# Isometry from Half-Plane metric to Poincare Disk metric

If I take the metric of the upper half-plane model $$(ds)^2=\frac{(dx)^2+(dy)^2}{y^2}$$ there is obviously an isometry which should provide the metric of the Poincare disk model $$(ds)^2= \frac{4\left((dx)^2+(dy)^2\right)}{(1-(x^2+y^2))^2}$$ Now, I understand this is achieved via a Möbius transform, with with $$z=x+iy$$, $$z \to \frac{z-i}{z+i}$$ but what I can't see is how this transforms one metric into the other.

Is this a standard calculation? I suppose it is just a change of variables from $$(x,y)$$ to $$(u(x),v(y))$$, and that with calculation of $$u'(x)$$ and $$v'(y)$$ from these coordinate transformations, one can just transform via the change of variables.

But is this how its done? What exactly are the coordinate transformation functions in this case?

Do you just use the Jacobian $$\frac{\partial(x,y)}{\partial(u,v)}$$ and transform?

• Similar to the maths below Example 5.5.2 here, but for the metric rather than the measure?
– apkg
Jun 16 at 20:01
• Yes, looks like you’ve got it right. Jun 16 at 21:24

Let $$\mathbb C = \mathbb R \oplus i \mathbb R$$, where $$x$$ is the real and $$y$$ is the imaginary coordinate. Then the transformation $$T : \mathbb C \to \mathbb C, T(z) = \frac{z -i}{z + i}$$ can be given in terms of the $$(x, y)$$ coordinates just by writing out the real and imaginary parts. That is, $$\frac{z -i}{z + i} = \frac{x + iy - i}{x + iy +i} = \frac{1}{x^2 + (y + 1)^2}\left(x^2 + y^2 - 1 + 2ix \right)$$ Now, if you restrict the domain to just the unit disk, then this map is smooth, and the image is not the upper half plane - note that $$x^2 + y^2 - 1$$ is always negative since we've restricted the domain, so the actual image is $$\{(x, y) : x < 0\}.$$ So I suppose the map you actually want is $$S = -i\frac{z - i}{z + i}.$$ Now a map $$T : (\mathbb C, g) \to (\mathbb C, h)$$ is an isometry if it preserves the metric, that is if the pullback of the metric in the image gives the metric in the domain. Written out this is just the condition $$g(z)(U, V) = h(T(z))(dT_z (U), dT_z(V)).$$ Here, $$U, V$$ are vectors and $$dT_z$$ is just the differential of T (the Jacobian). If you go through the calculation you'll find that the second metric is the pullback of the first, and so the map is an isometry with the given metrics.