2.1 Quotients in ${\mathbb{D}}{\mathcal{A}}$

From now on, we assume ${\mathcal{F}}\leq {\mathcal{A}}$ , so that all ${\mathcal{A}}$ -dioids are dioids. For a partial order $(D,\leq)$ , the down-closure of $U\subseteq D$ is $U^{\downarrow} := \left\{\,d\in D\,\mid\,d\leq u \textrm{ for some }u \in U\,\right\}$ .

If $D = (D,+^D,\cdot^D,0^D,1^D)$ is a dioid and $\rho$ a dioid-congruence on D, then $D/\rho$ , the set of congruence classes of $\rho$ , is a dioid under the operations

\begin{equation*} d/\rho + d'/\rho := (d+^Dd')/\rho, \quad d/\rho \cdot d'/\rho := (d\cdot^Dd')/\rho, \quad 0 := 0^D/\rho, \quad 1 := 1^D/\rho.\end{equation*}

The partial order $\leq$ on $D/\rho$ derived from $+$ is

\begin{equation*} d/\rho \leq d'/\rho :\iff (d+^Dd')/\rho = d'/\rho. \end{equation*}

An ${\mathcal{A}}$ -congruence on an ${\mathcal{A}}$ -dioid D is a dioid-congruence $\rho$ on D such that for all $U,U'\in {\mathcal{A}} D$ , if $(U/\rho)^{\downarrow} =(U'/\rho)^{\downarrow}$ , then $(\sum U)/\rho = (\sum U')/\rho$ .

If D is an ${\mathcal{A}}$ -dioid and $E\subseteq D\times D$ , there is a least ${\mathcal{A}}$ -dioid-congruence $\rho$ on D with $E\subseteq \rho$ , the intersection of all ${\mathcal{A}}$ -dioid-congruences on D above E.

The ${\mathcal{A}}$ -dioid ${\mathcal{A}}(X^*/\rho)$ of the quotient monoid $X^*/\rho$ is isomorphic to the quotient ${\mathcal{A}} X^*/\tilde\rho$ of the free ${\mathcal{A}}$ -dioid ${\mathcal{A}} X^*$ by a suitable ${\mathcal{A}}$ -congruence $\tilde\rho$ . If $\rho$ is determined by a set E of monoid equations, $\tilde\rho$ is determined by the corresponding set of dioid equations between singleton sets:

It is shown in Hopkins and Leiß (Reference Hopkins and Leiß2018) that ${\mathbb{D}}{\mathcal{A}}$ has coequalizers, and the quotient $D/\rho$ of an ${\mathcal{A}}$ -dioid D by an ${\mathcal{A}}$ -congruence $\rho$ on D is the coequalizer of two suitable ${\mathcal{A}}$ -morphisms $f,g:N\rightrightarrows D$ . The kernel $\ker(f)$ of an ${\mathcal{A}}$ -morphism $f:D\to D'$ between ${\mathcal{A}}$ -dioids is an ${\mathcal{A}}$ -congruence on D.

2.2 The polycyclic ${\mathcal{A}}$ -dioids $C_{n,{\mathcal{A}}}'$

Let $\Delta_n=\{p_0,\ldots,p_{n-1},q_0,\ldots,q_{n-1}\}$ be a set of n pairs of “brackets,” $p_0,q_0$ , …, $p_{n-1},q_{n-1}$ . We decompose $\Delta_n = P_n\cup Q_n$ into the set $P_n=\left\{\,p_i\,\mid\,i<n\,\right\}$ of “opening” brackets and the set $Q_n=\left\{\,\!q_i\,\mid\,i<n\!\,\right\}$ of “closing” brackets. We also use ${\langle{i}\!\mid}$ for $p_i$ and ${\mid\!{i}\rangle}$ for $q_i$ .

Let ${\mathcal{A}}$ be a monadic operator such that $\emptyset\in{\mathcal{A}} M$ for each monoid M. The polycyclic ${\mathcal{A}}$ -dioid $C_{n,{\mathcal{A}}}'$ is the quotient $C_{n,{\mathcal{A}}}' = {\mathcal{A}}\Delta_n^*/\rho_n$ of ${\mathcal{A}}\Delta_n^*$ by the ${\mathcal{A}}$ -congruence $\rho_n$ generated by the relations

(7) \begin{equation} p_iq_j = \delta_{i,j}\quad\textrm{ for }i,j<n,\end{equation}

where $\delta_{i,j}$ is the Kronecker $\delta$ . These equations allow us to algebraically distinguish matching brackets, where $p_iq_j=1$ , from non-matching ones, where $p_iq_j=0$ . Recall that in ${\mathcal{A}} \Delta^*_n$ , elements of $\Delta_n^*$ are interpreted by their singleton sets, 0 by the empty set, so that $\rho_n$ is the least ${\mathcal{A}}$ -congruence containing $(\{p_iq_i\},\{1\})$ and $(\{p_iq_j\},\emptyset)$ for $i,j<n, i\not=j$ . The bra-ket algebra $C_{n,{\mathcal{A}}}$ of Hopkins (Reference Hopkins2007) further assumes a “completeness” condition, $1 =\sum\left\{\,q_ip_i\,\mid\,i<n\,\right\}$ , i.e.

\begin{equation*} C_{n,{\mathcal{A}}} = {\mathcal{A}}\Delta_n^*/{\left\langle {\left\{\,p_iq_j=\delta_{i,j}\,\mid\,i,j<n\,\right\}\cup\left\{1 = \sum\left\{\,q_ip_i\,\mid\,i<n\,\right\}\right\}}\right\rangle}.\end{equation*}

We will only use the case ${\mathcal{A}} ={\mathcal{R}}$ and abbreviate $C_{n,{\mathcal{R}}}$ and $C_{n,{\mathcal{R}}}'$ by $C_n$ and $C_n'$ , respectively.

As we omit the completeness condition, we can view the semiring equations (7) as monoid equations, interpreted in monoids with an annihilating element 0, which leads to a different representation of $C_{n,{\mathcal{A}}}'$ that will be useful later.

The polycyclic monoid generated by $P_n$ is the quotient $P_n' := (\Delta_n\cup\{0\})^*/\rho_n$ , where $\rho_n$ now is the monoid congruence generated by

\begin{equation*} \left\{\,p_iq_j=\delta_{i,j}\,\mid\,i,j<n\,\right\}\cup \left\{\,x0=0=0x\,\mid\,x\in \Delta_n\cup\{0\}\,\right\}.\end{equation*}

Every string $w\in (\Delta_n\cup\{0\})^*$ can be reduced to a “normal form” $\textit{nf}_{n}(w)$ , using the equations generating $\rho_n$ as rules to shorten a string. The reduction either runs into a bracket mismatch and returns 0, or finds matching brackets only and returns 1, or ends in a string with all closing brackets in front of all opening brackets. Hence, we can identify $P_n'$ with $(Q_n^*P_n^*\cup\{0\},\cdot,1)$ , where

\begin{equation*} u\cdot v := \textit{nf}_{n}(uv), \qquad u,v\in Q_n^*P_n^*\cup\{0\}. \end{equation*}

The Cayley graph of the polycyclic monoid $P_n'$ is the graph whose nodes are the elements $x\in P_n'$ with edges $x{\,\stackrel{\delta}{\longrightarrow}\,} x\cdot\delta$ , for $\delta\in \Delta_n$ .

