The Riemann Sphere and Möbius Transformations

Direct algebra gives $\mathrm{\Sigma }(z)\in S^2$ for every finite $z$, and the displayed formulas satisfy $\mathrm{\Sigma }(\mathrm{\Pi }(x,y,t))=(x,y,t)$ for $(x,y,t)\mathrm{\ne }(0,0,1)$ and $\mathrm{\Pi }(\mathrm{\Sigma }(z))=z$ for finite $z$, with $\mathrm{\Sigma }(\mathrm{\infty })=(0,0,1)$ and $\mathrm{\Pi }(0,0,1)=\mathrm{\infty }$.

On $\mathbb{C}$ and on $S^2\setminus \{(0,0,1)\}$ the formulas are rational with nonzero denominator, so both restrictions are continuous; and for the cap $U_t=\{(x,y,s):s>t\}\cup \{(0,0,1)\}$ one has ${\mathrm{\Sigma }}^{-1}(U_t)=\{z:\mathrm{\mid }z\mathrm{\mid }>\sqrt{(1+t)\mathrm{/}(1-t)}\}\cup \{\mathrm{\infty }\}$, which is a neighbourhood of $\mathrm{\infty }$ by [L1].

If $V=\widehat{\mathbb{C}}\setminus K$ is a neighbourhood of $\mathrm{\infty }$, compactness of $K$ gives $R>0$ with $K\subseteq \bar{D}(0,R)$, so the cap $U_{(R^2-1)\mathrm{/}(R^2+1)}$ satisfies $\mathrm{\Pi }(U_{(R^2-1)\mathrm{/}(R^2+1)})\subseteq V$; therefore $\mathrm{\Pi }$ is continuous at the north pole.

The maps $\mathrm{\Sigma }$ and $\mathrm{\Pi }$ are continuous inverse bijections by the preceding three steps, so $\mathrm{\Sigma }$ is a homeomorphism.

By definition, $\chi (p,q)=\mathrm{\parallel }\mathrm{\Sigma }(p)-\mathrm{\Sigma }(q){\mathrm{\parallel }}_2$, so $\mathrm{\Sigma }$ is an isometry from $(\widehat{\mathbb{C}},\chi )$ onto $S^2$ with its Euclidean subspace metric.

The Euclidean subspace metric induces the usual topology of $S^2$, and [L1] already identifies that topology with the one-point-compactification topology on $\widehat{\mathbb{C}}$. Therefore $\chi$ induces the same topology.

For $A=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ and $B=\left(\begin{array}{cc}\alpha & \beta \\ \gamma & \delta \end{array}\right)$, direct algebra gives $M_A(M_B(z))=M_{AB}(z)$ wherever both sides are finite, hence on all of $\widehat{\mathbb{C}}$. So $A\mapsto M_A$ is a homomorphism, and it is surjective by the definition of Möbius transformation.

Because $A^{-1}\in G{L}_2(\mathbb{C})$, the identity $M_A\circ M_{A^{-1}}=M_I=M_{A^{-1}}\circ M_A$ shows that Möbius transformations form a group. The kernel condition $M_A(z)=z$ gives $c{z}^2+(d-a)z-b=0$ for all finite $z$, so $b=c=0$ and $a=d\mathrm{\ne }0$; hence the kernel is exactly ${\mathbb{C}}^{\times }I$.

Applying [L1] to the surjective homomorphism proved above and the kernel computation of step 1.2 gives $\mathrm{Mob}(\widehat{\mathbb{C}})\cong G{L}_2(\mathbb{C})\mathrm{/}({\mathbb{C}}^{\times }I)=PG{L}_2(\mathbb{C})$.

On every open set where $cz+d\mathrm{\ne }0$, the finite-chart expression $z\mapsto (az+b)\mathrm{/}(cz+d)$ is a rational function with nonvanishing denominator, hence holomorphic; and if $c\mathrm{\ne }0$, then near the finite pole $p=-d\mathrm{/}c$ the target infinity-chart expression is $(cz+d)\mathrm{/}(az+b)$, which is holomorphic because $az+b$ does not vanish at $p$.

