1.3 Spherical Representation

Spherical Representation

We have a sphere $S$ :

x_{1}^{2} + x_{2}^{2} + x_{3}^{2} = 1

Every point on $S$ except can associate a complex number

z = \frac{x_{1} + i x_{2}}{1 - x_{3}}

| z |^{2} = \frac{x_{1}^{2} + x_{2}^{2}}{(1 - x_{3})^{2}} = \frac{1 + x_{3}}{1 - x_{3}}

Mapping

⟹ {\begin{cases} x_{3} = \frac{| z |^{2} - 1}{| z |^{2} + 1} \\ x_{1} = \frac{z + \bar{z}}{1 + | z |^{2}} \\ x_{2} = \frac{z - \bar{z}}{i (1 + | z |^{2})} \end{cases}

Circles and Lines

z = x + i y \bar{z} = x - i y

where $x = x_{1}, y = x_{2} .$

Then:
$z + \bar{z} = 2 x$ , $z - \bar{z} = 2 i y$ , and $| z |^{2} = z \bar{z} = x^{2} + y^{2}$ .

We start with the plane on the unit sphere:

α_{1} x_{1} + α_{2} x_{2} + α_{3} x_{3} = α_{0}, α_{1}^{2} + α_{2}^{2} + α_{3}^{2} = 1, 0 \leq α_{0} < 1.

Under stereographic projection, write

x = \frac{z + \bar{z}}{| z |^{2} + 1}, y = \frac{z - \bar{z}}{i (| z |^{2} + 1)}, x_{3} = \frac{| z |^{2} - 1}{| z |^{2} + 1} .

Multiply the plane equation by $| z |^{2} + 1$ :

α_{1} (z + \bar{z}) - α_{2} i (z - \bar{z}) + α_{3} (| z |^{2} - 1) = α_{0} (| z |^{2} + 1) .

or equivalently,

(α_{0} - α_{3}) (x^{2} + y^{2}) - 2 α_{1} x - 2 α_{2} y + α_{0} + α_{3} = 0.

For $α_{0} \neq α_{3}$ this is a circle,
and for $α_{0} = α_{3}$ it reduces to a straight line.

Metric for Distance

We compute the distance $d (z, z^{'})$ between stereographic projections of $z$ and $z^{'}$ .
If the points on the sphere are $(x_{1}, x_{2}, x_{3})$ , $(x_{1}^{'}, x_{2}^{'}, x_{3}^{'})$ , then

(x_{1} - x_{1}^{'})^{2} + (x_{2} - x_{2}^{'})^{2} + (x_{3} - x_{3}^{'})^{2} = 2 - 2 (x_{1} x_{1}^{'} + x_{2} x_{2}^{'} + x_{3} x_{3}^{'}) .

From our mapping:

x_{1} x_{1}^{'} + x_{2} x_{2}^{'} + x_{3} x_{3}^{'} = \frac{(z + \bar{z}) (z^{'} + {\bar{z}}^{'}) - (z - \bar{z}) (z^{'} - {\bar{z}}^{'}) + (| z |^{2} - 1) (| z^{'} |^{2} - 1)}{(1 + | z |^{2}) (1 + | z^{'} |^{2})} = \frac{(1 + | z |^{2}) (1 + | z^{'} |^{2}) - 2 | z - z^{'} |^{2}}{(1 + | z |^{2}) (1 + | z^{'} |^{2})} .

Thus,

d (z, z^{'}) = \frac{2 | z - z^{'} |}{\sqrt{(1 + | z |^{2}) (1 + | z^{'} |^{2})}} .

For $z^{'} = \infty$ ,

d (z, \infty) = \frac{2}{\sqrt{1 + | z |^{2}}} .

Key point: $d (\frac{1}{z}, \frac{1}{w}) = d (z, w)$

Neighbourhood of Infinity

The neighbourhood $B (\infty, ε)$ is defined by the inequality:

d (z, \infty) < ε

Substituting the correct formula for chordal distance:

\frac{2}{\sqrt{1 + | z |^{2}}} < ε

Now, we solve for $| z |$ (assuming $ε > 0$ ):

\begin{array}{r} \frac{2}{ε} < \sqrt{1 + | z |^{2}} ⟹ {(\frac{2}{ε})}^{2} < 1 + | z |^{2} ⟹ \frac{4}{ε^{2}} - 1 < | z |^{2} \end{array}

So, the correct definition for the $ε$ -neighbourhood using chordal distance is:

B (\infty, ε) = {z \in C : | z | > \sqrt{\frac{4}{ε^{2}} - 1}}

Which is usually just considered as ${z \in C : | z | > 1 / ε}$