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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.

For a nonidentity Möbius transformation whose two fixed points are antipodal under stereographic projection, conjugate by a sphere rotation sending them to $0,\infty$. The conjugated map fixes both, hence is $z\mapsto az$ with $a\ne 0$. If $|a|=1$ it is an axial rotation; if $|a|<1$ every finite orbit tends to zero; if $|a|>1$ every nonzero orbit tends to infinity. Undoing the rotation gives either a sphere rotation or one attracting fixed point with the other as the sole exceptional initial point.

The Möbius transformation $M(z)=(3z-1)/(3-z)$ fixes the antipodal pair $1,-1$ under stereographic projection. But $M^{\prime }(1)=2$, so it multiplies infinitesimal spherical lengths at that fixed point by two, as follows from the round line element $2|dz|/(1+|z{|}^2)$. It is not an isometry and hence not a sphere rotation. Fixing an antipodal pair is therefore necessary for a nontrivial rotational Möbius transformation, but is not sufficient.

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.

A planispheric astrolabe uses a stereographic projection of the celestial sphere as a physical calculating device. A rotating star map is compared with a latitude-specific plate; a sighting rule measures an object's angular altitude. Aligning that measurement with the engraved geometry permits time and celestial-configuration calculations.^[ref]

For $E=\frac{1}{2}\int (|{\partial }_1\mathbf{n}{|}^2+|{\partial }_2\mathbf{n}{|}^2)$, completing the square gives $E=\frac{1}{2}\int |{\partial }_1\mathbf{n}\pm \mathbf{n}\times {\partial }_2\mathbf{n}{|}^2\pm 4\pi Q$, hence $E\ge 4\pi |Q|$. Saturation yields the Bogomolny equations of the sigma-model lump. With the oriented stereographic projection $\mathbf{n}=(2\mathrm{Re}w,2\mathrm{Im}w,1-|w{|}^2)/(1+|w{|}^2)$, holomorphic rational maps have positive degree and saturate the bound; antiholomorphic maps have negative degree.

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 knot is the image $K=f(S^1)$ of a smooth embedding $f:S^1\hookrightarrow {\mathbb{R}}^3$, considered up to ambient isotopy. The circle $S^1$ supplies a closed loop with no free ends; the smooth embedding condition rules out self-intersections and requires a nonzero tangent vector in a regular parametrization. The equivalent convention $S^1\hookrightarrow S^3$ uses the three-sphere: removing a point outside the knot identifies the remaining space with three-dimensional Euclidean space by stereographic projection.

More precisely, knots $K_0,K_1$ are equivalent if there is a smooth family of diffeomorphisms $F_s$ of the ambient space, for $0\le s\le 1$, such that $F_0$ is the identity and $F_1(K_0)=K_1$. Every intermediate $F_s(K_0)$ remains a smooth embedding. This ambient isotopy allows bending, stretching and moving a loop, but neither cutting it nor letting strands cross through each other. We compare the embedded images rather than their parametrizations; an oriented knot additionally remembers a direction of travel.

For a worked example, start with $U(t)=(\cos t,\sin t,0)$, where $t$ is taken modulo $2\pi$, and useBoth scale factors stay positive, so each $F_s$ is a diffeomorphism. At $s=1$ the circle becomes the ellipse $(2\cos t,\frac{1}{2}\sin t,0)$, yet its knot type is still the unknot. Exact lengths and angles are therefore irrelevant to knot type; self-intersection is the obstruction that the deformation must avoid.

In two spatial dimensions, a unit-vector field $\mathbf{n}:{\mathbb{R}}^2\to S^2$ has static energy $E=\int |\nabla \mathbf{n}{|}^2{d}^{2}x/2$. The sphere is also the complex projective line, so its stereographic projection gives the equivalent complex field description. A constant value at spatial infinity compactifies a regular configuration to a map $S^2\to S^2$, whose degree of a map between oriented manifolds is its topological charge.

Astronomical instruments do more than make an existing observation more accurate. They select measurable quantities, embody mathematical assumptions and create new classes of evidence. The astrolabe and the telescope illustrate two different relationships between a mathematical model and observation.