At $\mathrm{\infty }$, if $c\mathrm{\ne }0$ then the source infinity-chart expression is $(a+bw)\mathrm{/}(c+dw)$, which is holomorphic at $0$; if $c=0$, then $M(\mathrm{\infty })=\mathrm{\infty }$ and the target infinity-chart expression is $dw\mathrm{/}(a+bw)$, again holomorphic at $0$. Thus $M$ is holomorphic at every sphere point.

By [L2], the inverse map is again Möbius, so the same two chart computations apply to $M^{-1}$ as well. Therefore $M$ is a biholomorphism of the sphere.

Define $N(z):=\{\begin{array}{ll}{\displaystyle \frac{(z-b)(c-a)}{(z-a)(c-b)}},& a,b,c\in \mathbb{C},\\ {\displaystyle \frac{z-b}{c-b}},& a=\mathrm{\infty },\\ {\displaystyle \frac{c-a}{z-a}},& b=\mathrm{\infty },\\ {\displaystyle \frac{z-b}{z-a}},& c=\mathrm{\infty },\end{array}$ and define $N^{\mathrm{\prime }}$ by the same formula with $(a,b,c)$ replaced by $(a^{\mathrm{\prime }},b^{\mathrm{\prime }},c^{\mathrm{\prime }})$. In each case the displayed formula is Möbius and direct substitution gives $N(a)=\mathrm{\infty }$, $N(b)=0$, $N(c)=1$ and $N^{\mathrm{\prime }}(a^{\mathrm{\prime }})=\mathrm{\infty }$, $N^{\mathrm{\prime }}(b^{\mathrm{\prime }})=0$, $N^{\mathrm{\prime }}(c^{\mathrm{\prime }})=1$.

The composition $(N^{\mathrm{\prime }}{)}^{-1}\circ N$ is Möbius by [L1] and carries $(a,b,c)$ to $(a^{\mathrm{\prime }},b^{\mathrm{\prime }},c^{\mathrm{\prime }})$, so the required map exists.

If another Möbius map had the same three values, then composing with $N$ and $N^{\mathrm{\prime }}$ would produce a Möbius map fixing $0$, $1$, and $\mathrm{\infty }$; writing it as $(\alpha z+\beta )\mathrm{/}(\gamma z+\delta )$ forces $\beta =\gamma =0$ and then $\alpha =\delta$, so it is the identity. Thus the map is unique.

Let $N$ be the unique Möbius transformation sending $(z_2,z_3,z_4)$ to $(1,0,\mathrm{\infty })$, and let $N_M$ be the unique Möbius transformation sending $(M(z_2),M(z_3),M(z_4))$ to $(1,0,\mathrm{\infty })$. The defining formulas for the cross-ratio in its first variable give a Möbius map that sends $(z_2,z_3,z_4)$ to $(1,0,\mathrm{\infty })$, so uniqueness gives $N(z)=[z,z_2;z_3,z_4]$ for all $z$. Likewise $N_M(w)=[w,M(z_2);M(z_3),M(z_4)]$ for all $w$. In particular $N(z_1)=[z_1,z_2;z_3,z_4],N_M(M(z_1))=[M(z_1),M(z_2);M(z_3),M(z_4)]\mathrm{.}$

The maps $N\circ M^{-1}$ and $N_M$ have the same action on the triple $(M(z_2),M(z_3),M(z_4))$, so [L1] makes them equal. Evaluating at $M(z_1)$ yields $[M(z_1),M(z_2);M(z_3),M(z_4)]=[z_1,z_2;z_3,z_4]$.

