product defined by $$(\om_1\cdots \om_r) \sha (\om_{r+1}\cdots
 \om_{r+s}):=\sum_{\substack{\gs\in
 S_{r+s},\gs^{-1}(1)<\cdots<\gs^{-1}(r)\\
 \gs^{-1}(r+1)<\cdots< \gs^{-1}(r+s)}}
   \om_{\gs(1)}\cdots \om_{\gs(r+s)}.$$

For example, we have $$\begin{aligned}
\ &L_N(1|1)L_N(2,3|1,2)=Z(y_{1,1})Z(y_{2,1}y_{3,3})=Z(b_1\sha
(ab_1a^2b_3))\\
 =&Z(b_1ab_1a^2b_3+2ab_1^2a^2b_3+(ab_1)^2ab_3+ab_1a^2b_1b_3+ab_1a^2b_3b_1)\\
 =&Z(y_{1,1}y_{2,1}y_{3,3}+2y_{2,1}y_{1,1}y_{3,3}+y_{2,1}^2y_{2,3}
 +y_{2,1}y_{3,1}y_{1,3}+y_{2,1}y_{3,3}y_{1,1})\\
 =&L_N(1,2,3|1,0,2)+2L_N(2,1,3|1,0,2)+L_N(2,2,2|1,0,2)\\
 \ &\hskip3cm +L_N(2,3,1|1,0,2)+L_N(2,3,1|1,2,N-2).\end{aligned}$$

Let $\fA_\sha$ be the algebra of $\fA$ together with the multiplication defined by shuffle product $\sha$. Denote the subalgebra $\fA^0$ by $\fA_\sha^0$ when we consider the shuffle product. Then we can easily prove

\[shahomo\] The map $Z:  \fA_\sha^0\lra \C$ is an algebra homomorphism.

On the other hand, it is well known that MPVs also satisfy the series stuffle relations. For example $$L_N(2|5)L_N(3|4)=L_N(2,3|5,4)+L_N(3,2|4,5)+L_N(5|9).$$ because $$\sum_{j>0}\sum_{k>0}=\sum_{j>k>0}+\sum_{k>j>0}+\sum_{j=k>0}.$$ To study such relations in general we need the following definition.

Denote by $\fA^1$ the subalgebra of $\fA$ which is generated by words $y_{s,i}$ with $s\in \Z_{>0}$ and $i\equiv 0,\dots,N-1\pmod{N}$. Equivalently, $\fA^1$ is the subalgebra of $\fA$ generated by words not ending with $a$. For any word $w=y_{s_1,i_1}y_{s_2,i_2}\cdots
 y_{s_n,i_n}\in \fA^1$ and positive integer $j$ we define the exponent shifting operator $\tau_j$ by $$\tau_j(w)=y_{s_1,j+i_1}y_{s_2,j+i_2}\cdots y_{s_n,j+i_n}.$$ For convenience, on the empty word we have the convention that $\tau_j(\bfone)=\bfone.$ We then define a new multiplication $*$ on $\fA^1$ by requiring that $*$ distribute over addition, that $\bfone*w=w*\bfone=w$ for any word $w$, and that, for any words $\om_1,\om_2$, $$\begin{gathered}
  \label{equ:defnstuffle}
 y_{s,j}\om_1*y_{t,k}\om_2 = 