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Inspired by this question we ask:

Suppose that $G$ is an infinite group. Suppose that $X$ is a finite generating set of $G$. Let $\Gamma = \Gamma(G, X)$ be the resulting Cayley graph. Does $\Gamma$ contain a bi-infinite quasi-geodesic $L$ so that the closest points relation $\rho_L \colon \Gamma \to L$ is a quasi-retraction?

That is, so that the closest points relation $\rho_L$ has uniformly bounded images and is also (coarsely) Lipschitz.

Random thoughts:

For example, in $\mathbb{Z}^2$, we have "good" lines where the closest points relation is a retraction and "bad" lines where the closest points relation does not even have uniformly bounded images.

Perhaps (?) this question is related to groups which are wide: no asymptotic cone has a cut point. For example, $\mathbb{Z}^2$ is wide. A wide group has empty Morse boundary -- this is a necessary but not sufficient property for a "no" answer.

As Moishe Kohan suggests, the answer is certainly "no" if we try to generalise to constant valence infinite graphs. Here is a a four-valent example. We start with $\mathbb{Z}$ equipped with edges of the form $(i, i+1)$. We connect $2^k$ to $2^{k+1}$ by a path of length $2^k$, for all $k$ non-negative. We do the same between $-2^k$ to $-2^{k+1}$. Next we add the edge $(-1, 1)$. Now all vertices have valence four or two. To each of the latter we add a self loop to build the desired example.

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    $\begingroup$ Could you reformulate your last statement? Do you mean that every $G$ with nonempty Morse boundary has a positive answer to your question? $\endgroup$
    – YCor
    Commented Jul 29 at 21:44
  • $\begingroup$ A Morse geodesic $L$ gives a positive answer to my question. (Unless I have gotten confused...) Everything after the horizontal rule is commentary, and not part of the question proper. I'll edit to make that clear. $\endgroup$
    – Sam Nead
    Commented Jul 29 at 21:47
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    $\begingroup$ You may want to add that this is false for bounded degree graphs containing complete geodesics. $\endgroup$
    – Moishe Kohan
    Commented Jul 30 at 1:22

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The lamplighter group $L_2:= \mathbb{Z}_2 \wr \mathbb{Z}$ (and other wreath products) should provide a negative answer to your question. If I guess right, the following statement should be true:

Claim. In $L_2$, for every bi-infinite geodesic $\gamma$ and all integers $n \geq 1$, there is an isometrically embedded cycle with a subpath of length $n$ in $\gamma$.

The following argument just gives the idea. Precise computation has to be done, and maybe there are some additional cases to consider.

Sketch of proof. For convenience, assume that $(0,0) \in \gamma$. Let $(c,z) \in \gamma$ be a point at distance $n$. Fix a point $p \in \mathbb{Z}$ very far away from $z$ and $\mathrm{supp}(c)$.

Case 1: along $\gamma$, the last move when going from $(0,0)$ to $(c,z)$ is switching on a lamp. From $(c,z)$, you move the arrow to $p$, you switch on the lamp at $p$; then you come back to $0$, your arrow follows the same path when going from $(0,0)$ to $(c,z)$ but you undo what has been done, terminating at $(0,z)$; then move the arrow to $p$ and switch off the lamp at $p$; finally, you move the arrow at $0$. All the lamps are down, so your final point is $(0,0)$.

Case 2: along $\gamma$, the last move when going from $(0,0)$ to $(c,z)$ is moving the arrow. Then do essentially the same thing as above, but, when moving from $(c,z)$, switch on the lamp at $z$ first. $\square$

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    $\begingroup$ Here is a small observation to avoid cases: In the Diestel-Leader generating set (so $S=\{(0,1),(\delta_0,1)\}^{\pm 1}$), any geodesic can do at most 2 turnovers in the $\mathbb Z$ component, and a bi-infinite geodesic can do at most 1. Actually, up to isometries of the Cayley graph, these are the only two cases. And the geodesic with one turnover has long segments that look like geodesic of the other type. So we may suppose we work with a specific geodesic $((0,n))_{n\in\mathbb Z}$ $\endgroup$
    – Corentin B
    Commented Jul 30 at 9:09
  • $\begingroup$ What is a "turnover"? Is it a place where the lamplighter reverses direction? $\endgroup$
    – Sam Nead
    Commented Jul 31 at 14:59
  • $\begingroup$ Please forgive my ignorance - is this behaviour (arbitrarily large isometrically embedded loops) "the same as" the failure of finite presentability? More concretely: does the lamplighter group quasi-isometrically embed in some finitely presented group? $\endgroup$
    – Sam Nead
    Commented Jul 31 at 15:26
  • $\begingroup$ @SamNead The property implies not being finitely presentable, but it is not clear to me whether the converse holds. (My first guess is that it's not the case, but I should check.) The lamplighter group does quasi-isometrically embed into some finitely presented group, as any recursively presented group (this is an improvement of Higman's embedding theorem obtained by Ol'shanskii, but here it is possible to find an explicit embedding, e.g. Houghton groups or Thompson's group V). $\endgroup$
    – AGenevois
    Commented Jul 31 at 17:56
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    $\begingroup$ @SamNead Yes, the lamplighter moving along the street changing directions. Passing twice at the same place is not efficient, unless you need to switch some lamps in front of you, but your end destination is behind you. Passing three time at the same place never make sense (unless it is the starting and end point, depending how you understand « passing by ») $\endgroup$
    – Corentin B
    Commented Jul 31 at 18:21

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