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The map $X(\Gamma )\to X(\Gamma (1))\cong {\mathbb{P}}^1$ has degree $d$, the effective projective index. Ramification occurs only over the two elliptic points and the modular cusp. Over the order-two elliptic point, $r_2$ points have index one and the other $(d-r_2)/2$ points have index two, contributing $(d-r_2)/2$ to the ramification divisor. The analogous order-three contribution is $2(d-r_3)/3$.

The local degree at a modular cusp is its effective cusp width. The cusp widths sum to $d$, so the modular cusp contribution is $d-r_{\infty }$. The Riemann-Hurwitz formula, $2g(X)-2=d(2g(Y)-2)+{\sum }_P(e_P-1)$, now yieldsTherefore the genus formula for a modular curve isUsing the index in ${\mathrm{SL}}_2$ without accounting for its center would give the wrong degree when $-I\notin \Gamma$.

Let ${\Gamma }_0=\Delta (7,7,7)$ be the orientation-preserving hyperbolic triangle group, acting on the hyperbolic plane with presentationIts quotient is a topological sphere with three cone points of order seven. Define the surjective group homomorphismThe stated order conditions make this well defined, and the generation hypothesis makes it surjective. Set $K=\ker \varphi$. Every finite-order element of a hyperbolic triangle group is conjugate to a power of a cone-point generator. None of its nontrivial such powers lies in $K$, since all three images have order seven. Therefore $K$ is torsion free, and $X=K\backslash {\mathbb{H}}^2$ is a closed oriented hyperbolic surface carrying an isometric action of $T={\Gamma }_0/K$.

The orbifold Euler characteristic of the base isThe covering degree is $168$, so $\chi (X)=168(-4/7)=-96$ and $X$ has genus $49$. The subgroups have order $|U_i|=168/7=24$. Point stabilizers for the $T$ action are trivial or cyclic of order seven. Since seven does not divide $24$, each $U_i$ acts freely. Thusare smooth closed hyperbolic surfaces, andEquivalently, each $S_i$ is a seven-sheeted orbifold cover of the base with three branch values, and the Riemann-Hurwitz formula gives the same genus.

Gassmann equivalence means that $U_1,U_2$ meet each conjugacy class of $T$ in the same number of elements. To prove isospectrality directly, let $E_{\lambda }(X)$ be a Laplacian eigenfunction space on $X$, viewed as a representation of $T$. Pullback identifies $E_{\lambda }(S_i)$ with its $U_i$-invariant subspace. Averaging over $U_i$ is the projection onto this subspace, and henceThe character of a representation in this sum is constant on conjugacy classes. The sums are equal by Gassmann equivalence, so the eigenvalues and their multiplicities agree. This is the mechanism of the Sunada theorem. The constructed genus-three hyperbolic surfaces are isospectral.

Nonconjugacy of $U_1,U_2$ in $T$ does not prove nonisometry. For example, an isometry of $X$ outside the given subgroup $T$ might conjugate $U_1$ to $U_2$ and descend to an isometry of the quotients. More generally, an isometry between the quotients lifts to an isometry of ${\mathbb{H}}^2$ conjugating the two groups ${\varphi }^{-1}(U_i)$; there is no requirement that this lift come from an element of $T$. Additional symmetries of the three-cone-point construction are a possible source of this coincidence.

One can prevent this by modifying the common metric, while keeping the $T$ action isometric. Here is a geometric way to make that step precise. On the regular part of the base orbifold $O={\Gamma }_0\backslash {\mathbb{H}}^2$, choose a sufficiently small smooth conformal rescaling of a Riemannian metric of its metric whose local geometry distinguishes base points and local directions. One may use a generic smooth conformal factor with separate curvature features on a countable collection of small disks. Let these disks approach the cone points, with amplitudes decaying sufficiently rapidly that the lifted conformal factor and all its derivatives vanish at their centres. This gives smooth metrics on the covering surface, without retaining an open homogeneous region near a cone point. The purpose is to ensure that a local isometry in the regular part covering a quotient is the identity on the base: the curvature features identify its base point and local frame. Pull the metric back to $X$ and then descend it to both $S_i$. The only points with the cone-point rotational symmetry in their metric germs are the lifts of the cone points; the perturbation removes such symmetries in the regular part. Any quotient isometry must therefore preserve this finite distinguished set.

