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The projection $[f{]}_z\mapsto z$ is a local biholomorphism for the sheet topology on the space of germs of holomorphic functions. It need not be a topological covering map: branches may end at singularities even when another branch remains over that base point.

Every identity-preserving biholomorphism of $\mathbb{C}/\Lambda$ lifts to multiplication by a nonzero complex number $\zeta$ satisfying $\zeta \Lambda =\Lambda$. Relative to a $\mathbb{Z}$-basis of $\Lambda$, multiplication by $\zeta$ is represented by a unimodular matrix $A\in {\mathrm{GL}}_2(\mathbb{Z})$. Its real determinant is $|\zeta {|}^2$, while complex multiplication preserves orientation, so $\det A=1$ and $|\zeta |=1$. The Cayley-Hamilton theorem givesConsequently $2\mathrm{Re}\zeta =\mathrm{tr}(A)\in \{-2,-1,0,1,2\}$, and $\zeta$ is a root of unity of order $1$, $2$, $3$, $4$, or $6$.

A biholomorphism of the unit disk to itself. Every such map has the form $e^{i\theta }(z-a)/(1-\overline{a}z)$ with $|a|<1$. These maps normalize an extremal univalent function in the proof of the Riemann mapping theorem.

The equation $f_{\bar{z}}=\mu f_z$ prescribing the Beltrami coefficient of a quasiconformal map. Two homeomorphic solutions on the same source differ by a biholomorphism between their targets.

Exhaust a simply connected hyperbolic Riemann surface in potential theory by compact bordered Jordan domains $R_n$. Their Green functions on a Riemann surface $G_n(\cdot ,p)$ have logarithmic poles and zero boundary values. Exponentiating their harmonic conjugates produces proper degree-one maps $f_n:R_n\to \mathbb{D}$ with $f_n(p)=0$. If $G_n=-\log |\zeta |+c_n+o(1)$, normalize $f_n^{\prime }(p)=e^{-c_n}>0$. Domain monotonicity and a global positive Green function bound $c_n$ increasingly from above. Thus Montel theorem gives a nonconstant limit $F$, which is injective by the locally uniform limit of univalent functions theorem. The Schwarz lemma makes $F^{\prime }(p)$ extremal among normalized disk-valued maps. If $F$ omitted $a\ne 0$, taking a global holomorphic square root of a disk automorphism applied to $F$, then renormalizing at $p$, would multiply its derivative by $(1+|a|)/(2\sqrt{|a|})>1$. Thus $F$ is onto and gives the desired biholomorphism.

A holomorphic coordinate chart on a complex manifold is a biholomorphism from an open subset of the manifold to an open subset of ${\mathbb{C}}^n$. Its component functions are holomorphic coordinates.

A holomorphic embedding is a holomorphic map that is also a smooth embedding. Its image is a complex submanifold and the map is a biholomorphism onto that image.

An island over a Jordan domain $D$ is a relatively compact component $U$ of its inverse image on which the meromorphic function is a proper map onto $D$. A component reaching the boundary of the source is instead a peninsula. The island has an integer mapping degree; degree one means that the map is a biholomorphism. A simply connected island can have larger degree and need not be a simple island.

A local coordinate at a point $p$ of a Riemann surface is a biholomorphism from a neighbourhood of $p$ to an open subset of $\mathbb{C}$. After translating the image, one may require that the coordinate sends $p$ to zero.

A simply connected plane domain carrying a hyperbolic metric is proper. By the Riemann mapping theorem, choose a biholomorphism $\varphi :\Omega \to \mathbb{D}$ sending a basepoint to zero. The same distance formula now givesThus the Blaschke condition holds. In particular the $a_n$ have no interior accumulation point: near any such point the quantities $1-|a_n|$ would stay bounded away from zero.

Construct the Blaschke productHere $m=1$ precisely when the sequence contains zero. For $|\zeta |\le r<1$, its normalized Blaschke factor $b_a$ satisfiesThe Blaschke condition therefore makes the infinite product converge uniformly on every compact subdisk. Away from the prescribed zeros its tail factors tend uniformly to one with summable errors, so their logarithms converge and the product is nonzero. It has exactly the prescribed zeros, and the maximum modulus principle bounds every finite partial product by one, hence $|B|\le 1$. Thus $f=B\circ \varphi$ is the required bounded analytic function, with no extra zeros.

Simple connectivity cannot be dropped. On the punctured disk chooseThe hyperbolic metric on the punctured disk is $|dz|/(|z|\log (1/|z|))$. Its universal covering map $\tau \mapsto e^{i\tau }$ pulls it back to the curvature-minus-one upper-half-plane metric. The imaginary axis is a geodesic; the lift from $i$ to $i{n}^2$ has length $\log n^2$, and any other lift differs by a real deck translation and has at least that distance. ThereforeThe triangle inequality again makes the sum finite for every basepoint. However, any bounded holomorphic function on the punctured disk extends over zero by the Riemann removable singularity theorem. These zeros accumulate at zero, so the identity theorem makes the extension identically zero. It cannot have exactly the prescribed zero set. This example of hyperbolic zero sequences on a punctured disk satisfies the distance condition but not the analytic zero-set condition.