Example 3. A drawing of a part of the Cayley graph of $P_2'$ is given in Figure 1, not showing the sink node 0 and edges connected to it. For example, there is no edge $q_0{\,\stackrel{{p_0}}{\longrightarrow}\,}1$ since $q_0p_0\not=1$ . As the picture indicates, the subgraph with nodes in $P_2^*$ corresponds to a stack $P_2^*\simeq \{0,1\}^*$ , with $p_i$ for $\textit{push}(i)$ and $q_i$ for $\textit{pop}(i)$ . ^{Footnote 4} For $u\in\Delta_2^*$ , an equation $u=1$ holds in $P_2'$ iff u is the label of a path in the Cayley graph starting and ending in the node 1. The set of all those u is of course the pure Dyck language $D_2\in {\mathcal{C}} \Delta_2^*$ of balanced strings over $\Delta_2$ , i.e. the least solution in ${\mathcal{P}}\Delta_2^*$ of

\begin{equation*} y \geq 1 + p_0 y q_0 + p_1 y q_1 + yy. \end{equation*}

It is easy to define a pushdown automaton accepting $\left\{\,u\in\Delta_2^*\,\mid\,u=1 \textrm{ in }P_2'\,\right\}$ , keeping on its stack the $\rho$ -normal form of the input sequence read. $\diamond$

Figure 1.

Part of the Cayley graph of $P_2'$ .

Nivat and Perrot (Reference Nivat and Perrot1970) initiated the study of the connection between polycyclic monoids and formal languages.

Remark 8. The polycyclic dioid $C_n'$ has an interpretation in the algebra of binary relations on the Cayley graph of $P_n'$ , arising from the interpretation of $p_i$ and $q_i$ as the transition relations ${\,\stackrel{{p_i}}{\longrightarrow}\,}$ and ${\,\stackrel{{q_i}}{\longrightarrow}\,}$ shown in the graph. Under this interpretation, the completeness condition $1= \sum_{i<n} q_ip_i$ is not valid, as node 1 is not related to itself by any of the relations ${\,\stackrel{q_ip_i}{\longrightarrow}\,}$ . We will later use a restriction of this interpretation to the “stack part” $P_n^*$ of $P_n'$ . It may seem that a constant $\pi$ representing an “empty stack” test is needed to model a stack properly by

\begin{equation*} 1=\pi + \sum{\left\{\,q_ip_i\,\mid\,i<n\,\right\}},\quad \pi\pi=\pi,\quad \bigwedge_{i<n}(\pi q_i=0=p_i\pi).\end{equation*}

But in our case, this turns out to be unnecessary, since computations popping from the empty stack can be interpreted by the empty transition relation (cf. the proof of Theorem 17).

The map from $P_n'$ to ${\mathcal{A}}\Delta_n^*$ given by $0 \mapsto \emptyset$ , $1\mapsto\{1\}$ , and $p_i\mapsto\{p_i\}$ , $q_i \mapsto \{q_i\}$ for $i<n$ , extends to a homomorphism from $P_n'$ to the multiplicative monoid of $C_{n,{\mathcal{A}}}'$ . Since $P_n'$ has no non-trivial congruences (by Nivat and Perrot Reference Nivat and Perrot1970), this is an embedding of $P_n'$ into $C_{n,{\mathcal{A}}}'$ , for ${\mathcal{F}}\leq{\mathcal{A}}$ .

In the original definition $C_n' = {\mathcal{R}} \Delta_n^*/\rho_n$ , we take the quotient of the regular sets ${\mathcal{R}}\Delta_n^*$ under the semiring congruence $\rho_n$ , where 1 stands for $\{1\}$ and 0 for $\emptyset$ . We now take the regular sets ${\mathcal{R}} P_n'$ of $P_n'$ and remove the annihilating element:

Proof. We first prove the second statement. $\Leftarrow$ : Since $\{0\}/\nu=\emptyset/\nu$ is the least element of ${\mathcal{R}} P_n'/\nu$ , for $B\in {\mathcal{R}} P_n'$ we have

\begin{equation*} B/\nu = \sum\left\{\,\{w\}/\nu\,\mid\,w\in B\,\right\} = \sum\left\{\,\{w\}/\nu\,\mid\,w\in B\setminus\{0\}\,\right\}.\end{equation*}

Hence, if $B,B'\in{\mathcal{R}} P_n'$ with $B\setminus\{0\}=B'\setminus\{0\}$ , then $B/\nu=B'/\nu$ .

$\Rightarrow$ : One shows by induction on the regular operations that

\begin{equation*} \nu_0:=\left\{\,(B,B')\in {\mathcal{R}} P_n'\times {\mathcal{R}} P_n'\,\mid\,B\setminus\{0\} = B'\setminus\{0\}\,\right\}\end{equation*}

is an ${\mathcal{R}}$ -congruence containing $(\{0\},\emptyset)$ , and obviously $\nu\subseteq \nu_0$ .

For the first statement, notice that a congruence class $A/\rho_n$ of $C_n'= {\mathcal{R}}\Delta_n^*/\rho_n$ can be represented by a set of reduced strings, $\left\{\,\textit{nf}_{n}(w)\,\mid\,w\in A\,\right\}\setminus\{0\}$ , and $A/\rho_n\mapsto\left\{\,\textit{nf}_{n}(w)\,\mid\,w\in A\,\right\}/\nu$ constitutes an isomorphism between $C_{n,{\mathcal{R}}}'$ and ${\mathcal{R}} P_n'/\nu$ .

The $n>2$ pairs of brackets $p_i,q_i$ , $i<n$ of $\Delta_n$ can be coded by two pairs, say b,d and p,q of $\Delta_2$ , via $p_i:=bp^i$ and $q_i := q^id$ . This extends to an embedding of $P_n'$ in $P_2'$ .

2.3 The tensor product of ${\mathcal{A}}$ -dioids

In a category whose objects have a monoid structure, two morphisms $F_1:M_1\to M$ and $F_2:M_2\to M$ are relatively commuting, if for all $m_1\in M_1$ and $m_2\in M_2$ , $F_1(m_1)\cdot F_2(m_2) = F_2(m_2) \cdot F_1(m_1)$ .

The tensor product of $M_1$ and $M_2$ is an object $M_1\otimes M_2$ with two relatively commuting morphisms $\top_1 : M_1\to M_1\otimes M_2$ and $\top_2:M_2\to M_1\otimes M_2$ such that for any pair of relatively commuting morphisms $f:M_1\to M$ and $g:M_2\to M$ there is a unique morphism $h_{f,g}: M_1\otimes M_2\to M$ with $f=h_{f,g}\circ \top_1$ and $g=h_{f,g}\circ \top_2$ :

By the universal property, the tensor product of $M_1$ and $M_2$ is unique up to isomorphism.

