# Free groups - extending quasi isometry to commuting square

Let's assume one has a finitely generated free group on $n$- generators, and two homomorphisms : $F \rightarrow F/k_1$, $F \rightarrow F/k_2)$ (for $k_i$ normal subgroup ) such that $F/k_1$ is quasi-isometric to $F/k_2$ and we have one fixed QI $G: F/k_1 \rightarrow F/k_2$. Does it mean that there is a quasi isometry $H:F \rightarrow F$ giving a commuting square up to a bounded distance ? Since $F$ is a tree it is easy to show that we can inductively find quasi isometric embedding $H:F \rightarrow F$ which gives a commuting square but is it possible to extend it somehow to full quasi isometry?

Edit:

We have one fixed quasi-isometry $G: F/k_1 \rightarrow F/k_2$ . For anyone reading this post the comments are now irrelevant - I've updated it all to the post itself. Here's how I do my lift.

We will denote the lift $H$. Let's denote projection $F \rightarrow F/k_1$ with $p$ and projection $F \rightarrow F/k_2$ with $q$. We equip $F$ with standard symmetric generating set $<a_1,..a_n>$ and we fix their projection images as generating sets for $F/k_i$. Additionaly let $c_i$ be a reduced word in $k_2$ beggining with letter $a_i$ and of minimal norm (it always can be found if $k_2$ is not trivial). $e$ denotes the neutral element. We build our lift inductively. Start with an injective lift of $\lbrace{ e,a_1,..,a_n \rbrace}$ and assume that $H(e) =e$ (for technical reasons and also I am specifically interested in a QI $G$ which sends neutral element to itself). By injective I mean the following : if$|b| < |d|$ then $|H(b)| < |H(d)|$. Now assume we have an injective lift of elements from the ball of radius $n$ -$B(e,n) = \lbrace{ d \in F : |d| \le n \rbrace}$. Take any vertex $v$ of norm $n$ - we will describe lifting of elements $\gamma \in F$ such that $|\gamma^{-1} \cdot v| =1$ and $|\gamma| > |v|$ (i.e. vertices which share one edge with $v$ but are further from the neutral element).

Let's denote such vertices $\gamma_1,..,\gamma_{n-1}$. $G$ is a QI hence $d_{F/k_2}(G(p(v)),G(p(\gamma_{i})) \le K$ for some uniform constant $K$.

As a result $G(p(v))^{-1} \cdot G(p(\gamma_i)) = q(a_{i_{1}}) \cdot ... \cdot q(a_{i_{k_{i}}})$ for $i = \lbrace{1,..,n-1 \rbrace}$ and for all $i$ we have that $i_{k_{i}} \le K$.( it is the definition of the word metric $G(p(v))^{-1} \cdot G(p(\gamma_i))$ can be written as a word shorter then $K$ but for each $i$ word's letters and its lenght may differ -but we know they are uniformly bounded!).

Since we have a lift of $v$ - denoted $H(v)$ , we can lift $\gamma_i$ to $H(\gamma_i) = H(v) \cdot a_{i_{1}} \cdot ... \cdot a_{i_{k_{i}}} \cdot c_{j_{i}}$.

So far the choice of $c_{j_{i}}$ may be arbitrary - the lift will give us commuting square and will be coarse Lipschitz. But it should additionaly be QI embedding hence we choose $c_{j_{i}}$ in a special way - forcing the lifts to "spread".

For given $i$ we choose $c_{j_{i}}$ in such a way that:

i) $j_i \neq i_{k_{i}}$ hence word $H(v) \cdot a_{i_{1}} \cdot ... \cdot a_{i_{k_{i}}} \cdot c_{j_{i}}$ is a reduced word.

ii) we choose them in such a way that if $G(p(\gamma_m)) = G(p(\gamma_r))$ then $c_{j_{m}} \neq c_{j_{r}}$ hence lifts $H(\gamma_m),H(\gamma_r$ separate at vertex $H(v)$

iii) and we even choose them so that $c_{j_{i}} \neq c_{j_{r}}$ if $i \neq r$.

Such choices can obviously be made. It ends the construction.