We use the complete hyperbolic metric of curvature $-1$; on the unit disc its length element is $2|dz|/(1-|z{|}^2)$. This normalization matters in the exponential summability criterion. A nonzero bounded holomorphic function on the entire plane is constant by Liouville's theorem, so the prescribed infinite zero set forces $\Omega \ne \mathbb{C}$. The Riemann mapping theorem supplies a biholomorphism $\varphi :\Omega \to \mathbb{D}$, and transports the disk hyperbolic metric to $\Omega$.

Fix $w\in \Omega$ and choose $\varphi (w)=0$. Put $a_n=\varphi (z_n)$ and $F=f\circ {\varphi }^{-1}$. First suppose $F(0)\ne 0$. For radii avoiding its zeros, Jensen's formula givesThe monotone convergence theorem as $r\uparrow 1$ gives ${\sum }_n-\log |a_n|<\infty$, hence ${\sum }_n(1-|a_n|)<\infty$. If $F$ has a zero at zero, factor off its finite order first; the resulting holomorphic function is still bounded, since division by that power is bounded away from zero and extends near zero. Thus the Blaschke condition holds in either case, with the possible origin zero treated as one additional finite term.

The Hyperbolic distance in the Poincare disc satisfiesConsequently the exponential distances are summable. This holds for every basepoint: the triangle inequality bounds the ratio of $e^{-\rho (w,z_n)}$ and $e^{-\rho (w_0,z_n)}$ between $e^{-\rho (w,w_0)}$ and $e^{\rho (w,w_0)}$.

Use the curvature-minus-one normalization of the hyperbolic metric:The associated distance is $d(z,w)=2\mathrm{artanh}|(z-w)/(1-\overline{w}z)|$. A connected Riemann surface has its complete canonical Poincare metric on a Riemann surface precisely when its universal cover is conformally the unit disc. By the uniformization theorem, the other simply connected covering types are the complex plane and Riemann sphere. Disc automorphisms preserve $\lambda (z)|dz|$, as direct differentiation of $(z-a)/(1-\overline{a}z)$ shows; therefore the metric descends through the deck group. This is the uniformization notion of hyperbolicity, not the different potential-theoretic Green-function classification.

First prove the needed Schwarz lemma. If $F:\mathbb{D}\to \mathbb{D}$ is holomorphic and $F(0)=0$, then $F(z)/z$ extends holomorphically at zero. On $|z|=r<1$ its modulus is at most $1/r$. The maximum modulus principle and $r\uparrow 1$ give $|F(z)|\le |z|$ and $|F^{\prime }(0)|\le 1$.

Write ${\psi }_a(z)=(z-a)/(1-\overline{a}z)$. Conjugating $f$ by the source and target disc automorphisms, ${\psi }_{f(a)}\circ f\circ {\psi }_a^{-1}$ fixes zero. Applying the derivative bound givesIntegrate along any piecewise smooth path and then take the infimum of lengths: $d(f(z),f(w))\le d(z,w)$. This proves the Schwarz-Pick theorem. Contraction means nonexpansion, not a uniform Lipschitz constant strictly smaller than one; automorphisms preserve all distances.

If both $R$ and $S$ have disc universal covers, lift $g\circ {\pi }_R$ to $\tilde{g}:\mathbb{D}\to \mathbb{D}$ through ${\pi }_S$. The disc is simply connected, so the lift exists, and it is holomorphic because the covering maps are local biholomorphisms. Both covering maps are local hyperbolic isometries. The disc contraction therefore descends to $g^{\ast }d{s}_S\le d{s}_R$ and, by the path-length argument, $d_S(g(p),g(q))\le d_R(p,q)$. If only $S$ is known to have a hyperbolic metric, a nonconstant $g$ forces $R$ to have one too: otherwise lift from the plane or sphere universal cover of $R$ to the disc cover of $S$. A bounded entire function is constant by the Liouville theorem, and a holomorphic map from the compact sphere into the disc is constant by the maximum modulus principle. Thus a nonconstant map into a hyperbolic surface is distance-nonincreasing between the canonical hyperbolic metrics; a nonhyperbolic source admits only constant such maps.

For the prescribed point, Schwarz-Pick theorem gives $|{\psi }_{w_0}(f(z))|\le |{\psi }_{z_0}(z)|$. Thereforeis holomorphic after filling the removable singularity at $z_0$, and $|h|\le 1$. Inverting the target automorphism provesConversely every holomorphic $h$ with this bound gives an admissible disc map: $|{\psi }_{z_0}(z)h(z)|<1$, its inverse target automorphism remains in the disc, and the denominator cannot vanish.

Fix $z$ and put $r=|{\psi }_{z_0}(z)|<1$. The possible intermediate values fill $|v|\le r$, since constant choices $h\equiv c$, $|c|\le 1$, attain them all. Thus the one-point value region of a holomorphic disc map is $|{\psi }_{w_0}(w)|\le r$. Completing the square in $|w-w_0{|}^2\le r^2|1-{\overline{w}}_{0}w{|}^2$ gives its Euclidean centre and radius:This closed disc lies strictly inside the unit disc because $r<1$. For $z=z_0$ it degenerates to the single point $w_0$. Its boundary is attainable by unimodular constant $h$; requiring $f$ to map into the open disc does not remove that boundary.