Theorem 10 (Hopkins and Leiß Reference Hopkins and Leiß2018) A tensor product of ${\mathcal{A}}$ -dioids $D_1$ and $D_2$ exists,

\begin{equation*} {\top_1} : {D_1} \longrightarrow {D_1\mathop{\otimes_{\mathcal{A}}} D_2} \longleftarrow {D_2} : {\top_2}, \end{equation*}

consisting of $D_1\mathop{\otimes_{\mathcal{A}}} D_2 := {\mathcal{A}}(M_1\times M_2)/_\equiv$ , where $M_i =\, \widehat{\!{\mathcal{A}}} D_i$ is the monoid underlying $D_i$ and $\equiv$ is the ${\mathcal{A}}$ -congruence generated by the “tensor product relations”

\begin{equation*} \left\{\left(\sum A,\sum B\right)\right\} = A\times B \qquad (A\in{\mathcal{A}}{M_1}, B\in {\mathcal{A}}{M_2}),\end{equation*}

together with the embeddings $\top_1,\top_2$ defined by

\begin{equation*} \top_1(d_1) := \{(d_1,1)\}/_\equiv \,\textrm{ for }d_1\in D_1\textrm{ and }\top_2(d_2) := \{(1,d_2)\}/_\equiv\, \textrm{ for }d_2\in D_2.\end{equation*}

The liftings of the embeddings of $M_i$ into $M_1\times M_2$ map $A\in {\mathcal{A}}M_1$ to $A\times\{1\}\in {\mathcal{A}}(M_1\times M_2)$ and $B\in {\mathcal{A}} M_2$ to $\{1\}\times B \in {\mathcal{A}}(M_1\times M_2);$ hence, $A\times B =(A\times\{1\})(\{1\}\times B)\in {\mathcal{A}}(M_1\times M_2)$ . The tensor product relations are needed to make the $\top_i:D_i \to D_1\mathop{\otimes_{\mathcal{A}}} D_2$ be ${\mathcal{A}}$ -morphisms.

For $U\in{\mathcal{A}}(M_1\times M_2)$ , we write [U] for $U/_\equiv$ . By the definition of $\sum$ on $D_1\mathop{\otimes_{\mathcal{A}}} D_2$ ,

\begin{eqnarray*} [U] &=& \left[\bigcup\left\{\,\{(a,b)\}\,\mid\,(a,b)\in U\,\right\}\right]= \sum \left\{\,[\{(a,b)\}]\,\mid\,(a,b)\in U\,\right\}= \sum \left\{\,a\otimes b\,\mid\,(a,b)\in U\,\right\},\end{eqnarray*}

using $a\otimes b$ for $\top_1(a)\cdot \top_2(b)=[\{(a,b)\}]\in D_1\mathop{\otimes_{\mathcal{A}}} D_2$ . In particular, for $A\in {\mathcal{A}} M_1, B\in {\mathcal{A}} M_2$ , via $A \times B \in {\mathcal{A}}(M_1\times M_2)$ we have, by the tensor product relations,

\begin{equation*} \left(\sum A\right)\otimes \left(\sum B\right) = \sum\left\{\,a\otimes b\,\mid\,a\in A, b\in B\,\right\}.\end{equation*}

Since $D_1\mathop{\otimes_{\mathcal{A}}} D_2 $ is the quotient of ${\mathcal{A}}(M_1\times M_2)$ by the ${\mathcal{A}}$ -congruence $\equiv$ , its partial order is the dioid-order on the quotient, i.e. $[U]\leq [V]$ iff $[U\cup V] = [V]$ for $U,V\in{\mathcal{A}}(M_1\times M_2)$ , which is coarser than the order on ${\mathcal{A}}(M_1\times M_2)$ among the representatives U,V.

Proposition 11 (Hopkins and Leiß Reference Hopkins and Leiß2018) For monoids $M_1$ and $M_2$ , the tensor product of the ${\mathcal{A}}$ -dioids ${\mathcal{A}} M_1$ and ${\mathcal{A}} M_2$ is isomorphic to the ${\mathcal{A}}$ -dioid ${\mathcal{A}}(M_1\times M_2)$ :

\begin{equation*} {\mathcal{A}} M_1\mathop{\otimes_{\mathcal{A}}} {\mathcal{A}} M_2 \ \simeq\ {\mathcal{A}}(M_1\times M_2).\end{equation*}

Since $(A,B)\mapsto A\times B$ is a homomorphism from ${\mathcal{A}} M\times {\mathcal{A}}N$ to ${\mathcal{A}}(M\times N)$ , its lifting is a homomorphism from ${\mathcal{A}}({\mathcal{A}}M\times{\mathcal{A}} N)$ to ${\mathcal{A}}({\mathcal{A}}(M\times N))$ , so for $R\in{\mathcal{A}}({\mathcal{A}} M\times {\mathcal{A}} N)$ ,

\begin{equation*} S_R : = \bigcup\left\{\,A\times B\,\mid\,(A,B)\in R\,\right\} \in {\mathcal{A}}(M\times N). \end{equation*}

By the isomorphism, we have $[R]=[R']$ iff $S_R = S_{R'}$ . Conversely, for $S\in{\mathcal{A}}(M\times N)$ ,

\begin{equation*} R_S := \left\{\,(\{a\},\{b\})\,\mid\,(a,b)\in S\,\right\} \in {\mathcal{A}}({\mathcal{A}} M\times {\mathcal{A}} N); \end{equation*}

hence, $[R_S] = \sum\left\{\,\{a\}\otimes\{b\}\,\mid\,(a,b)\in S\,\right\} \in {\mathcal{A}} M\mathop{\otimes_{\mathcal{A}}} {\mathcal{A}}N$ . Notice that $S_{(R_S)}=S$ and $[R_{(S_R)}]=[R]$ .

We can push this a bit further:

Theorem 12. Let M be a monoid and N a monoid with annihilating element 0. Then

(8) \begin{equation}{\mathcal{A}} M\mathop{\otimes_{\mathcal{A}}} ({\mathcal{A}} N/\nu) \simeq {\mathcal{A}}(M\times N)/{\tilde\nu}\end{equation}

where $\nu$ is the least ${\mathcal{A}}$ -congruence on ${\mathcal{A}} N$ containing $(\{0\},\emptyset)$ and ${\tilde\nu}$ is the least ${\mathcal{A}}$ -congruence on ${\mathcal{A}}(M\times N)$ containing $(\{(1,0)\},\emptyset)$ .

Putting $R_\nu := \left\{\,(A,B/\nu)\,\mid\,(A,B)\in R\,\right\}$ for $R\in {\mathcal{A}}({\mathcal{A}}M\times {\mathcal{A}} N)$ , the isomorphism is given by

\begin{eqnarray*} [R_\nu] &\mapsto& (S_R)/{\tilde\nu}, \quad\textrm{where } S_R := \bigcup\left\{\,A\times B\,\mid\,(A,B)\in R\,\right\} \textrm{ for }R\in {\mathcal{A}}({\mathcal{A}} M\times {\mathcal{A}} N),\\ S/{\tilde\nu} &\mapsto& [(R_S)_\nu],\quad\textrm{where }R_S := \left\{\,(\{m\},\{n\})\,\mid\,(m,n)\in S\,\right\} \textrm{ for }S\in {\mathcal{A}}(M\times N).\end{eqnarray*}

To prove ${\mathcal{A}} M\mathop{\otimes_{\mathcal{A}}} ({\mathcal{A}} N/\nu) \simeq {\mathcal{A}}(M\times N)/{\tilde\nu}$ , it is sufficient to show that

\begin{equation*} {\top_1'} : {{\mathcal{A}} M} \longrightarrow {{\mathcal{A}}(M\times N)/{\tilde\nu}} \longleftarrow {{\mathcal{A}} N/\nu} : {\top_2'} \end{equation*}

form a tensor product of ${\mathcal{A}} M$ and ${\mathcal{A}} N/\nu$ in ${\mathbb{D}}{\mathcal{A}}$ , where $\top_1'$ and $\top_2'$ are defined by

\begin{equation*} \top_1'(A) = (A\times\{1\})/{\tilde\nu} \quad\textrm{ and }\quad \top_2'(B/\nu) = (\{1\}\times B)/{\tilde\nu}.\end{equation*}

This will be done in the Appendix.

