How to prove $\mathbb Z^2 \ast \mathbb Z$ is quasi isometric to $\mathbb Z^2 \ast \mathbb Z^2$?

Question

How to prove $\mathbb Z^2 \ast \mathbb Z$ is quasi isometric to $\mathbb Z^2 \ast \mathbb Z^2$? Here $\ast$ is the free product of groups.

I am thinking of proving they are commensurable. In other words, proving they share a finite index subgroup. $\mathbb Z^2 \ast \mathbb Z$ is the fundamental group of $T \vee S^1$, and $\mathbb Z^2 \ast \mathbb Z^2$ is fundamental group of $T \vee T$ (here $\vee$ is wedge sum). So we can also turn the problem into finding compact cover of each space such that the two covers have the same fundamental group. My problem right now is the only compact cover of a torus is torus itself, so I'm not sure how to find such a compact cover?

Answer 1

We can show that these groups are quasi-isometric by showing that these are both q.i. to $\mathbb{Z}^2*\mathbb{Z}^2*\mathbb{Z}$.

1. Consider, as you mention in the question, the space $T\vee S^1$, whose fundamental group is $\mathbb{Z}^2 * \mathbb{Z}$. Using the standard double cover of this space, $T\vee T\vee S^1$, we see that $\mathbb{Z}^2 * \mathbb{Z}$ is quasi-isometric to $\mathbb{Z}^2*\mathbb{Z}^2*\mathbb{Z}$.

2. Express $\mathbb{Z}^2 * \mathbb{Z}^2$ as $\langle a,b,c,d~|~aba^{-1}b^{-1},cdc^{-1}d^{-1}\rangle$. Then $\mathbb{Z}^2 * \mathbb{Z}^2 * \mathbb{Z}$ is realized as the subgroup $\langle a^2,b,c^2,d,ac^{-1}\rangle$. (The map is the obvious one: $a^2$ and $b$ generate the first $\mathbb{Z}^2$, and so on.) Then you can show this has index 2 by showing that $1$ and $a^{-1}$ are the only left cosets.

Answer 2

$\mathbb Z^2 * \mathbb Z^2$ is indeed commensurable to $\mathbb Z^2 * \mathbb Z$, in fact they each have an index 2 subgroup isomorphic to $\mathbb Z^2 * \mathbb Z^2 * \mathbb Z$.

You can see this with covering spaces if you choose the correct spaces.

For the first group $\mathbb Z^2 * \mathbb Z \approx \pi_1(X)$, take $X$ to be a torus $T$ with an arc $A$ attached by identifying its endpoints to two different points of $T$.

For $\mathbb Z^2 * \mathbb Z^2 \approx \pi_1(Y)$, take $Y$ to be two disjoint toruses $T_i$ ($i=1,2$), and an arc $B$ with endpoints attached to $p_i \in T_i$ ($i=1,2$).

Define a double covering space $Z$ of $Y$ consisting of disjoint toruses $\widetilde T_i \subset Z$ ($i=1,2$), such that $\widetilde T_i$ double covers $T_i$ with two points $\tilde p'_i,\tilde p''_i \in \widetilde T_i$ covering $p_i$, and with two arcs $\widetilde B', \widetilde B'' \subset Z$ that each cover the arc $B$, where $\widetilde B'$ has endpoints attached to $p'_1,p'_2$, and $\widetilde B''$ has endpoints attached to $p''_1,p''_2$.

But there is also a double covering map $Z \to X$ taking each $\widetilde T_i$ homeomorphically to $T$ and taking each of $\widetilde B', \widetilde B''$ homeomorphically to $A$.

By the way, as a consistency check one can do a quick Euler characteristic calculation: $\chi(X)=\chi(Y)=-1$, and $\chi(Z)=-2$. My first, now overwritten construction failed this check :-/

Notice finally that $\pi_1(Z) \approx \mathbb Z^2 * \mathbb Z^2 * \mathbb Z$.