Let $\mathbb{H}^n$ be the usual hyperbolic space $\mathbb{H}^n := \{(x_1,\ldots,x_n) : x_n > 0\}$ with metric $g = \frac{1}{x_n^2}\sum_{i=1}^ndx_i^2.$ Lets construct another model: $H_{-1}^n := \{(x_0,\ldots,x_n) : x_0 = \sqrt{1 + \sum_{i=1}^nx_i^2}\}$ whose metric is the restriction to $H_{-1}^n$ of the metric $h = - dx_0^2 + \sum_{i=1}^ndx_i^2.$
How can I show that these two are isometric?
I am having trouble about construct the function that gives the isometry.
Thanks in advance.
$$H_1=\{ (x_1,x_2)| x_2>0 \}\ {\rm with}\ g_1= \frac{dx_1^2+dx_2^2}{x_2^2} $$
$$ H_2=\{ (x,y,z)\in \mathbb{E}^3| z=\sqrt{1+x^2+y^2} \}\ {\rm with}\ g_2 $$
If $p=(0,1)\in H_1,\ q=(0,0,1)\in H_2$ then define $$ i : S^1(1)\subset T_pH_1\rightarrow S^1(1)\subset T_q H_2 $$ : Assume that $i(v)=w$ If $\angle (v,v_1)=\angle (w,w_1)$ then $i(v_1)=w_1$
Then we have $$ f(\exp_p\ tv)=\exp_q\ t\cdot i(v),\ |v|=1$$
At $p,\ q$ $g_1=dt^2+ h(t)^2 d\theta^2=g_2$ since $H_i$ has constant Gaussian curvature Hence $f$ is isometry
