Quasi-automatic groups are asynchronously automatic

TL;DR

This paper proves that quasi-automatic groups are also asynchronously automatic, establishing an equivalence between these two classes of groups under certain rationality conditions.

Contribution

It demonstrates that quasi-automatic groups are asynchronously automatic, clarifying the relationship between these two notions in group theory.

Findings

Quasi-automatic groups are asynchronously automatic.

The equivalence holds specifically for groups, not just semigroups.

Provides a characterization of automaticity in certain algebraic structures.

Abstract

A quasi-automatic semigroup is a finitely generated semigroup with a rational set of representatives such that the graph of right multiplication by any generator is a rational relation. A asynchronously automatic semigroup is a quasi-automatic semigroup for which these rational relations are also recognisable by two-tape automata. We show that when such a semigroup happens to be a group, the converse actually holds, meaning quasi-automatic groups are asynchronously automatic.

Figures

Equations

$$pfs\in H,|f|>k⟹\alpha (f)\ne \epsilon \ne \beta (f)$$

$$L(q)=\{x\in B^{+}|q_{0}x=q\}\{y\in B^{+}|qy=q,\alpha (y)=\epsilon ,|y|\le k\}\{z\in B^{+}|qz\in F\}.$$

$$\beta (xyz)\sim \alpha (xyz)=\alpha (x)\alpha (y)\alpha (z)=\alpha (x)\epsilon \alpha (z)=\alpha (xz)\sim \beta (xz)$$

$$(\alpha \times \beta )(H)=(\alpha \times \beta )((\alpha \times \beta {)}^{-1}(K\times K)\cap H_1)=(K\times K)\cap (\alpha \times \beta )(H_1)=(K\times K)\cap R_{\epsilon }^L=R_{\epsilon }^K$$

$$(u,v)\in R_{\epsilon }^K⟹|u|\le \ell |v|.$$

$$D(n)=2c\ell +\ell \text{ max}\{m(q,q^{\prime },g)|q,q^{\prime }\in Q,|g{|}_G\le n\}$$

$$|y|\le |x^{\prime }y{z}^{\prime }|\le \ell |x^{\prime }{y}^{\prime }{z}^{\prime }|=\ell (|x^{\prime }|+|y^{\prime }|+|z^{\prime }|)\le \ell (c+|y^{\prime }|+c)=2c\ell +\ell |y^{\prime }|\le D(n)$$

Full text

For semigroups in general, asynchronously automatic dictionaries are quasi-automatic (see [2], Proposition 3.4.2). This implies the if part. For the only if part, we use a theorem due to Epstein and Levy (see [1], theorem 7.2.8) which goes as follows: a group is asynchronously automatic if it has a Lipschitz Hausdorff dictionary along with a departure function for that dictionary. This means it suffices to show there exist such a dictionary for any quasi-automatic group. This will be Proposition 4; we first need a couple of lemmas.

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Proof By definition, $R_{\varepsilon}^{L}$ is rational and by Nivat’s theorem there exist some finite $B$, rational langage $H_{1}\subseteq B^{*}$ and alphabetic morphisms $B^{*}\to A^{*}$ such that $(\alpha\times\beta)(H_{1})=R_{\varepsilon}^{L}$. Let $W=(Q,T,q_{0},F)$ be a finite state deterministic automaton which recognizes $H_{1}$ and let $k$ be the number of states of $W$. For each state $q\in Q$, we consider the set of circuits $y$ from $q$ to $q$ such that $\alpha(y)=\varepsilon$ and $0<|y|\leq k$. Note that this set finite. Let $L(q)$ be the set of accepted words of $W$ with paths including such a circuit. Explicitly, we have

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It is rational by concatenation of rational langages. Symmetrically, we define $R(q)$ to be the set of accepted words whose path contains a non-empty circuit $y$ of length at most $k$ such that $\beta(y)=\varepsilon$. It is rational by the same reasoning.

We claim $K=L-\bigcup\limits_{q\in Q}\beta(L(q))\cup\alpha(R(q))$ is a rational dictionary of $S$. The relations ${R_{a}^{K}=R_{a}^{L}\cap(K\times K)}$ are rational by intersection. We show that $p(K)=S$. Consider some $s\in S$ and $w\in L$ a word of minimal length such that $p(w)=s$. Suppose for a contradiction that $w\notin K$. Then either $w\in\beta(L(q))$ or $w\in\alpha(R(q))$. Suppose the former; the other case is identical. Then $w=\beta(xyz)$ with $y$ a short circuit around $q$ with $\alpha(y)=\varepsilon$, $q_{0}x=q$ and $qz\in F$. Then $q_{0}xz=qz\in F$, indicating $xz\in H_{1}$. This implies $(\alpha\times\beta)(xz)\in R_{\varepsilon}^{L}$, which implies $\alpha(xz)\sim\beta(xz)$. Similarly, $\alpha(xyz)\sim\beta(xyz)$. Then

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We conclude that $\beta(xz)$ and $\beta(xyz)$ represent the same element of $S$. Since $\alpha\times\beta$ is a Nivat bimorphism, $\alpha(y)=\varepsilon$ implies $\beta(y)\neq\varepsilon$. This means $\beta(xz)$ is shorter than $w=\beta(xyz)$, a contradiction to the minimality of $w$.

