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Write $g_2=60G_4$ and $g_3=140G_6$, with the lattice sums taken over nonzero lattice elements. Reindexing the normally convergent series shows that the Weierstrass elliptic function $\wp$ is an even function, and consequently ${\wp }^{\prime }$ is an odd function. Expanding $(z-w{)}^{-2}$ near zero and cancelling the odd lattice sums gives the Laurent coefficients of the Weierstrass elliptic functionThus the elliptic function ${\wp }^{\prime 2}-4{\wp }^3+60G_4\wp$ has no pole at zero and its constant term is $-140G_6$. By periodicity it has no poles anywhere. An entire elliptic function is bounded on a fundamental parallelogram and hence everywhere, so Liouville theorem makes it constant. This proves the Weierstrass elliptic differential equation

We first verify that the cubic is smooth, rather than presupposing it when using its group law. There are three nonzero two-torsion points of a complex torus $h_j\in (\frac{1}{2}\Lambda )/\Lambda$. Oddness and periodicity imply ${\wp }^{\prime }(h_j)=0$. Since ${\wp }^{\prime }$ has a triple pole and no other poles on the torus, value multiplicity of an elliptic function says that these three distinct zeros exhaust its zeros and are simple. Put $e_j=\wp (h_j)$. Each $\wp -e_j$ has a double zero at $h_j$, since ${\wp }^{\prime }(h_j)=0$ and the zero of ${\wp }^{\prime }$ is simple. If $e_i=e_j$ with $i\ne j$, this same function would have at least four zeros counted with multiplicity, although its only pole is double. Therefore the half-period values of the Weierstrass elliptic function are distinct. They are roots of $4x^3-g_{2}x-g_3$, so all its roots are distinct and the affine curve is a smooth algebraic curve. Its projective closure isIt has the single point $O=(0:1:0)$ at infinity. The derivative with respect to $Z$ of the defining homogeneous polynomial is nonzero at $O$, proving smoothness there too.

For any finite value $a$, value multiplicity of an elliptic function gives exactly two solutions of $\wp (z)=a$, counting multiplicities. Evenness supplies the pair $z,-z$. At a nonzero half-period these coincide and the zero is double; elsewhere they are distinct. An even elliptic function $f$ therefore descends to a meromorphic function of $x=\wp (z)$ on the Riemann sphere. To justify descent at a half-period, put $z=h+u$: both $f(h+u)$ and $\wp (h+u)$ are even in $u$, and $\wp (h+u)-\wp (h)$ is $u^2$ times a nonvanishing even analytic function. The local Laurent series of $f$ is consequently meromorphic in the coordinate $x-e$. At zero the same argument uses $1/x=z^2+O(z^6)$. Every meromorphic function on the Riemann sphere is a rational function, proving that even elliptic functions are rational in the Weierstrass function.

For an arbitrary elliptic function $f$, its even part $(f(z)+f(-z))/2$ is $A(\wp )$. Its odd part divided by ${\wp }^{\prime }$ is an even meromorphic function, hence is $B(\wp )$ with $B$ rational. We have therefore proved the elliptic function-field decompositionUniqueness follows by taking even and odd parts. In particular the field of elliptic functions is $\mathbb{C}(\wp ,{\wp }^{\prime })$.

Define $\Phi (0)=O$ and $\Phi (z)=(\wp (z),{\wp }^{\prime }(z))$ elsewhere on $\mathbb{C}/\Lambda$. Periodicity makes the map well defined. Near zero, the projective chart $Y\ne 0$ has coordinates $X/Y=\wp /{\wp }^{\prime }=-z/2+O(z^5)$ and $Z/Y=1/{\wp }^{\prime }=-z^3/2+O(z^7)$, so the extension is holomorphic. The fiber description above proves injectivity: the two candidates $z,-z$ are distinguished by ${\wp }^{\prime }$ unless they coincide at a half-period. It also proves surjectivity, since every $x$ occurs and the two derivatives are the two allowed $y$ values, with $y=0$ at a branch value.

