I've been reading about uniform Roe algebras lately and in particular the rigidity problem. I know that in general $C_u^*(X)$ and $C_u^*(Y)$ being *-isomorphic doesn't necessrily imply that $X$ and $Y$ are isomorphic as coarse spaces, but rather only coarsely equivalent. I'm thus looking for an example of coarse spaces $X$ and $Y$ which have isomorphic associated Roe algebras but are not themeselves isomorphic as coarse spaces. Any help would be appreciated!
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What do you mean by isomorphic as coarse spaces? – Aweygan Jan 23 '24 at 14:13
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Sorry for the confusion. I mean so that there are $f:X \to Y$ and $g:Y \to X$ coarse maps which are genuinely inverses in the set-theoretic sense. I've learned in the meantime that the category of coarse spaces is so defined so that two spaces are isomorphic in $\mathbf{Coarse}$ iff they are coarsely equivalent so I can see that my formulation was poorly chosen, but I think the question still makes sense! – ham_ham01 Jan 23 '24 at 17:30
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Then how about $X$ being the discrete metric space ${1,2}$ and $Y$ being the space ${1}$. These are not "isomorphic" in your sense (there can be no bijection between the two) but if i'm not mistaken their uniform Roe algebras should be both just $\mathbb{C}$, no? – Just dropped in Feb 13 '24 at 01:10
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@Justdroppedin The uniform Roe algebra of $Y$ would be $\mathbb{C}$, but the uniform Roe algebra of $X$ would be $\mathbb{M}_2(\mathbb{C})$ – Aweygan Feb 18 '24 at 15:29
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@Aweygan Oh, I have no idea about roe algebras, i just assumed that since the spaces are coarsely equivalent their roe algebras should be isomorphic. Isn't this the case? – Just dropped in Feb 18 '24 at 18:22
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The Roe algebras are isomorphic (both are isomorphic to the algebra of compact operators on a separable infinite-dimensional Hilbert space). Their uniform Roe algebras, on the other hand, are not isomorphic. – Aweygan Feb 18 '24 at 22:10