If the four points lie on one circline, then taking that circline to be $C(z_1,z_2,z_3)$ forces $z_4\in C(z_1,z_2,z_3)$. Because the four points are distinct, $z_4\mathrm{\notin }\{z_1,z_2,z_3\}$, so [L1] gives $[z_1,z_2;z_3,z_4]\in \mathbb{R}$.

Conversely, if $[z_1,z_2;z_3,z_4]\in \mathbb{R}$, then $z_4\mathrm{\notin }\{z_1,z_2,z_3\}$ and [L1] says exactly that $z_4\in C(z_1,z_2,z_3)$. Hence the four points lie on a single circline.

Writing $C=C(a,b,c)$, the defining points $a,b,c$ map to $M(a),M(b),M(c)\in C(M(a),M(b),M(c))$. If $z\mathrm{\notin }\{a,b,c\}$, then [L2] says $z\in C$ exactly when $[a,b;c,z]\in \mathbb{R}$. By [L1] this is equivalent to $[M(a),M(b);M(c),M(z)]\in \mathbb{R}$, which is exactly the condition $M(z)\in C(M(a),M(b),M(c))$. Thus $M(C)=C(M(a),M(b),M(c))$ is again a circline.

If $C=C(a,b,c)$, then [L3] gives a Möbius map $N$ with $N(a)=\mathrm{\infty }$, $N(b)=0$, and $N(c)=1$. Step 1.1 makes $N(C)$ a circline. Its defining triple $\mathrm{\infty },0,1$ already lies in $C(\mathrm{\infty },0,1)$, and for every $w\mathrm{\notin }\{\mathrm{\infty },0,1\}$ one has $[\mathrm{\infty },0;1,w]=w$, so [L2] makes $w\in C(\mathrm{\infty },0,1)$ exactly when $w\in \mathbb{R}$. Hence $N(C)=C(\mathrm{\infty },0,1)=\widehat{\mathbb{R}}$, so every circline admits a Möbius normalization to the standard real circline.

Let $N$ and $N^{\mathrm{\prime }}$ be two normalizing maps for $C$, and put $H:=N^{\mathrm{\prime }}\circ N^{-1}$. By [L4], $H$ is Möbius and preserves $\widehat{\mathbb{R}}$. The map $\tilde{H}:={\sigma }_{\widehat{\mathbb{R}}}\circ H\circ {\sigma }_{\widehat{\mathbb{R}}}$ is Möbius by conjugating the coefficients of a fractional-linear formula. Moreover, $H$ and $\tilde{H}$ agree at $0$, $1$, and $\mathrm{\infty }$, because these points and their $H$-images lie in $\widehat{\mathbb{R}}$. By [L3], $\tilde{H}=H$, equivalently $H\circ {\sigma }_{\widehat{\mathbb{R}}}={\sigma }_{\widehat{\mathbb{R}}}\circ H$.

Since $N^{\mathrm{\prime }}=H\circ N$, step 3.1 gives $(N^{\mathrm{\prime }}{)}^{-1}\circ {\sigma }_{\widehat{\mathbb{R}}}\circ N^{\mathrm{\prime }}=N^{-1}\circ H^{-1}\circ {\sigma }_{\widehat{\mathbb{R}}}\circ H\circ N=N^{-1}\circ {\sigma }_{\widehat{\mathbb{R}}}\circ N\mathrm{.}$ Thus the reflection ${\sigma }_C$ is independent of the normalizing map.

Choose a normalizing map $N$ for $C$. Since step 1.1 makes $M(C)$ a circline, $N\circ M^{-1}$ normalizes $M(C)$ to $\widehat{\mathbb{R}}$. Using the well-defined reflection from step 4.1 gives ${\sigma }_{M(C)}=(N\circ M^{-1}{)}^{-1}\circ {\sigma }_{\widehat{\mathbb{R}}}\circ (N\circ M^{-1})=M\circ {\sigma }_C\circ M^{-1}\mathrm{.}$

If $R=P\mathrm{/}Q$ is rational with coprime polynomials, then it is holomorphic on $\mathbb{C}$ away from the zeros of $Q$, those zeros are poles of finite order, and in the infinity chart the expression $R(1\mathrm{/}w)$ is meromorphic at $0$. So every rational map is meromorphic on the sphere.