The Blaschke product here is taken over distinct points of the orbit; its existence is part of the hypothesis. It is not necessary to assert that every discrete orbit satisfies the Blaschke condition.

The modulus of a Blaschke factor is the pseudohyperbolic distance. Thus, away from the orbit,The logarithms converge absolutely at such a point, by the compact convergence estimate for the Blaschke product. Every $T\in G$ preserves the hyperbolic metric and permutes the orbit. Reindexing this convergent product yields $|B(Tz)|=|B(z)|$.

The functions $B\circ T$ and $B$ have the same simple zero set, since $T$ is a biholomorphism. Their quotient therefore extends holomorphically and without zeros across the orbit. It has modulus one everywhere. The open mapping theorem makes it a constant, denoted $\chi (T)$. FinallySince $B$ is not identically zero, cancellation givesThis is the automorphy character of an orbit Blaschke product. The equal-modulus argument is essential: equality of zero sets alone would not make the quotient of two bounded holomorphic functions constant.

A hyperbolic Riemann surface in potential theory is a noncompact Riemann surface admitting a positive Green function on a Riemann surface. With pole $p$, this is a positive function harmonic off $p$, having local form $-\log |\zeta |$ plus a harmonic function in a coordinate $\zeta (p)=0$, and obtained as the minimal such positive function by exhaustion. This is a noncircular definition for the present proof. For a simply connected surface it is equivalent to disk conformal type; for arbitrary surfaces potential-theoretic hyperbolicity and disk universal-cover type should be distinguished.

The unit disk is an example. At pole zero its function is $G(z,0)=-\log |z|$: it is positive in the punctured disk, harmonic there, has the correct logarithmic singularity, and tends to zero at the boundary. If another positive function has the same singularity and is harmonic elsewhere, subtract $G$ and apply the maximum principle for harmonic functions on disks of radius $r<1$; the boundary lower bound is $\log r$. Letting $r\uparrow 1$ shows that the other function dominates $G$. Thus $G$ is indeed the positive Green function on a Riemann surface. At a general pole it is

Here is a Green-function exhaustion proof of disk uniformization. The main analytic idea is to build conformal maps on bordered disks from their Green functions and pass to an extremal limit. The positive global Green function prevents that limit from becoming constant.

First exhaust the noncompact simply connected surface $R$ by relatively compact domains $R_n$ containing $p$, each with a smooth Jordan boundary. This topological step does not use conformal uniformization: take regular neighborhoods of finitely many coordinate disks covering successive compact sets and fill bounded complementary components. Simple connectedness and the Jordan curve theorem make the filled neighborhoods topological disks; enlarge them to be nested.

Each $R_n$ has a zero-boundary Green function on a Riemann surface $G_n(\cdot ,p)$. One analytic construction, independent of conformal uniformization, is as follows. Equip the compact bordered domain with a smooth conformal metric. Multiply $-\log |\zeta |$ by a smooth cutoff function equal to one near $p$ and zero near the boundary; call the result $s$. Distributionallywhere $k$ is smooth and supported away from $p$. Solve $\Delta v=k$ with zero boundary data. This auxiliary Dirichlet problem can be obtained by minimizing $\frac{1}{2}\int |\nabla v{|}^2+\int kv$ over functions with zero boundary trace: the Poincaré inequality gives coercivity, minimization in a Hilbert space gives a weak solution, and interior and smooth-boundary elliptic regularity give a smooth solution. Then $G_n=s+v$ is harmonic away from $p$, has the logarithmic pole, and vanishes at the boundary. It is positive by the maximum principle for harmonic functions, applied after removing a sufficiently small circle around its positive pole. This construction uses ordinary linear elliptic existence, not the desired uniformization theorem. Equivalently, the same bordered-domain Green functions can be constructed by the Perron method for the Dirichlet problem using local logarithmic barriers.

On $R_n\setminus \{p\}$ choose local harmonic conjugates $G_n^{\ast }$. Their period around the pole is $-2\pi$, and every closed loop is homologous to an integer multiple of that loop. Thereforeis a single-valued holomorphic function. It extends across $p$ with one simple zero and has no others. Since $G_n>0$, $|f_n|<1$; since $G_n=0$ on the boundary, $|f_n|\to 1$ there. Thus $f_n:R_n\to \mathbb{D}$ is proper. For any $a\in \mathbb{D}$, choose a smooth interior contour sufficiently close to the boundary that $|f_n|>|a|$ everywhere on it. The contour surrounds $p$ and all possible preimages of $a$, because $|f_n|>|a|$ throughout a thin boundary collar. By Rouché's theorem, $f_n-a$ has the same number of zeros inside as $f_n$, namely one. Hence $f_n$ is onto and one-to-one, with nonzero derivative: it is a biholomorphism from $R_n$ to the disk.

