have $$\exp(du)(w)=(1-b_0u)\left(\big(1-b_0u\big)^{-1}\sha w\right)\qquad (w\in
    \fA^1).$$ On the generators we have $$\label{equ:dgens}
 \exp(du)(a)=a\big(1-b_0u\big)^{-1},\qquad\exp(du)(b_j)=b_j\big(1-b_0u\big)^{-1},\
 j=0,\dots,N-1.$$

In fact, we can replace the whole $\fA$ by $\fA^1$ in the first part of Prop. \[prop:d\]. We can do the same in the next corollary. However, in the proof of Theorem \[thm:RDS\] we only need this weaker version.

\[cor:Delta\] Let $u$ be a formal parameter. Let $\gD_u=\exp(-du)\circ \Phi_{b_0u}\in {\rm Aut}(\fA)[\![u]\!]$ (here $\circ$ means the composition). Then $$(1-b_0u)^{-1}*w=(1-b_0u)^{-1} \sha \gD_u(w), \qquad \forall
w\in\fA^1.$$ In particular, for $w\in \fA^0$ by taking $\operatorname{{reg}}$ on both sides of the above equation we get $$\operatorname{{reg}}_\sha\Big((1-b_0u)^{-1}*w\Big)=  \gD_u(w).$$

\[prop:reg\] For $\om_0=a\om_0' \in\fA^0$ we have $$\operatorname{{reg}}_\sha^T\left(\big(1-b_0u\big)^{-1} \om_0\right)=
   \exp(-du)(\om_0)e^{Tu}=a\left(\big(1+b_0u\big)^{-1}\sha \om_0'\right)e^{Tu}.$$

Theorem \[thm:RDS\] now follows easily from a detailed computation as in [@IKZ]. As a matter of fact, the same argument shows that [@IKZ Prop. 10] and its Cor. are both valid in our general setup if we replace $\HH^0$ there by $\fA^0$.

Seeded (or weight one) relations {#sec:seed}
================================

When $N\ge 4$ there exist linear relations among MPVs of weight one by a theorem of Bass [@Bass]. These relations are important because by multiplying any MPV of weight $w-1$ by such a relation we can get a relation between MPVs of weight $w$ which we 