If the perturbed quotients were isometric, their local maps on the dense regular part would preserve the projection to $O$. They would therefore give equivalent connected covers of the punctured base; the correspondence between connected covering spaces and subgroups would conjugate ${\varphi }^{-1}(U_1)$ to ${\varphi }^{-1}(U_2)$ inside ${\Gamma }_0$. Applying $\varphi$ would conjugate $U_1,U_2$ inside $T$, contrary to hypothesis. Thus the perturbed quotients are nonisometric. The same averaging argument still proves isospectrality, because it requires a common $T$-invariant Riemannian metric, not constant curvature. The genus remains three.

This last step produces nonisometric isospectral metrics of variable curvature on genus-three surfaces. An oriented metric defines a Riemann surface through its conformal structure; using a conformal perturbation even preserves the original conformal structures. If “Riemann surface” means specifically its canonical curvature $-1$ metric, the last step should instead be described as producing Riemannian surfaces. The $\Delta (7,7,7)$ orbifold has no hyperbolic deformation parameters: uniformizing the perturbed metrics does not preserve their Laplace-Beltrami operator spectra, and cannot be used to claim nonisometric hyperbolic examples from this argument.

Consider the singularity of the injective entire function $f$ at infinity. If removable, it is bounded outside a disk and therefore bounded globally; Liouville theorem makes it constant, a contradiction. If essential, the Casorati-Weierstrass theorem says its values outside every large disk are dense. Choose a small neighbourhood $U$ of a finite point; the open mapping theorem puts a nonempty open disk inside $f(U)$. A point outside a disk containing $U$ would map into that open disk, duplicating a value already attained in $U$. This contradicts injectivity. Thus infinity is a pole, so $f$ is a polynomial. If its degree were greater than one, its derivative would have a root; the local power-series form at that point has local degree at least two and cannot be injective. HenceFor a nonconstant holomorphic map of connected compact Riemann surfaces of degree $d$, the Riemann-Hurwitz formula iswhere $g_R,g_S$ are genera and $e_p$ is the local mapping degree, the multiplicity in the target fibre. There are finitely many points with $e_p>1$.

For a degree-two sphere map, the formula gives total ramification two. Since $e_p\le 2$, there are exactly two distinct ramification points, each of local degree two. Their images are distinct, since a fibre has total multiplicity two. Choose a Möbius transformation $T$ taking $0,\infty$ to these domain points and $S_0$ taking their images to $0,\infty$. The resulting rational map has its only zero at zero of order two and its only pole at infinity of order two, hence is $c{z}^2$ with $c\ne 0$. Replace $S_0$ by $S(z)=c^{-1}{S}_0(z)$, obtaining

Set $F(z,w)=w^2-z^n+1$. Its complex gradient is $(-n{z}^{n-1},2w)$, never zero on $F=0$: simultaneous zero would give $z=w=0$, which is not on the curve. The complex implicit function theorem therefore gives a one-dimensional complex manifold, hence a Riemann surface. Where $w\ne 0$ use $z$ as coordinate, and where $z\ne 0$ use $w$; the transition maps and both projections are holomorphic.

For $\pi$, a ramification point of a holomorphic map has $w=0$ and $z=\zeta$ with ${\zeta }^n=1$. In the coordinate $w$, $z-\zeta =w^2/(n{\zeta }^{n-1})+O(w^4)$, giving ramification index $2$. Thus there are $n$ such points, and their branch points are the $n$th roots of unity.

For $\tau$, a ramification point of a holomorphic map has $z=0,w=\pm i$. In the coordinate $z$, $w-w_0=z^n/(2w_0)+O(z^{2n})$, giving ramification index $n$ at both points and branch points $\pm i$.