Write near $p$Multiply $f_n$ by a unit constant so that, in the fixed coordinate, $f_n^{\prime }(p)=e^{-c_n}>0$. Domain monotonicity gives $G_n\le G_{n+1}$: their difference is harmonic even at $p$ and nonnegative on $\partial R_n$. The global positive Green function on a Riemann surface $G_R$ similarly majorizes every $G_n$. Comparing the regular parts shows that $c_n$ increases and is bounded above by the finite regular part of $G_R$ at $p$. Consequently

On each fixed compact subdomain, the $f_n$ are eventually bounded holomorphic functions. Montel theorem and a diagonal subsequence therefore give a locally uniform convergence limit $F:R\to \mathbb{D}$ with $F(p)=0$ and $F^{\prime }(p)=d>0$. The maximum modulus principle keeps its values strictly inside the disk. It is injective: for fixed $q$, the functions $f_n-f_n(q)$ have no zeros away from $q$; Hurwitz's theorem implies that their nonconstant limit has no other zeros. This is the locally uniform limit of univalent functions argument.

There is also an extremality property. For any holomorphic function $g:R\to \mathbb{D}$ with $g(p)=0$, the function $g\circ f_n^{-1}$ is a disk self-map fixing zero. The Schwarz lemma gives $|g^{\prime }(p)|\le f_n^{\prime }(p)$. Passing to the limit yields $|g^{\prime }(p)|\le d=F^{\prime }(p)$.

Finally suppose $F$ omits $a\in \mathbb{D}$. Necessarily $a\ne 0$. Apply the disk Möbius transformation ${\varphi }_a(w)=(w-a)/(1-\overline{a}w)$. The nowhere-zero function ${\varphi }_a\circ F$ has a global holomorphic logarithm because $R$ is simply connected, hence a holomorphic square root $h$. It maps into the disk and $|h(p)|=\sqrt{|a|}$. Normalize it by setting $g={\varphi }_{h(p)}\circ h$, so $g(p)=0$. Direct differentiation givesThe strict inequality follows from $(1-\sqrt{|a|}{)}^2>0$. This contradicts extremality. Therefore $F$ omits no point, and we conclude

Choose holomorphic coordinates centered at the point $p$ and a polydisc $U\subset {\mathbb{C}}^n$ about the origin. The local blowup of a complex manifold at a point is the incidence manifoldAway from $z=0$ the direction is forced to be $[\ell ]=[z]$, giving a biholomorphism with $U\setminus \{0\}$. Glue this punctured part to $X\setminus \{p\}$. The charts below make the result a complex manifold. The projection is proper near $p$, since it is the restriction of projection from a closed subset of $U\times {\mathbb{CP}}^{n-1}$; globally the modification is proper and is an isomorphism off $p$. Its exceptional divisor is

On ${\ell }_i\ne 0$, let $t=z_i$ and $u_j={\ell }_j/{\ell }_i$ for $j\ne i$. ThenThese are coordinates on the open domain where $t$ and all $t{u}_j$ lie in the original polydisc; at $t=0$ all finite $u_j$ are allowed. Their inverse reads the direction ratios and $z_i$ from the incidence point. In this chart $E$ is the nonsingular hypersurface $t=0$. On overlap with chart $k\ne i$, the holomorphic coordinate changes areThey are holomorphic with holomorphic inverses where $u_k\ne 0$, and the $n$ charts cover every point of $E$. Intrinsically, changing the original coordinates by $F$ lifts via $[\ell ]\mapsto [A(z)\ell ]$, where $F(z)=A(z)z$, $A(0)=d{F}_0$, and $A$ is holomorphic and invertible near zero. One can take $A(z)={\int }_0^{1}d{F}_{sz}ds$ on a sufficiently small polydisc. Away from the origin the lift is uniquely forced; this proves coordinate-independence of the gluing near $E$.

The canonical bundle is $K_X={\Lambda }^n{\Omega }_X^1$. In the $i$th blowup chart the top-form Jacobian iswhere the $u_j$ occur in increasing order. Every term with a second $dt$ vanishes in the wedge product of differential forms. Thus the differential gives a map ${\sigma }^{\ast }{K}_X\to K_{\tilde{X}}$ vanishing to order $n-1$ along $E$ and nowhere else. To express this through the supplied meromorphic-section hypothesis, take a nonzero meromorphic section $s$ of $K_X$ and put $D=\mathrm{div}(s)$. Locally $s=f(z)d{z}_1\wedge \cdots \wedge d{z}_n$, soHere ${\sigma }^{\ast }D$ is the total divisor pullback, including any additional exceptional multiplicity from $f$; it is not merely the strict transform. A line bundle with a nonzero meromorphic section is isomorphic to the line bundle of that section's divisor, as its local coefficient ratios are its frame transitions. Pullback of divisor line bundles and additivity of analytic divisor line bundles therefore give the canonical bundle formula for a point blowup:For $n=1$ the blowup is already an isomorphism, consistently with the zero exponent.

For the projective-plane construction, an invertible linear change of homogeneous coordinates puts $x=[1:0:0]$. Lines through $x$ are parametrized by $[a:b]\in {\mathbb{CP}}^1$, with equation $b{Z}_1-a{Z}_2=0$. Thus the required map on the punctured plane isIt is well-defined under rescaling and holomorphic on the covering sets $Z_1\ne 0$ and $Z_2\ne 0$, where its target coordinates are $Z_2/Z_1$ and $Z_1/Z_2$. These sets cover the complement of $x$.