Finally, a nonvertical line $y=ax+b$ pulls back to the elliptic function ${\wp }^{\prime }-a\wp -b$, whose only pole is triple at zero. Its three zeros $z_1,z_2,z_3$, counted with intersection multiplicity, satisfy $z_1+z_2+z_3=0$ modulo $\Lambda$ by the zero-pole sum of an elliptic function. On the cubic, reflection $y\mapsto -y$ corresponds to $z\mapsto -z$. The chord-and-tangent group law therefore gives $\Phi (z_1)+\Phi (z_2)=\Phi (-z_3)=\Phi (z_1+z_2)$. Vertical lines give inverse pairs; tangencies are included through repeated zeros, and the identity cases follow from the extension at zero. This completes the Weierstrass uniformization by half-period values: $\Phi$ is a group isomorphism.

For $\mathrm{Re}s>1$, define the Riemann zeta function by the absolutely convergent Dirichlet seriesThe Euler product follows from unique prime factorization and absolute convergence. The counting-function integral givesBecause the fractional part is bounded, the last integral, and its derivatives with respect to $s$ on compact subsets, converge uniformly for $\mathrm{Re}s>0$. It is a holomorphic function there. This proves the Meromorphic continuation of the Riemann zeta function to the right half-plane, with a simple pole at one of residue one and no other singularity in that half-plane.

The Gamma function isThe Gamma function recurrence extends it meromorphically; its poles are at the nonpositive integers and it has no zeros. The functional equation of the Riemann zeta function can be stated asor, equivalently,These are identities of meromorphic functions; the second also extends the Riemann zeta function to the remaining half-plane. The critical strip is $0<\mathrm{Re}s<1$; its boundary lines will also be treated below, rather than included among possible exceptional zeros.

For $\mathrm{Re}s>1$, the absolutely convergent reciprocal Euler product gives $\zeta (s)\ne 0$. For $\mathrm{Re}s<0$, the factors $2^s$, ${\pi }^{s-1}$, $\Gamma (1-s)$ and $\zeta (1-s)$ in the second functional equation of the Riemann zeta function are finite and nonzero. Hence the only zeros there are the zeros of the sine factor:Each is simple. At zero, the sine zero cancels the pole of $\zeta (1-s)$; using its residue one gives $\zeta (0)=-1/2$, not zero. Nonvanishing on the rest of $\mathrm{Re}s=0$ follows from the boundary-line proof below and the functional equation of the Riemann zeta function. Together these facts prove that the trivial zeros of the Riemann zeta function are the only zeros outside the open critical strip.

Here is a Jensen disk proof of the zeta zero-count bound which avoids any unproved left-half-plane growth estimate. Set $F(s)=(s-1)\zeta (s)$, a holomorphic function throughout $\mathrm{Re}s>0$, including at one. The integral continuation formula givesFor each integer $j$, apply Jensen's formula in the disk with centre $2+ij$, outer radius $R=7/4$ and inner radius $r=8/5$. The outer disk stays in $\mathrm{Re}s\ge 1/4$, so its maximum modulus is $O((|j|+2{)}^2)$. At its centre,since $|1/\zeta (2+ij)|\le {\sum }_n|\mu (n)|n^{-2}\le \zeta (2)$. Therefore the number of zeros in its inner disk, counted with multiplicity, is at mostThe rectangle $1/2\le \mathrm{Re}s\le 1$, $|\mathrm{Im}s-j|\le 1/2$ fits in the inner disk because its farthest point has distance $\sqrt{(3/2{)}^2+(1/2{)}^2}=\sqrt{5/2}<8/5$. Summing over $O(T)$ such disks counts $O(T\log T)$ zeros in the right half of the critical strip up to height $T$. The first functional equation of the Riemann zeta function bijects zeros, with multiplicities, in the left half with their reflections $s\mapsto 1-s$ in the right half; its Gamma function factors are finite and nonzero in the strip. ThusThe pole of zeta function at one does not count as a zero: $F(1)=1$.