A planispheric astrolabe represents the celestial sphere on a flat plate by stereographic projection. A rotating star map is fitted over a plate appropriate to the observer's latitude; the back carries a sighting rule, or alidade, and an angular scale. A measured stellar altitude can be matched to the engraved geometry to infer the celestial configuration and hence the time. The instrument combines an observation with a physical analogue calculation. Its geometry is indispensable to the inference: the sighting alone does not tell the time.^[ref]

This also explains why manufacture, calibration and users' knowledge mattered. An unsuitable latitude plate or an inaccurate sighting produces a systematic error rather than a philosophical refutation of celestial geometry. Conversely, engraved scales and repeatable settings allow predictions and observations to be compared in an organized way. The astrolabe belongs to a history of practical calculation as well as cosmological ideas. Its use to represent apparent sky rotation does not establish that its user had independently proved every physical claim of a geocentric cosmology.

The early telescope changed what could be included in an astronomical description. In Sidereus Nuncius by Galileo Galilei, the Moon's changing light and shadow supports an interpretation of surface relief, the Milky Way resolves into many stars, and repeated observations distinguish small bodies accompanying Jupiter from ordinary fixed stars. Temporal sequences were crucial: a few points in one image are ambiguous, whereas their changing configurations around the same bright body support an orbital interpretation.^[ref]

There were physical reasons not to accept every magnified appearance immediately. Lens defects, limited resolution, stray light and imperfect alignment can create or distort features. Consistency under repeated observations, comparison with familiar terrestrial objects and a successful account of how the instrument forms images help distinguish an optical artefact from a celestial object. An instrument extends sight only through a chain of physical inferences; it does not remove that chain.

The discoveries also had discriminating, rather than unlimited, theoretical force. A visibly irregular Moon challenged the doctrine of perfectly smooth celestial bodies. Moons accompanying Jupiter showed that not every visible revolution had the Earth as its immediate centre. Neither observation alone proved that the Earth moves around the Sun: a different system could allow Jupiter's moons while retaining a stationary Earth. Later observations such as the phases of Venus must also be distinguished chronologically from the observations published in the March 1610 book.

The two instruments thus shaped astronomy in complementary ways. The astrolabe made a mathematical sky model operational, whereas the telescope exposed phenomena that demanded changes to parts of the accepted model. In both cases, reliable knowledge depended on geometry, material workmanship, repeated observations and explicit limits on what the observations entailed.

Project from the north pole $N=(0,0,1)$ onto the equatorial plane, identifying that plane with the complex plane. The line from $N$ to $P=(X,Y,Z)$ meets the plane inThis is stereographic projection. Its inverse isRotation through $\theta$ about the $z$-axis changes $X+iY$ to $e^{i\theta }(X+iY)$ and preserves $Z$, so it becomes the Möbius transformation $u\mapsto e^{i\theta }u$.

Let $B$ be the rotation matrix supplied in the question. It sends the $z$-axis to the $x$-axis, and therefore $R_x(\theta )=B{R}_z(\theta )B^{-1}$. If $b=\varphi B{\varphi }^{-1}$ is its given Möbius transformation, thenwhich is again a Möbius transformation. Such $z$- and $x$-axis rotations can send any chosen point of the sphere to the south pole: first rotate its horizontal projection onto the $y$-axis, then rotate in the $yz$-plane. This supplies the rotations needed below.

The antipodal stereographic coordinate relation is $\varphi (-P)=-1/\overline{\varphi (P)}$, as follows immediately from the inverse formula. Fix the cross-ratio conventionWe claim the required ordering isFor the antipodal cross-ratio and spherical distance identity, rotate $P$ to the south pole. Its coordinate becomes zero and its antipode becomes infinity. A point at angular spherical distance $d$ from the south pole has vertical coordinate $-\cos d$ and horizontal magnitude $\sin d$, so its new projected coordinate $v^{\prime }$ has modulus $\sin d/(1+\cos d)=\tan (d/2)$. By Möbius invariance of the cross-ratio, the displayed expression equalsFor finite nonsingular coordinates the same ratio is $-|u-v{|}^2/|1+u\bar{v}{|}^2$. Zero or infinite antipodal coordinates are interpreted by limits. When $Q=-P$, $d=\pi$ and the ratio has the extended value infinity; the finite real formula applies for $0<d<\pi$.

With determinant-one normalization, the necessary and sufficient condition for sphere rotations as special-unitary Möbius transformations isEquivalently, the matrix $M=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ belongs to the special unitary group $SU(2)$. The two determinant-one representatives $M$ and $-M$ induce the same Möbius transformation, and both satisfy this condition.