The graph closure isOver the complement of $x$ it is the graph of $f$. In the affine chart $Z_0\ne 0$, its equation is exactly the two-dimensional local incidence model for the blowup of a complex manifold at a point, so its first projection identifies it with the required blowup globally. The second projection is a holomorphic map $\tilde{f}:S\to {\mathbb{CP}}^1$ extending $f\circ \sigma$. Explicitly, the two blowup charts giveThese formulas agree when $v=1/u$ and remain holomorphic at $t=0$. Hence the pencil extends over the exceptional divisor, where it records the projective tangent direction. This is the projective pencil resolved by a point blowup.

A Riemann surface is a connected one-complex-dimensional complex manifold: it has local complex coordinates with holomorphic transition maps. These coordinates induce an orientation and a conformal structure. Fix a closed oriented topological surface $S_g$ of genus $g$. A point of Teichmüller space is a marked Riemann surface $(X,f)$, where $f:S_g\to X$ is an orientation-preserving marking. Two pairs are equivalent if there is a biholomorphism $h:X\to X^{\prime }$ with $h\circ f$ isotopic to $f^{\prime }$. Thus the marking remembers how curves on the reference surface sit in $X$; forgetting it gives the moduli space of Riemann surfaces, obtained by quotienting by the mapping class group.

For $g\ge 2$, the uniformization theorem supplies a unique curvature-$-1$ hyperbolic metric in each conformal class. A pants decomposition has $3g-3$ cuffs, and their positive lengths together with their real twists give Fenchel–Nielsen coordinates. HenceFor completeness, ${\mathcal{T}}_0$ is a point, whereas ${\mathcal{T}}_1$ is the upper half-plane, parametrizing marked complex tori $\mathbb{C}/(\mathbb{Z}+\tau \mathbb{Z})$ with $\mathrm{Im}\tau >0$.

The Wolpert generic spectral rigidity theorem says that, for fixed $g\ge 2$, there is a closed proper real-analytic exceptional subset of ${\mathcal{T}}_g$ such that a hyperbolic surface outside that subset is determined up to isometry by its unmarked length spectrum, equivalently its Laplacian spectrum. This is an unoriented isometry statement: reversing orientation cannot be detected by either spectrum.

The Huber spectral equivalence theorem says that two closed curvature-$-1$ hyperbolic surfaces have the same Laplacian spectrum, with multiplicities, if and only if they have the same unmarked length spectrum, with multiplicities. One may consistently count unoriented closed geodesics including their iterates; alternatively, the primitive multiset determines the multiset with iterates and conversely. This equivalence comes from the Selberg trace formula.

The Buser finite length spectrum theorem is the finite-determination ingredient: for each $g\ge 2$ and $\epsilon >0$ there is a finite number $L(g,\epsilon )$ such that, for any two closed hyperbolic surfaces of genus $g$ with hyperbolic systole at least $\epsilon$, equality of their length multisets through $L(g,\epsilon )$ implies equality of their entire length spectra. The finite-length statement is used in Buser's chapter on Wolpert's theorem. It determines the full spectrum, not necessarily the isometry class.

Here is a proof of the Buser finite length spectrum theorem through finite spectral matching by polynomial invariants. We use the following subsidiary results, stating the required content. First, Mumford's compactness theorem makes the part of the genus-$g$ moduli space of Riemann surfaces with hyperbolic systole at least $\epsilon$ compact. Second, marked hyperbolic holonomy representations of a closed oriented surface lift from ${\mathrm{PSL}}_2(\mathbb{R})$ to ${\mathrm{SL}}_2(\mathbb{R})$, and normalized representatives can be chosen continuously on sufficiently small parameter charts. Their hyperbolic translation length satisfiesThird, the length spectrum of a compact hyperbolic surface is locally finite, and metrics on compact marked parameter charts are uniformly bilipschitz equivalent. Finally, the Hilbert basis theorem implies that a descending sequence of algebraic sets in a finite-dimensional affine space stabilizes. These are compactness, holonomy, elementary geometric finiteness and algebraic finiteness results; none is the finite unmarked-spectrum assertion we are proving.

By compactness, finitely many relatively compact marked charts cover the thick moduli space of Riemann surfaces. Choose normalized lifted hyperbolic holonomy representations on these charts and take their compact closures. Their finite union $K$ is a compact set of determinant-one matrix tuples representing every surface in the thick part, possibly more than once. Each tuple records the images of the $2g$ generators of the fundamental group. Thus $K$ sits in a fixed finite-dimensional real matrix space. Every group word is a polynomial in these matrix entries: multiplication is polynomial, and for a determinant-one two-by-two matrix its inverse is its adjugate, also polynomial. Consequently each $q_{\gamma }$ is a polynomial on the ambient space. Squared traces remove the sign ambiguity of the lift, and $4{\cosh }^2(\ell /2)$ is strictly increasing for positive $\ell$.