Write $n=2m$. At infinity set $u=1/z$, $v=w{u}^m$; then $v^2=1-u^n$. At $u=0$ there are two smooth points, $v=\pm 1$, and $u$ is a local coordinate. The extended $\pi$ has a simple pole at each, so neither is ramified. The degree-two Riemann-Hurwitz formula for the map to the Riemann sphere gives $2g-2=2(-2)+n$, henceThe list for $\tau$ above concerns the original affine Riemann surface; on the compactification its poles at the two added points have order $m$ and are additionally ramified if $m>1$.

There is an exact rational-map description of sigma-model lumps and a restricted variational rational-map description of Skyrmions. The first follows from a Bogomolny equation; the second separates angular and radial dependence in a field that generally does not saturate a Bogomolny bound.

For the exact example, take the two-dimensional O3 nonlinear sigma model with a unit field $\mathbf{n}$, energy $E=\frac{1}{2}\int |\nabla \mathbf{n}{|}^2{d}^{2}x$, and a fixed limit at infinity. Regular fields then have a one-point compactification to maps $S^2\to S^2$. Choose the orientation so that the stereographic fieldhas positive charge when it is holomorphic. Equivalently, this is the CP1 nonlinear sigma model since the target sphere is the complex projective line. Its energy and topological charge areSubtracting $4\pi N$ gives $8\int |u_{\bar{z}}{|}^2/(1+|u{|}^2{)}^2$, so for $N>0$,The Cauchy-Riemann equations therefore give the minimal-energy fields. A holomorphic map from the compactified domain Riemann sphere to the target Riemann sphere is a rational map $u=p(z)/q(z)$, with common polynomial factors cancelled. Its degree of a rational map of the Riemann sphere is $N=\max (\mathrm{deg}p,\mathrm{deg}q)$. Poles of $u$ are coordinate singularities, not singularities of $\mathbf{n}$. For instance, $u=(z-a)/\lambda$ with $\lambda \ne 0$ is an exact unit lump with energy $4\pi$, arbitrary centre $a$, and scale $|\lambda |$. Its density is $4|\lambda {|}^2/(|z-a{|}^2+|\lambda {|}^2{)}^2$, whose plane integral is $4\pi$. For negative charge use antiholomorphic maps. Scale invariance allows arbitrarily small lumps and does not by itself prevent a singular concentration limit in the time-dependent theory.

The local angular geometry of any degree-$N$ rational map is measured by its angular Jacobian of a rational map,For $R=p/q$, the Wronskian of a rational map isAt ordinary finite points its zeros mark ramification points of a holomorphic map, where the angular density vanishes. At a pole use $q/p$ as the target coordinate, and at infinity use $1/z$ as the domain coordinate. A pole of order $m$ is ramified by $m-1$. The Riemann-Hurwitz formula counts total ramification $2N-2$, including infinity, even if the affine Wronskian has smaller degree. For $R=z^N$, $W=N{z}^{N-1}$ accounts for $N-1$ zeros at zero; the reciprocal coordinate shows the other $N-1$ at infinity.

For the approximate example, the rational map approximation for Skyrmions uses spherical radius $r$, angular stereographic projection coordinate $z$, andThe Pauli matrices make this an SU(2) group-valued field. The boundary conditions give a continuous centre and the vacuum value at infinity; the profile must also make the energy finite. Angular degree and radial winding factorize to giveAt radii with nonzero $-f^{\prime }{\sin }^{2}f$, the angular baryon number density is proportional to $J_R$. Thus the Wronskian of a rational map locates zero-density directions and reveals the holes or face directions in shell-like Skyrmion configurations.