We here show that the isomorphism ${{\mathcal{A}} M\mathop{\otimes_{\mathcal{A}}} ({\mathcal{A}} N/\nu)}\simeq {{\mathcal{A}}(M\times N)/{\tilde\nu}}$ consisting of the ${\mathcal{A}}$ -morphisms ${h_{\top_1',\top_2'}}$ and ${h'_{\top_1,\top_2}}$ induced by the universal properties of the two tensor product constructions

is as claimed in the theorem. The induced ${\mathcal{A}}$ -morphisms are defined by

\begin{eqnarray*} {h_{\top_1',\top_2'}}([R_\nu]) &:=& \sum\left\{\,\top_1'(A)\cdot\top_2'(B/\nu)\,\mid\,(A,B)\in R\,\right\}\\ &=& \sum\left\{\,(A\times\{1\})/{\tilde\nu}\cdot(\{1\}\times B)/{\tilde\nu}\,\mid\,(A,B)\in R\,\right\}\\ &=& \sum\left\{\,(A\times B)/{\tilde\nu}\,\mid\,(A,B)\in R\,\right\}\\ &=& \left(\bigcup\left\{\,A\times B\,\mid\,(A,B)\in R\,\right\}\right)/{\tilde\nu}= (S_R)/{\tilde\nu}\end{eqnarray*}

and

\begin{eqnarray*} {h'_{\top_1,\top_2}}(S/{\tilde\nu}) &:=& \sum\left\{\,\top_1(\{m\})\cdot\top_2(\{v\}/\nu)\,\mid\,(m,v)\in S\,\right\}\\ &=& \sum\left\{\,\{m\}\otimes \{v\}/\nu\,\mid\,(m,v)\in S\,\right\}\\ &=& [(R_S)_\nu].\end{eqnarray*}

Since $S_{(R_S)} = S$ , we have

\begin{eqnarray*} ({h_{\top_1',\top_2'}}\circ {h'_{\top_1,\top_2}})(S/{\tilde\nu}) &=& {h_{\top_1',\top_2'}}([(R_S)_\nu]) = S_{(R_S)}/{\tilde\nu} = S/{\tilde\nu},\end{eqnarray*}

and since the tensor product embeddings $\top_1,\top_2$ are ${\mathcal{A}}$ -morphisms,

\begin{eqnarray*} ({h'_{\top_1,\top_2}}\circ{h_{\top_1',\top_2'}})([R_\nu]) &=& {h'_{\top_1,\top_2}}(S_R/{\tilde\nu}) = [(R_{(S_R)})_\nu]\\ &=& \sum\left\{\,A\otimes B/\nu\,\mid\,(A,B)\in R_{(S_R)}\,\right\}\\ &=& \sum\left\{\,\{m\}\otimes \{v\}/\nu\,\mid\,(m,v)\in S_R\,\right\}\\ &=& \sum\left\{\,\{m\}\otimes \{v\}/\nu\,\mid\,(m,v)\in A\times B, (A,B)\in R\,\right\}\\ &=& \sum\left\{\,A\otimes B/\nu\,\mid\,(A,B)\in R\,\right\} = [R_\nu]. \\[-28pt]\end{eqnarray*}

With Proposition 9, Theorem 12 provides us with a representation of ${\mathcal{R}} M \mathop{\otimes_{\mathcal{R}}} C_2'$ as a quotient of a free ${\mathcal{R}}$ -dioid extension:

Corollary 13. Let $\nu$ be the least ${\mathcal{R}}$ -congruence on ${\mathcal{R}} P_2'$ containing $(\{0\},\emptyset)$ and ${\tilde\nu}$ the least ${\mathcal{R}}$ -congruence on ${\mathcal{R}}(M\times P_2')$ containing $(\{(1,0)\},\emptyset)$ . Then

\begin{equation*} {\mathcal{R}} M\mathop{\otimes_{\mathcal{R}}} C_2' = {\mathcal{R}} M\mathop{\otimes_{\mathcal{R}}} ({\mathcal{R}} P_2'/\nu) \simeq {\mathcal{R}}(M\times P_2')/{\tilde\nu}. \end{equation*}

The centralizer $Z_C(D)$ of a set C under an embedding $i:C\to D$ in an ${\mathcal{A}}$ -dioid D is the set

\begin{equation*} \left\{\,d\in D\,\mid\,d\cdot i(c) = i(c)\cdot d \textrm{ for each }c\in C\,\right\}\end{equation*}

of all elements of D that commute elementwise with the image of C in D under i.

In the following, we are concerned with $Z_{C_2'}({\mathcal{R}} M\mathop{\otimes_{\mathcal{R}}} C_2')$ , the centralizer of $C_2'$ in ${\mathcal{R}} M\mathop{\otimes_{\mathcal{R}}} C_2'$ under the tensor product embedding $\top_2 : C_2'\to {\mathcal{R}} M\mathop{\otimes_{\mathcal{R}}} C_2'$ .

Proof. It is clear that $Z_{C_2'}({K} \mathop{\otimes_{\mathcal{R}}} {C_2'})$ is closed under $0,1,+$ and $\cdot$ , so it is a sub-dioid of $K\mathop{\otimes_{\mathcal{R}}} C_2'$ . It is also closed under the restriction of $\sum:{\mathcal{R}}(K\mathop{\otimes_{\mathcal{R}}} C_2')\to K\mathop{\otimes_{\mathcal{R}}} C_2'$ to ${\mathcal{R}}(Z_{C_2'}({K} \mathop{\otimes_{\mathcal{R}}} {C_2'}))$ : if $c\in C_2'$ and $U\in {\mathcal{R}}(Z_{C_2'}({K} \mathop{\otimes_{\mathcal{R}}} {C_2'}))\subseteq {\mathcal{R}}(K\mathop{\otimes_{\mathcal{R}}} C_2')$ , then since $K\mathop{\otimes_{\mathcal{R}}}{C_2'}$ is ${\mathcal{R}}$ -distributive,

\begin{equation*} (1\otimes c)\sum U = \sum\left\{\, (1\otimes c)u\,\mid\,u\in U\,\right\} = \sum\left\{\, u(1\otimes c)\,\mid\,u\in U\,\right\} = \left(\sum U\right)(1\otimes c),\end{equation*}

so $\sum U\in Z_{C_2'}({K} \mathop{\otimes_{\mathcal{R}}} {C_2'})$ .

