# explicit isometry between metric spaces

Let $X=\mathbb{H}^2\times\left[0,1\right]$ (just $\left\{ (x,y,z)\big\vert y>0\right\}$ as a set) and consider the following two metrics: $$ds_1=\frac{dx+dy}{y}+dz$$ and $$ds_2=\frac{dx+dy+dz}{z}$$ which are (or should be, I hope!) the product metric and the metric induced from $\mathbb{H}^3\cong_{Top}\mathbb{H}^2\times\mathbb{R}\supseteq\mathbb{H}^2\times\left[0,1\right]$, respectively. I would like to find a map $f\colon (X,ds_1)\to (\mathbb{H}^3,ds_2)$ such that its restriction to $\mathbb{H}^2\times(0,1)$ is an isometry onto the image $\mathbb{H}^2\times (n,n+1)$. The problem is to find a map which balances somehow the fact that the 'length' with respect to the third coordinate is 'homogeneous' in only one of the two metrics, namely in $ds_1$. I thought that, since $(\mathbb{H}^3,ds_1)$ and $(\mathbb{H}^3, ds_2)$ are not isometric spaces (only the latter being hyperbolic) I should exploit some property given by the boundedness of the interval $[0,1]$, like multiplying for a factor $(1-z)$ or something similar, and 'moving' the stretching problem on the boundary, where the restriction is not defined; anyway, I wasn't able to write down a map explicitly, or at least to prove theoretically that such a map exists. Could you help me? Thank you.

-
I think you mean $ds_1^2$, $dx^2$ etc. – Robert Israel Jan 3 '12 at 20:34

There is no isometric embedding of $\mathbb H^2\times (0,1)$ into $\mathbb H^3$. Indeed, $\mathbb H^2$ contains geodesic lines (isometric images of $\mathbb R$), and the product of such a line $L$ with $(0,1)$ is a flat infinite strip. Which is a non-hyperbolic space: it contains triangles with angle sum 180 degrees (imagine that!) and a pair of geodesic lines $L\times \{1/3\}$, $L\times \{2/3\}$ which do not diverge exponentially in either direction, but stay at constant distance from each other (the horror!). No such ugly things can ever happen in the nice hyperbolic space $\mathbb H^3$.