# 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 [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.

• 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 '20 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 '20 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. – Benjamin Steinberg Mar 26 '20 at 21:22
• That should be 2n+1 not 2n. – Benjamin Steinberg Mar 26 '20 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. – user6976 Mar 26 '20 at 22:31
• @MarkSapir any quotient of an E-unitary inverse semigroup that is above the maximal group image is E-unitary – Benjamin Steinberg Mar 26 '20 at 22:59
• Yes, and this is, I think, mentioned in one of the MM papers. – user6976 Mar 26 '20 at 23:08
• @MarkSapir, likely. It is elementary and probably in Petrich's book too. – Benjamin Steinberg Mar 26 '20 at 23:11