Now let's take any two points $u,t \in F$ such that $d(u,t) = n$ it means that they are connected by a unique shortest path $d_0,..,d_n$ such that $d_0 = u,d_n =t$ and $d(d_i,d_{i+1})=1$. This path divides into two parts:

$d_0,...,d_{s},d_{s+1},..,d_n$

For $i \le s$ $|d_{i}| <|d_{i-1}|$ and for $i \ge s$ we have that $|d_{i+1}| >|d_{i}|$ .

The lift of the first part of the path path gives us path $H(d_0),..,H(d_s)$ such that $1 \le d(H(d_i),H(d_{i+1})) \le K'$ ($K'$ is a little bit bigger then $K$ because we have added $c_j$'s but nonetheless it is uniform. Now the very construction of the lift assures us that $H(d_{i})$ lies in the complement of $F - H(d_{i+1})$ which does not contain lift of neutral element - $H(e)$ hence If you take the shortest path connecting $H(d_0),H(d_s)$ - $f_0,..,f_k$ ($d(f_j,f_{j+1}) =1$) then $H(d_i)$ lay on this path and if $f_{j}$ is a vertex equal to $H(d_i)$ and $f_l$ is a vertex equal to $H(d_{i+1})$ then $l > j$ which means that $d(H(d_0),H(d_s)) = k \ge \frac{1}{M+1} \cdot s = \frac{1}{M+1} \cdot d(d_0,d_s)$ ( informally speaking path $H(d_0,...,H(d_s)$ always flows forward from $H(e)$, never backwards )

We can conduct the same reasoning with the second part - $d_s,...,d_n$ - hence $d(H(d_s),H(d_n)) = k \ge \frac{1}{M+1} \cdot (n-s) = \frac{1}{M+1} \cdot d(d_s,d_n)$

All that is left is to analyse the critical three points - $d_{s-1},d_{s},d_{s+1}$.

We know that $|H(d_{s-1})| < |H(d_{s})|$ and that $|H(d_{s+1}) < |H(d_{s})|$

But we chose $c_j$ in such a way that points $H(d_{s-1}),H(d_{s}),H(d_{s+1})$ do not belong to one geodesic (. Hence if $r_0,..,r_m$ is the shortest path connecting $d_{s-1}$ with $d_{s+1}$ then $1 \le m \le K'$ hence path $H(d_0),..,H(d_s),r_0,..,r_m,H(d_{s+1}),..,H(d_n)$ is path joining $H(d_0)$ with $H(d_n)$ and it only "flows" in one direction hence $d(H(d_0),H(d_n)) \ge \frac{1}{M+1} \cdot (s + n-s)$

• In the question you fix a quasi-isometry $F/k_1\to F/k_2$ and you're asking whether you can "lift" it to a quasi-isometry? (this is not exactly what you write as you just say that $F/k_1$ and $F/k_2$ are QI which is not the same as fixing a QI between them, so I'm confused) – YCor Apr 21 '17 at 21:37
• Yes,sorry that's what I meant. We have one fixed quasi-isometry between $F/k_1$ and $F/k_2$ and we want to lift/extend it to $F$ – Kat Apr 22 '17 at 10:41
• Please, explain your construction of a lift: What I see is a lift which is coarse Lipschitz and coarsely surjective. – Moishe Kohan Apr 22 '17 at 19:11
• Let the QI betwen $F/k_1$ and $F/k_2$ be named $G$.Let's give $F/k_1$ and $F/k_2$ generators which are projections of generators from $F$. Let's lift $e \in F$ to any element so that the diagram commutes. Now we do it inductively. Suppose some word $a$ has its lift and $b$ is a word such that $|a^{-1}b| =1$. Then $d_{F/k_1}(p_i(a),p_i(b)) \le K$ for some uniform $K$ ($p_i:F \rightarrow F/k_i$). – Kat Apr 22 '17 at 19:50
• Now we send $p(a),p(b)$ to $F/k_2$ - $G(p(a))$ has a lift to one vertex belonging to $p_{2}^{-1}(p(a))$ and $d_{F/k_2}(G(p(a)),G(p(b)) \le M$ for some uniform constant $M$ hence $G(p(a))^{-1} \cdot G(p(b)) = p(\gamma_1) \cdot ... \cdot p(\gamma_n)$ where $n \le M$ and $\gamma$ are generators of $F$. Let's define the lift of $b$ to be $H(a) \cdot \gamma_1 \cdot ... \cdot \gamma_{n} \cdot k$ where $H(a)$ is a lift of $a$ and $k$ is one fixed element from $k_2$ which is not equal to zero – Kat Apr 22 '17 at 19:52

