Let $S = \langle K\rangle$ be a finitely generated inverse semigroup, where $K \subset S$ is a fixed, finite and symmetric set of generators.


**Preliminaries:** Recall that we say that $s, t \in S$ are $\mathcal{L}$-related if $s^{-1}s = t^{-1}t$. Given an $\mathcal{L}$-class $L \subset S$, we may construct its' *Schützenberger graph* $\Lambda(L, K)$, whose vertices are the points of $L$ and where $x, y \in L$ are connected by an edge labeled by $k \in K$ if $kx = y$. We consider $L$ equipped with the natural path metric $d\_L$ via $\Lambda(L, K)$. Another studied congruence in $S$ is $\sigma$, where $s \sigma t$ if $sx = tx$ for some $x \in S$. The quotient $S/\sigma$ is a group $G$ known as the *maximal homomorphic image of $S$*. Moreover, we say that $S$ is E-unitary if whenever $s \sigma t$ and $s \mathcal{L} t$ then $s = t$, i.e. the quotient map embeds every $\mathcal{L}$-class into $G$.


**Question:** Let $S = \langle K \rangle$ be a fin. gen. E-unitary inverse semigroup. Let $L \subset S$ be an $\mathcal{L}$-class. Is the quotient map $L \rightarrow G$ a quasi-isometric embedding? That is, are there constants M, C > 0 such that for all $x, y \in L$
$$ \frac{1}{M} d\_L(x, y) - C \leq d\_G(x\sigma, y\sigma) \leq M d\_L(x, y) + C$$
where $d\_G$ is the path metric in the left Cayley graph of $G$ with respect to the generating set $K \sigma$. Observe that the right inequality above is true for any $M \geq 1$, since any geodesic between $x, y$ *falls down* to a path between $x\sigma, y\sigma$.


**Partial results/remark:** it's clear that if $L$ has only finitely-many $\mathcal{R}$-classes then the quotient map is going to be a quasi-isometry. Indeed, the $\mathcal{H}$-class of the idempotent of $L$ is a group included in $G$, and that inclusion of groups is a quasi-isometry. Since $L$ has only finitely-many $\mathcal{R}$-classes, then so is the *inclusion* of $L$ into $G$, i.e., the quotient map.


**Motivation:** In [1] quasi-isometries in monoids are studied, from the point of view of the Cayley graph. However, little is said about my inquiry, and I haven't been able to find any reference on this in the literature. My guess is the answer should be yes, but any help is greatly appreciated.


[1] Gray and Kambites, *Groups acting on semimetric spaces and quasi-isometries of monoids*, Trans. Ame. Math. Soc. **365** (2013) 555--578.



Quasi-isometries and E-unitary inverse semigroups
The answer is no. Let $G$ be a free abelian group of rank 2 generated by $x,y$. Let $S$ be the Meakin-Margolis expansion of $G$. It consists of all pairs $(X,g)$ with $X$ a finite connected subgraph of the Cayley graph of $G$ containing the origin and $g$. The product is $(X,g)(Y,h)=(X\cup gY,gh)$. The projection to $G$ is an idempotent pure homomorphism, so $S$ is E-unitary, and $S$ is generated by the edge from the origin to (1,0) and the edge from the origin to (0,1). Call these generators $x,y$ respectively. Now let $T$ be the quotient of $S$ by the relations $xx^{-1}=1=x^{-1}x$. It is not hard to see that $T$ is E-unitary since it is sandwiched between $S$ and $G$. Its elements can be viewed as pairs $(X,g)$ with $X$ a connected subgraph of the Cayley graph which contains the origin and $g$ with only finitely many vertical $y$ edges and containing the horizontal line through any vertex of $X$. Such graphs are precisely the Schutzenberger graphs of $T$. 


These graphs in general do not quasi-isometrically embed as soon as they have a $y$ edge. For example of you take the Schutzenberger graph of $y$ you have the lines $x=0$ and $x=1$ and the edge from $(0,0)$ to $(0,1)$. So the distance from $(n,0)$ to $(n,1)$ in this graph is $2|n|+1$ while in the Cayley of $G$ the distance is $1$. So the embedding is not a quasi-isometry.