Use north-pole stereographic projection in the direction from the plane to the sphere:Its chord-distance identity is $|\phi (z)-\phi (w)|=2\chi (z,w)$, where $\chi$ is the chordal metric. Direct subtraction, using $ad-bc=1$, givesThis is first calculated away from poles and then extended continuously to the Riemann sphere. A sphere rotation preserves chord distances, so for distinct finite nonpole pointsUsing three distinct points shows all these positive ratios equal one. Hence $H(z)=1+|z{|}^2$ everywhere. Comparing its coefficients givesThus $M$ is a unitary matrix. Comparing $M^{-1}=M^{\ast }$ with $M^{-1}=\left(\begin{array}{cc}d& -b\\ -c& a\end{array}\right)$ proves the boxed relations.

Conversely, these relations give $H(z)=1+|z{|}^2$, so the induced sphere map $T=\phi g{\phi }^{-1}$ preserves all Euclidean chord distances. Such a bijection of the unit sphere is the restriction of an orthogonal matrix: let $v_j=T(e_j)$; preservation of distances makes these an orthonormal basis, and $T(p)\cdot v_j=p\cdot e_j$ forces $T(p)={\sum }_j{p}_j{v}_j$. Finally the Möbius transformation is holomorphic with positive real Jacobian at every ordinary coordinate point. Conjugating by stereographic projection preserves orientation, so this orthogonal matrix has determinant $+1$. It is therefore a genuine rotation in three dimensions, proving sufficiency rather than just spherical isometry.

For the circle assertion, a nondegenerate generalized circle has an equationFor $A\ne 0$ this is an ordinary circle, and for $A=0$ it is a line with infinity included. On the sphere write coordinates $(p_1,p_2,p_3)$, with $z=(p_1+i{p}_2)/(1-p_3)$ and $|z{|}^2=(1+p_3)/(1-p_3)$. Multiplication by $1-p_3$ gives the planeIf $n=(2\mathrm{Re}B,-2\mathrm{Im}B,A-C)$ and $h=-(A+C)$, then $|n{|}^2-h^2=4(|B{|}^2-AC)>0$. Thus the plane intersects the unit sphere in a proper circle. It contains the north pole exactly when $A=0$, accounting for the point at infinity on a line.

Conversely, a circle on the sphere is the intersection with a plane $n\cdot p=h$, $|h|<|n|$. Set $A=(n_3-h)/2$, $C=(-n_3-h)/2$ and $B=(n_1-i{n}_2)/2$. The preceding equation is recovered and has $|B{|}^2-AC=(|n{|}^2-h^2)/4>0$. Its inverse stereographic image is an ordinary circle or a line with infinity attached. This proves both directions of the circle-plane relation under stereographic projection.

Project from the north pole $N=(0,0,1)$ along the line through a sphere point onto the equatorial plane $Z=0$, identified with $\mathbb{C}$. Parameterizing this line givesThe inverse stereographic projection isNegating the three sphere coordinates consequently sends $z$ to $-1/\bar{z}$, with $0,\infty$ interchanged. This proves the antipodal stereographic coordinate relation.

A nonidentity Möbius transformation $T(z)=(az+b)/(cz+d)$ has $ad-bc\ne 0$. For $c\ne 0$, its fixed points are the roots of $c{z}^2+(d-a)z-b=0$, giving one or two distinct points. If $c=0$ and $a\ne d$, there is one finite root and the fixed point infinity. If $c=0$, $a=d$, and $b\ne 0$, it is a translation and only infinity is fixed. The remaining case is the identity, already excluded. Thus the claim includes fixed points at infinity, not just finite solutions of the quadratic.

A nontrivial rotation of the sphere fixes exactly the two poles on its axis, which are antipodal. It therefore gives exactly two fixed stereographic points with $z_2=-1/{\bar{z}}_1$. Here the nonzero rotation angle is understood modulo $2\pi$; an angle that is a whole multiple of $2\pi$ would be the identity.

Conversely, use a sphere rotation whose Möbius map $S$ sends the two antipodal fixed points to $0,\infty$. The conjugate $ST{S}^{-1}$ fixes both, hence equals $z\mapsto az$ for some $a\ne 0$. If $|a|=1$, write $a=e^{i\theta }$; the inverse stereographic formula shows that this rotates $X+iY$ while leaving $Z$ fixed. Conjugating back by the rotation $S$ gives a sphere rotation. If $|a|<1$, $a^{n}z\to 0$ for every finite $z$. If $|a|>1$, $a^{n}z\to \infty$ for every $z\ne 0$, in the sphere topology. Pulling these conclusions back proves the antipodal fixed-point classification of Möbius transformations: either the map is a sphere rotation or one fixed point attracts every point except the other fixed point.

