Skip to main content
8 events
when toggle format what by license comment
Mar 27, 2020 at 8:45 vote accept Diego Martinez
Mar 26, 2020 at 21:57 comment added Benjamin Steinberg That should be 2n+1 not 2n.
Mar 26, 2020 at 21:48 answer added Benjamin Steinberg timeline score: 3
Mar 26, 2020 at 21:22 comment added Benjamin Steinberg 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 20:34 comment added user6976 The firsr semigroup I would check is the free e-unitary semigroup with cover thw free abelian group of rank 2.
Mar 26, 2020 at 18:04 comment added user6976 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.
Mar 26, 2020 at 17:38 history edited YCor
edited tags
Mar 26, 2020 at 17:17 history asked Diego Martinez CC BY-SA 4.0