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There is a small domain issue in the printed notation. If ${\mathbb{R}}^3/{\mathbb{R}}^2$ means a quotient vector space by the horizontal plane, the displayed formula does not descend to that quotient vector space: $(0,0,0)$ and $(1,0,0)$ represent the same class but give different values. The intended construction is stereographic projection, restricted to the unit sphere with the north pole removed. Interpreting the slash as removal of the plane $z=1$ also supplies a suitable ambient domain. The holomorphic stereographic atlas of the sphere is obtained as follows.

Write $N=(0,0,1)$ and $S=(0,0,-1)$. On $U_N=S^2\setminus \{N\}$ use $\zeta =(x+iy)/(1-z)$. Its inverse, with $\zeta =u+iv$, isThese formulas give a smooth manifold chart from $U_N$ onto $\mathbb{C}$. On $U_S=S^2\setminus \{S\}$ choose the second manifold chartIts inverse is $x=2\mathrm{Re}\eta /(1+|\eta {|}^2)$, $y=-2\mathrm{Im}\eta /(1+|\eta {|}^2)$ and $z=(1-|\eta {|}^2)/(1+|\eta {|}^2)$, so this is also a smooth manifold chart onto $\mathbb{C}$. The conjugation in this second stereographic projection is essential. On the overlap, $x^2+y^2=1-z^2$, soBoth directions of this transition are holomorphic maps on ${\mathbb{C}}^{\times }$, with nonzero derivative. The two manifold charts cover the sphere, hence define a holomorphic atlas, giving precisely the Riemann sphere. If one instead used $x+iy$ in both manifold charts, the transition would be $1/\bar{\zeta }$ and would not be holomorphic.

Orient the sphere by this holomorphic atlas. The given volume form is smooth at infinity: replacing $\zeta$ by $1/\eta$ in its exterior product gives the same expressionMoreover, $id\zeta \wedge d\bar{\zeta }=2du\wedge dv$, so the normalization of this volume form isThus the printed volume form has half the area of the standard round unit sphere; replacing its integral by $4\pi$ would introduce an erroneous factor of two.

For $k\ge 1$, the holomorphic map $f(\zeta )={\zeta }^k$ extends over infinity, since the target reciprocal coordinate is ${\eta }^k$ when the source reciprocal coordinate is $\eta$. Its pullback of a differential form isUsing $s=r^{2k}$ givesThe degree of a map between oriented manifolds is thereforeFor $k=0$, the formula on the finite manifold chart is the constant $1$. Its unique continuous extension is also $1$ at infinity, rather than an undefined expression ${\infty }^0$. Its pullback of a differential form is zero and its degree of a map between oriented manifolds is zero.

For the preimage calculation when $k\ge 1$, choose a regular value $w\in {\mathbb{C}}^{\times }$. There are exactly $k$ distinct roots of ${\zeta }^k=w$. At each root the real Jacobian determinant is $|k{\zeta }^{k-1}{|}^2>0$, so every local contribution to the degree of a map between oriented manifolds is $+1$. Their sum is $k$, agreeing with the integral. The exceptional values $0$ and infinity are avoided because they are branch values when $k>1$. For the constant map, any $w\ne 1$ is a regular value with no preimages, giving the same answer zero. This establishes the degree of a power map of the Riemann sphere for every allowed $k$.

Take an oriented closed string in flat, critical bosonic string theory, with ${\alpha }^{\prime }=1/(2\pi T)$ and a mostly-plus target Minkowski metric. After continuation of the worldsheet to Euclidean signature, the Polyakov path integral sums over embeddings $X$ and worldsheet metrics, divided by worldsheet diffeomorphisms and Weyl transformations. In a flat target its kinetic action isTarget-time continuation or analytic continuation of external momenta defines the Lorentzian scattering amplitude; a naive real Euclidean Gaussian for timelike $X^0$ would not be convergent.

An external tachyon is represented by the tachyon vertex operator $V_p(z,\bar{z})=:e^{ip\cdot X}:$, with conformal weights $h=\bar{h}={\alpha }^{\prime }{p}^2/4$. The physical integrated vertex has $(h,\bar{h})=(1,1)$, so $p^2=4/{\alpha }^{\prime }$. Schematically its tree amplitude isThe Faddeev-Popov determinant from conformal gauge is represented by the worldsheet ghost fields. At tree level the worldsheet is a Riemann sphere, whose unpunctured complex structure has no moduli. Its residual conformal automorphisms are the Möbius transformations, $PSL(2,\mathbb{C})$. Fix three insertion points, accompanying their unintegrated vertices by the required $c\tilde{c}$ ghost factors. The remaining $N-3$ complex insertion positions are integrated over the sphere; equivalently one integrates all positions and divides by the residual conformal group.

The embedding fields are free, withTheir zero-mode integral gives momentum conservation, and their nonzero-mode Gaussian integral gives the Koba-Nielsen factorThis explains the sphere tachyon position integral without needing its evaluation. For four tachyons the resulting Virasoro–Shapiro amplitude displays the exchanged string spectrum directly.