In dimensionless Skyrme model units, insert this ansatz into the quadratic and quartic derivative energies. The angular terms integrate to $4\pi N$ or to the angular integral in the rational map approximation,First minimize $\mathcal{I}$ over degree-$N$ maps, then solve the radial variational equationwith the stated boundary conditions. The Cauchy-Schwarz inequality gives $\mathcal{I}\ge N^2$ because $J_R$ has sphere average $N$. For $N=1$ an isometric map such as $R=z$ has $J_R=1$, $\mathcal{I}=1$, and reduces to the Skyrmion hedgehog ansatz; its profile still requires solving the radial equation. For general $N$, angular and radial separation restrict the allowed fields. The minimum within this class is an upper bound on the full sector minimum, not a claim that every exact Skyrmion has this separated form. The original construction is Houghton, Manton and Sutcliffe's arxiv.org/abs/hep-th/9705151.

A rotational symmetry of a rational map must satisfy $R(gz)=hR(z)$, where $g,h$ are the domain and target rotations written as Möbius transformations from SU(2) matrices. Since a target rotation is an isometry, this identity implies $J_R(gz)=J_R(z)$. For $R=z^N$, $R(e^{i\alpha }z)=e^{iN\alpha }R(z)$, giving axial spatial symmetry accompanied by an internal rotation. This explains the axial symmetry of the degree-two toroidal ansatz.

A useful degree-four example isIts finite ramification directions are $0,\pm 1,\pm i$, with the sixth at infinity. They are the six coordinate-axis directions under stereographic projection, so they are the face normals of a cube. The full map has octahedral rotational equivariance: for example $R(iz)=1/R(z)$, and the cyclic-axis generator $g(z)=(iz+1)/(1-iz)$ obeys $R(gz)=e^{2\pi i/3}R(z)$. These generate the cube's rotational group and give a cubic angular density. The associated profile then produces the familiar cubic charge-four approximation. Symmetry of the Wronskian is a useful necessary diagnostic but is not sufficient: replacing this map by $cR$, $c>0$, leaves the same ramification directions, while the identity under $z\mapsto iz$ becomes $cR(iz)=c^2/(cR(z))$, whose target transformation is a sphere rotation only when $c=1$. Equivariance of the entire rational map, rather than symmetry of its critical-point set alone, determines the physical rotational symmetry.

For a nonconstant holomorphic map $f:X\to Y$ between compact connected Riemann surfaces, its degree iswhere $e_x$ is the ramification index of a holomorphic map at $x$. The valency theorem makes this sum independent of $y\in Y$. The Riemann-Hurwitz formula is

Write $E=\mathbb{C}/\Lambda$. The map $\psi (z+\Lambda )=-z+\Lambda$ is well-defined and biholomorphic, with ${\psi }^2=\mathrm{id}$. Its fixed points satisfyThey are therefore indexed by $\frac{1}{2}\Lambda /\Lambda \cong (\mathbb{Z}/2\mathbb{Z}{)}^2$, so there are four, as in the negation map on a one-dimensional complex torus.

For each $x\in E^{\prime }$ choose a small open disc $U_x$ for which $U_x\cap \psi (U_x)=\varnothing$. The projection restricts to a homeomorphismUse $(p{|}_{U_x}{)}^{-1}$ followed by a holomorphic coordinate on $U_x$ as a chart on $p(U_x)$. Choosing $\psi (U_x)$ instead changes this coordinate by the holomorphic map induced by $\psi$, so the transition functions are holomorphic. These charts make $S^{\prime }$ a Riemann surface and $p:E^{\prime }\to S^{\prime }$ a holomorphic double covering map.

At each omitted fixed point, a centred local coordinate changes by $z\mapsto -z$, so the invariant quotient coordinate is $w=z^2$. Hence the extension has degree two and exactly four ramification points, each with $e_x=2$. Since the complex torus $E$ has genus one, Riemann--Hurwitz givesThus $\boxed{g_S=0}$, in agreement with the general quotient of a one-dimensional complex torus by negation.

