Riemann sphere ($\hat{\mathbb{C}}=\mathbb{C}\cup \{\infty \}$)

The Riemann sphere is $\mathbb{C}\cup \{\infty \}$ with two charts related by reciprocal coordinates.

The extra point completes the complex plane into a compact Riemann surface. It is one point shared by every direction of escape: $n$, $-n$ and $in$ all tend to $\infty$ as $n\to \infty$. A basis of neighbourhoods at $\infty$ isThis is the one-point compactification of the complex plane; convergence to $\infty$ means that the modulus tends to infinity.^[ref]

For a concrete picture, use the unit sphereand identify $z=x+iy$ with $(x,y,0)$ in its equatorial plane. The north pole $N=(0,0,1)$ represents $\infty$ and the south pole $S=(0,0,-1)$ represents $0$. The stereographic projection below gives the identification. This convention fixes the scale: the equator corresponds to $|z|=1$.^[ref]

Why add this point? The Möbius transformation $z\mapsto 1/z$ then exchanges $0$ and $\infty$ and becomes a holomorphic map everywhere on the Riemann sphere. Its apparent blow-up at zero disappears when its output is expressed in the reciprocal chart at infinity: $w=1/(1/z)=z$. This is a concrete instance of a meromorphic function as a holomorphic map to the Riemann sphere. The additional point makes the complex plane compact, while its second manifold chart lets us do ordinary complex analysis near infinity.

The chordal metric on the Riemann sphere is $\chi (z,w)=|z-w|/\sqrt{(1+|z{|}^2)(1+|w{|}^2)}$ for finite points, with $\chi (z,\infty )=(1+|z{|}^2{)}^{-1/2}$. It is half the Euclidean chord distance between stereographic images on the unit sphere, so the triangle inequality follows from the Euclidean one. If $\rho$ is round spherical geodesic distance, then $\rho =2\arcsin \chi$ and $2\chi \le \rho \le \pi \chi$. These inequalities transfer uniform-continuity estimates between spherical and chordal distances, including at poles of meromorphic functions.

A rational map is smooth as a map between compact round spheres, including at its poles in reciprocal coordinates. Its spherical differential norm therefore has a finite supremum $L$. Integrating along a shortest spherical geodesic gives $\rho (Rz,Rw)\le L\rho (z,w)$. Since $2\chi \le \rho \le \pi \chi$, we obtain $\chi (Rz,Rw)\le (\pi L/2)\chi (z,w)$. In finite coordinates the differential norm is $R^{\#}(z)=|R^{\prime }(z)|(1+|z{|}^2)/(1+|R(z){|}^2)$, extending continuously to the whole sphere. This uniform bound is useful for passing equicontinuity through compositions.

Stereographic projection identifies a sphere minus one pole with the plane and supplies the standard charts of the Riemann sphere.

In the equatorial-plane convention of the Riemann sphere, draw the straight line through $N=(0,0,1)$ and a point $P=(X,Y,Z)\ne N$ on the unit sphere. Its intersection with the plane $Z=0$ has complex coordinateThe inverse function, for $z=x+iy$ and $r^2=x^2+y^2$, isHere positive $x$ gives positive $X$, and positive $y$ gives positive $Y$. Thus $1$ and $i$ lie on the positive $X$ and $Y$ ends of the equator. The omitted north pole is assigned $z=\infty$.^[ref]

For original worked points, the real numbers $1/2$ and $2$ lift to points with the same $X$ coordinate but opposite $Z$ coordinates:Projecting back gives $(4/5)/(1+3/5)=1/2$ and $(4/5)/(1-3/5)=2$. A point off the real axis checks the other sign:The coordinates satisfy $4/9+1/9+4/9=1$, as a point on the unit sphere must. Also $P(0)=S$, $P(-1)=(-1,0,0)$ and $P(i)=(0,1,0)$.

A short derivation starts with the line $N+t(P-N)=(tX,tY,1+t(Z-1))$. Setting its last coordinate to zero gives $t=1/(1-Z)$, hence the projection formula. Conversely the line from $N$ to $(x,y,0)$ is $(tx,ty,1-t)$. Substitution in the unit sphere equation givesThe root $t=0$ is $N$; the other root $t=2/(1+r^2)$ gives $P(z)$. These line intersections establish both the formula and its bijection property without guessing the inverse.

The same algebra explains how large planar distances become small spherical ones:Consequently $|z|\to \infty$ brings $P(z)$ to $N$ independently of direction. The last expression is twice the chordal metric distance $\chi (z,\infty )$. A whole unbounded exterior in the complex plane therefore becomes a small cap on the unit sphere.

For north-pole stereographic projection $z=(X+iY)/(1-Z)$ on the unit sphere, substitute $|z{|}^2=(1+Z)/(1-Z)$ into a planar circle equation. The displayed plane equation follows, and its intersection with the sphere is a circle. The plane contains the sphere's centre exactly when $r^2=1+|c{|}^2$, giving a great circle.