Use all-incoming external momenta and introduce $u=-(p_1+p_4{)}^2/(8\pi T)$ alongside the two printed dimensionless Mandelstam variables. Since $p_a^2=8\pi T$ and ${\sum }_a{p}_a=0$, one obtains $s+t+u=-4$. The physical center-of-mass energy squared in the $s$ channel is $s_{\mathrm{phys}}=8\pi Ts=4s/{\alpha }^{\prime }$. The $t$ channel measures the analogous crossed momentum transfer, with sign determined by the mostly-plus convention. Rewriting the Gamma factors in a symmetric form givesThe Gamma function poles imply, at generic fixed values of the other invariant,The denominator Gamma factors can remove residues at special intersecting channel kinematics; the statement concerns a generic single-channel limit. An $s$-channel pole occurs when the intermediate momentum $p_1+p_2$ satisfies the mass-shell condition for a closed-string state:The pole at $s=-1$ exchanges the ground-state tachyon, the pole at $s=0$ exchanges massless states, and the positive integer poles exchange the infinite massive tower. The $t$-channel interpretation is the crossed version. Factorization means that each residue is a sum of products of couplings to intermediate physical states that couple to the chosen external particles. It need not expose every representation at that mass.

For example, the Gamma function recurrence and Gamma function residue at a nonpositive integer give the dimensionless Virasoro–Shapiro amplitude pole residueThe residue polynomial has degree $2n$, consistent with maximum spin $2n$ in the exchanged level. The amplitude also has the corresponding $u$-channel poles by crossing symmetry.

The massless fields can be treated as target backgrounds rather than separate asymptotic insertions. Write the target metric as $G_{mn}={\eta }_{mn}+h_{mn}$, introduce a Kalb–Ramond field $B_{mn}$, and a dilaton $\Phi$. In conventional Euclidean signs their string nonlinear sigma model action isHere ${ϵ}^{ab}$ is the antisymmetric tensor density. The three backgrounds correspond to the graviton, antisymmetric tensor and scalar states at closed-string level one. Expanding the vacuum functional in $h,B,\Phi$, then Fourier expanding the backgrounds, produces exactly their integrated string vertex operators. Its functional derivatives therefore generate the amplitudes with massless external strings. The connected vacuum functional organizes connected amplitudes; the spacetime effective action organizes the corresponding vertices after treating massless propagation consistently.

At momenta small compared with $1/\sqrt{{\alpha }^{\prime }}$, massive string propagators can be expanded in powers of momenta over their masses. The analytic part of the amplitudes consequently determines local higher-derivative interactions, ordered by powers of ${\alpha }^{\prime }$. Massless exchange poles are retained through propagation of the massless fields, rather than expanded into local contact terms. Up to field redefinitions, the leading massless-sector action in the string-frame metric isThe terms denoted $O({\alpha }^{\prime })$ contain additional derivatives, including curvature-squared terms in the bosonic theory. The expansion concerns the massless sector around the perturbative bosonic background; the tachyon instability remains and is not cured by omitting its field from this displayed action. Thus this is a formal perturbative effective description, not a claim of a stable bosonic vacuum.

Finally split the dilaton into a constant and its variation, $\Phi ={\Phi }_0+\varphi$. By the Gauss-Bonnet theorem, its dilaton Euler-characteristic weighting on a connected closed oriented surface follows fromDefine the string coupling by $g_s=e^{{\Phi }_0}$. A genus-$g$ path integral is weighted byThe sphere carries $g_s^{-2}$, the torus carries $g_s^0$, and each extra handle adds $g_s^2$. At higher genus one integrates over complex-structure moduli as well as insertion points, with the associated antighost insertions supplying the correct moduli measure. With canonically normalized external vertices an $N$-point genus-$g$ amplitude scales as $g_s^{2g-2+N}$.

Summing connected worldsheets of every genus of a surface yields the string-loop effective action expansionEach coefficient has its own low-energy ${\alpha }^{\prime }$ expansion. The two parameters have different roles: ${\alpha }^{\prime }$ resolves finite string size through higher derivatives, while $g_s^2$ counts additional string loops.

The degree of a map between oriented manifolds measures how many times the domain covers the target, with signs recording the local local orientation of a manifold. Let $M,N$ be connected, oriented closed manifolds of the same dimension $n>0$. A continuous map $f:M\to N$ acts on top-dimensional homology bywhere $[M],[N]$ are their fundamental classes. Connectedness and the choices of local orientation of a manifold identify $H_n(N;\mathbb{Z})$ with $\mathbb{Z}$. Reversing the orientation of either manifold changes the sign; reversing both does not.