Let $H$ denote a hyperplane-section divisor of a projective plane curve. The Adjunction formula for a smooth degree-$d$ projective plane curve statesSince $\mathrm{deg}(H{|}_C)=d$, this gives the canonical degree of a smooth plane curveCombining this with $\mathrm{deg}{K}_C=2g-2$ gives the genus of a smooth plane curve

Homogenization of the affine equation gives the projective completionThis is the plane model y plus x cubed plus xy cubed equals zero of the Klein quartic: after the coordinate relabelling $[A:B:C]=[Y:X:Z]$, its equation is $A^{3}B+B^{3}C+C^{3}A=0$.

Its first partial derivatives areIf one of $X,Y,Z$ vanishes at a common zero of these three derivatives, the displayed equations successively force all three coordinates to vanish, which is impossible in projective space. If $XYZ\ne 0$, multiplying the three derivative equations givesagain a contradiction. The Jacobian criterion therefore proves that $C$ is smooth.

The rational function $y=Y/Z$ defines a rational map of projective varieties $C⇢{\mathbb{P}}^1$. Because $C$ is a smooth projective curve, it extends uniquely to a morphismFor a generic finite value of $y$, its fibre is given by the cubicso the degree of a holomorphic map is $3$. The discriminant of a depressed cubic is

At $y=0$ the fibre consists of $(0,0)$. Since $F_y(0,0)=1$, the holomorphic implicit function theorem givesso $x$ is a local coordinate and $\pi$ has ramification index of a holomorphic map $3$ there. Each of the seven distinct roots of $4y^7+27$ gives one double, but not triple, root of the cubic in $x$, hence seven further ramification points of index $2$.

It remains to inspect the points at infinity. They areNear $Q$, set $u=X/Y$ and $v=Z/Y$. The equation becomes $u^{3}v+u+v^3=0$, so $u=-v^3+O(v^{10})$, while $y=1/v$ has a simple pole. Thus $e_Q=1$. Near $P$, set $s=Y/X$ and $t=Z/X$. Now $t+s^3+s{t}^3=0$, so $t=-s^3+O(s^{10})$ andThus $y$ has a double pole at $P$ and $e_P=2$. These calculations are the ramification of the x-coordinate on the Klein quartic after the coordinate relabelling above.

The total ramification contribution isThe Riemann-Hurwitz formula for the degree-three map to the projective line now givesand therefore

Homogenizing the affine equation gives the projective closureThis is the Klein quartic. It is smooth: if all three projective partial derivatives vanished, thenMultiplying them gives $28X^3{Y}^3{Z}^3=0$. If one coordinate is zero, the displayed equations force all three to vanish, which is impossible in projective space. Thus $C$ is a smooth compactification of $X^{\prime }$. The smooth compactification compatible with the extended coordinate functions is its normalization, and since $C$ is already smooth it is biholomorphic to $C$.

For a direct genus computation, consider the extended coordinate map ${\pi }_x:X\to {\mathbb{C}}_{\infty }$. A generic $x$ leaves the cubic equationso $\mathrm{deg}{\pi }_x=3$. In the affine part, ramification occurs where $F_y=0$. Solving $F=F_y=0$ gives $(0,0)$ and the seven pointsAt the origin, $x=-y^3+O(y^6)$, so the ramification index of a holomorphic map is three and the contribution is two. At each of the other seven points, $F_x=-7/2\ne 0$, $F_y=0$, and $F_{yy}=6y\ne 0$, so the ramification index is two and each contributes one.

There are two points at infinity. Near $P=[1:0:0]$, in coordinates $u=Y/X$, $v=Z/X$, the equation isand $1/x=v$, so $P$ is unramified. Near $Q=[0:1:0]$, in coordinates $r=X/Y$, $s=Z/Y$, the equation isThus $s=-r^3+O(r^{10})$ and $1/x=s/r=-r^2+O(r^9)$, so $Q$ has ramification index two and contributes one. The total ramification is thereforeThe Riemann-Hurwitz formula for the degree-three map to the sphere givesand henceThis also agrees with the genus of a smooth plane curve of degree four.