Use the unit sphere $X^2+Y^2+Z^2=1$, its north pole $(0,0,1)$, and the equatorial complex plane $Z=0$. The line from the north pole to a sphere point meets that plane atThe north pole maps to $\infty$. The inverse stereographic projection isThe antipodal point has coordinate $w=-(X+iY)/(1+Z)$. Since $X^2+Y^2=(1-Z)(1+Z)$, this is $-1/\bar{z}$ when both sides are finite and nonzero. Zero and infinity correspond to the two poles and are exchanged by the same extended convention. The inverse formula also proves the converse, giving the antipodal stereographic coordinate relation

The requested rotation result is that sphere rotations correspond exactly to Möbius transformationsThe matrices $\left(\begin{array}{cc}a& b\\ -\bar{b}& \bar{a}\end{array}\right)$ form $SU(2)$, and opposite matrices induce the same transformation. Thus the rotation group $SO(3)$ is identified with $SU(2)/\{\pm I\}$. This is the sphere rotations as special-unitary Möbius transformations result, which need not be proved here.

For a planar circle $|z-c|=r$, $c=a+ib$ and $r>0$, substitute $|z{|}^2=(1+Z)/(1-Z)$ and $\Re (\bar{c}z)=(aX+bY)/(1-Z)$. Multiplication by $1-Z$ gives the circle-plane relation under stereographic projection:The lift lies in this plane, and conversely its intersection with the sphere projects onto the circle. It is a nondegenerate circle on the sphere. The north pole is not on the displayed plane, since substitution there gives two rather than zero.

A great circle is the intersection with a plane through the sphere's centre. Here this holds exactly whenThe unit circle lifts to the equator. If a different circle lifts to a great circle, its plane and the equatorial plane are distinct planes through the origin, so their intersection cuts the sphere at two antipodal equatorial points. Their planar coordinates are $z$ and $-z$ on $|z|=1$.

Conversely, if the planar circle meets the unit circle in $z$ and $-z$, their lifts are antipodal equatorial points. The plane of the lifted circle contains their midpoint, the origin, so that circle is great. Equivalently, adding $|z-c{|}^2=r^2$ and $|-z-c{|}^2=r^2$ gives $r^2=1+|c{|}^2$. The exclusion of the unit circle ensures the two intersection points are a genuine pair rather than coincident circles. This proves the antipodal equatorial intersection criterion for a great circle.

Use the normalized chordal metricUnder stereographic projection it is half the Euclidean chord distance on the unit sphere, which proves the metric axioms and the agreement with the usual sphere topology. If $\rho$ is spherical geodesic distance, then $\rho =2\arcsin \chi$, so $2\chi \le \rho \le \pi \chi$.

A rational map is holomorphic, hence smooth, as a map of compact spheres. At a pole use $1/R$ as target coordinate, and at infinity use $1/z$ as source coordinate; neither is a singularity of the sphere map. Its differential in the round metric has a finite maximum norm $L$. Integrating its norm along a minimizing spherical geodesic gives $\rho (Rz,Rw)\le L\rho (z,w)$. The preceding comparisons implyThis proves that the rational map is Lipschitz in the chordal metric. For a constant map take constant zero; for a nonconstant map the bound is finite by compactness. In finite coordinates the differential norm is $R^{\#}(z)=|R^{\prime }(z)|(1+|z{|}^2)/(1+|R(z){|}^2)$, with its continuous sphere extension.

Now suppose $R,S$ are commuting maps. We first prove $S(F(R))\subseteq F(R)$. The family $\{R^n\}$ is locally equicontinuous on $F(R)$, by the normality equivalence in question 1. Commutation and the Lipschitz estimate give, for all $n$,For $x_0\in F(R)$ choose a source ball within that open set, small enough to make the right side less than any prescribed $ϵ$ uniformly in $n$. Holomorphic openness makes its image contain a ball about $S(x_0)$. Every point $w$ of that target ball has some preimage $x$ in the chosen source ball, so the same inequality proves equicontinuity of the iterates of $R$ at $S(x_0)$. Repeating at each point of the image gives local normality on that image. Thus the stated inclusion holds even where $S$ has a rational critical point.