For a smooth map, Sard theorem supplies a regular value $y$. Its inverse image is discrete and, by compactness, finite. At each $x\in f^{-1}(y)$ the differential is an isomorphism; let its sign be $+1$ or $-1$ according to whether it preserves or reverses the chosen local orientations. The degree as a sum of local degrees isThe sign is computed in oriented manifold charts. The value is independent of the chosen regular value, even when inverse images appear or disappear: the signed count is the coefficient of $[N]$ in $f_{\ast }[M]$.

There is a useful local-density expression for the same topological degree. If $\omega$ is a volume form with ${\int }_N\omega =1$, thenThis degree by integration of a pullback volume form follows first by choosing a smooth top-form supported in a small neighborhood of a regular value, where the inverse branches contribute their orientation signs. Any other normalized top-form differs from it by an exact form: integration identifies $H_{\mathrm{dR}}^n(N)$ with $\mathbb{R}$. The integral of its pullback difference vanishes by Stokes theorem. In particular, for any top-form $\eta$, ${\int }_M{f}^{\ast }\eta =\mathrm{deg}(f){\int }_N\eta$.

A homotopy $H:M\times [0,1]\to N$ preserves this integral, since $d\omega =0$ and Stokes theorem givesThus topological degree is a homotopy invariant. It is multiplicative under composition, because the induced maps on homology compose: $\mathrm{deg}(g\circ f)=\mathrm{deg}(g)\mathrm{deg}(f)$. The identity has degree one, a constant map has degree zero for $n>0$, and an orientation-reversing diffeomorphism has degree minus one. An orientation-preserving finite covering map has degree equal to its number of sheets. Nonzero topological degree forces surjectivity, since an omitted point would be a regular value with an empty inverse image.

For the circle, $e^{i\theta }\mapsto e^{ik\theta }$ has degree $k$, positive or negative. This is its winding number, computable as $(2\pi i{)}^{-1}\int f^{-1}df$. The antipodal map of $S^n$ has degree $(-1{)}^{n+1}$: its extension $-I$ on the ambient $(n+1)$-dimensional vector space has that determinant sign and respects the outward-normal convention. A holomorphic map $z\mapsto z^k$, $k\ge 1$, on the Riemann sphere has degree $k$, whereas its complex conjugate has degree $-k$. These examples show how orientation, rather than simply the number of inverse images, determines the integer.

For maps $S^n\to S^n$, topological degree gives the complete homotopy classification ${\pi }_n(S^n)\cong \mathbb{Z}$. The degree does not classify general manifold maps: the identity of the torus and the map induced by the integer matrix $\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)$ both have degree one, but have different induced maps on $H_1(T^2;\mathbb{Z})$ and so are not homotopic. A nonzero-degree map $S^n\to S^n$ cannot extend continuously to $B^{n+1}$, because such an extension would make the boundary map null-homotopic. In the smooth setting, Stokes theorem gives the same obstruction by applying it to the pulled-back normalized volume form.

The hypotheses can be adjusted, but must be stated. For connected oriented noncompact manifolds, a proper map has a degree defined using compactly supported top-forms, and it is invariant under proper homotopies. For manifolds with boundary one uses relative fundamental classes and maps of pairs, or fixes appropriate boundary conditions. Without an integral orientation one can still count inverse images modulo two, obtaining a mod-two degree. The integer integral formula used below assumes the oriented setting.

In classical field theory, these ideas turn continuous fields into quantized topological charges. Suppose a field on ${\mathbb{R}}^d$ approaches one fixed target value at spatial infinity. The one-point compactification makes it a map $\varphi :S^d\to \mathcal{V}$. When the target $\mathcal{V}$ is an oriented closed $d$-manifold, its topological degree labels topological sectors. More generally the sectors are described by homotopy groups; an integer degree is available only when the domain and target have the appropriate dimensions and orientations. Smooth time evolution preserving the boundary condition is a homotopy, so it cannot change the integer. A change requires a singular field, escape from the allowed target, or a change at the boundary.

A normalized closed target $d$-form gives the pullback-volume representation of a topological current. On spacetime, put $\alpha ={\varphi }^{\ast }\omega$. Since $d\alpha ={\varphi }^{\ast }(d\omega )=0$, its dual current is identically conserved, andis independent of time when there is no flux at infinity. This conservation law follows from geometry without using the field equations; it need not arise from a continuous symmetry through Noether theorem.

A concrete example is the O3 nonlinear sigma model in two spatial dimensions. Its unit-vector field $\mathbf{n}$ approaches a constant at infinity, defining $S^2\to S^2$. The normalized area form of the target gives the degree charge of an O3 sigma-model lump:For the energy normalization $E=\frac{1}{2}\int (|{\partial }_1\mathbf{n}{|}^2+|{\partial }_2\mathbf{n}{|}^2)$, the identities $\mathbf{n}\cdot {\partial }_i\mathbf{n}=0$ giveThis is the Bogomolny degree bound for the O3 sigma model. Choosing the sign appropriate to $Q$ makes the square nonnegative; vanishing of the square gives first-order Bogomolny equations and a sigma-model lump saturating the bound. With the oriented stereographic projectionthe maps $w=z^k$ have $Q=k$ and $E=4\pi k$. Their conjugates have $Q=-k$ with the same energy. Holomorphic rational maps have positive degree equal to their degree as rational maps; taking a reciprocal does not reverse the orientation. Antiholomorphic dependence reverses it.

