Isometry of Riemann sphere?

Question

The complex metric on the Riemann sphere is given in the Wikipedia article to be

$$ds^2=\frac{4}{(1+\zeta\bar \zeta)^2}d\zeta d\bar \zeta$$

while the sphere should be mapped to itself under $SL(2,\mathbb{C})$ Moebius transformations

$$\zeta\to\frac{a \zeta + b}{c \zeta + d}~,\qquad \bar \zeta\to\frac{\bar a \bar \zeta + \bar b}{\bar c \bar \zeta + \bar d}$$

with $$a d-b c=\bar a\bar d-\bar c\bar b=1.$$ However, when I explicitly plug these transformations into $ds^2$ it does not reduce back to itself, as it should under an isometry. Is there a mistake somewhere, or did I get myself confused? What is the correct Riemann sphere metric with complex coordinates that is isometric to itself under Moebius transformations?

Answer 1

By proposition 8.4 in these lecture notes:

https://www.dpmms.cam.ac.uk/~tkc/GeometryandGroups/GeometryandGroups.pdf

a Moebius transformation is only an isometry of the Riemann sphere if

$$d=\bar a~~~,~~~ c=-\bar b~~~,~~~\bar d= a~~~,~~~ \bar c=- b~~~,~~~a\bar a+b\bar b=1 .$$

Plugging these parameters into the Moebius transformation and applying it to $ds^2$ indeed returns the same $ds^2$ back, so that the isometry is correct for the metric given on Wikipedia.