Conversely, assume $f$ is sphere-meromorphic. Meromorphy at $\mathrm{\infty }$ gives a radius beyond which there are no finite poles, and [L3] makes the remaining finite pole set discrete. Covering that pole set by isolating discs and using compactness of the containing closed disc shows that only finitely many finite poles occur.

Fact [L1] gives a finite principal part at each finite pole and a finite principal part in the infinity chart. Subtracting all of those principal parts leaves an entire function that is bounded near $\mathrm{\infty }$, hence bounded on all of $\mathbb{C}$; [L2] therefore makes the remainder constant. So $f$ is a rational function.

The first step proves the rational-to-meromorphic direction and the latter two steps prove the converse, so sphere-meromorphic functions are exactly rational functions.

For a finite value $w$, the finite preimages of $w$ are exactly the roots of $P-wQ$. If $\mathrm{deg}(P-wQ)=d$, then [L1] gives exactly $d$ such roots. If $\mathrm{deg}(P-wQ)=m<d$, then $\mathrm{\infty }$ is also a preimage and its multiplicity is exactly $d-m$ in the infinity chart, so the total multiplicity is still $d$.

For $w=\mathrm{\infty }$, the finite preimages are the roots of $Q$ with multiplicity. If $\mathrm{deg}P\le \mathrm{deg}Q=d$, then [L1] gives all $d$ preimages in the finite chart; if $\mathrm{deg}P>\mathrm{deg}Q$, then $\mathrm{\infty }$ contributes the remaining multiplicity $\mathrm{deg}P-\mathrm{deg}Q$. So the total multiplicity is again $d$.

Every sphere value is either finite or $\mathrm{\infty }$, and both cases give total fibre multiplicity $d$.

Because $F$ is holomorphic on the sphere, [L1] makes it a rational map. Bijectivity means every sphere value has exactly one preimage, and that preimage has multiplicity $1$.

Applying [L2] to any fibre forces the degree of $F$ to be $1$, and a degree-$1$ rational self-map is exactly a Möbius transformation.

Since $f$ and $f^{-1}$ are homeomorphisms of $\mathbb{C}$, [L2] extends $f$ to a sphere homeomorphism $F$ with $F(\mathrm{\infty })=\mathrm{\infty }$. In the infinity chart, the reciprocal expression $1\mathrm{/}f(1\mathrm{/}w)$ is bounded near $0$, so the removable-singularity theorem makes $F$ meromorphic at $\mathrm{\infty }$. Thus $F$ is a meromorphic self-map of the sphere.

Fact [L1] makes $F$ a rational sphere map of degree $1$, hence Möbius. Because $F(\mathrm{\infty })=\mathrm{\infty }$, its denominator has zero $z$-coefficient, so restricting back to $\mathbb{C}$ gives $f(z)=az+b$ with $a\mathrm{\ne }0$.

If $z_n\to 0$ in ${\mathbb{C}}^{\times }$ and $f(z_n)\to a\in {\mathbb{C}}^{\times }$, continuity of $f^{-1}$ on ${\mathbb{C}}^{\times }$ forces $z_n=f^{-1}(f(z_n))\to f^{-1}(a)\in {\mathbb{C}}^{\times }$, a contradiction. Hence every cluster value of $f(z)$ as $z\to 0$ lies in $\{0,\mathrm{\infty }\}$. If both $0$ and $\mathrm{\infty }$ were cluster values, then every sufficiently small punctured disc would have image meeting both $\{\mathrm{\mid }w\mathrm{\mid }<1\}$ and $\{\mathrm{\mid }w\mathrm{\mid }>1\}$; connectedness of that image would then produce points with $\mathrm{\mid }f(z)\mathrm{\mid }=1$ approaching $0$, giving a finite nonzero cluster value after all. Therefore $f(z)$ tends to a single limit ${\mathrm{\ell }}_0\in \{0,\mathrm{\infty }\}$ as $z\to 0$. The same argument applied to $z\mapsto f(1\mathrm{/}z)$ shows that $f(z)$ tends to a single limit ${\mathrm{\ell }}_{\mathrm{\infty }}\in \{0,\mathrm{\infty }\}$ as $z\to \mathrm{\infty }$.