The Skyrme model supplies a three-dimensional example. A field $U:{\mathbb{R}}^3\to \mathrm{SU}(2)$ with $U\to I$ at infinity is a map $S^3\to \mathrm{SU}(2)\cong S^3$. Take $T_i=-i{\tau }_i$ and $U^{-1}dU={\theta }^i{T}_i$, with ${\theta }^1\wedge {\theta }^2\wedge {\theta }^3$ positive. Since $\mathrm{tr}(T_i{T}_j{T}_k)=-2{ϵ}_{ijk}$, the normalized target volume form isThe integral is one on the unit three-sphere. Consequently the Skyrme baryon number as a mapping degree isThis is the topological baryon number in the Skyrme model; the sign has been fixed by the stated orientation and anti-Hermitian generator convention.

A topological charge alone does not guarantee a stable finite-size solution. The degree and energetic stability of a field configuration concern different properties. For a three-dimensional configuration of size $R$, the two-derivative energy scales as $R$, so it can decrease by shrinking while the topological degree remains fixed for every $R>0$. The limit can be singular. The Skyrme term, with four derivatives, scales as $R^{-1}$ and can balance the shrinking tendency. This is the role of Derrick theorem in distinguishing topological obstruction from energetic stability.

For defects, the relevant boundary map can instead be the sphere surrounding a core. A vacuum manifold equal to $S^1$ gives the integer winding number of a vortex; a vacuum manifold $S^2$ gives the degree of a surrounding $S^2$ for a magnetic monopole. This vacuum-boundary degree as a defect charge obstructs extending the normalized vacuum field through the enclosed ball. A nonzero integer therefore forces the field to leave the vacuum manifold somewhere in the core. This construction does not require the field to take one constant value in every direction at infinity.

Degree also appears in four-dimensional gauge theory through a boundary transition function. For an anti-Hermitian SU(2) gauge connection on ${\mathbb{R}}^4$, write $F=dA+A\wedge A$ and assume finite-action boundary behavior $A\to g^{-1}dg$ on the large bounding three-sphere. In the second-Chern conventionthe identity $d\mathrm{tr}(A\wedge dA+\frac{2}{3}{A}^3)=\mathrm{tr}(F\wedge F)$ and the Maurer-Cartan equation giveThis boundary winding representation of Yang-Mills topological charge relates the Second Chern number to the degree of $g:S^3\to \mathrm{SU}(2)$. The Chern-Simons 3-form turns the bulk integral into the boundary winding integral. Conventions which define the instanton number with the opposite trace sign reverse $k$; the integer quantization is unchanged. A Yang-Mills theta term weights a sector by $e^{i\vartheta k}$, giving periodicity $\vartheta \mapsto \vartheta +2\pi$. Thus the same topological degree that counts oriented inverse images also labels field sectors and expresses their quantized charges as integrals of local densities.

On the complex torus $\mathbb{C}/\Lambda$, the Weierstrass elliptic function $\wp$ is a degree-two map to the Riemann sphere, and its fibres are the pairs $\{z,-z\}$, with coalescence at the four branch points. An even elliptic function is constant on every fibre, so away from those branch points it descends to a meromorphic function $Q$ of $\wp$.

The descent also holds at a branch point. At a half-period $h$, evenness and periodicity give $f(h+u)=f(h-u)$, so its local Laurent series uses only even powers of $u$. Also $\wp (h+u)-\wp (h)$ is a local coordinate proportional to $u^2$. At $0$, use $1/\wp (z)\sim z^2$ instead. Thus $Q$ extends meromorphically across every branch point and infinity. Every meromorphic function on the Riemann sphere is a rational function, proving $f=Q\circ \wp$.

For a nonconstant rational function $Q=p/q$ in coprime form, $\mathrm{deg}Q=\max (\mathrm{deg}p,\mathrm{deg}q)$: a generic equation $p-wq=0$ has that many roots on the sphere. Degrees of holomorphic maps multiply under composition. Hence the degree isConstant functions have degree zero. Degree two therefore occurs exactly when $Q$ is a Möbius transformation:Conversely every such function is even, elliptic and of degree two.

The Papperitz symbol describes a homogeneous second-order Fuchsian differential equation on the Riemann sphere, with three regular singular points and the two indicial exponents at each. At a finite point $z_j$, an exponent $r$ corresponds locally to a factor $(z-z_j{)}^r$; at infinity it corresponds to $z^{-r}$. Repeated or resonant exponents can require logarithms. The six exponents obey the Fuchs relation that their sum is one. With three singularities there is no independent accessory parameter, so these data determine the normalized equation.