We will later see that for monoids M, $Z_{C_2'}(\mathcal{R} \mathop{\otimes_{\mathcal{R}}} {C_2'})$ is a ${\mathcal{C}}$ -dioid and isomorphic to ${\mathcal{C}} M$ . The reason to care about this is that it provides us with regular expressions as a notation system for context-free sets. Every $L\in {\mathcal{C}} M$ is, under ${\mathcal{C}} M \simeq Z_{C_2'}(\mathcal{R} \mathop{\otimes_{\mathcal{R}}} {C_2'})$ , an element of the Kleene algebra ${\mathcal{R}} M\mathop{\otimes_{\mathcal{R}}} C_2'$ and therefore denoted by a regular expression in the generators of ${\mathcal{R}} M\mathop{\otimes_{\mathcal{R}}} C_2'$ .

Example 5. Let $\Delta_2=\{b,d,p,q\}$ and M be a monoid. For $x,y\in M$ , consider $L = \left\{\,x^ny^n\,\mid\,n\in{\mathbb{N}}\,\right\}\in{\mathcal{C}} M$ and the regular expression $b(px)^*(yq)^*d$ . Interpreted in ${\mathcal{R}} M\mathop{\otimes_{\mathcal{R}}} C_2'$ , elements $m\in M$ mean $\top_1(\{m\}) = \{m\}\otimes \{1\}/\nu$ and elements $\delta\in\Delta_2$ mean $\top_2(\{\delta\})=\{1\}\otimes \{\delta\}/\nu$ . By ${}^*$ -continuity and since m and $\delta$ commute with each other in ${\mathcal{R}} M\mathop{\otimes_{\mathcal{R}}} C_2'$ , we obtain

\begin{eqnarray*} b(px)^*(yq)^*d &=& b\left(\sum\left\{\,(px)^k\,\mid\,k\in{\mathbb{N}}\,\right\}\right)\left(\sum\left\{\,(yq)^n\,\mid\,n\in{\mathbb{N}}\,\right\}\right)d\\ &=& \sum\left\{\,b(px)^k(yq)^nd\,\mid\,k\in{\mathbb{N}},n\in{\mathbb{N}}\,\right\}\\ &=& \sum\left\{\,x^ky^nbp^kq^nd\,\mid\,k\in{\mathbb{N}},n\in{\mathbb{N}}\,\right\}\\ &=& \sum\left\{\,x^ny^n\,\mid\,n\in{\mathbb{N}}\,\right\},\end{eqnarray*}

where in the final step, $bp^kq^nd = \delta_{k,n}$ since $pq=1=bd$ and $bq=0=pd$ . We will see that for any $L\in {\mathcal{C}} M$ , its image $\widehat L := \sum\left\{\,\{m\}\otimes 1\,\mid\,m\in L\,\right\}$ in ${\mathcal{R}} M\mathop{\otimes_{\mathcal{R}}} C_2'$ belongs to $Z_{C_2'}(\mathcal{R} \mathop{\otimes_{\mathcal{R}}} {C_2'})$ . $\diamond$

To prove that $Z_{C_2'}(\mathcal{R} \mathop{\otimes_{\mathcal{R}}} {C_2'})$ , together with the embedding $\top_1$ , is the ${\mathcal{C}}$ -closure of ${\mathcal{R}} M$ , we want to show that any ${\mathcal{R}}$ -morphism $f:{\mathcal{R}} M\to C$ to a ${\mathcal{C}}$ -dioid C has a unique extension to a ${\mathcal{C}}$ -morphism $\bar f: Z_{C_2'}(\mathcal{R} \mathop{\otimes_{\mathcal{R}}} {C_2'})\to C$ , which can be defined by

\begin{equation*} \bar f(\widehat L) = \sum\left\{\,f(\{m\})\,\mid\,m\in L\,\right\}\qquad\textrm{ for }L\in{\mathcal{C}} M.\end{equation*}

At least this works if for each element $e\in Z_{C_2'}(\mathcal{R} \mathop{\otimes_{\mathcal{R}}} {C_2'})$ there is a unique $L\in {\mathcal{C}} M$ with $e=\widehat L$ . This uniqueness condition will be established in the following two sections: by Theorem 17 of Section 3, there is at most one such $L\in {\mathcal{C}} M$ , by Theorem 26 of Section 4, there is at least one such L.

We remark that the category ${\mathbb{D}}{\mathcal{A}}$ has coproducts and free extensions (Hopkins and Leiß Reference Hopkins and Leiß2018). In particular, the free ${\mathcal{A}}$ -dioid extension of ${\mathcal{A}} M$ by the set $\Delta$ , $({\mathcal{A}} M)[\Delta]$ , is isomorphic to ${\mathcal{A}}(M[\Delta])$ , whence we simply write ${\mathcal{A}} M[\Delta]$ below.

3.1 The CST for monoids

For a finite set X– and $\Delta_2=\{p_0,q_0,p_1,q_1\}$ –, Dyck’s language $D_2(X)\in{\mathcal{C}} X^*[\Delta_2]$ of balanced strings over $X^*[\Delta_2]$ is the least solution of

\begin{equation*} y \geq 1 + X + yy + p_0yq_0 + p_1yq_1 \end{equation*}

in ${\mathcal{P}}(X^*[\Delta_2])$ . By the classical theorem of Chomsky and Schützenberger (Reference Chomsky and Schützenberger1963),

\begin{equation*} {\mathcal{C}} X^* \subseteq\left\{\,h_{X^*}(R\cap D_2(X))\,\mid\,R\in {\mathcal{R}} X^*[\Delta_2]\,\right\}. \end{equation*}

Since, for any monoid M, ${\mathcal{C}} M$ is defined inductively as an extension of ${\mathcal{F}} M$ by least solutions (in ${\mathcal{P}} M$ ) of polynomial systems with parameters from ${\mathcal{C}} M$ , one can reduce the parameters in the polynomial systems to the base case ${\mathcal{F}} M$ , so that all elements of M that occur in the parameters come from a finite set $X\subseteq M$ . Therefore,

\begin{equation*} {\mathcal{C}} M = \bigcup\left\{\,{\mathcal{C}} {\left\langle {X}\right\rangle}\,\mid\,X\in{\mathcal{F}} M\,\right\}, \quad \textrm{likewise }\quad {\mathcal{R}} M = \bigcup\left\{\,{\mathcal{R}} {\left\langle X\right\rangle}\,\mid\,X\in {\mathcal{F}} M\,\right\},\end{equation*}

where ${\left\langle X\right\rangle}\subseteq M$ is the submonoid of M generated by X. We identify a system $y_1\geq p_1,\ldots,y_n\geq p_k$ with polynomials $p_i\in{\mathcal{C}} {\left\langle {X}\right\rangle}[y_1,\ldots,y_k]$ with a context-free grammar with nonterminal symbols $y_1,\ldots,y_k$ , start symbol $y_1$ and terminal symbols from X. Since M resp. $D_2(M)$ belong to ${\mathcal{C}} M$ resp. ${\mathcal{C}} M[\Delta_2]$ only if M is finitely generated, we first state the CST for finitely generated monoids:

Theorem 15 (CST for finitely generated monoids) For any finitely generated monoid M,

\begin{equation*} {\mathcal{C}} M \subseteq \left\{\,h_M(R\cap D_2(M))\,\mid\,R\in {\mathcal{R}} M[\Delta_2]\,\right\}\end{equation*}

where $D_2(M)\in {\mathcal{C}} M[\Delta_2]$ is the set of elements of $M[\Delta_2]$ with balanced brackets.