Carefully state the permitted Montel theorem: a family of meromorphic maps on an open domain, all omitting the same three distinct values of the Riemann sphere, is normal for locally uniform spherical convergence. Since $\mathrm{deg}R\ge 2$, question 1 shows that $J(R)$ contains at least three distinct points. The forward inclusion just proved means every iterate of $S$ maps $F(R)$ into itself and therefore omits those three points there. Montel gives $F(R)\subseteq F(S)$. Exchanging the roles, using $\mathrm{deg}S\ge 2$, gives the opposite inclusion. ConsequentlyThis proves that commuting rational maps of degree at least two have the same Julia set without assuming either is an iterate of the other.

For commuting maps of degree one with different Julia sets, takeTheir two compositions are the identity, and the Julia sets follow from the expanding and contracting Möbius cases in question 1.

For equal Julia sets without commutation, takeBoth affine maps contract toward their finite fixed points and have a repelling periodic point at infinity. But $R\circ S=z/4+1/2$, while $S\circ R=z/4+1$, so they do not commute.

Finally the distinct Chebyshev polynomialshave degrees two and three and neither is a pure power of $z$. Direct expansion gives both compositions equal to $32z^6-48z^4+18z^2-1$. Equivalently $T_m(T_n(z))=T_{mn}(z)$, first seen at $z=\cos \theta$ and then as a polynomial identity. This provides the requested higher-degree pair.

A nonlinear sigma model describes a field taking values in a Riemannian target rather than an unconstrained linear space. Its static configurations are harmonic maps, critical points of the Dirichlet energy. The standard example for sigma-model lumps is the O3 nonlinear sigma model in two spatial dimensions: a unit vector $\mathbf{n}(x_1,x_2)\in S^2$ hasThe same target is the complex projective line, giving the CP1 nonlinear sigma model description. A lump is a localized finite-energy static solution, whose existence and minimum energy are governed by the degree of its field map.

Impose a fixed angle-independent value ${\mathbf{n}}_{\infty }$ at spatial infinity and require the field to extend smoothly there. The one-point compactification of the plane is a sphere, so the configuration becomes a map $S^2\to S^2$. Its topological degree is the integerThis is the degree charge of an O3 sigma-model lump: the integrand is the pullback of the oriented unit-sphere area form, normalized by its area $4\pi$. Equivalently, the degree counts inverse images of a regular target point with orientation signs. It is unchanged under smooth homotopy with the prescribed boundary value. Consequently fields of different degree belong to distinct topological sectors.

Topology also determines an energy lower bound. The derivatives are tangent to the sphere, so $|\mathbf{n}\times {\partial }_2\mathbf{n}|=|{\partial }_2\mathbf{n}|$. Since
${\partial }_1\mathbf{n}\cdot (\mathbf{n}\times {\partial }_2\mathbf{n})=-\mathbf{n}\cdot ({\partial }_1\mathbf{n}\times {\partial }_2\mathbf{n})$, square completion givesChoosing the sign of $N$ proves the Bogomolny degree bound for the O3 sigma modelFor positive degree, equality holds precisely when ${\partial }_1\mathbf{n}+\mathbf{n}\times {\partial }_2\mathbf{n}=0$; for negative degree the opposite sign applies. These are the Bogomolny equations for the lump. A saturating field minimizes energy in its degree sector and hence solves the static field equation. Explicitly, that equation is $\Delta \mathbf{n}+|\nabla \mathbf{n}{|}^2\mathbf{n}=0$, obtained by varying under ${\mathbf{n}}^2=1$.

The first-order equations become especially simple in stereographic projection. Use $z=x_1+i{x}_2$ andThe round target metric is $4dwd\bar{w}/(1+|w{|}^2{)}^2$. In the one-half energy normalization used here, the stereographic energy of the O3 sigma model and degree areThus $E-4\pi N=8\int |w_{\bar{z}}{|}^2/(1+|w{|}^2{)}^2$. Positive-degree lumps are holomorphic maps, $w_{\bar{z}}=0$, and negative-degree lumps are antiholomorphic maps, $w_z=0$.

A holomorphic sphere map is meromorphic in a stereographic coordinate and hence a rational map $w=p(z)/q(z)$ with coprime polynomials. To see why the rational form suffices, its finitely many coordinate poles can be removed by subtracting their principal parts, leaving a holomorphic function on the compact sphere and therefore a constant. The degree of a rational map of the Riemann sphere isafter canceling common factors. A generic target value $c$ has inverse images solving $p-cq=0$, with the root at infinity counted in homogeneous coordinates when necessary; the number is $N$, and holomorphic local degrees are positive. Poles of $w$ do not make the physical field singular: the reciprocal target coordinate $1/w$ is regular there. Antiholomorphic rational maps have the opposite topological degree and the same minimum energy.

