$\DeclareMathOperator\SL{SL}$Let $d$ be a metric on the upper-half plane $\mathbb H = \{(x,y) : y > 0\}$ which is invariant with respect to the action of $\SL(2, \mathbb R)$ to $\mathbb H$ which is defined by
$$A \cdot z = \frac{az+b}{cz+d}, \ \ A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \in \SL(2, \mathbb R), \ z \in \mathbb H. $$
Specifically,
$$ d(A \cdot z, A \cdot w) = d(z,w), \ \ z, w \in \mathbb H, A \in \SL(2, \mathbb R).$$
Then, without assuming that $(\mathbb H,d)$ is geodesic, can we say that $d$ is Gromov-hyperbolic? Our original motivation comes from the square roots of some $f$-divergences between univariate Cauchy distributions which are non-geodesic metrics on $\mathbb H$.



Gromov hyperbolicity for (non-geodesic) metrics on the upper-half plane invariant with respect to SL(2, R) action
(Recall that an arbitrary metric space is Gromov-hyperbolic if
$$\sup\_{a,b,c,d}\Big((ab+cd)-\max(ac+bd,ad+bc)\Big)<\infty;$$
where $ab$ is the distance between $a$ and $b$. Note that this passes to subspaces.)


The answer is no: a counterexample being the square root of the usual distance on $\mathbf{H}^2$, which is invariant under the whole isometry group. For $\mathbf{R}\_{\ge 0}$ embeds isometrically into $\mathbf{H}^2$, and just the square root of the usual distance on $\mathbf{R}\_{\ge 0}$ is not hyperbolic:


Choose $a=0$, $b=2t$, $c=t$, $d=3t$. Then $ab+cd=2\sqrt{2}t$, $ac+bd\le ad+bc=(\sqrt{3}+1)\sqrt{t}$, so $(ab+cd)-\max(ac+bd,ad+bc)=(2\sqrt{2}-\sqrt{3}-1)\sqrt{t}$ is unbounded when $t\to\infty$.