Now let $H=(\alpha\times\beta)^{-1}(K\times K)\cap H_{1}$. Reverse morphisms preserve recognisability in any monoid; $H$ is therefore recognisable. But as a subset of the free monoid $B^{*}$, this means it is rational. Furthermore, we have

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We now show $H$ satisfies the lemma. Let $w=pfs\in H$ with $|f|>k$. Suppose for a contradiction that $\alpha(f)=\varepsilon$; here again, the symmetric case is identical. Since $H\subseteq H_{1}$, $pfs$ is recognised by $W$. Since the length of $f$ is greater than the number of states of $W$, its path during the reading of $w$ must contain some circuit $y$ of length at most $k$ around some $q$. Also, as $y$ is a factor of $f$, we know $\alpha(y)=\varepsilon$. As $w$ is an accepted word with path including an appropriate circuit $y$, $w\in L(q)$ and in turn $\beta(w)\not\in K$. This is a contradiction because $w\in H$ implies $(\alpha\times\beta)(w)\in R_{\varepsilon}^{K}$ and thus $\beta(w)\in K$.

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Proof Take $w\in H$ such that $(\alpha\times\beta)(w)=(u,v)$. Decompose $w$ into factors of length $k$. This gives us $w=f_{1}...f_{h+1}$ with $|f_{i}|=k$ for all $i\leq h$ and $|f_{h+1}|\leq k$. By the previous lemma, $\alpha(f_{i})\neq\varepsilon\neq\beta(f_{i})$ for all $i\leq h$. We have $|u|=|\alpha(f_{1}...f_{h+1})|=|\alpha(f_{1})|+...+|\alpha(f_{h+1})|$ and thus $h\leq|u|\leq kh+k$. Symmetrically, we have $h\leq|v|\leq kh+k$. Then $|u|\leq kh+k\leq k|v|+k\leq 2k|v|$ and $\ell=2k$ satisfies the condition of the lemma.

Proof Start by taking $K\subseteq L$ like in the previous lemma and let $W=(Q,T,q_{0},F)$ be a new finite state automaton recognising $K$; assume that $W$ is accessible and coaccessible. Let $c$ be the number of states of $W$ and $\ell$ be the constant given by the previous lemma. For all triplets $(q,q^{\prime},g)$ with $q,q^{\prime}\in Q$ and $g\in G$, set ${m(q,q^{\prime},g)=\text{min}\{|w|\mathrel{\mathop{\mathchar 58\relax}}qw=q^{\prime},p(w)=g\}}$ if there exist some $w$ with $qw=q^{\prime}$ and $p(w)=g$, and $m(q,q^{\prime},g)=0$ otherwise. We define $D\mathrel{\mathop{\mathchar 58\relax}}\mathbb{N}\to\mathbb{N}$ by setting

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We claim this is a departure function for $K$. Choose $xyz\in K$ such that $|y|>D(n)$ and suppose for a contradiction that $|y|_{G}\leq n$. Consider the path $q_{0}\xrightarrow{x}q_{1}\xrightarrow{y}q_{2}\xrightarrow{z}q_{3}\in F$. Choose a word $y^{\prime}$ such that $q_{1}y^{\prime}=q_{2}$ and $y^{\prime}\sim y$ of minimal length $m(q_{1},q_{2},p(y))$. Since $W$ is accessible, coaccessible and has $c$ states, we can choose $x^{\prime}$ and $z^{\prime}\in A^{*}$ such that $|x^{\prime}|,|z^{\prime}|\leq c$ and $q_{0}x^{\prime}=q_{1}$ and $q_{2}z^{\prime}\in F$. This implies $x^{\prime}yz^{\prime}$ and $x^{\prime}y^{\prime}z^{\prime}$ are in $K$. Also, since $y\sim y^{\prime}$, we have $x^{\prime}yz^{\prime}\sim x^{\prime}y^{\prime}z^{\prime}$. These are sufficient conditions to know that $(x^{\prime}yz^{\prime},x^{\prime}y^{\prime}z^{\prime})\in R_{\varepsilon}^{K}$. The previous lemma then tells us that $|x^{\prime}yz^{\prime}|\leq\ell|x^{\prime}y^{\prime}z^{\prime}|$. Then,

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which is a contradiction to the choice of $y$.

Proof of Theorem 1 Let $G$ be a quasi-automatic group and $K$ be the quasi-automatic dictionary with departure function built in Lemma 2. By a theorem of Blanchette et al. (see [2], Theorem 4.5), as with all quasi-automatic dictionaries, it is weak Lipschitz. It is therefore Lipschitz Hausdorff by the remark at the end of the introduction. Applying the theorem of Epstein and Levy let us conclude that $G$ is asynchronously automatic.

Conclusion

While we conjecture that quasi-automatic semigroups strictly contains the class of asynchronously automatic semigroups, we don’t have an explicit example of a quasi-automatic semigroup which is not asynchronously automatic. On the level of relations, the inclusion is known to be strict. Rational relations are stable under union while relations recognised by two-tape automata are not; this indicates how different they are. Building examples that are asynchronously automatic but not automatic is far from trivial (see [1] example 7.4.1, the Baumslag–Solitar group). As a similar problem, it hints us that finding an appropriate quasi-automatic semigroup and showing it is not asynchronously automatic could be a hard task.