For example, the unit sigma-model lump and size modulus isIts degree is one and radial integration gives $E=4\pi$. The complex center $z_0$ supplies two position parameters, while the magnitude and phase of $\rho$ give size and internal orientation. More generally, fixing $w(\infty )=0$ allows a monic degree-$N$ denominator and a numerator of degree at most $N-1$, with no common roots. Their $2N$ complex coefficients give the based rational-map moduli of sigma-model lumps.

Two-dimensional Derrick scaling leaves this pure-gradient energy unchanged, so degree fixes the minimum energy but not a preferred size. As $\rho \to 0$, the unit-lump energy concentrates and the field limit is singular at its center. A degree cannot change through a smooth boundary-preserving deformation, but concentration can leave that smooth configuration space. Thus topological degree protects the smooth sector and determines the Bogomolny energy, while not by itself stabilizing the lump size. There is no contradiction with the scalar obstruction in Question 1: a curved-target harmonic map obeys a nonlinear constrained field equation, rather than having harmonic unconstrained components.

The kink panel uses the normalization of Question 1. The lump panel has $z_0=0$ and $|\rho |=1$ in the unit sigma-model lump and size modulus; changing its size redistributes the density while keeping its total energy $4\pi$.

Use stereographic projection charts $\phi (X,Y,Z)=(X+iY)/(1-Z)$ away from the north pole and $\psi (X,Y,Z)=(X-iY)/(1+Z)$ away from the south pole. On their overlap $\psi =1/\phi$, a holomorphic transition function, defining the Riemann sphere.

Where $F$ avoids the north pole, $\phi \circ F$ is an ordinary holomorphic function exactly when $F$ is holomorphic. Near a north-pole value, the other chart gives $\psi \circ F=1/(\phi \circ F)$; a zero of this holomorphic coordinate becomes a pole of $\phi \circ F$. This proves the correspondence with meromorphic functions. On a disconnected domain one must also allow components mapped constantly to infinity, or state the correspondence on each nonconstant component. Applying the same reciprocal charts at source and target infinity shows that every nonconstant rational function defines a holomorphic sphere map.

The multiplicity of the value $a$ at infinity is the zero order at $w=0$ of $f(1/w)-a$ when $a$ is finite, and of $1/f(1/w)$ when $a=\infty$. For coprime polynomials $P,Q$, the degree of a rational map of the Riemann sphere is $n=\max (\mathrm{deg}P,\mathrm{deg}Q)$, also its total number of poles with multiplicities.

The quotient rule $f^{\prime }=(P^{\prime }Q-P{Q}^{\prime })/Q^2$ gives the upper bound $\mathrm{deg}{f}^{\prime }\le 2n$. Each finite pole of order $m$ contributes order $m+1$ to $f^{\prime }$. If $\mathrm{deg}P>\mathrm{deg}Q$, infinity contributes an additional order $\mathrm{deg}P-\mathrm{deg}Q-1$ when this is positive. Counting gives the degree bounds for the derivative of a rational functionThe lower bound is attained by $f=z^n$; the upper by $f=1/(z^n-1)$, whose $n$ simple finite poles become double poles. Product degrees are not additive: $f=z$ and $g=(z+1)/z$ both have degree one, but $fg=z+1$ also has degree one because the zero and pole cancel.

For stereographic projection, choose the north pole $N=(0,0,1)$ of the unit sphere and identify the plane $Z=0$ with the complex plane by $z=X+iY$. Project a point $P=(X,Y,Z)\ne N$ along the line joining it to $N$; adjoin $z=\infty$ for the north pole to identify the whole sphere with the Riemann sphere. A point of that line is $N+t(P-N)$. Its last coordinate vanishes when $t=1/(1-Z)$, givingUsing $X^2+Y^2+Z^2=1$, or solving this line's second intersection with the sphere, gives the inverseA great circle lies in a plane $h_{1}X+h_{2}Y+h_{3}Z=0$ through the origin. Its projected equation isIt is a straight line only if $h_3=0$, meaning the plane contains the north pole and the south pole. Thus if all three sides of a nondegenerate spherical triangle projected to straight sides of a Euclidean triangle, all three distinct side planes would contain both poles. Each pair of side planes could then meet on the sphere only at those same two poles, which cannot supply three distinct triangle vertices. A nondegenerate spherical triangle cannot project to a Euclidean triangle. If a side passes through the projection pole, its image is unbounded and also cannot be a finite triangle side.