Enumerate the unoriented nontrivial free homotopy classes as ${\gamma }_1,{\gamma }_2,\dots$, including the iterates, and write ${\ell }_j(x)$ for their geodesic lengths at $x\in K$. The uniform finiteness of short geodesic classes follows here directly: on each of the finitely many compact marked charts, choose a fixed reference metric $h$ and a uniform bilipschitz constant $C$. A class with ${\ell }_j(x)\le R$ somewhere on that chart has reference length at most $CR$. Only finitely many classes have that reference length, and a finite union of these finite collections is finite. In particular, for every $j$ the candidate setis finite. If two parameters in $K$ have equal lengths for classes $j,k$, then $k\in F_j$. These candidate sets depend on the fixed compact set $K$ and the source label $j$, not on the number of labels currently being matched.

For every $N$, define $E_N$ in the full ambient space of pairs of matrix tuples $(x,y)$ as follows. Require an injective choice $\sigma :\{1,\dots ,N\}\to \mathbb{N}$ with $\sigma (j)\in F_j$ andand independently require an injective choice $\tau :\{1,\dots ,N\}\to \mathbb{N}$ with $\tau (j)\in F_j$ andThere are finitely many possible pairs of choices $(\sigma ,\tau )$ because every $F_j$ is finite. For each choice these are finitely many polynomial equations. Therefore $E_N$ is a finite union of algebraic sets, hence an algebraic set. Restricting an injective matching to its first $N$ entries provesThe Hilbert basis theorem now supplies $N_0$ such that $E_N=E_{N_0}$ for all $N\ge N_0$. This step takes place in a fixed affine space and uses polynomial equations, so it gives stabilization of the actual sets, not just stabilization of germs at individual parameters.

For $(x,y)\in E_{N_0}\cap (K\times K)$, stabilization gives injective partial length matchings in both directions for every $N$. Fix any length $a>0$. Only finitely many labels have length $a$ at $x$ or at $y$, by local finiteness. Take $N$ large enough to contain every such label at $x$. Its injective matching into the labels at $y$ preserves $q_j$, and hence preserves length, so the multiplicity of $a$ at $x$ is at most its multiplicity at $y$. The reverse matching gives the opposite inequality. Thus the two multiplicities are equal for every $a$, proving equality of the entire unmarked length spectra. No assumed global matching or unjustified exchange of infinitely many neighborhoods is needed.

Finally choose a numberSuppose the two length multisets agree through $L$. For each of the first $N_0$ labels at $x$, select a distinct equal-length label at $y$, which is possible precisely because the truncated multisets agree with multiplicity. Each selected label belongs to the corresponding $F_j$. Do the same in the reverse direction. Then $(x,y)\in E_{N_0}$, and the previous paragraph proves equality of the full length spectra. Since $K$ was selected using only the fixed thick genus-$g$ parameter space, the resulting finite $L$ depends only on $g,\epsilon$. This proveswith each intersection interpreted as a multiset. By the Huber spectral equivalence theorem, the corresponding entire Laplacian spectra also agree. This is the finite matching reduction needed before the analytic and geometric rigidity steps in the Wolpert generic spectral rigidity theorem.

The Schwarz-Pick lemma says that every holomorphic self-map $F$ of the unit disc contracts pseudohyperbolic distance and therefore hyperbolic distance:Its infinitesimal form isEquality for distinct points in the distance inequality, or at one point in the derivative inequality, holds exactly when $F$ is a disc Möbius transformation.

For completeness, first prove the Schwarz lemma. If $H:\mathbb{D}\to \mathbb{D}$ is holomorphic and $H(0)=0$, the quotient $H(z)/z$ extends holomorphically at zero. On $|z|=r<1$ its modulus is at most $1/r$. The maximum modulus principle, followed by $r\uparrow 1$, gives $|H(z)|\le |z|$ and $|H^{\prime }(0)|\le 1$. Equality at a nonzero point or in the derivative forces that quotient to be a constant of modulus one, again by the maximum modulus principle; hence $H(z)=e^{i\theta }z$.

Now let ${\varphi }_a(z)=(z-a)/(1-\overline{a}z)$ and apply this result to $H={\varphi }_{F(w)}\circ F\circ {\varphi }_w^{-1}$, which fixes zero. The modulus inequality is exactly the displayed distance inequality, and $\rho =\log ((1+\delta )/(1-\delta ))$ is increasing in $\delta$, giving hyperbolic contraction. Also $|{\varphi }_a^{\prime }(a)|=(1-|a{|}^2{)}^{-1}$ and $|({\varphi }_w^{-1}{)}^{\prime }(0)|=1-|w{|}^2$, so the chain rule proves the derivative inequality. The equality statement follows from the rotation case of the Schwarz lemma after undoing the two automorphisms. Conversely a disc automorphism preserves the distance and attains equality.

There is a maximum of the derivative modulus in the given annulus family, which is nonempty because it contains the constant function one. Indeed the functions are bounded by $e$, and Cauchy estimates bound their derivatives at zero. Take a sequence whose derivative moduli approach the finite supremum. By Montel theorem it has a subsequence converging locally uniformly to a holomorphic function $f$, with $f(0)=1$ and $e^{-1}\le |f|\le e$. The upper equality cannot occur at an interior point, because the maximum modulus principle would force a constant of modulus $e$, contrary to $f(0)=1$. The lower equality is excluded by applying the same argument to $1/f$, which is holomorphic because $|f|\ge e^{-1}$. Thus the limit still maps into the open annulus. Locally uniform convergence gives convergence of derivatives by the Cauchy integral formula, so this limit attains the supremum.

