Smooth transition from Euclidean plane to hyperbolic plane

Question

If I have a Poisson point process $\mathcal{X}$ of density $\lambda$ on the Euclidean plane $\mathbb{R}^2$, with the Euclidean metric taking pairs of points to the Euclidean distance,

$$ \operatorname{dist} (\langle x_1, y_1 \rangle, \langle x_2, y_2 \rangle) = \sqrt{(x_1 -x_2)^2-(y_1-y_2)^2} $$

can I allow the metric to depend on time, in such a way that it smoothly (or otherwise) approaches the hyperbolic metric as time traverses the closed unit interval $[0,1]$?

So, initially, $\mathcal{X}$ sees Euclidean space, and finally hyperbolic space,

$$ \operatorname{dist} (\langle x_1, y_1 \rangle, \langle x_2, y_2 \rangle) = \operatorname{arcosh} \left( \cosh y_1 \cosh (x_2 - x_1) \cosh y_2 - \sinh y_1 \sinh y_2 \right) $$ with a number (perhaps a finite set) of "negatively curved" spaces in between. Only a small region would need this property, not the whole of $\mathbb{R}^2$.

The reason for this is it would affect the structure of a random graph or simplical complex built on the points (e.g. the degree distribution). Is this possible?

Note: I understand Ricci flow is similar to this, but where the metric satisfies a partial differential equation.

Answer 1

Yes, this is possible with a simple formula (although with an infinite family of spaces in between; see below).

As $t \in [0,1]$ varies, just use the family of metrics $$e^{2ty} dx^2 + dy^2 $$ When $t=1$ one gets the metric $e^{2y}dx^2 + dy^2$ which is isometric to the upper half plane metric $\frac{dx^2 + dz^2}{z^2}$ using the transformation $z=e^y$ (or perhaps $z=e^{-y}$).

As $t$ decreases from $1$ to $0$ the curvature $\kappa(t)$ varies from $\kappa(1)=-1$ to $\kappa(0)=0$.

Notice that one gets an infinite family of hyperbolic metric spaces between $t=0$ and $t=1$. In fact its not reasonable to expect only a finite family, if one wants the metrics to vary smoothly, because the curvature will then be a smooth function of $t$ and it must therefore decrease smoothly from $\kappa(1)=-1$ to $\kappa(0)=0$.