A nonidentity rotation in three dimensions has an axis through the origin: an orthogonal three-by-three matrix of determinant one has an eigenvalue one, since nonreal eigenvalues pair as conjugates, real eigenvalues are $\pm 1$, and their product is one. Its axis meets the unit sphere in an antipodal pair, which is fixed. In the identity case every point is fixed, so an antipodal pair can again be selected. Projecting $P$ and $-P$ giveswhenever both points are finite. Thus the associated Möbius transformation fixes an antipodal pair satisfying the claimed relation. The universally valid relation is $q=-1/\overline{p}$ on the Riemann sphere, with $0$ and $\infty$ paired. A rotation about the projection axis fixes exactly that pair, so the PDF's literal finite product needs this extended-coordinate convention. This is the antipodal stereographic coordinate relation.

Write $w=u+iv$ and $R=1+u^2+v^2$. Inverting the specified stereographic projection givesThe coordinate takes the value infinity at the south pole, so a general map is described using the extended complex plane and a second chart there. A pole of $w$ is not necessarily a singularity of $\varphi$.

Differentiate the inverse formulas:On squaring and summing, the terms containing $(dR{)}^2$ cancel, leaving $|d\varphi {|}^2=4(d{u}^2+d{v}^2)/R^2$. Thus the stereographic energy of the O3 sigma model isThe same derivatives give $\varphi \cdot ({\partial }_u\varphi \times {\partial }_v\varphi )=4/R^2$, with positive sign: at $u=v=0$ the ordered tangent vectors are $(2,0,0)$ and $(0,2,0)$ at the north pole. The chain rule therefore givesThis is the topological degree for the orientation in the question.

Put $z=x_1+i{x}_2$, $w_z=(w_{x_1}-i{w}_{x_2})/2$ and $w_{\bar{z}}=(w_{x_1}+i{w}_{x_2})/2$. Expanding the squares givesFor $N\ge 0$, subtracting the second expression times $8\pi$ givesfor $N\le 0$, adding $8\pi N$ gives the analogous square with $w_z$. ConsequentlyThe factor is $8\pi$, not $4\pi$, because the printed energy has no prefactor $1/2$. Equality requires a holomorphic map for positive degree, an antiholomorphic map for negative degree, or a constant map for zero energy.

Explicit examples for every integer degree areTo verify both degree and finite energy for $w=z^m$, $m>0$, use polar coordinates: $|w_z{|}^2=m^2{r}^{2m-2}$ and $w_{\bar{z}}=0$. Thenwhere $s=r^{2m}$. Its energy is $8\pi m$. Complex conjugation changes the degree's sign while preserving the energy, and the constant example has degree and energy zero. These sigma-model lumps are smooth sphere-valued maps; the coordinate infinity at spatial infinity is regular in the other target chart.

The starting point is the Grassmannian of complex two-planes $W\subset \mathbb{T}\cong {\mathbb{C}}^4$. Choose a basis $u,v$ and form the simple bivector $P=u\wedge v\in {\Lambda }^2\mathbb{T}$. A change of basis multiplies $P$ by a nonzero determinant, so its projective class is independent of that choice. The Plücker embedding is thereforeIt is injective because $W=\{z:z\wedge P=0\}$. In a basis of $\mathbb{T}$, its six homogeneous coordinates $p_{ij}=u_i{v}_j-u_j{v}_i$ satisfyThis is the condition $P\wedge P=0$. Conversely, a nonzero antisymmetric $4\times 4$ matrix satisfying it has Pfaffian zero, hence rank at most two; its nonzero rank is even, so it has rank two and is a simple bivector. Therefore the equation characterizes the image, the Klein quadric $Q\subset {\mathbb{CP}}^5$. Its gradient cannot vanish at a projective point, since its six partial derivatives are the six coordinates up to permutation and sign. Thus $Q$ is smooth and has complex dimension four, agreeing with $\mathrm{Gr}(2,4)$.

The Klein quadric is the complexified compactified Minkowski space of the twistor correspondence. A volume form on $\mathbb{T}$ turns $P\wedge P^{\prime }$ into a scalar symmetric bilinear form $B(P,P^{\prime })$. It is zero exactly when the two underlying two-planes have nontrivial intersection: the four spanning vectors are then linearly dependent. In projective twistor space this says that the two twistor lines meet. In the affine graph chart supplied by the twistor incidence relation, it is $\det (x-y)=0$, so it is precisely null separation.