The formula $[X:Y:Z]\mapsto [X:Z]$ is initially undefined at the unique point $P_{\infty }=[0:1:0]$ of $X$ at infinity. Since $X$ is a smooth projective curve, the extension of a rational map from a smooth projective curve extends it uniquely to a morphismOn the affine chart it is the rational function $x$, and the equation $y^3=x^4+1$ shows that $\phi$ has degree three.

For each of the four roots $\alpha$ of $x^4+1$, let $P_{\alpha }=(\alpha ,0)$. Since the root is simple, $t=y$ is a uniformizer at $P_{\alpha }$, and the equation givesThus the ramification index of a holomorphic map at every $P_{\alpha }$ is three.

At $P_{\infty }$, use the chart $Y=1$ and put $u=X/Y$, $v=Z/Y$. The equation becomesThe implicit function theorem makes $u$ a uniformizer and gives $v=u^4$ times a local unit. Henceso $P_{\infty }$ also has ramification index three. These are all the ramification points of a holomorphic map because the resulting ramification degree is $5(3-1)=10$. Indeed, the Riemann-Hurwitz formula givesTherefore $g_X=3$, in agreement with the genus of a smooth plane curve, and the Euler characteristic of the underlying compact orientable surface is

Choose a triangulation of the sphere having every branch point as a vertex. Its lift has twice every edge and face, and twice every nonbranch vertex, but only one vertex above each of the $n$ branch points. HenceSince $\chi ({\Sigma }_g)=2-2g$, this gives $n=2g+2$, the degree-two Riemann-Hurwitz formula.

The Riemann-Hurwitz formula for a nonconstant degree-$n$ morphism
$f:C\to C^{\prime }$ of smooth projective connected curves in characteristic zero iswhere $R_f$ is the effective ramification divisor. If $g(C^{\prime })\ge 2$ and
$g(C)<g(C^{\prime })$, the right side is at least $2g(C^{\prime })-2>2g(C)-2$, impossible. If $g(C^{\prime })=1$, then $g(C)=0$ and the left side is $-2$ while the right side is nonnegative. Thus every morphism is constant.

Choose $p\in C_d$ and project from $p$ to the pencil of lines through it. Bézout's theorem says that a general such line meets $C_d$ in $p$ plus $d-1$ further points, so the resolved projectionhas degree $d-1$. Since the genus of a smooth plane curve is
$g=(d-1)(d-2)/2$, Riemann--Hurwitz givesEvery branch point receives at least one ramification point, hence

The adjunction formula givesSince $\mathrm{deg}D=d$ and $\mathrm{deg}{K}_{C_d}=d(d-3)$, linear equivalence
$D\sim K_{C_d}$ would force $d=d(d-3)$, hence $d=4$. It is therefore impossible for $d\ge 5$.

For a smooth plane quartic, projection from a point of the curve gives a degree-three map, so its gonality is at most three. Its genus is three and
$K_C\sim D$, so its plane embedding is the canonical map. A genus-three curve with a degree-two map to ${\mathbb{P}}^1$ would be hyperelliptic, and its canonical map would factor through that double cover and map onto a conic rather than embed the curve. Thus no degree-two map exists, and a degree-one map is excluded by the genus. The quartic's gonality isFor a genus-one curve, Riemann--Roch supplies a degree-two map to ${\mathbb{P}}^1$, while degree one would make it isomorphic to ${\mathbb{P}}^1$. Its gonality is therefore

A germ of a holomorphic function at $z\in D$ is an equivalence class $[f{]}_z$ of pairs consisting of a neighbourhood $U$ of $z$ and a holomorphic function $f:U\to \mathbb{C}$. Two pairs represent the same germ if their functions agree on some neighbourhood of $z$.

The space of germs of holomorphic functions isFor every holomorphic $f$ on an open $U\subseteq D$, the setis declared open; these sets form a basis. The forgetful mapmaps each ${\tilde{U}}_f$ homeomorphically onto $U$. The inverses of these restrictions give the charts defining the complex structure, so $\pi$ is a local biholomorphism.

