Question: Determine whether the following spaces are quasi-isometric: $\mathbb{H}^m$, $\mathbb{H}^n$, trees, $\mathbb{R}^m$, $\mathbb{R}^n$ with $m\neq n$.
What I know: $\mathbb{H}^m$, $\mathbb{H}^n$ and trees are quasi-isometry as they belong to hyperbolic plane but $\mathbb{R}^m$, $\mathbb{R}^n$ are not quasi- isometry with other spaces, am I right? I am not sure.Thanks for your help.
Among your spaces:
All are Gromov-hyperbolic except $\mathbf{R}^d$ for $d\ge 2$; these are pairwise non-QI since they have growth $\simeq n^d$.
Among the Gromov-hyperbolic ones, $\mathbf{H}^d$ for $d\ge 2$ can be distinguished since their boundary is a $(d-1)$-sphere (while others have a totally disconnected boundary).
Remain trees and $\mathbf{R}^d$ for $d\le 1$; the latter two can be identified to trees (a point and a line). There are plenty of QI-types of trees; for instance for every totally disconnected compact metrizable topological space $Y$ there's a tree of valency $\le 3$ whose boundary is homeomorphic to $Y$ (usually far from unique up to isometry). Among them, of special importance are regular trees of valency $d\ge 3$, which are all quasi-isometric (boundary is then a Cantor space).
One difference between the spaces is their curvature. $\mathbb{H}^n$ is $\delta$-hyerbolic for every $n$. Trees are even $0$-hyperperbolic, where $\mathbb R^n$ is not $\delta$-hyperbolic (for $n \geq 3$). $\delta$-hyperbolicity and $0$-hyperbolicity are quasi-isometrie invariants, so we only have to distinguish $\mathbb H^n$ from $\mathbb H^m$ and $\mathbb R^n$ from $\mathbb R^m$, asides from the trees.
I don't know what knowledge you want to use, but a good way to distinguish the hyperbolic spaces is the boundary. There are some diverse definitions of boundary, one of them is for example the Gromov boundary $\partial X$ for a $\delta$-hyperbolic space $X$. The Gromov boundary is a quasi-isometry-invariant and $\partial \mathbb H^n \cong S^{n-1}$. Since $S^n \ncong S^m$, we can distinguish $\mathbb H^n$ and $\mathbb H^m$.
$\mathbb R^n$ is quasi-isometric to $\mathbb Z^n$ for all $n$. So the question is, is $\mathbb Z^n$ quasi isometric to $\mathbb Z^m$? Consider the growth function of a finitely presented group. It's growth rate is a quasi-invariant. One can calculate that the growth rate of $\mathbb Z^n$ is $n$ (see wikipedia).
Therefore we concluded that all your spaces are not quasi isometric. There could be easier ways to get those results.
Addendum: Forgot to mention trees. All finite trees are qi to a point. For infinite trees, this is much harder. The topology of the boundary is a quasi-invariant, as is the asymptotic growth of the connected components if you take the complement of exhausting compacta. But it won't be good enough.
Addendum2: Forgot that $\mathbb{R}$ is $0$-hyperbolic (or, equivalently, an $\mathbb{R}$-tree). $Cay(\mathbb Z,\{\,1\,\}) \underset{QI}{\cong} \mathbb R$, so $\mathbb R$ is QI to an infinite tree. This is the only example where objects of your list coincide up to QI.