This also determines a conformal geometry on $Q$. At $[P]$, the tangent space is $P^{\perp }/⟨P⟩$ for $B$. The restriction of $B$ descends to a nondegenerate quadratic form on this four-dimensional quotient. Rescaling the representative $P$ changes the identification by a scalar, so its invariant content is a conformal class. In a graph chart its null tangent vectors are the rank-one matrices in $\mathrm{Hom}(W,\mathbb{T}/W)$, equivalently $d{x}^{A{A}^{\prime }}=a^A{b}^{A^{\prime }}$.

The compact complex geometry does not yet specify a real spacetime. The Lorentzian slice comes from the twistor Hermitian form $h$ of signature $(2,2)$: real points are maximal totally isotropic two-planes. In the incidence chart, isotropy of $W_x=\{(ix\pi ,\pi )\}$ is equivalent to $x=x^{†}$, the usual real Minkowski condition. Its twistor lines lie wholly in projective null twistor space. Diagonalizing $h$ as positive and negative two-dimensional blocks also gives a useful global description: an isotropic two-plane projects isomorphically to the positive block, so it is a graph $(z,Uz)$, and isotropy is $U^{†}U=1$. The real compact space is therefore $U(2)$, equivalently $(S^1\times S^3)/{\mathbb{Z}}_2$.

The same result is especially transparent in coordinates of the Plücker embedding after a complex linear change of coordinates on ${\Lambda }^2\mathbb{T}$. The real Lorentzian cone has equationIts nonzero real generators represent points. Normalize the lengths of its positive and negative parts to one; real projectivization identifies their simultaneous antipodes. ConsequentlyThis is the Projective Minkowski compactification. Its universal cover is $\mathbb{R}\times S^3$, with deck generator $(s,v)\mapsto (s+\pi ,-v)$, so its fundamental group is $\mathbb{Z}$. The projective version includes periodic identifications; an unwrapped Einstein-cylinder or Penrose diagram description uses the appropriate cover and affine domain instead.

For example, use homogeneous cone coordinates $[U,V,t,\mathbf{x}]$ with quadratic equation $2UV+t^2-|\mathbf{x}{|}^2=0$. The affine patch $U\ne 0$ sets $U=1$ and then $V=-(t^2-|\mathbf{x}{|}^2)/2$. The complement $U=0$ is the null cone $t^2-|\mathbf{x}{|}^2=0$ at infinity. It is much larger than a single point: different null directions retain different limits there.

For the Euclidean slice, choose an antilinear quaternionic structure on a complex vector space $J$ on twistor space, for exampleThe Euclidean reality structure on twistor space selects $J$-invariant complex two-planes. Such a plane is a quaternionic line in ${\mathbb{H}}^2$, givingThere are no fixed projective twistors for $J$: if $JZ=cZ$, applying $J$ again would give $-Z=|c{|}^{2}Z$. Every projective twistor instead lies on the unique invariant line $\mathbb{P}\mathrm{span}(Z,JZ)$. Thus the Euclidean correspondence becomes a fibration ${\mathbb{CP}}^3\to S^4$ with fiber ${\mathbb{CP}}^1$, in contrast to the Lorentzian correspondence where only null twistors represent real null geodesics. Distinct Euclidean real points have disjoint twistor lines, since the line through any $Z$ and $JZ$ is unique. This reflects the absence of nonzero real null vectors for a positive-definite metric.

On the Klein quadric the Euclidean real cone has signature $(1,5)$; normalize its timelike coordinate to one and obtain the unit sphere $S^4$. Its affine Euclidean chart omits one point. Inverse stereographic projection explicitly gives the round conformal metricwhich extends smoothly across that point. This proves the one-point conformal compactification of Euclidean space. The distinction between the Lorentzian Hermitian and Euclidean quaternionic reality structures is also described in Adamo's lectures on twistor theory; the global statements here follow from the explicit models just derived.

Finally, Wick rotation $t\mapsto it$ compares affine real slices through their complexification. Locally it changes $d{t}^2-|d\mathbf{x}{|}^2$ to $-d{\tau }^2-|d\mathbf{x}{|}^2$, the negative of the Euclidean metric; the overall sign does not change its null cone in the complexified setting. It does not identify the two real conformal compactifications globally. The Lorentzian projective space has fundamental group $\mathbb{Z}$, while $S^4$ is simply connected. Its causal universal cover $\mathbb{R}\times S^3$ is noncompact, so passing to that cover does not repair the mismatch with the compact Euclidean sphere.