Our proof is essentially a proof of the classical Chomsky–Schützenberger theorem, with a perhaps more intuitive construction of R from L than can be found in textbooks; it may be helpful to first consider Example 6 below. We use the grammatical definitions of ${\mathcal{C}} M$ and ${\mathcal{R}} M[\Delta_2]$ .

Approximate $L_1'$ from above by a regular set $R_1\in {\mathcal{R}} M[\Delta_n]$ as follows. To each monomial $\alpha$ of $p_j'$ (i.e. grammar rule $(y_j,\alpha)$ or $y_j\to\alpha$ ) attach a “follow”- or “continuation”-variable $y_j^F$ and split $\alpha y_j^F$ into right-linear factors m’y, where $m'\in M[\Delta_n]$ and $y\in\{y_1,\ldots,y_k,y_j^F\}$ . Let $G^F$ be the right-linear polynomial system

\begin{equation*} y_1 \geq p_1'',\ldots,y_k\geq p_k'',\quad y_1^F \geq p_1^F + 1,\ y_2^F \geq p_2^F,\ldots,y_k^F\geq p_k^F,\end{equation*}

where the monomials of $p_j''$ are the initial factors m’y of the monomials of $p_j'y_j^F$ and those of $p_j^F$ are the factors m’y that follow the occurrences of $y_j$ in $p_1'y_1^F,\ldots,p_k'y_k^F$ . Let $R_1$ be the first component of the least solution of $G^F$ in ${\mathcal{R}} M[\Delta_n]$ .

Proof of Claim 1. By induction on the derivation length $r\geq 1$ in $y_j\Rightarrow^r_{G'} w$ . If $r=1$ , then $y_j\to w$ is a rule of G’, hence $y_j\to wy_j^F$ is a rule of $G^F$ . If $y_j \Rightarrow_{G'}^1 \alpha \Rightarrow_{G'}^r w$ , and the rule $y_j\to\alpha$ of G’ is $y_j\to m_1{\langle{i_1}\!\mid}y_{j,1}{\mid\!{i_1}\rangle}\ldots m_s{\langle{i_s}\!\mid}y_{j,s}{\mid\!{i_s}\rangle}m_{s+1}$ , then $G^F$ has the rules

\begin{equation*} y_j \to m_1{\langle{i_1}\!\mid}y_{j,1},\quad y_{j,1}^F\to {\mid\!{i_1}\rangle}m_2{\langle{i_2}\!\mid}y_{j,2},\quad\ldots,\quad y_{j,s}^F\to {\mid\!{i_s}\rangle}m_{s+1}y_j^F,\end{equation*}

and so $y_j \Rightarrow^*_{G^F} \alpha$ . By the induction hypothesis, we get $\alpha \Rightarrow^*_{G^F} wy_j^F$ ; hence, $y_j \Rightarrow^*_{G^F} wy_j^F$ . $\triangleleft$

Using $y_1^F \geq 1$ , Claim 1 implies $L_1'\subseteq R_1$ ; hence, $L_1'\subseteq R_1\cap D_n(M)$ .

Proof of Claim 2. By induction on the derivation length $r\geq 1$ in $y_j\Rightarrow_{G^F}^rwy_j^F$ . If $r=1$ , then $y_j\to w$ is a rule of G’; hence, $y_j\Rightarrow^*_{G'} w$ . For length $r+1 > 1$ , suppose $y_j\Rightarrow_{G^F}^1 \alpha \Rightarrow_{G^F}^r wy_j^F$ . By the choice of $y_j\geq p_j''$ of $G^F$ , there is $m_1\in M$ such that $\alpha =m_1{\langle{i_1}\!\mid}y_{j,1}$ for some $i_1$ and $y_{j,1}\in \{y_1,\ldots,y_k\}$ , or $\alpha = m_1 y_{j}^F$ . The second case cannot occur, since $m_1y_j^F\Rightarrow_{G^F}^r wy_j^F$ with $r\geq 1$ is impossible: all monomials of $p_j^F$ either begin with a closing bracket ${\mid\!{i}\rangle}$ for some i, or are 1. In the first case, we have $m_1{\langle{i_1}\!\mid}y_{j,1} \Rightarrow_{G^F}^r wy_j^F$ . As $w\in D_n(M)$ , there are $u_1,w_1\in D_n(M)$ such that $w =m_1{\langle{i_1}\!\mid}u_1{\mid\!{i_1}\rangle}w_1$ and $y_{j,1} \Rightarrow_{G^F}^r u_1{\mid\!{i_1}\rangle}w_1y_j^F$ . By the construction of $G^F$ , ${\mid\!{i_1}\rangle}$ occurs only in rules $y_{j,1}^F\to{\mid\!{i_1}\rangle}m_2y_{j}^F$ or $y_{j,1}^F\to {\mid\!{i_1}\rangle}m_2{\langle{i_2}\!\mid}y_{j,2}$ with $m_2\in M$ and $y_{j,2}\in \{y_1,\ldots,y_k\}$ . So we must have

\begin{equation*} y_{j,1} \Rightarrow_{G^F}^{r_1} u_1y_{j,1}^F \Rightarrow_{G^F}^{r_2} u_1 {\mid\!{i_1}\rangle}w_1 y_j^F \quad\textrm{ with }r = r_1+r_2.\end{equation*}

Since $r_1\leq r$ , we get $y_{j,1} \Rightarrow_{G'}^* u_1$ by induction. If the part $y_{j,1}^F \Rightarrow^{r_2}_{G^F} {\mid\!{i_1}\rangle}w_1y_j^F$ is

\begin{equation*} y_{j,1}^F\Rightarrow_{G^F} {\mid\!{i_1}\rangle}m_2 y_{j}^F \Rightarrow^{r_2-1}_{G^F} {\mid\!{i_1}\rangle}w_1y_j^F,\end{equation*}

we must have $r_2=1$ and $w_1=m_2$ , so $y_j \to m_1{\langle{i_1}\!\mid}y_{j,1}{\mid\!{i_1}\rangle}m_2$ is a rule of G’, and since $w=m_1{\langle{i_1}\!\mid}u_1{\mid\!{i_1}\rangle}m_2$ , the claim $y_j\Rightarrow_{G'}^* w$ is shown. If the part $y_{j,1}^F \Rightarrow^{r_2}_{G^F} {\mid\!{i_1}\rangle}w_1y_j^F$ is

\begin{equation*} y_{j,1}^F\Rightarrow_{G^F} {\mid\!{i_1}\rangle}m_2{\langle{i_2}\!\mid}y_{j,2} \Rightarrow^{r_2-1}_{G^F} {\mid\!{i_1}\rangle}w_1y_j^F,\end{equation*}

there are $u_2,w_2\in D_n(M)$ such that $w_1=m_2{\langle{i_2}\!\mid}u_2{\mid\!{i_2}\rangle}w_2$ and $y_{j,2}\Rightarrow_{G^F}^{r_2-1}u_2{\mid\!{i_2}\rangle}w_2$ . As ${\mid\!{i_2}\rangle}$ occurs only in rules $y_{j,2}^F\to{\mid\!{i_2}\rangle}m_3y_j^F$ or $y_{j,2}^F\to{\mid\!{i_2}\rangle}m_3{\langle{i_3}\!\mid}y_{j,3}$ with $m_3\in M$ and $y_{j,3}\in\{y_1,\ldots,y_k\}$ , we can continue this way and get $1\leq i_1,\ldots,i_s\leq n$ , $r_1,\ldots,r_s\leq r$ , $m_1,\ldots,m_{s+1}\in M$ and $u_1,\ldots,u_s\in D_n(M)$ such that $w=m_1{\langle{i_1}\!\mid}u_1{\mid\!{i_1}\rangle}\ldots m_s{\langle{i_s}\!\mid}u_{s}{\mid\!{i_s}\rangle}m_{s+1}$ and