A Möbius transformation of a Papperitz symbol moves the singular points without changing their exponents. Multiplication of the dependent variable by a local power shifts the two exponents by that power, using the reciprocal local coordinate at infinity.

Set $w=z/(z-1)$ and $v(w)=F(a,c-b,c;w)$. At $w=0,1,\infty$, the exponent pairs for $v$ are respectively $(0,1-c)$, $(0,b-a)$, and $(a,c-b)$. The Möbius transformation sends $z=0,1,\infty$ to $w=0,\infty ,1$. Before rescaling, $v(w(z))$ therefore has exponent pairs $(0,1-c)$ at zero, $(a,c-b)$ at one, and $(0,b-a)$ at infinity. Multiplication by $(1-z{)}^{-a}$ subtracts $a$ at one and adds $a$ at infinity. The resulting pairs areexactly those of the original Gauss hypergeometric equation. Near zero choose the branch with $(1-z{)}^{-a}=1$ at zero. The resulting solution is analytic there and has value one. Uniqueness of the normalized analytic solution yieldsThis is the Pfaff transformation. The normalization presupposes the usual admissible parameters, in particular $c\notin \{0,-1,-2,\dots \}$. Elsewhere the identity holds by analytic continuation with compatible branches, or by parameter limits wherever those are defined.

The Möbius group consists of all Möbius transformations of the Riemann sphere ${\mathbb{C}}_{\infty }=\mathbb{C}\cup \{\infty \}$, with operation function composition. A nonzero-determinant matrix $M=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ represents the map $z\mapsto (az+b)/(cz+d)$. If $c\ne 0$, its pole $-d/c$ maps to $\infty$ and $\infty$ maps to $a/c$; if $c=0$, the map is affine and fixes $\infty$. Nonzero scalar multiples of $M$ represent the same transformation.

Define $\varphi (M)$ by this action for $M\in S{L}_2(\mathbb{C})$. Substitution of fractional-linear expressions shows $\varphi (MN)=\varphi (M)\circ \varphi (N)$, so $\varphi$ is a group homomorphism. Any invertible matrix can be rescaled into the special linear group: choose $\lambda \in {\mathbb{C}}^{\times }$ with ${\lambda }^2\det M=1$, which is possible because every nonzero complex number has a square root. This does not change the transformation, proving surjectivity. A transformation represented by $M$ is the identity only if fixing $0$, $\infty$ and $1$ forces $b=c=0$ and $a=d$. The determinant-one condition then gives $a^2=1$. ThusFor the fixed-point claim, a nonidentity Möbius transformation has at least one fixed point of a Möbius transformation: when $c\ne 0$ the finite fixed points solve a quadratic, and when $c=0$ infinity is fixed. A quadratic shows there can be no more than two distinct fixed points unless the transformation is the identity. Suppose there is exactly one. Conjugate it to infinity by a Möbius transformation. The resulting map has form $z\mapsto az+b$, since it fixes infinity. If $a\ne 1$ it also has the finite fixed point $b/(1-a)$, a contradiction. Hence it is $z\mapsto z+b$ with $b\ne 0$. Its $m$th power is $z\mapsto z+mb$, never the identity for $m>0$. Conjugation preserves order, so a nonidentity finite-order Möbius transformation has exactly two fixed points.

Now let $K$ be a finite abelian subgroup of the Möbius group. First suppose it contains an element $g$ of order greater than two. Conjugate the two fixed points of $g$ to $0$ and $\infty$, so $g(z)=\zeta z$, with $\zeta$ a root of unity of order greater than two. Every element $h$ commuting with $g$ permutes its fixed-point set, because $g(h(p))=h(g(p))=h(p)$. A Möbius transformation preserving each of $0,\infty$ is $h(z)=az$; one exchanging them is $h(z)=a/z$. In the second case $hg(z)=a/(\zeta z)$ and $gh(z)=\zeta a/z$, so commuting would force ${\zeta }^2=1$, impossible. Thus all elements of $K$ are scalings. A finite subgroup of a field multiplicative group is cyclic; here one can see this directly by choosing the least common multiple $m$ of all element orders and observing that the scalars lie among the $m$th roots of unity, a cyclic group. Hence $K$ is cyclic.

Otherwise every nonidentity element of $K$ has order two. If $K$ is nontrivial, choose $g$ and conjugate it to $z\mapsto -z$. The action of $K$ on $\{0,\infty \}$ has image of order at most two. Its kernel consists of scalings $z\mapsto az$, and the order-two condition forces $a^2=1$, so the kernel has at most two elements. Hence $|K|\le 4$. Orders one and two give cyclic groups; order four with all nonidentity elements involutions gives $C_2\times C_2$, generated by any two distinct nonidentity elements. Both possibilities occur: scalings by roots of unity give every finite cyclic group, and $\{z,-z,1/z,-1/z\}$ gives the Klein four group. The classification is

At the origin, the Taylor series $\sin z=z+O(z^3)$ and $1-\cos z=z^2/2+O(z^4)$ giveThus $f_1$ has a pole of order three and $f_2$ a pole of order four at zero. For the third function, the Laurent serieshas infinitely many nonzero negative-power terms, so zero is an essential singularity of $f_3$.