Finally, for real $\sigma >1$ and $t\ne 0$, the Euler-product logarithms and the nonnegative trigonometric polynomialgive the three-four-one product proof of zeta boundary nonvanishing:If $\zeta (1+it)$ had a zero of order $a\ge 1$, its factor would be $O((\sigma -1{)}^{4a})$ as $\sigma \downarrow 1$. The real zeta function factor has pole order three, and the factor at $1+2it$ stays bounded because $t\ne 0$. The product would tend to zero as $O((\sigma -1{)}^{4a-3})$, contradicting that it is at least one. ThereforeAt $t=0$ there is a pole, not a zero. Applying the sine-form functional equation of the Riemann zeta function at $s=it\ne 0$ now proves nonvanishing on the remaining imaginary axis, completing the earlier assertion about all zeros outside the critical strip.

The weights of $E_4^3$ and $\Delta$ agree, so their ratio is invariant under the modular group. Nonvanishing of the modular discriminant on the half-plane makes the ratio holomorphic there. At an elliptic point, an invariant holomorphic function has a power series in the quotient coordinate, so the ratio descends holomorphically. At the unique modular cusp its expansion isThus it is a meromorphic function on the compact Riemann surface with precisely one pole, of order one. The degree of a nonconstant meromorphic map to the sphere equals its total pole order. It therefore has degree one, hence is a biholomorphism:This is the single-pole criterion for a spherical coordinate.

The matrix $\left(\begin{array}{cc}0& -1\\ 1& 1\end{array}\right)$ fixes $\omega$. Weight-four transformation gives $E_4(\omega )=(\omega +1{)}^4{E}_4(\omega )$, and $(\omega +1{)}^4\ne 1$. Hence $E_4(\omega )=0$ and $j(\omega )=0$. Since $j$ is a degree-one coordinate, this is a simple zero on the quotient surface. Pulled back to the half-plane it has order three, by the analytic ramification index in part (a).

View $j$ as a meromorphic function on $X({\Gamma }_0(2))$, using its coordinate $t=j_2$. At the infinity modular cusp its pole order is one. At the zero modular cusp, $j(S\tau )=j(\tau )=q^{-1}+O(1)=q_0^{-2}+O(1)$, so its pole order is two. Part (c) identifies these poles with $t=\infty$ and $t=0$.

The zero of $j$ on the full modular curve lies at its order-three elliptic point. There are no effective order-three stabilizers in ${\Gamma }_0(2)$: their lifts have trace $\pm 1$, whose characteristic polynomial modulo two is $X^2+X+1$, whereas an upper triangular matrix over ${\mathbb{F}}_2$ has both diagonal entries one. Therefore the degree-three covering has exactly one point over that elliptic point, with analytic ramification index three. Its $t$-coordinate is $a=j_2(\omega )$, and the zero divisor of the pulled-back $j$ is $3[a]$.

A rational function with these zeros and poles must be $C(t-a{)}^3/t^2$. At the infinity modular cusp both $j$ and $t$ have leading coefficient one times $q^{-1}$, giving $C=1$. HenceThe proof fixes the coefficient and the powers using modular cusp cusp widths and elliptic ramification, rather than assuming a formula for the level-two coordinate.

