# Quasi-isometries and E-unitary inverse semigroups

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  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.

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

• I doubt it is always a quasi-isometry. Look at papers by Margolis and Meakin. Also in our paper with Meakin about e-unitary semigrous with Abelian covers the Schutzenberger graphs of our semigroup and the distance functions are described.
– user6976
Mar 26, 2020 at 18:04
• The firsr semigroup I would check is the free e-unitary semigroup with cover thw free abelian group of rank 2.
– user6976
Mar 26, 2020 at 20:34
• What if you take the free E-unitary cover of the free abelian group of rank 2 generated by x,y and add the idempotent relations $xx^{-1}=1=x^{-1}x$. This should give an E-unitary inverse semigroup where maximal group image is the free abelian group where Schutzenberger graphs have finitely many y edges but all horizontal x edges through any vertex. Then the Schutzenberger graph of y does not quasisometrically embed because the distance from (n,0) to (n,1) is 2n in the Schutzenberger graph and is 1 in the group. Mar 26, 2020 at 21:22
• That should be 2n+1 not 2n. Mar 26, 2020 at 21:57

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.
• The fact that your semigroup $T$ is e-unitary should follow from another paper by Margolis and Meakin.