Let $F$ be the extension of $f$ to $\widehat{\mathbb{C}}$ with $F(0)={\mathrm{\ell }}_0$ and $F(\mathrm{\infty })={\mathrm{\ell }}_{\mathrm{\infty }}$, and let $G$ be the analogous extension of $f^{-1}$. Step 1.1 gives continuity of both extensions at the added points, and on the dense subset ${\mathbb{C}}^{\times }$ one has $G\circ F=\mathrm{id}=F\circ G$. By continuity the same identities hold on all of $\widehat{\mathbb{C}}$, so $F$ is a sphere homeomorphism and $F(\{0,\mathrm{\infty }\})=\{0,\mathrm{\infty }\}$.

Near each of $0$ and $\mathrm{\infty }$, the homeomorphism $F$ lands either in a bounded finite chart or in a neighbourhood of $\mathrm{\infty }$. In the first case the corresponding chart expression is bounded near the puncture and extends holomorphically by [L2]; in the second case its reciprocal is bounded and again extends holomorphically by [L2]. Thus $F$ is meromorphic at both added points, and therefore on the whole sphere.

Fact [L1] makes $F$ a rational sphere map of degree $1$, hence Möbius. A Möbius map preserving the set $\{0,\mathrm{\infty }\}$ is either $z\mapsto az$ or $z\mapsto a\mathrm{/}z$ with $a\mathrm{\ne }0$, and restricting back to ${\mathbb{C}}^{\times }$ gives exactly the claimed automorphisms.

Writing $M(z)=(az+b)\mathrm{/}(cz+d)$, the fixed-point equation is $c{z}^2+(d-a)z-b=0$ in the finite chart, together with the possibility that $\mathrm{\infty }$ is fixed. Therefore a nonidentity Möbius transformation has at most two fixed points.

If $M$ has exactly one fixed point $p$, then [L1] provides a Möbius map $T$ sending $p$ to $\mathrm{\infty }$. The conjugate $TM{T}^{-1}$ fixes $\mathrm{\infty }$, so it has the form $z\mapsto \alpha z+\beta$; uniqueness of the fixed point forces $\alpha =1$ and $\beta \mathrm{\ne }0$, so a further scaling conjugates it to $z\mapsto z+1$.

If $M$ has two fixed points $p\mathrm{\ne }q$, then [L1] provides a Möbius map sending them to $0$ and $\mathrm{\infty }$. The conjugate therefore fixes both $0$ and $\mathrm{\infty }$, hence has the form $z\mapsto \lambda z$ with $\lambda \in {\mathbb{C}}^{\times }\setminus \{1\}$.

Replacing a representing matrix $A$ by $tA$ multiplies both $\mathrm{tr}(A{)}^2$ and $\det A$ by $t^2$, so $\tau (M)=\mathrm{tr}(A{)}^2\mathrm{/}\det A$ is well defined on the projective class. For $z\mapsto \lambda z$ it equals $(\lambda +1{)}^2\mathrm{/}\lambda =\lambda +2+{\lambda }^{-1}$, and for $z\mapsto z+1$ it equals $4$.

Step 1.1 leaves only the one-fixed-point and two-fixed-point cases, and the preceding three steps identify those cases with the parabolic and dilation normal forms, the standard dilation-branch terminology, and the projective trace invariant.