The evaluation map isIt is well-defined by the germ equivalence relation. On the chart ${\tilde{U}}_f$, its coordinate expression iswhich is holomorphic. Hence $\mathcal{E}$ is analytic, as stated in evaluation map on a space of germs.

Now putThis gives the germ surface of the square root of z to the eighth minus one. An explicit gluing description is obtained by pairing the eight roots into four adjacent pairs and cutting the plane along four disjoint arcs joining the members of each pair. On the complement of the cuts choose one branch $q(z)$ of $\sqrt{z^8-1}$. Take two copies, labelled by $q$ and $-q$, and glue the upper bank of each cut in one copy to the lower bank in the other, and conversely. The cut interiors are restored by the gluing, while the eight endpoints remain absent because they are not in $D$.

On this surface defineFor each $(z,w)\in R$, the holomorphic function $z^8-1$ is nonzero near $z$, so it has a unique local square root $q$ with $q(z)=w$. DefineThis is well-defined, injective, and analytic in the displayed local sheets; its inverse on its image is $[q{]}_z\mapsto (z,q(z))$. Thus $\Phi$ is the required analytic embedding, and it intertwines both the forgetful and evaluation maps.

To compactify, first add one point above each of the eight roots of unity. These are simple branch points of the double cover. Since the polynomial has even degree eight, the even-degree hyperelliptic model has two distinct, unbranched points above infinity. Consequently

Finally apply the Riemann-Hurwitz formula to the degree-two meromorphic map $\bar{\pi }:\overline{R}\to {\mathbb{C}}_{\infty }$. Its only ramification consists of the eight simple finite branch points, soThereforeThis is the compactification of y squared equals x to the eighth minus one.

For a divisor on an algebraic curve $D$ on a smooth projective curve $X$ of genus $g$, the Riemann-Roch theorem stateswhere $K_X$ is a canonical divisor. Taking $D=0$ gives $1-\ell (K_X)=1-g$, so $\ell (K_X)=g$. Taking $D=K_X$ then gives

To obtain a uniform projective embedding, choose a divisor $D$ of degree $2g+1$. Since every divisor appearing below has degree greater than $2g-2$, Riemann--Roch givesThe first two equalities show that the complete linear system of a divisor $|D|$ has no base point. The strict drops in the last two comparisons show respectively that its sections separate distinct points $P,Q$ and tangent directions at $P$. Thus $D$ is a very ample divisor, in accordance with the general fact that a high-degree divisor is very ample on a smooth projective curve, and its sections define a closed embeddingThe ambient dimension therefore depends only on $g$.

The Riemann-Hurwitz formula for a nonconstant morphism $h:C_1\to C_2$ of degree $n$ isChoose a smooth plane quartic $C$, so $g(C)=3$, and form the product of projective varieties $S=C\times C$. This is a smooth projective variety of dimension two. If $B\subset S$ is an irreducible curve, pass to its normalization $\tilde{B}$. At least one coordinate projection $\tilde{B}\to C$ is nonconstant, since otherwise $B$ would be a point. For that projection, Riemann--Hurwitz givesso the geometric genus of $B$ is at least three. Hence $S$ is the required surface; this is the product surface without low-genus curves construction.

Finally let $X\subset {\mathbb{P}}^2$ be a smooth plane curve of degree $d$ and let $p\notin X$. After a projective change of coordinates, take $p=[0:0:1]$. Projection away from $p$ isIts two homogeneous coordinate functions cannot vanish simultaneously on $X$, because their common zero in ${\mathbb{P}}^2$ is $p$. The criterion for a morphism of algebraic varieties therefore shows that the restrictionis a morphism. A fibre is the intersection with a line through $p$, and a general such line meets $X$ in $d$ points counted with multiplicity. Thus the projection of a plane curve from an exterior point has degree $d$.