\begin{eqnarray*}y_j\Rightarrow_{G^F} m_1{\langle{i_1}\!\mid}y_{j,1}, && \quad y_{j,1}\Rightarrow_{G^F}^{r_1} u_1 y_{j,1}^F,\quad y_{j,1}^F \Rightarrow_{G^F} {\mid\!{i_1}\rangle}m_2{\langle{i_2}\!\mid} y_{j,2}, \quad\ldots\\ & &{j,s}\Rightarrow_{G^F}^{r_s} u_s y_{j,s}^F,\quad y_{j,s}^F\Rightarrow_{G^F} {\mid\!{i_s}\rangle}m_{s+1} y_{j}^F.\end{eqnarray*}

By construction of $G^F,$ there is a rule $y_j\to m_1{\langle{i_1}\!\mid}y_{j,1}{\mid\!{i_1}\rangle}\ldots{\langle{i_s}\!\mid}y_{j,s}{\mid\!{i_s}\rangle}m_{s+1}$ in G’, and by induction, $y_{j,1}\Rightarrow_{G'}^* u_1,\ \ldots,\ y_{j,s}\Rightarrow^*_{G'} u_s$ . The claim $y_j\Rightarrow^*_{G'} w$ follows. $\triangleleft$

Using $y_1^F\Rightarrow_{G^F} 1$ , Claim 2 implies $R_1\cap D_n(M)\subseteq L_1'$ . So $L_1'=R_1\cap D_n(M)$ , and $L_1= h_M(L_1')=h_M(R_1\cap D_n(M))$ . Finally, $D_n(M)$ can be replaced by $D_2(M)$ , since two bracket pairs b,d and p,q can be used to code ${\langle{i}\!\mid}$ by $bp^i$ and ${\mid\!{i}\rangle}$ by $q^id$ for $i< n$ .

Example 6. Let $a,b,c\in M$ and let $G = {\left\langle {y\geq p_y(y,z), z\geq p_z(y,z)}\right\rangle}$ be the following grammar

\begin{equation*} \begin{array}{rcl@{\qquad}rcl} y &\geq& a + bzc, & z &\geq& yy,\end{array}\end{equation*}

which defines sets $L_y,L_z\in{\mathcal{C}} M$ . Adding brackets around the three variable occurrences on the right turns G into the grammar $G' = {\left\langle {y\geq p_y', z\geq p_z'}\right\rangle}$

\begin{equation*} \begin{array}{rcl@{\qquad}rcl} y &\geq& a + b{\langle{l}\!\mid}z{\mid\!{1}\rangle}c, & z &\geq& {\langle{2}\!\mid}y{\mid\!{2}\rangle}{\langle{3}\!\mid}y{\mid\!{3}\rangle},\end{array}\end{equation*}

defining $L_y',L_z'\in{\mathcal{C}} M[\Delta_3]$ . It is clear that $L_y = h_M(L_y')$ and $L_y'\subseteq D_4(M)$ . Now add follow- or continuation-variables to the summands of $p_y'$ resp. $p_z'$

\begin{equation*} \begin{array}{rcl@{\qquad}rcl} y &\geq& ay^F + b{\langle{l}\!\mid}z{\mid\!{1}\rangle}cy^F, & z &\geq& {\langle{2}\!\mid}y{\mid\!{2}\rangle}{\langle{3}\!\mid}y{\mid\!{3}\rangle}z^F;\end{array}\end{equation*}

now split the monomials of $p_y'y^F$ and $p_z'z^F$ into right-linear factors mx with $m\in M[\Delta_3]$ and $x\in \{y,z,y^F,z^F\}$ to build the right-linear grammar $G^F = {\left\langle {y\geq p_y'', z\geq p_z'', y^F\geq p_y^F + 1,z^F\geq p_z^F}\right\rangle}$ where $p_y''$ resp. $p_z''$ are the sums of the initial factors from $p_y'y^F$ resp. $p_z'z^F$ , and $p_y^F$ resp. $p_z^F$ are the sums of factors following an occurrence of y resp. z in $p_y'y^F$ and $p_z'z^F$ :

(9) \begin{equation} \begin{array}{rcl@{\qquad}rcl@{\qquad}rcl@{\qquad}rcl} y &\geq& a y^F + b{\langle{l}\!\mid}z, & z &\geq& {\langle{2}\!\mid}y,& y^F &\geq& {\mid\!{2}\rangle}{\langle{3}\!\mid}y + {\mid\!{3}\rangle}z^F + 1, & z^F &\geq& {\mid\!{1}\rangle}c y^F.\end{array}\end{equation}

We need $y^F\geq 1$ , since y is the start symbol of G. Let $R_y\in {\mathcal{R}}M[\Delta_3]$ be the y-component of the least solution of $G^F$ . Intuitively, $L_y'\subseteq R_y$ since $y^F$ collects what follows any occurrence of y. A finite automaton accepting $R_y$ is shown in Figure 1 (with initial node y and accepting node $y^F$ ). A regular expression for $R_y$ can be obtained by solving the inequation system in Kleene algebra (i.e. replace $x \geq Ax+B$ with constant A,B by $x=A^*B$ ) with respect to y. This leads via $z^F={\mid\!{1}\rangle} cy^F$ ,

\begin{eqnarray*} y^F &=& {\mid\!{2}\rangle}{\langle{3}\!\mid}y + {\mid\!{3}\rangle}{\mid\!{1}\rangle}c y^F +1\\ &=& ({\mid\!{3}\rangle}{\mid\!{1}\rangle}c)^*({\mid\!{2}\rangle}{\langle{3}\!\mid}y+1)\\ &=& ({\mid\!{3}\rangle}{\mid\!{1}\rangle}c)^*{\mid\!{2}\rangle}{\langle{3}\!\mid}y + ({\mid\!{3}\rangle}{\mid\!{1}\rangle}c)^*\end{eqnarray*}

and $z = {\langle{2}\!\mid} y$ to the least solution in y by

\begin{eqnarray*} y &=& a y^F + b{\langle{l}\!\mid}{\langle{2}\!\mid}y\\ &=& a ({\mid\!{3}\rangle}{\mid\!{1}\rangle}c)^*{\mid\!{2}\rangle}{\langle{3}\!\mid}y + a ({\mid\!{3}\rangle}{\mid\!{1}\rangle}c)^* + b{\langle{l}\!\mid}{\langle{2}\!\mid}y\\ &=& (b{\langle{l}\!\mid}{\langle{2}\!\mid} + a ({\mid\!{3}\rangle}{\mid\!{1}\rangle}c)^*{\mid\!{2}\rangle}{\langle{3}\!\mid})y + a({\mid\!{3}\rangle}{\mid\!{1}\rangle}c)^*\\ &=& (b{\langle{l}\!\mid}{\langle{2}\!\mid}+ a ({\mid\!{3}\rangle}{\mid\!{1}\rangle}c)^*{\mid\!{2}\rangle}{\langle{3}\!\mid})^* a({\mid\!{3}\rangle}{\mid\!{1}\rangle}c)^*.\end{eqnarray*}

For example, $R_y$ contains the description $b{\langle{l}\!\mid}{\langle{2}\!\mid}a{\mid\!{2}\rangle}{\langle{3}\!\mid}a{\mid\!{3}\rangle}{\mid\!{1}\rangle}c$ of a parse tree of $baac\in L_y$ .