To classify the point at infinity on the Riemann sphere, use the coordinate $w=1/z$. Thenso $f_3$ has a simple pole at infinity.

For the other two functions, an important qualification is necessary. The poles of $f_1$ at nonzero integer multiples of $\pi$, and those of $f_2$ at nonzero integer multiples of $2\pi$, approach infinity. Equivalently, their images in the $w$ coordinate accumulate at zero. Hence infinity is a non-isolated singularity of $f_1$ and $f_2$. Neither function is holomorphic on any punctured neighbourhood of infinity, so the classification of isolated singularities into removable singularities, poles and essential singularities does not apply there.

Write the period lattice as $\Lambda =\mathbb{Z}{\omega }_1+\mathbb{Z}{\omega }_2$, with ${\omega }_1,{\omega }_2$ real-linearly independent and $\mathrm{Im}({\omega }_2/{\omega }_1)>0$. A meromorphic function $f$ is an elliptic function for this lattice if $f(z+\lambda )=f(z)$ for every $\lambda \in \Lambda$. If it has no poles, it is an entire function bounded on the closure of a fundamental parallelogram. Periodicity makes it bounded on all of $\mathbb{C}$, so the Liouville theorem proves that an elliptic function without poles is constant.

Here are the two contour identities needed for counting its zeros and their sum. Choose a positively oriented fundamental parallelogram $P$ whose boundary avoids all zeros and poles, and put $h=f^{\prime }/f$. The argument principle givesbecause $h$ is periodic and opposite edges cancel. Here $N,P_f$ count zeros and poles with multiplicity. For the weighted integral, let $J_i$ be the integral of $h$ along the edge from a vertex $c$ to $c+{\omega }_i$. Translating the opposite edges givesThe endpoint values of $f$ agree along either edge. A continuous logarithm along the edge therefore changes by $2\pi i{n}_i$, with $n_i\in \mathbb{Z}$, and $J_i=2\pi i{n}_i$. On the other hand, the residue theorem evaluates the left side as the sum of the zero positions minus the sum of the pole positions, both with multiplicity. Thus the elliptic divisor-sum identity isUnder the stated pole hypothesis there is just one pole class, represented by zero, of order $m$. ConsequentlyChanging representatives of the zero classes changes their weighted sum by an element of $\Lambda$, so the congruence is well defined.

The Weierstrass elliptic function isOn a bounded set of $z$, the summand is $O(|\omega {|}^{-3})$ for large $|\omega |$. The lattice sum of $|\omega {|}^{-3}$ converges in real dimension two, proving Normal convergence of the Weierstrass elliptic-function series on compact sets away from $\Lambda$. Hence $\wp$ is a holomorphic function there, and at zero its principal part is $z^{-2}$, so it has a double pole. Replacing $\omega$ by $-\omega$ shows that $\wp$ is even. Differentiating the normally convergent series giveswhose absolute convergence permits reindexing by any $\lambda \in \Lambda$. Thus ${\wp }^{\prime }$ is periodic and $\wp (z+\lambda )-\wp (z)$ is constant. Substituting $-z-\lambda$ for $z$ and using evenness changes that constant to its negative, so it is zero. This proves that $\wp$ is an elliptic function, with double poles precisely at the lattice points.

The preceding zero count applied to $\wp -a$ says that it has exactly two zeros modulo $\Lambda$, counted with multiplicity. Evenness pairs any zero $z$ with $-z$. If these are distinct classes they must both be simple. If they coincide, $2z\in \Lambda$, and locally $\wp (z+t)=\wp (z-t)$: the zero has even order, which the total count forces to be two. In particular such a zero cannot be a lattice point, where $\wp$ has a pole. Every finite fibre of the Weierstrass function consists of two opposite simple points or one double nonzero half-period point.

To obtain the rational representation, first consider an even elliptic function $g$. The preceding description proves that $\wp :\mathbb{C}/\Lambda \to {\mathbb{P}}^1$ has exactly the fibres $\{z,-z\}$, so $g$ defines a function $A$ of $w=\wp (z)$. Away from the branch values a local inverse of $\wp$ makes $A$ meromorphic. At a half-period $z_0$, the local expansion is $w-w_0=c{t}^2+O(t^4)$, with $c\ne 0$ because the zero has order exactly two. The Laurent series of $g(z_0+t)$ contains only even powers, and $t^2$ is a holomorphic local coordinate as a function of $w-w_0$. Thus $A$ is meromorphic at that value too. At zero the same argument uses $1/w=t^2+O(t^6)$, proving meromorphy at infinity. A meromorphic function on the Riemann sphere is a rational function: subtract its finitely many principal parts and use compactness to make the remainder constant. This proves that even elliptic functions are rational in the Weierstrass function.

