{\mathrm{\ell }}$ acts by ${\zeta }_H^{j\mathrm{\ell }}$, so $R_H$ acts by $r_{H,j}$.