The Riemann Sphere and Möbius Transformations — Examples

The determinant condition is $ad-bc=2i\mathrm{\ne }0$, so [L1] makes $C$ Möbius. For real $x$, one has $\mathrm{\mid }C(x)\mathrm{\mid }=1$, and for $z=x+iy$ with $y>0$ the identity $\mathrm{\mid }z+i{\mathrm{\mid }}^2-\mathrm{\mid }z-i{\mathrm{\mid }}^2=4y>0$ gives $\mathrm{\mid }C(z)\mathrm{\mid }<1$.

The inverse formula is $C^{-1}(w)=i(1+w)\mathrm{/}(1-w)$, and the same calculation in reverse shows that $\mathrm{\mid }w\mathrm{\mid }<1$ implies $\mathrm{Im}{C}^{-1}(w)>0$. Hence $C$ carries the upper half-plane biholomorphically onto the unit disc.

The formula $M(z)=(1-i)(z-i)\mathrm{/}(z-1)$ satisfies $M(1)=\mathrm{\infty }$, $M(i)=0$, and $M(-1)=1$ by direct substitution.

Fact [L1] makes a Möbius transformation with those three prescribed values unique, so this formula is exactly the desired map.

The map $z\mapsto z+1$ fixes only $\mathrm{\infty }$, so [L1] classifies it as parabolic; the map $z\mapsto 2z$ fixes $0$ and $\mathrm{\infty }$ and is already the dilation normal form with multiplier $2$, so [L1] classifies it as hyperbolic.

The map $z\mapsto 1\mathrm{/}z$ fixes $1$ and $-1$, and conjugating by $(z-1)\mathrm{/}(z+1)$ turns it into $w\mapsto -w$. Since that multiplier has modulus $1$, [L1] places it in the elliptic branch.

Fact [L1] already gives the inverse formulas. Substituting $z=0$ gives the south pole $(0,0,-1)$, and the defining clause sends $\mathrm{\infty }$ to the north pole $(0,0,1)$.

Substituting $(0,0,-1)$ into the inverse formula returns $0$, while the exceptional north-pole clause returns $\mathrm{\infty }$. Thus the coordinate formulas behave exactly as claimed.

Substituting the stereographic coordinates of $z$ and $w$ into the Euclidean distance formula on $S^2$ simplifies to $\chi (z,w{)}^2=4\mathrm{\mid }z-w{\mathrm{\mid }}^2\mathrm{/}((1+\mathrm{\mid }z{\mathrm{\mid }}^2)(1+\mathrm{\mid }w{\mathrm{\mid }}^2))$, and taking square roots gives the finite-point formula.

Using $\mathrm{\Sigma }(\mathrm{\infty })=(0,0,1)$ in the same calculation gives $\chi (z,\mathrm{\infty }{)}^2=4\mathrm{/}(1+\mathrm{\mid }z{\mathrm{\mid }}^2)$, so $\chi (z,\mathrm{\infty })=2\mathrm{/}\sqrt{1+\mathrm{\mid }z{\mathrm{\mid }}^2}$.

The function $e^z$ is entire on $\mathbb{C}$, so it is meromorphic on $\mathbb{C}$. If it were meromorphic on $\widehat{\mathbb{C}}$, then [L1] would make it rational.

A rational function with no finite poles is a polynomial, but $e^{z+2\pi i}=e^z$ whereas no nonconstant polynomial is $2\pi i$-periodic. Therefore $e^z$ is not rational, so it cannot be meromorphic on the sphere.

If $w=r(\cos \theta +i\sin \theta )$ with $r>0$, [L1] gives a real $x$ with $e^x=r$, and then $\exp (x+i\theta )=w$. So the exponential map is surjective onto ${\mathbb{C}}^{\times }$.

Fact [L2] gives $\exp (0)=\exp (2\pi i)=1$, so the exponential map is not injective. It is therefore a holomorphic surjection onto ${\mathbb{C}}^{\times }$ that is not an automorphism.

The map $\kappa$ is continuous, involutive, and fixes $\mathrm{\infty }$, so it is a self-homeomorphism of $\widehat{\mathbb{C}}$; under [L1] it is the reflection of the unit sphere across the $xz$-plane.

At $0$, the complex difference quotient along real increments equals $1$ while along imaginary increments $it$ it equals $-1$, so the complex derivative does not exist there. Thus this sphere homeomorphism is not holomorphic.

Fact [L1] makes $M(z)=1\mathrm{/}z$ a biholomorphic self-map of the sphere.

As a sphere map, $M$ is defined at $0$ and satisfies $M(0)=\mathrm{\infty }$. Therefore its restriction to the finite plane does not even take values in $\mathbb{C}$, so it is not an entire map $\mathbb{C}\to \mathbb{C}$, let alone an entire biholomorphism.

The identity map is Möbius and has the same values as $M$ on those three fixed points.

Fact [L1] makes a Möbius transformation with those three prescribed values unique, so $M$ must already be the identity. Hence the statement is false.

Fact [L1] provides a sphere homeomorphism that sends the cross-ratio value $i$ to the different value $-i$ on an explicit quadruple.

Therefore not every sphere homeomorphism preserves cross-ratios, so the statement is false.

Fact [L1] makes $\widehat{\mathbb{C}}$ compact, while the open cover $\{B(0,n):n\ge 1\}$ of $\mathbb{C}$ has no finite subcover, so $\mathbb{C}$ is not compact.

Homeomorphisms preserve compactness, so a compact space cannot be homeomorphic to a noncompact one. Hence the statement is false.