For an arbitrary elliptic function, split $f=f_{+}+f_{-}$, where $f_{\pm }(z)=(f(z)\pm f(-z))/2$. The even part is $A(\wp )$. Since ${\wp }^{\prime }$ is odd and not identically zero, $f_{-}/{\wp }^{\prime }$ is an even meromorphic elliptic function, including at the zeros of ${\wp }^{\prime }$ where the quotient may have poles. It is therefore $B(\wp )$ for a rational function $B$. The elliptic function-field decomposition isThe representation is unique: its even and odd parts determine $A$ and $B$, and $\wp$ takes every value on the Riemann sphere.

Use the Riemann sphere convention for evaluating a Möbius transformation at a pole and at infinity. When the three prescribed points are finite, takeIts zero is $w_2$, its pole is $w_1$, and substitution at $w_3$ gives one. The remaining three cases areAll coefficient determinants are nonzero because the points are distinct. There is only one such Möbius transformation: the composite of any two candidate maps, one inverted, fixes $\infty ,0,1$; fixing infinity makes it affine, and fixing zero and one makes it the identity.

For this question use the cross-ratio normalizationIn the finite case this is $(w_3-w_1)(w_4-w_2)/[(w_3-w_2)(w_4-w_1)]$, with the corresponding limits if a point is infinite. It is finite and distinct from zero and one, since $w_4$ is distinct from the other three points and $f$ is a bijection. Ordering conventions for the cross-ratio differ; the definition here is fixed by the specified images, rather than by importing another ordering formula.

A generalized circle means either a circle or a straight line completed by infinity. Let $K$ be the unique generalized circle through the first three points. Its image under $f$ is the unique generalized circle through $\infty ,0,1$, namely $\mathbb{R}\cup \{\infty \}$. Therefore $w_4\in K$ exactly when $f(w_4)$ is real. Conversely a real $f(w_4)$ lies on that extended real line, whose inverse image is $K$. This proves the real cross-ratio criterion for a generalized circle in both directions.

Write $w$ for the complex coordinate. The inverse stereographic projection iswith $\infty$ mapped to the north pole. A rotation $R_z(\theta )$ about the third axis sends $w$ to $e^{i\theta }w$, represented byFor the rotation $R_y(\theta ):(x,y,z)\mapsto (x\cos \theta +z\sin \theta ,y,-x\sin \theta +z\cos \theta )$, substitution in the stereographic projection givesBoth displayed matrices are unitary with determinant one, hence belong to the special unitary group.

Use the rotation-generation result that every $R\in SO(3)$ can be written $R_z(\alpha )R_y(\beta )R_z(\gamma )$, with angles chosen suitably. This is the Euler-angle decomposition, which is allowed here. Composition of Möbius transformations corresponds to multiplication of their matrix representatives, soThe formulas extend at poles in the Riemann sphere. This realizes sphere rotations as special-unitary Möbius transformations. The two representatives $U$ and $-U$ define the same Möbius transformation, as expected for the double covering of the special orthogonal group.

The prescribed-pole form of Runge theorem is this: let $K\subset \mathbb{C}$ be compact, and let $E\subset \hat{\mathbb{C}}\setminus K$ meet every connected component of the complement in the Riemann sphere. If $f$ is holomorphic on a neighbourhood of $K$, then for every $ϵ>0$ there is a rational function with all its poles in $E$ whose uniform distance from $f$ on $K$ is less than $ϵ$. A polynomial is regarded as having its only possible pole at infinity. The empty compact set is trivial, so assume $K\ne \varnothing$.

First suppose $\infty \in E$. Let $A\subset C(K)$ be the uniform rational approximation algebra with prescribed poles, the uniform closure of rational functions whose finite poles lie in $E$. It is a commutative unital Banach algebra, and contains $u(z)=z$. Setwhere the inverse in this definition is the pointwise continuous function on $K$. The set $S$ is relatively open: if the inverse at $\lambda$ belongs to $A$, the Neumann series gives inverses at nearby points. It is relatively closed: when ${\lambda }_n\to \lambda \notin K$, the corresponding scalar functions converge uniformly on $K$, and $A$ is closed.

Each bounded complementary component meets $E$ in a finite point $a$, and $(u-a1{)}^{-1}$ is an allowed rational function. The unbounded component meets $S$ because for $|\lambda |>\Vert u\Vert$ the geometric series for $(u-\lambda 1{)}^{-1}$ belongs to $A$. Being both open and closed in the complement, $S$ therefore contains every component, so it is the entire complement. Conversely, evaluation at any $z\in K$ prevents $u-z1$ from being invertible. HenceApply the holomorphic functional calculus in $A$ to $f$ near $K$. For every $z\in K$, the evaluation character of an algebra and the contour formula from (b) give $f(u)(z)=f(z)$. Thus $f{|}_K\in A$. By the definition of this uniform closure, the required rational approximations exist.