A planar circle other than the unit circle lifts to a great circle exactly when it meets the unit circle at two opposite points. A great circle other than the equator intersects the equator in an antipodal pair. Conversely the plane containing a lifted circle and an antipodal pair contains their midpoint, the sphere's centre. Algebraically, adding the equations $|z-c{|}^2=r^2$ and $|-z-c{|}^2=r^2$ with $|z|=1$ gives the displayed condition.

For north-pole stereographic projection $p=(X+iY)/(1-Z)$ on the unit sphere, the antipodal pair point has coordinate $q=-(X+iY)/(1+Z)$. Since $X^2+Y^2=1-Z^2$, this gives $p\overline{q}=-1$ when both coordinates are finite. On the whole Riemann sphere, write $q=-1/\overline{p}$, pairing zero with infinity. The two axis endpoints fixed by a sphere rotation obey this relation.

Use the north-pole stereographic projection and the cross-ratio convention $[a,b;c,d]=(a-c)(b-d)/[(a-d)(b-c)]$. Antipodes have coordinates $-1/\bar{u}$. Rotate the first point to the south pole, whose coordinate is zero: its antipode becomes infinity, while a point at spherical distance $d$ has projected modulus $\tan (d/2)$. Möbius invariance of the cross-ratio then gives the displayed identity. Antipodal endpoints require the extended-value interpretation.

On the unit sphere, use $\zeta =(x+iy)/(1-z)$ away from the north pole and $\eta =(x-iy)/(1+z)$ away from the south pole. Their overlap relation $\eta =1/\zeta$ is a holomorphic map, giving a holomorphic atlas for the Riemann sphere. The opposite sign of $iy$ in the second manifold chart ensures a holomorphic rather than antiholomorphic transition.

Writing the sphere coordinates as $(X,Y,Z)$ avoids confusing their height with the complex coordinate $z$. The two manifold charts cover the whole unit sphere:On their overlap, $X^2+Y^2=1-Z^2$ givesThe coordinate transformation and its inverse are holomorphic, so these manifold charts make the sphere a Riemann surface. The sign in $X-iY$ is essential. Geometric projection from the south pole with the same planar axis labels would give $v=(X+iY)/(1+Z)=1/\overline{z}$. The complex conjugation in that transition prevents it from being holomorphic. For $z=i$, the correct second coordinate is $w=-i$: positive $Y$ corresponds to negative imaginary part in this manifold chart.

The reciprocal chart makes infinity an ordinary coordinate origin. Its domain is the Riemann sphere minus the point $z=0$; $w(\infty )=0$ and $w(z)=1/z$ elsewhere. A neighbourhood $U_R$ of infinity becomes the open disc $|w|<1/R$. In sphere coordinates this is exactly the capFor $R=2$, the exterior $|z|>2$ together with infinity becomes $Z>3/5$ and $|w|<1/2$. The boundary itself is excluded from each open set.

The point $z=2+i$ lies in this neighbourhood because $|z|=\sqrt{5}>2$. Its reciprocal coordinate is $w=(2-i)/5$, with $|w|=1/\sqrt{5}<1/2$, and its sphere height is $Z=2/3>3/5$. The negative imaginary part of $w$ is the axis reversal required by the holomorphic stereographic atlas of the sphere.

For a complex-valued holomorphic function $f$ defined outside a large disc, having a removable singularity at infinity means that $g(w)=f(1/w)$ extends to a holomorphic function at $w=0$. This definition reduces questions at infinity to familiar questions at a finite point.^[ref] For example,The geometric series gives a Taylor series at the new origin, so $f(\infty )=1$ and $g^{\prime }(0)=-5$. The coefficient $-5$ is the derivative in the specified $w$ coordinate. In contrast, $f(z)=z^2$ becomes $g(w)=w^{-2}$, displaying a pole of order two at infinity when the target is the complex plane. As a holomorphic map to the Riemann sphere, it uses the target reciprocal coordinate $1/f(1/w)=w^2$, which is regular at zero.

Under the stereographic projection $w=(x+iy)/(1-z)$, rotations about the third axis have representatives $\mathrm{diag}(e^{i\theta /2},e^{-i\theta /2})$, and rotations about the second axis have representativesThese belong to the special unitary group $SU(2)$. Euler-angle generation of the special orthogonal group $SO(3)$ shows that every sphere rotation becomes a Möbius transformation represented in $SU(2)$. Representatives $U$ and $-U$ induce the same transformation.

Let a finite group $H$ of conformal automorphisms of the Riemann sphere fix $p$. The derivative representation at $p$ embeds $H$ into ${\mathbb{C}}^{\times }$, so $H$ is cyclic. Averaging a local coordinate linearizes the action to $z\mapsto \zeta z$, and the quotient has coordinate $u=z^{|H|}$.

For a finite conformal group action, choose a disc around each point that meets only its stabilizer translates. The local cyclic quotient chart $z\mapsto z^e$, where $e$ is the stabilizer order, gives the orbit space a Riemann-surface structure and makes the quotient map holomorphic.

For $r(z)=\zeta z$ and $s(z)=1/z$, where ${\zeta }^n=1$, the rational functionis invariant under $D_{2n}$. Equality of two values factors as $(u-v)(uv-1)=0$ for $u=z_1^n$ and $v=z_2^n$, so every fibre is exactly one dihedral orbit.