# Stereographic projection is conformal --- from the line element

I'm looking over some fairly basic stuff on complex methods and the book I'm using takes the formula for the stereographic projection:

$$z = \cot(\beta/2)e^{i\phi}$$

as well as the line element on the sphere:

$$ds^2 = d\beta^2 + \sin^2 \beta d\phi^2$$

(where $\beta$ is the polar angle and $\phi$ the azimuthal angle) and uses this to derive

$$\frac{4 dz d \bar{z}}{(1 + z\bar{z})^2} = d\beta^2 + \sin^2 \beta d\phi^2 \,.$$

It then says "from this we deduce that the stereographic projection is conformal". I don't understand this step. At the moment the only mention made of conformal maps is that holomorphic maps from C to C are conformal.

• Do you assert that the textbook says “the stereographic projection is conformal” before any definition of a conformal map? Commented Nov 11, 2014 at 19:58
• No, the first definition is that a conformal map is one which preserves angles. The book then shows that holomorphic maps from C to C are conformal. Then it goes on to state that the stereographic projection is conformal. Thanks. Commented Nov 16, 2014 at 13:41

A map is conformal if it induces a rescaling of the metric, ie. for $f: (M,g_M) \rightarrow (N,g_N)$, $f^* g_N = \phi \; g_M$ for some $\phi \in C^\infty(M)$.
If above is not helpful, then you can picture it in the following way: pick a point on the sphere $(\beta, \phi )$ on the sphere, and take two tiny vectors tangent to the surface at this point, say $v_1 = (d \beta_1, d \phi_1 )$ and $v_2 ( d \beta_2, d \phi_2 )$. The line element $ds^2 = d\beta^2 + \sin^2 \beta d\phi^2$ tells you how to take the dot product of the vectors, eg. $v_1 \cdot v_2 = (d \beta_1 d \beta_2 + \sin(\beta)^2 d \phi_1 d \phi_2)$, and also the angle between $v_1$ and $v_2$, eg. $\frac{v_1 \cdot v_2 }{ |v_1| |v_2| }$.
The stereographic projection map takes the point $(\beta, \phi)$ to the point $(z, \bar{z}) = (\cot(\beta/2)e^{i\phi}, \cot(\beta/2)e^{-i\phi})$, and sends the tiny tangent vectors $v_1$ and $v_2$ to tiny tangent vectors starting at $(z, \bar{z})$, say $v_1' = (dz_1, d\bar{z}_1)$ and $v_2'$. In the complex plane, the line element $dz \; d \bar{z}$ again tells you how to take inner product, eg $v_1' \cdot v_2' = dz_1 d\bar{z}_2+dz_2 d\bar{z}_1$. (You can check that if you write $z = x + i y$, then it gives you $x_1 x_2 + y_1 y_2$, or times some factor of 2).
The calculation he does shows that the the dot product $v_1' \cdot v_2' = (1 + z \bar{z})^2 \; v_1 \cdot v_2$. So in particular, the angle between $v_1$ and $v_2$, is the same as $v_1'$ and $v_2'$.