If $\infty \notin E$, choose a finite point $a\in E$ in the component containing infinity. The Möbius transformation $w=(z-a{)}^{-1}$ sends $K$ to a compact subset $K^{\prime }$ of the plane and sends $E$ to a set $E^{\prime }$ containing infinity. It preserves complementary components, so $E^{\prime }$ meets every component of the complement of $K^{\prime }$. The transformed function $F(w)=f(a+1/w)$ is holomorphic near $K^{\prime }$, since $0\notin K^{\prime }$. The case already proved approximates $F$ by rational functions with poles in $E^{\prime }$. Pulling them back gives rational approximations to $f$ on $K$ with poles only in $E$: a pole at infinity in the $w$-plane becomes a pole at $a$, and every other pole has its prescribed preimage. In particular the pullbacks do not acquire a pole at the original infinity, since $0\notin E^{\prime }$.

This proves the full prescribed-pole theorem, including the case where a pole at infinity is forbidden. If $\mathbb{C}\setminus K$ is connected, take $E=\{\infty \}$ to obtain the polynomial Runge theorem as a corollary.

Interpret the printed congruence entrywise in the integer lattice:It then defines the usual level-two principal congruence subgroup of ${\mathrm{SL}}_2(\mathbb{Z})$, despite the PDF's ambient ${\mathrm{SL}}_2(\mathbb{R})$. A literal ideal congruence inside $\mathbb{R}$ would be vacuous because $2\mathbb{R}=\mathbb{R}$, and would make the asserted conclusion false. The integer-lattice interpretation is essential.

Pass to $\overline{\Gamma (2)}=\Gamma (2)/\{\pm I\}$, which has exactly the same action. Reduction modulo two maps the modular group onto ${\mathrm{SL}}_2({\mathbb{F}}_2)$, a group of order six: the reductions of $S$ and $T$ generate it. Its kernel is $\overline{\Gamma (2)}$, so the index of a subgroup is six. The standard fundamental domain of the modular group has hyperbolic area $\pi /3$, and hence the quotient has hyperbolic area $2\pi$.

There are no nonidentity elliptic Möbius transformations in $\overline{\Gamma (2)}$. An integral matrix representing an elliptic Möbius transformation has trace $0$ or $\pm 1$. Here the trace is even, excluding $\pm 1$; trace zero would give $d=-a$ and $1=-a^2-bc\equiv -1(\mathrm{mod}4)$, impossible. Thus the effective action is a free properly discontinuous group action, and the quotient is a Riemann surface.

A cusp of a modular group is represented by a rational boundary point. Their orbits correspond towhich has $6/2=3$ elements. They are represented by $\infty ,0,1$, or by the three nonzero parity vectors of a primitive numerator-denominator pair. Each width of a cusp is two. A union of six copies of the standard fundamental domain of the modular group gives a fundamental region for this subgroup. Removing small horocycle neighbourhoods of its cusps leaves a compact core. Adding one point at each cusp of a modular group, using the local parameter $e^{\pi iz}$ after moving that cusp to infinity, gives a compact Riemann surface $\overline{X}$.

For a finite-area hyperbolic surface of genus $g$ with $r$ cusps, the Gauss-Bonnet theorem gives area $2\pi (2g-2+r)$. Thus $2\pi =2\pi (2g-2+3)$ and $g=0$. A compact genus-zero Riemann surface is the Riemann sphere, by the uniformization theorem. A Möbius transformation sends the three added points to $0,1,\infty$. Restricting it gives

By the argument principle, the number of zeros minus poles, counted with multiplicity, in the fundamental parallelogram is $(2\pi i{)}^{-1}\oint f^{\prime }/fdz$. The logarithmic derivative is periodic with the same lattice, so the integrals over opposite sides cancel. Thus the numbers of zeros and poles are equal.

The Weierstrass elliptic function gives a degree-two map $\wp :\mathbb{C}/\Lambda \to {\mathbb{P}}^1$ whose fibers identify $z$ and $-z$. An even elliptic function is constant on these fibers, so it descends to a single-valued function $Q$ on the sphere. This is meromorphic even at a branch value: near a half-period the involution is $t\mapsto -t$, the Laurent series of $f$ has only even powers, and $\wp -{\wp }_0$ has a double zero and provides a local coordinate in $t^2$. At the image of $0$, $\wp \sim t^{-2}$ gives the same conclusion at infinity. A meromorphic function on the Riemann sphere is a rational function, so $\boxed{f=Q(\wp )}$.

A meromorphic antiderivative need not be elliptic. Take $f=\wp$. Its double poles have zero residue, so integrating on the punctured plane gives a single-valued primitive, since every closed-contour integral is a sum of zero residues. The primitive extends meromorphically over each pole, with local principal part $-1/z$. If $F$ were elliptic, it would have precisely one pole modulo the lattice, with residue $-1$, contradicting the zero sum of residues in a fundamental parallelogram. This is a nonconstant counterexample satisfying the stated premise; the preceding descent proves that even elliptic functions are rational in the Weierstrass function.