Let us add a preview to the algebraic version of the theorem and its reverse. Those elements of $R_y$ that are not descriptions of parse trees of words in $L_y$ contain (modulo commuting brackets with elements of M) bracket mismatches or have initial closing brackets or final opening brackets, like $w=a{\mid\!{3}\rangle} {\mid\!{1}\rangle} c{\mid\!{2}\rangle}{\langle{3}\!\mid}a\in R_y$ . With the reserved pair ${\langle{0}\!\mid},{\langle{0}\!\mid}$ of brackets, we can go from $R_y$ to $\{{\langle{0}\!\mid}\}R_y\{{\langle{0}\!\mid}\}$ and thereby turn the latter kind of elements into further ones with bracket mismatches, like ${\langle{0}\!\mid} w{\langle{0}\!\mid} = {\langle{0}\!\mid} a{\mid\!{3}\rangle}{\mid\!{1}\rangle} c{\mid\!{2}\rangle}{\langle{3}\!\mid} a{\langle{0}\!\mid}$ . The modified automaton shown in Figure 2 therefore is a description of $L_y$ : if we only extract sequences without bracket mismatches, we obtain all parse trees of words in $L_y$ . An algorithm to extract only bracket mismatch-free strings from a similar automaton is given by Hulden (Reference Hulden2011) for a CFG-parser based on the Chomsky–Schützenberger theorem.

Figure 2.

automaton for $R_y\in{\mathcal{R}} M[\Delta_3]$ .

Figure 3.

automaton for $L_y\in {\mathcal{C}} M \simeq Z_{C_4'}(\mathcal{R} \mathop{\otimes_{\mathcal{R}}} {C_4'})$ .

A version without the assumption that M be finitely generated follows:

Corollary 16 (CST for monoids). Let M be a monoid M and $D_2 \in {\mathcal{C}}\Delta_2^*$ the pure Dyck language over $\Delta_2$ . For any $L\in {\mathcal{C}} M,$ there is some $R\in {\mathcal{R}}(M\times\Delta_2^*)$ such that

\begin{equation*} L = \left\{\,m\,\mid\,(m,d)\in R, d\in D_2\,\right\}. \end{equation*}

Moreover, for $(m,d)\in R$ the normal form $\textit{nf}_{2}(d)$ in $P_2'=Q_2^*P_2^*\cup\{0\}$ of d belongs to $\{0,1\}$ .

The normal form $\textit{nf}_{2}:(\Delta_2\cup\{0\})^*\to P_2'=Q_2^*P_2^*\cup\{0\}$ maps those $d\in\Delta_2^*$ with $d\in D_2$ to $1\in Q_2^*P_2^*$ and those containing a bracket mismatch to 0. To obtain $\textit{nf}_{2}(d)\in \{0,1\}$ for all $(m,d)\in R$ , the idea is to wrap all strings of $\Delta_2^*$ with a third pair of brackets; this turns strings with normal form 0 or 1 into strings with the same normal form and those with normal form in $Q_2^*P_2^*\setminus\{1\}$ into strings with normal form 0, because of a mismatch with the new wrapping brackets; finally, the three bracket pairs are coded by two.

To carry this out, suppose $\Delta_2=\{{\langle{0}\!\mid},{\langle{0}\!\mid},{\langle{1}\!\mid},{\mid\!{1}\rangle}\}$ . The embedding of $\Delta_2$ in $\Delta_2^*$ given by ${\langle{i}\!\mid} \mapsto{\langle{0}\!\mid}{\langle{1}\!\mid}^{i+1}$ and ${\mid\!{i}\rangle} \mapsto {\mid\!{1}\rangle}^{i+1}{\langle{0}\!\mid}$ for $i<2$ extends to a homomorphism $':\Delta_2^*\to\Delta_2^*$ , such that $d \in D_2$ iff $d'\in D_2$ for each $d\in \Delta_2^*$ . Then, $(m,d) \mapsto(m,d')$ also is a homomorphism, which lifts to a homomorphism from ${\mathcal{R}}(M\times \Delta_2^*)$ to ${\mathcal{R}}(M\times\Delta_2^*)$ . From $R\in{\mathcal{R}}(M\times \Delta_2^*),$ we therefore obtain $R':= \left\{\,(m,d')\,\mid\,(m,d)\in R\,\right\}\in{\mathcal{R}}(M\times\Delta_2^*)$ ; hence, also

\begin{equation*} \widetilde R := \{(1,{\langle{0}\!\mid})\}\cdot R'\cdot \{(1,{\mid\!{0}\rangle})\}\in {\mathcal{R}}(M\times\Delta_2^*). \end{equation*}

For $d\in\Delta_2^*,$ we have $d\in D_2$ iff $d'\in D_2$ iff ${\langle{0}\!\mid} d'{\langle{0}\!\mid}\in D_2$ , so $\textit{nf}_{2}(d)=1$ iff $\textit{nf}_{2}{{\langle{0}\!\mid}d'{\langle{0}\!\mid}}=1$ , and

\begin{equation*} L=\left\{\,m\,\mid\,(m,d)\in R, d\in D_2\,\right\} = \left\{\,m\,\mid\,(m,\tilde d)\in \widetilde R, \tilde d\in D_2\,\right\}. \end{equation*}

Moreover, if $\textit{nf}_{2}(d)=0$ , then $\textit{nf}_{2}{{\langle{0}\!\mid} d'{\langle{0}\!\mid}}=0$ . Finally, if $\textit{nf}_{2}(d)\in Q_2^*P_2^*\setminus\{1\}$ , then $\textit{nf}_{2}{d'}$ begins with ${\mid\!{1}\rangle}^{i+1}{\langle{0}\!\mid}$ or ends in ${\langle{0}\!\mid}{\langle{1}\!\mid}^{i+1}$ , for some $i<2$ , so that $\textit{nf}_{2}{{\langle{0}\!\mid}d'{\mid\!{0}\rangle}}=0$ . Hence, with $\widetilde R$ instead of R both claims in the statement of the corollary hold.

3.2 Algebraic version of the CST for monoids

In the algebraic formulation, the Chomsky–Schützenberger theorem relates the objects ${\mathcal{R}} M$ and ${\mathcal{C}} M$ of the Kleisli subcategories of ${\mathbb{D}}{\mathcal{R}}$ and ${\mathbb{D}}{\mathcal{C}}$ . While in the classical formulation, each $L\in{\mathcal{C}} M$ has a regular approximation $R\in{\mathcal{R}} M[\Delta_2]$ such that $L=h_M(R\cap D_2(M))$ , one can now perform both the intersection with $D_2(M)$ and the removal of brackets by $h_M$ algebraically in ${\mathcal{R}} M\mathop{\otimes_{\mathcal{R}}} C_2'$ , and the relation between L and R becomes one of representing the same tensor.