To compute it and all extremizers, the nonvanishing function $f$ on the simply connected disc has a holomorphic logarithm $h$ with $h(0)=0$. One can construct it by integrating $f^{\prime }/f$ from zero; differentiation shows $e^h/f$ is constant, hence one. Its real part obeys $-1<\mathrm{Re}h<1$. The mapis a biholomorphism from this vertical strip onto the disc. To check the inverse explicitly, for $|z|<1$ the ratio $(1+iz)/(1-iz)$ has positive real part, so its logarithm with value zero at zero has imaginary part in $(-\pi /2,\pi /2)$. Thereforehas real part between minus one and one, and direct substitution gives $T(s(z))=z$. The strip restriction removes the period ambiguity of the tangent, so these are inverse maps.

Applying the Schwarz lemma to $T\circ h$ gives $(\pi /4)|h^{\prime }(0)|\le 1$. Since $f^{\prime }(0)=h^{\prime }(0)$, the exact maximum isIt is attained by $f(z)=\exp ((4/\pi )\arctan z)$. The equality case requires $T(h(z))=e^{i\theta }z$, and hence all maximizing functions areTheir derivatives are $(4/\pi )e^{i\theta }$, so the extremizer is not unique: there is exactly this circle of distinct functions, with the parameter considered modulo $2\pi$. This is the extremal derivative of a disc map into a symmetric annulus for $L=1$.

A rational function is a quotient $r=P/Q$ of polynomials with $Q\ne 0$. Cancel common factors. It determines a map on the Riemann sphere by assigning infinity at poles and assigning at infinity the limit in the sphere. Equivalently homogenize $P,Q$ to the same degree $d=\max (\mathrm{deg}P,\mathrm{deg}Q)$ and set $f([Z:W])=[P_h(Z,W):Q_h(Z,W)]$. They have no common projective zero: finite ones would contradict coprimality, and at infinity at least one has a nonzero leading coefficient. This also proves the map is holomorphic.

The degree of the nonconstant rational function is $d$. The degree of a holomorphic map of compact Riemann surfaces is the number of preimages of any target point, counted with local mapping multiplicity. For target $[a:b]$, the homogeneous equation $b{P}_h-a{Q}_h=0$ is a nonzero homogeneous polynomial of degree $d$. Over $\mathbb{C}$ it factors into $d$ linear factors, with multiplicity, giving exactly $d$ projective zeros. Locally its zero order is precisely the map's multiplicity, since the other target-coordinate denominator is nonzero. The two degrees are equal.

There is a Möbius transformation sending any ordered triple of distinct points to $0,1,\infty$. When all are finite one can use $T_A(z)=(z-a_1)(a_2-a_3)/[(z-a_3)(a_2-a_1)]$; the formulas for infinity are its limits. Define $T_B$ in the same way. Then $T_B^{-1}\circ T_A$ sends $A$ to $B$ and restricts to a biholomorphism $X\setminus A\to X\setminus B$.

Two function elements $(U,f)$ and $(V,g)$ define the same germ of a holomorphic function at $p$ when $p\in U\cap V$ and they agree on some neighborhood of $p$. Write the germ as $[f{]}_p$. Basic open sets of the space of germs of holomorphic functions arefor open $U$ on which $f$ is holomorphic. If two such sets meet, equality of the germs gives a smaller open sheet contained in their intersection, so these sets form a basis. The projection $\pi ([f{]}_p)=p$ maps each sheet homeomorphically onto $U$. Compose this with a local coordinate on $X$ to define the complex charts on the germ space. Transition maps are the corresponding transition maps on $X$, hence holomorphic; $\pi$ is a local biholomorphism and is onto, since even constant functions provide germs over every point.

The germ space is Hausdorff. Germs over different points are separated by base neighborhoods. For distinct germs over the same point, choose representatives on one small connected disk. If their two sheets met anywhere in that disk, the representatives would agree on an open set and hence everywhere on the disk by the identity theorem, a contradiction. These sheets thus separate the germs. Together with the allowed second-countability fact, the charts make each component a Riemann surface.

Here “covering in the sense of complex analysis” means this locally biholomorphic, possibly incomplete analytic germ projection. It does not mean an evenly covered topological covering map: an analytic branch can cease to exist at a singularity, so the restriction to a component need not project onto all of $X$.

The evaluation map $E([f{]}_p)=f(p)$ is holomorphic, since on the sheet ${\tilde{U}}_f$ it is just $f$. Starting from one germ, continuation through overlapping disks produces exactly the germs in its connected component. Indeed such continuations give paths in the germ space; conversely, cover a path by finitely many successive sheets to obtain a chain of function elements. Components are path-connected because the surface is locally path-connected. No two distinct germs can be added to that component without being obtainable by continuation. It therefore represents the maximal analytic continuation, or complete holomorphic function, of any of its germs. Conversely every such maximal continuation embeds into the germ space by taking its local germs; completeness makes its image the entire component. These constructions are inverse, giving the requested bijection between complete holomorphic functions and connected components, with functions understood up to their natural equivalence as analytic continuations, rather than merely as single-valued functions defined on all of $X$.