By the genus of a smooth plane curve, $g(X)=(d-1)(d-2)/2$. Applying Riemann--Hurwitz to $\pi$ and its ramification divisor $R_{\pi }$ givesEvery ramification point contributes at least one to this degree, so

The uniformization theorem states that every simply connected Riemann surface is conformally equivalent to exactly one of the Riemann sphere ${\mathbb{C}}_{\infty }$, the complex plane $\mathbb{C}$, or the open unit disc $\mathbb{D}$. Equivalently, every connected Riemann surface is a quotient of one of these by a free properly discontinuous group of conformal automorphisms.

For ${\mathbb{C}}_{\infty }$, every nonidentity Möbius transformation has a fixed point. A free action is therefore trivial, so the only surface uniformized by the sphere is

Every conformal automorphism of $\mathbb{C}$ has the form $z\mapsto az+b$. If $a\ne 1$, it has a fixed point, so a group acting freely consists only of translations. A discrete subgroup of $(\mathbb{C},+)$ has rank zero, one, or two. The corresponding quotients arewhere $\Lambda$ is a rank-two lattice and $\mathbb{C}/\Lambda$ is a complex torus. This proves the completeness of the Riemann surfaces uniformized by the complex plane.

Now let $D\subseteq \mathbb{C}$ be a domain whose complement contains two distinct points. Its universal cover cannot be the sphere because no nonconstant holomorphic map from a compact Riemann surface has image in the noncompact proper domain $D$. If its universal cover were $\mathbb{C}$, the covering map followed by $D\hookrightarrow \mathbb{C}$ would be a nonconstant entire function omitting the two chosen complementary points, contradicting the Little Picard theorem. The uniformization theorem therefore leaves onlywhich is the plane domain with two omitted points is hyperbolic result.

Finally let $j:{\mathbb{C}}^{\ast }\hookrightarrow R$ be a holomorphic embedding with $R$ compact. At each puncture zero and infinity, an essential singularity would, by the Great Picard theorem in a coordinate chart, force values to be taken repeatedly and contradict injectivity. Thus $j$ extends holomorphically toA generic point of the open image $j({\mathbb{C}}^{\ast })$ has exactly one preimage, so $J$ has degree one. A degree-one holomorphic map between compact Riemann surfaces is a biholomorphism; equivalently, the Riemann-Hurwitz formula rules out branching and gives the same conclusion. Therefore the compact Riemann surface containing an embedded punctured plane is

For a nonconstant holomorphic map $f:X\to Y$ of degree $d$ between compact Riemann surfaces, Riemann-Hurwitz formula is $2g_X-2=d(2g_Y-2)+{\sum }_p(e_p-1)$. Triangulate $Y$ with branch values as vertices and lift the triangulation: faces and edges lift $d$ times, while the vertex deficit is exactly the ramification sum. Comparing Euler characteristics proves the formula.

If $C$ is a smooth projective curve of genus at least three, then the smooth projective surface $C\times C$ contains no curve of geometric genus below three. On the normalization of any curve in the product, at least one coordinate projection to $C$ is nonconstant, and the Riemann-Hurwitz formula cannot decrease genus.

Away from the four fixed points, negation acts freely and the quotient map is a holomorphic double covering. Near a fixed point, an invariant coordinate is $w=z^2$, so the quotient extends across the image as a Riemann surface and the map has ramification index two. The Riemann-Hurwitz formula giveshence the quotient has genus zero and is biholomorphic to the Riemann sphere.

For a finite morphism $f:X\to Y$ of smooth curves, the ramification divisor isThe Riemann-Hurwitz formula states $2g_X-2=(\mathrm{deg}f)(2g_Y-2)+\mathrm{deg}{R}_f$.

The coordinate function $x$ has degree three on the Klein quartic. It has ramification index three at $(0,0)$, index two at the seven affine points satisfying $y^7=-3/8$ and $x=2y^3$, and index two at $[0:1:0]$. The total ramification is ten, so the Riemann-Hurwitz formula gives genus three.