Isometries of hyperbolic plane

Question

How can i show that the metric of minkowski (or hyperboloid model) is the same that half plane model without use calculus of complex variables?

Answer 1

I will expand on the discussion in this post.

Consider the Lorentz metric $ds^2_L=dx^2+dy^2-dz^2$ on $3$-space and take the (half) hyperboloid $z=\sqrt{1+x^2+y^2}$. To map to the Poincaré disk, we send $(x,y,z)$ to $(\frac xz,\frac yz,1)$, then to $(\frac xz,\frac yz,0)$, then to the unit sphere at $(\frac xz,\frac yz,\frac 1z)$. Finally, we stereographically project from the south pole to the unit disk and map to $\frac1{z+1}(x,y)$. You can check that the inverse mapping — from the Poincaré disk to the hyperboloid — is $$f(u,v) = \frac1{1-(u^2+v^2)}(2u,2v,1+u^2+v^2).$$ (You should check that this point actually does lie on the hyperboloid.)

We have \begin{align*} x&=\frac{2u}{1-(u^2+v^2)},\\ y&=\frac{2v}{1-(u^2+v^2)},\\ z&=\frac{1+u^2+v^2}{1-(u^2+v^2)}, \end{align*} and so \begin{align*} f^*dx&=\frac{2(1+u^2-v^2)du + 4uv\,dv}{(1-(u^2+v^2))^2},\\ f^*dy&=\frac{4uv\,du + 2(1-u^2+v^2)dv}{(1-(u^2+v^2))^2},\\ f^*dz&=\frac{4(u\,du+v\,dv)}{(1-(u^2+v^2))^2}. \end{align*} Now let's compute the pullback $f^*ds^2_L$ of the Lorentz metric on the hyperboloid. We have \begin{align*} f^*(dx^2+dy^2-dz^2) &= (f^*dx)^2 + (f^*dy)^2 - (f^*dz)^2 \\ &= \frac {4(du^2+dv^2)}{(1-(u^2+v^2))^2}, \end{align*} which is the metric on the Poincaré disk with constant curvature $-1$.

You can convert this to the upper half-plane by pulling back by any standard conformal mapping from the half-plane to the disk.