What you're asking for is not possible in general. You can't extend the quasi-isometry of the quotient groups to a quasi-isometry of the free groups. Here is an example when it cannot be done: $$\newcommand{\lra}{\longrightarrow} \newcommand{\da}{\left\downarrow\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{ccccccccc} 1 & \lra & 1 & \lra & \langle x,y,z\rangle & \lra & \langle x,y,z\rangle & \lra & 1\\ \da & & \da & & \da & & \,\,\,\,\da\phi & & \da & \\ 1 & \lra & \langle a\rangle & \lra & \langle a,b,c\rangle & \lra & \langle b,c\rangle & \lra & 1 \end{array}$$ The map $\phi$ is defined not just as a quasi-isometry, but a group homomorphism: $\phi(x)=b$, $\phi(y)=cbc^{-1}$, and $\phi(z)=c^2$. This is an injective group homomorphism which embeds $F_3$ as a finite index subgroup of $F_2$. Since $\phi$ is an injective homomorphism, it's a quasi-isometric embedding of $\langle x,y,z\rangle$ into $F_2$. Since the image is finite index, $\phi$ is coarsely surjective and is therefore a quasi-isometry between $\langle x,y,z\rangle$ and $\langle b,c\rangle$.
Assume $\Phi$ is a quasi-isometry making the diagram commute. Let let $q_1$ and $q_2$ denote the quotient maps so that $q_2\circ\Phi$ is a bounded distance from $\phi\circ q_1$. Well $\phi\circ q_1$ is a QI, since $q1$ is the identity and $\phi$ is a QI. If you post-compose a quasi-isometry with function which is bounded distance from the identity, you get another QI. Thus, $q_2\circ \Phi$ must be a QI. But that's impossible since $\Phi$ was assumed to be a QI, and $q_2$ is most definitely NOT a QI. So $\phi$ cannot be lifted to a quasi-isometry.
As a note, if you dislike that one of the kernels is trivial, then you can stick an extra generator in each of the $F_3$'s (replace $\langle x,y,z\rangle$ with $\langle x,y,z,w\rangle$ and similarly for the other $F_3$). Then just kill that generator so that one of the quotients is $F_2$ and the other is $\mathbb{Z}$. It's not as clear that the quasi-isometry can't be extended, but I believe it's still true.
• I'm not convinced in the argument that $\phi$ can't be lifted to a QI (btw the word is "lift", not "extend"). "All I can do is change $\phi$ by a bounded amount": no, a priori you can change $\phi$ into, for instance, $\phi'(t)=a\phi(t)$; a priori you change it to $\phi(t)A(t)$ where $A(t)$ goes to the kernel (I'm a bit sloppy too), an in a way that it's still a large-scale Lipschitz map (which restricts the function A$). Possibly the approach might work, anyway. – YCor Apr 25 '17 at 17:41 • Thank you for your answer but I'm afraid I am not really convinced too- it's pretty obvious that such lift (assuming it exists) will not be close to a homorphism. – Kat Apr 25 '17 at 17:57 • @YCor Here's a more rigorous argument: Assume$\Phi$is a quasi-isometry making the diagram commute. Let let$q_1$and$q_2$denote the quotient maps so that$q_2\circ \Phi$is a bounded distance from$\phi\circ q_1$. Well$\phi\circ q_1$is a QI, since$q_1$is the identity. If you post-compose a quasi-isometry with function which is bounded dist. from the identity, you get another QI. Thus,$q_2\circ \Phi$must be a QI. That's impossible since$\Phi$was assumed to be a QI, and$q_2\$ is most definitely NOT a QI. – D Wiggles Apr 25 '17 at 18:00