The Riemann mapping theorem asserts that a nonempty simply connected domain $\Omega ⊊\mathbb{C}$ is biholomorphic to the unit disk $\mathbb{D}$. More precisely, for each $z_0\in \Omega$ there is a unique biholomorphism $f:\Omega \to \mathbb{D}$ satisfying $f(z_0)=0$ and $f^{\prime }(z_0)>0$. We prove existence through an extremal derivative, deriving the compactness and injectivity steps rather than assuming special lemmas for the theorem.

First construct a bounded univalent function on $\Omega$. Choose $a\notin \Omega$ and a holomorphic logarithm of $z-a$, which exists by the permitted logarithm result on the translated simply connected domain. PutThen $u(z{)}^2=z-a$, so $u$ is univalent and never zero. Its image $G=u(\Omega )$ is open, and $G\cap (-G)=\varnothing$: equality $u(x)=-u(y)$ would imply $x=y$ after squaring and then $u(x)=0$.

Fix $w_0\in G$ and choose $r>0$ with $B(w_0,r)\subset G$. It follows that $B(-w_0,r)\subset -G$ is disjoint from $G$, so $|u(z)+w_0|\ge r$ throughout $\Omega$. Henceis a univalent function with $|h|\le 1/2$. For $b\in \mathbb{D}$, defineThe identityand the inverse $T_b^{-1}(w)=(w+b)/(1+\overline{b}w)$ show directly that $T_b$ is an automorphism of the unit disk. Apply $T_{h(z_0)}$ to $h$ and then a rotation to make its nonzero derivative at $z_0$ positive. The family $\mathcal{A}$ of univalent maps $\Omega \to \mathbb{D}$ with value zero and positive derivative at $z_0$ is therefore nonempty.

LetIt is positive. Choose $\rho >0$ with $\overline{B(z_0,\rho )}\subset \Omega$. The Cauchy estimate gives $v^{\prime }(z_0)\le 1/\rho$ for every $v\in \mathcal{A}$, so $M$ is finite. Take a sequence $v_j\in \mathcal{A}$ with $v_j^{\prime }(z_0)\to M$.

We now justify a locally uniform convergence subsequence. On any compact subset of $\Omega$, the bound $|v_j|\le 1$ and the Cauchy estimate on slightly larger interior disks give a uniform derivative bound. They therefore give equicontinuity. Apply the Arzelà-Ascoli theorem on a compact exhaustion of $\Omega$, then take a diagonal subsequence for locally uniform convergence. Its limit $v$ is holomorphic: pass to the limit in the Cauchy integral formula on every small interior circle. The same formula for derivatives givesThus $v$ is nonconstant. Initially $|v|\le 1$, and the maximum modulus principle improves this to $|v|<1$ everywhere.

For completeness, the needed locally uniform limit of univalent functions argument is short. If $x\ne y$ and $v(x)=v(y)$, the nonconstant holomorphic function $v-v(y)$ has an isolated zero at $x$. Choose a small closed disk around $x$ excluding $y$, with no zero on its boundary. Uniform convergence there gives $v_j-v_j(y)\to v-v(y)$. The Rouche theorem then makes $v_j-v_j(y)$ have a zero inside that disk for all sufficiently large $j$. Its only possible zero, by injectivity of $v_j$, is $y$, outside the disk. This contradiction proves injectivity of $v$. Therefore $v\in \mathcal{A}$ and attains the supremum $M$.

It remains to prove surjectivity. Suppose $c\in \mathbb{D}$ is omitted by $v$; since $v(z_0)=0$, necessarily $c\ne 0$. The domain $T_c(v(\Omega ))$ is simply connected, because both constituent maps are homeomorphisms onto their images, and it excludes zero. The permitted holomorphic logarithm gives a holomorphic square rootThis maps $\Omega$ into $\mathbb{D}$ and is univalent, since equality of its values implies equality after squaring and hence equality of the original arguments. Write $\gamma =g(z_0)$. Then ${\gamma }^2=T_c(0)=-c$ and $|\gamma |=\sqrt{|c|}$. Normalize again with $V=T_{\gamma }\circ g$ and a rotation. Its derivative magnitude isThe strict inequality follows from $(1-\sqrt{|c|}{)}^2>0$. This square-root improvement of a normalized conformal map contradicts extremality. Thus $v(\Omega )=\mathbb{D}$. An injective holomorphic map has nonzero derivative and a holomorphic local inverse; these inverses agree, so the resulting bijection is biholomorphic. We have proved existence.

For uniqueness, two normalized maps give an automorphism of the unit disk $v_2\circ v_1^{-1}$ fixing zero. Apply the Schwarz lemma, proved in the next part, to this automorphism and its inverse. Both inequalities force equality, so the equality case makes it a rotation. Its derivative at zero is the positive real number $v_2^{\prime }(z_0)/v_1^{\prime }(z_0)$, forcing that rotation to be the identity. ThereforeThe proper-domain hypothesis is necessary: a bounded entire holomorphic function has derivative at most $1/R$ on disks of arbitrarily large radius by the Cauchy estimate, and hence is constant. Conversely, a domain biholomorphic to the disk is simply connected because the map is a homeomorphism.