The effective group is torsion free and every modular cusp is regular in the preceding sense, so orders of a meromorphic weight-$k$ form are integers. At interior points use a local automorphy trivialization; at a modular cusp use the Fourier order of the appropriate slash transform in its cusp width coordinate. Let $C$ be the sum of all modular cusp points, each once, and setA meromorphic function $\phi$ belongs to the Riemann-Roch space $\mathcal{L}(D(f))$ exactly when $\mathrm{div}\phi +D(f)\ge 0$. In the interior this requires $f\phi$ to have no pole; at a modular cusp it requires order at least one. Conversely, the quotient of any weight-$k$ cusp form by $f$ is a meromorphic weight-zero function satisfying precisely those inequalities. This proves the cusp-form divisor presentationTo compute the degree without imposing a valence formula as an extra assumption, use the meromorphic tensor differential $f^{12}(d\tau {)}^{6k}$. Its automorphy factors cancel. Its order at an interior point is $12{\mathrm{ord}}_{P}f$; at a modular cusp it is $12{\mathrm{ord}}_{P}f-6k$, since $d\tau$ is a nonzero constant times $d{q}_c/q_c$. A meromorphic section of the $6k$th tensor power of the canonical bundle has total divisor degree $6k(2g-2)$. The regular-cusp valence formula on a torsion-free modular curve is thereforeFor $k\ge 3$, $\mathrm{deg}D(f)-(2g-2)=(k-2)d/12>0$. The Riemann-Roch theorem says $\ell (D)-\ell (K-D)=\mathrm{deg}D+1-g$, and a divisor of negative degree has no nonzero sections. Thus $\ell (K-D)=0$ andUsing $d=(p^2-1)/2$ and $r_{\infty }=p-1$ givesThe canonical-degree and Riemann-Roch facts used here are general results for compact Riemann surfaces, as permitted.

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.

Let $\mathcal{P}={\mathcal{M}}_X/{\mathcal{O}}_X$, the sheaf of meromorphic principal parts. Exactness of taking stalks givesChoose a holomorphic coordinate $t$ vanishing at $x$. Every meromorphic germ has a finite negative tail in its Laurent series; the holomorphic tail vanishes in the quotient. HenceThis coordinate description is a vector-space identification; the intrinsic space is the quotient of germs, so a coordinate change need not preserve the displayed basis.

For each point inclusion $i_x:\{x\}\to X$, a principal part defines a local meromorphic germ near $x$, whose quotient class is zero off $x$. Gluing with zero on the complement gives a map from its skyscraper sheaf to $\mathcal{P}$. These maps induce an isomorphism on every stalk and therefore an isomorphism of sheaves:The direct sum of sheaves means the sheafification of the sectionwise direct-sum presheaf. Its sections can have infinitely many nonzero components globally, but their point supports form a locally finite family of subsets. This matches the fact that poles of a meromorphic function are locally finite. On a compact Riemann surface only finitely many points can occur; on a noncompact one a discrete infinite family is allowed.

For a global principal-part family $P$, choose local meromorphic lifts $m_i$ on a sufficiently small open cover. The differences $m_j-m_i$ are holomorphic functions and form a Čech cocycle. The connecting homomorphism sends $P$ to the resulting class in $H^1(X,{\mathcal{O}}_X)$. If this class vanishes, after refining the cover write $m_j-m_i=h_j-h_i$; then the functions $m_i-h_i$ glue to a global meromorphic function. Conversely any global lift makes the class zero. Equivalently, exactness of the given long exact sequence in sheaf cohomology saysThe connecting class is the obstruction to the Mittag-Leffler problem on a Riemann surface. It rules on the specified poles and their finite negative Laurent tails, with no additional poles allowed. When a solution exists, any two solutions differ by a global holomorphic function.

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.

The residue theorem states that a meromorphic function with finitely many poles inside a positively oriented simple closed contour and none on it satisfies ${\oint }_{C}F(z)dz=2\pi i\sum \mathrm{Res}(F;a)$. The function must be holomorphic on a neighbourhood of the contour and its interior away from those poles.

Write $J={\int }_0^{\infty }{x}^{1/2}\log x/(1+x^2)dx$ and $I={\int }_0^{\infty }{x}^{1/2}/(1+x^2)dx$. Both integrals converge absolutely. Use the principal complex logarithm in the upper half-plane, with $z^{1/2}=e^{\log z/2}$, and an upper semicircle of radius $R>1$ indented above zero by a clockwise semicircle of radius $ϵ<1$. The branch values on the negative real side are upper limits; equivalently use contours arbitrarily slightly above that side before taking a limit.

