Lipschitz retractions on finitely generated groups

Question

Let $G$ be a finitely generated group and let $|\cdot|$ be a word norm (word length function) on $G$ with respect to some finite generating set. For every $n\in\mathbb{N}$, let $B_n$ denote the ball of radius $n$ around $1_G$, i.e. the set $\{g\in G\colon |g|\leq n\}$.

Does there exist a positive constant $K>0$ such that for every $n$ there is a $K$-Lipschitz retraction $P_n: G\rightarrow B_n$? That is, a $K$-Lipschitz map from $G$ onto $B_n$ which is identity on $B_n$.

There are particular examples for which it is clear. For instance, if $G$ is a free group, then $P_n$ for every $g\in G$, written as a reduced word $a_1a_2\ldots a_n\ldots a_k$, produces the reduced word $a_1\ldots a_n$. That can be checked to be $1$-Lipschitz for every $n$. With a little bit more care, a similar argument works for any hyperbolic group.

A bit different (but easy) argument works for $\mathbb{Z}^n$. I believe I have also an argument for two-step nilpotent groups, but that's already a bit messy. It's plausible it may work for all nilpotent (torsion-free) groups.

But perhaps I am missing something and by some general argument this is actually true for all finitely generated groups. Is it possible? Or is there some obvious counter-example (or a candidate for a counter-example)?

EDIT: I forgot to add that ideally I would prefer these retractions to commute, i.e. $P_n\circ P_m=P_m\circ P_n$, for all $n,m\in\mathbb{N}$.

Answer 1

My guess in the comments was somewhat wrong, you can get a combing by quasigeodesics, but it is not necessarily synchronous.

A (asynchronous) combing of a group $G$, with finite generating set $S$, is a choice of paths $\sigma_g$ from the identity to $g \in G$ with the property that if $d_S(g,h)=1$ then the paths $\sigma_g,\sigma_h$ $K'$-fellow travel, up to reparametrization, for some fixed $K'$. If they $K'$-fellow travel without reparametrization we say that we have a synchronous combing. Recall the fellow travel property says that the paths stay close if the endpoints are close―depending on $K'$.

If you have a sequence of commuting $K$-Lipschitz retractions, $P_n: G \to B_n$, then you can construct a combing on $G$ like so:

• First $$P_n\circ P_m =P_m \circ P_n=P_m$$ for $n>m$, which can be seen by just plugging in elements and using that they are retracts.

• Take $g \in G$ distance $n$ from the origin. $$P_n(g)=g,P_{n-1}(g),\dots P_0(g)=e$$ gives a list of elements with the property that $$d(P_k(g),P_{k+1}(g)) \leq K+1:$$ If $P_{k+1}(g)$ is in $B_k$ then the distance is $0$, otherwise $P_{k+1}(g)$ is adjacent to $x$ in $B_k$ and $d(P_k(x),P_k(g)) \leq K$.

• Now define $\sigma_g$ by choosing geodesics, $\gamma:[0,d(P_k(g),P_{k+1}(g)] \to G$, between $P_k(g),P_{k+1}(g)$. Note that it is not necessarily true that $P_k(g)$ is on the $k$th shell of the ball $B_n$. $\sigma_g$ is a $(K+1,K+1)$-quasigeodesic, independent of $g$: Each geodesic segment is at most $K+1$ making the upperbound obvious; the lower bound is found by noticing if there is a geodesic between $\sigma_g(t)$ and $\sigma_g(s)$ of length $k$ then the path there used at most $k$ geodesic segments of length less than or equal to $K+1$.

• We need to verify that we have the (asynchronous) $K'$-fellow travel property, for some $K'$. Assume that $d(g,e)\geq d(h,e)$. By $K$-Lipschitz, if $d(g,h)=1$ then $d(P_k(g),P_k(h) \leq K$, and the sequence $P_n(h),\dots, P_0(h)$ will be the defining sequence for $\sigma_h$ except possible repeating $h$ at $P_n$ and $P_{n-1}$. Arranging "squares" for the $P_k(g),P_k(h),P_{k+1}(g),P_{k+1}(h)$ gives that any two points on them is at most $4(K+1)$ away from any other. Reparametrize on of the quasigeodesics so that you are traveling through the sides of the squares at roughly the same rate. The thing preventing this from being synchronized is that maybe for $h$ you have an actual geodesic and for $g$ you have an as-bad-as-it-gets quasigeodesic and it takes about $(K+1)n$ time to transverse vs just $n$ time.

Having a combing implies that you have solvable word problem and is finitely presented. I personally don't know how much more restrictive adding quasigeodesics is (probably a lot more). There is a nice survey on the subject Hairdressing in groups: a survey of combings and formal languages by Sarah Rees.