Exam

Prove that the mapping

ψ (x) = x / x_{n + 1}

maps the hemisphere

H : = S^{n} \cap [x_{n + 1} < 0]

onto

R^{n}

. Is

ψ : H \to R^{n}

conformal? Determine the inverse of

ψ

.

hemisphere

The case

n = 2

suffices. The matrices of

D ψ (x)

and

D ψ (x)^{*} D ψ (x)

are given by:

\frac{1}{x_{3}} (\begin{array}{ccc} 1 & 0 & - \frac{x_{1}}{x_{3}} \\ 0 & 1 & - \frac{x_{2}}{x_{3}} \end{array}) and \frac{1}{x_{3}^{2}} (\begin{array}{ccc} 1 & 0 & - \frac{x_{1}}{x_{3}} \\ 0 & 1 & - \frac{x_{2}}{x_{3}} \\ - \frac{x_{1}}{x_{3}} & - \frac{x_{2}}{x_{3}} & \frac{x_{1}^{2} + x_{2}^{2}}{x_{3}^{2}} \end{array})

Thus for

x \in S^{2}

and

⟨ u, x ⟩ = 0

we obtain for

⟨ D ψ^{*} D ψ (x) u, u ⟩

(without the factor

x_{3}^{- 2}

):

\begin{array}{rcl} u_{1}^{2} + u_{2}^{2} - 2 \frac{x_{1} u_{1} + x_{2} u_{2}}{x_{3}} u_{3} + \frac{x_{1}^{2} + x_{2}^{2}}{x_{3}^{2}} u_{3}^{2} \\ = & u_{1}^{2} + u_{2}^{2} + 2 u_{3}^{2} + \frac{1 - x_{3}^{2}}{x_{3}^{2}} u_{3}^{2} \\ = & u_{1}^{2} + u_{2}^{2} + u_{3}^{2} + \frac{1}{x_{3}^{2}} u_{3}^{2} \end{array}

i.e.

ψ

is not conformal!
From

y_{j} = x_{j} / x_{n + 1}

,

x_{n + 1} < 0

and

\sum x_{j}^{2} = 1

we infer that

1 - x_{n + 1}^{2} = \sum x_{j}^{2} = x_{n + 1}^{2} \sum y_{j}^{2} i.e. x_{n + 1} = - \frac{1}{1 + ‖ y ‖^{2}}

and for

j = 1, \dots, n

:

x_{j} = y_{j} x_{n + 1}

.