For $F(z)=z^{1/2}\log z/(1+z^2)$, the large arc is $O(R^{-1/2}\log R)$ and the small arc is $O({ϵ}^{3/2}|\log ϵ|)$, so both vanish. On $z=-x$ approached from above, $z^{1/2}=i\sqrt{x}$ and $\log z=\log x+i\pi$. The oriented negative segment therefore contributes $iJ-\pi I$, and the positive segment contributes $J$. The only enclosed pole is $i$, with residueConsequentlyComparing imaginary and real parts yields the upper-half-plane contour for square-root logarithmic integrals evaluation:

Use the closed genus $g\ge 2$ version of Teichmüller's uniqueness theorem. Let $f_0:X\to Y$ be a Teichmüller map: for a nonzero holomorphic quadratic differential $q$, normalized by ${\int }_X|q|=1$, and $0<k_0<1$,The value at a zero of $q$ is irrelevant to the Beltrami coefficient, which is defined almost everywhere. For every quasiconformal map $f:X\to Y$ preserving orientation and homotopic to $f_0$, the conclusion isThe normalized $q$ is also uniquely determined when $k_0>0$.

We use the following analytic input, stated with its hypotheses. The Reich–Strebel inequality says that if $f_0$ has the displayed Beltrami coefficient and a competitor $f$ has the same target and homotopy class, then, for $\mu ={\mu }_f$,This is the standard fundamental inequality for an integrable holomorphic quadratic differential; it also holds on a finite-type punctured surface with the homotopy fixing the punctures. Its formulation is given in Gardiner and Hu, §5. We quote this analytic inequality as the lecture result used in the proof.

Put $k=\Vert \mu {\Vert }_{\infty }$. Pointwise, away from the isolated zeros of $q$,Integrating against the probability density function $|q|$ proves $K_0\le K(f)$. If $K(f)=K_0$, all these inequalities are equalities almost everywhere. The strictly increasing last function forces $|\mu |=k_0$ almost everywhere; equality in the triangle inequality then forces $\mu q/|q|=k_0$. ThereforeTwo quasiconformal maps with the same Beltrami coefficient differ by postcomposition with a biholomorphism, by the local chain rule for the Beltrami equation. Consequently $h=f\circ f_0^{-1}$ is a biholomorphism from $Y$ to itself homotopic to the identity.

For completeness, such a biholomorphism is the identity when $g\ge 2$. Apply the uniformization theorem and choose the lift of the homotopy to $\mathbb{H}$ starting at the identity. Its endpoint lift $\tilde{h}$ commutes with every deck transformation. It is a real Möbius transformation. The compact quotient has no parabolic Möbius transformations in its deck group, and freeness excludes elliptic Möbius transformations. Two distinct hyperbolic axes exist: a discrete free group preserving just one axis would be cyclic, contradicting the fundamental group of a closed surface of genus at least two. Commutation with two deck transformations represented by hyperbolic Möbius transformations having distinct axes makes it fix their boundary endpoints individually; there are at least three such endpoints. A Möbius transformation fixing three points is the identity. Hence $h=\mathrm{id}$ and $f=f_0$.

If $q_1$ is another unit-area differential for the same map, then $|q_1|/q_1=|q|/q$, so $q_1/q$ is positive real wherever defined. This meromorphic function is constant by the open mapping theorem, and normalization makes the constant one. Without normalization, positive multiples of $q$ describe the same map. At $k_0=0$, uniqueness of the map still holds in genus at least two, but there is no distinguished $q$. In genus one, translations supply nontrivial biholomorphisms homotopic to the identity, so equality determines the map only up to those biholomorphisms; the genus hypothesis cannot be omitted.