# Existence of a quasi-isometric residually finite group?

It's, by now, more or less well known that residual finiteness is not a quasi-isometry invariant for finitely generated groups (see here for an example). Thus the following question makes sense:

Question: given a finitely generated group $$G$$, is there a finitely generated group $$H$$ such that $$H$$ is residually finite and $$G$$ and $$H$$ are quasi-isometric? That is, does the class of groups quasi-isometric to $$G$$ always contain residually finite groups?

Acknowledging this question is fairly open-ended, I would be glad with any reference in this direction. Thanks in advance. The following tries to motivate the above via some C*-algebraic questions.

Motivation (and guess): Should the above be known true, then one would be able to manually build so-called quasi-diagonalizing projections for the reduced group C*-algebra of $$G$$, that is, finite rank projections $$p_n \in \mathcal{B}(\ell^2 G)$$, with $$p_n \rightarrow 1$$ in the strong operator topology and $$||p_n \lambda_g - \lambda_g p_n|| \rightarrow 0$$ for every $$g \in G$$, where $$\lambda$$ is the left regular representation of $$G$$. Indeed, note that if $$H$$ is QI to $$G$$ and res. finite and amenable, then, by work of Orfanos there are projections $$q_n$$ for $$H$$ as above, and those can be pullback-ed to projections $$p_n$$ for $$G$$. Thus, since this C*-question is still open, my guess is that the original question is open as well.

• The answer is certainly no, and the question is whether a counterexample (a f.g. group not QI to any RF f.g. group) is already known. The limit is that QI rigidity is known for quite few groups. Possibly Kac-Moody groups (these are simple finitely presented CAT(0) groups) or non-RF Baumslag-Solitar groups would be workable candidates. Also for most non-RF groups I can imagine, there's no natural candidate for a RF group QI to it. – YCor Apr 15 '20 at 9:39
• The quasi-isometry class of higher Baumslag-Solitar groups is known (White, Papasoglu and others) and contains no r.f. groups. – user6976 Apr 15 '20 at 10:34
• @MarkSapir I googled on White's paper before posting my first comment. Could you provide a reference? I'm aware of an unpublished paper of Whyte, which is known to contain mistakes in its claim on QI-classification (I don't know whether it affects your assertion). – YCor Apr 15 '20 at 11:09
• PS it's Whyte, not White, we both misspelled... it's Kevin Wkyte, Coarse bundles arXiv link. Nevertheless I can find in this paper a statement describing arbitrary groups quasi-isometric to $\mathrm{BS}(2,3)$, so I'd be happy to hear about a reference, if any. – YCor Apr 15 '20 at 20:54
• @YCor I think you're looking for [Whyte, K. The large scale geometry of the higher Baumslag-Solitar groups. Geom. Funct. Anal. 11 (2001), no. 6, 1327–1343.]. Theorem 5.1 in there is the following: a f.g. group $\Gamma$ is qi to $BS(2,3)$ iff it is a graph of virtual $\mathbb{Z}$s which is neither commensurable to $F_n \times \mathbb{Z}$ nor virtually solvable. – Carl-Fredrik Nyberg Brodda Apr 16 '20 at 9:10

## 1 Answer

Take any finitely-presented group $$G$$ with undecidable word problem. Then $$G$$ is not quasi-isometric to any finitely generated group with decidable word problem, in particular, to any residually-finite group. (Note that finite presentability and decidability of the WP are quasi-isometry invariant. The latter is because quasi-isometries preserve the equivalence class of the Dehn function and WP is for a finitely-presented group decidable iff the Dehn function is recursive.)

• Since it's implicit, you're using that finitely presented residually finite groups have solvable word problem, and that among finitely presented groups, to have solvable word problem is QI-invariant. Nice answer! – YCor Apr 15 '20 at 16:26