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The printed assertion for every cohomological degree is false. The valid conclusion supplied by the stated Hartogs extension theorem is the degree-zero isomorphism. Indeed, around each removed point choose a small coordinate ball on which the holomorphic vector bundle $E$ is trivial. A holomorphic section on the punctured ball has finitely many holomorphic coefficient functions, each of which extends by Hartogs extension theorem when the complex dimension is at least two. Uniqueness of holomorphic extension makes these local extensions agree with the original section on overlaps, givingThis also identifies the direct image sheaf $j_{\ast }\mathcal{O}(E{|}_{X_0})$ with $\mathcal{O}(E)$ for the inclusion $j:X_0\hookrightarrow X$. It does not identify higher sheaf cohomology.

Here is an explicit higher-degree obstruction. Take $X={\mathbb{P}}^2$, remove $[0:0:1]$, and take the trivial line bundle. Cover $X_0$ by $U=\{x\ne 0\}$ and $V=\{y\ne 0\}$. On $U$ use coordinates $t=y/x$, $s=z/x$; on $V$ use $u=x/y=1/t$, $v=z/y=s/t$. Both charts are ${\mathbb{C}}^2$, and the overlap is ${\mathbb{C}}^{\ast }\times \mathbb{C}$. For every integer $m\ge 2$, the holomorphic functiondefines a Čech cocycle. It cannot be a Čech coboundary $b(1/t,s/t)-a(t,s)$ with $a,b$ entire on their charts. Taking the coefficient of $s^m$, the term from $a$ has only nonnegative powers of $t$, whereas the term from $b$ has the form $t^{-m}{b}_m(1/t)$ and only powers at most $-m$. Neither can provide the coefficient $t^{-1}$. Uniqueness of Laurent series proves the contradiction. The same argument applied to each fibre degree proves that all the classes $[g_m]$, $m\ge 2$, are linearly independent.

These classes remain nonzero in Čech cohomology of $X_0$, rather than merely of this cover. The low-degree Mayer-Vietoris sequence for sheaf cohomology injects the quotient of overlap sections by the two chart-section groups into $H^1(X_0,\mathcal{O})$. Concretely, a partition of unity gives smooth $f_U,f_V$ with $f_V-f_U=g$; their common Dolbeault operator defines a global $(0,1)$-form. If that form were $\bar{\partial }$-exact, subtracting its global smooth primitive would make $f_U,f_V$ holomorphic and split $g$. The nonsplitting just proved therefore gives the same obstruction in Dolbeault cohomology, and the Dolbeault theorem identifies it with sheaf cohomology. Thus $H^1(X_0,\mathcal{O})$ is infinite dimensional. In contrast, $H^1({\mathbb{P}}^2,\mathcal{O})$ is finite dimensional by compact Dolbeault Hodge decomposition for the Fubini-Study form. Hence the two degree-one groups cannot be isomorphic, disproving the printed all-degree request already in complex dimension two.

In complex dimension one, even the degree-zero conclusion fails. Take $X={\mathbb{P}}^1$, remove its point at infinity and use the trivial line bundle. Holomorphic functions on compact connected ${\mathbb{P}}^1$ are constant by the maximum modulus principle, whereas $X_0\cong \mathbb{C}$ has nonconstant entire functions such as $z$. Therefore the restriction map is not onto even in degree zero.

Any nonsingular holomorphic function $f({\Phi }_1,\dots ,{\Phi }_n)$ of chiral superfields is itself a chiral superfield. The graded Leibniz rule and the chain rule giveThus sums, products, convergent power series and inverses on patches where the denominator is nonzero preserve chirality. An ordinary function involving ${\Phi }^{†}$ is generally not chiral, because $\bar{D}{\Phi }^{†}$ need not vanish. This is why a superpotential is holomorphic. Spacetime derivatives also commute with $\bar{D}$, so derivatives of chiral superfields remain chiral; chirality by itself does not impose renormalizability.

Use the Wirtinger derivative $q_z=(q_x-i{q}_y)/2$ and write $z=x+iy$. Since $q$ solves the Laplace equation, $q_z$ is a holomorphic function. The transforms in the question select solutions with sufficient decay at infinity. In this class the solution, when it exists, is unique; hence reflection of the symmetric data about $y=\ell /2$ givesFor $\gamma >0$, this uniqueness follows directly from Green's first identity: the homogeneous problem has integral $\int |\nabla q{|}^2+\gamma {\int }_{y=0,\ell }{q}^2=0$. For arbitrary real $\gamma$, the decaying-class qualifications are explained in part (c). Without a condition at infinity, symmetry of the data alone would not force symmetry of every solution.

Put $c=q(0,0)$, $f(x)=q(x,0)$, and introduce a known transformThe bottom Robin boundary condition gives $q_y=\gamma f$, while the top one gives $q_y=-\gamma f$. The bottom side is traversed from infinity to zero. Integration by parts therefore givesThe top side is traversed from $i\ell$ towards infinity, and its exponential supplies $e^{k\ell }$, givingOn the vertical side, $dz=idy$ and the prescribed Neumann boundary condition is $q_x=g$. ThusBoth the side orientations and the factor $1/2$ from the Wirtinger derivative are essential to these signs.

Two holomorphic function elements represent the same germ of a holomorphic function at $z$ if they agree on some neighbourhood of $z$. For a holomorphic $f$ on an open $D\subseteq G$, set $U(f,D)=\{[f{]}_z:z\in D\}$. These sets form a basis of a topology on the space of germs of holomorphic functions: every germ lies in one, and if a germ lies in two, equality of the representatives on a smaller neighbourhood supplies a basis of a topology set in their intersection. Declaring arbitrary unions of basis of a topology sets open therefore gives a topology.

Different base points are separated by disjoint small discs in $G$. If two different germs have the same base point, choose representatives on one connected small disc. Their sheets are disjoint: any common germ elsewhere on that disc would force equality throughout the disc by the identity theorem. This proves the Hausdorff property.

The projection $\pi ([f{]}_z)=z$ maps each sheet $U(f,D)$ homeomorphically onto $D$. Use these maps as complex charts. On overlaps their transition functions are the identity, hence holomorphic; thus each connected component is a Riemann surface, and $\pi$ is a local biholomorphism, an analytic germ projection. For completeness, second countability of a component follows by following continuation chains through a countable basis of a topology of rational discs: starting at one germ, finite chains of such discs give only countably many continuation sheets, by uniqueness on overlaps, and they cover every germ in that component.

A complete analytic function consists of all germs reachable from one initial germ by analytic continuation. The supplied path-lifting equivalence identifies these with its path component. Since the germ space is locally path connected, connected components and path components agree. Thus its continuation surface is exactly one component ${\mathcal{G}}_{\mathcal{F}}$.

A local biholomorphism need not evenly cover the whole base; the germ projection is an analytic covering in this local sense. The examples below distinguish a genuine regular covering map from a surjective nonregular projection, with the meaning of regular stated explicitly.

Analytic continuation extends a holomorphic function by holomorphic elements agreeing on overlaps; along a path, a chain of overlapping neighbourhoods transports the initial element. The identity theorem makes compatible continuation unique on connected overlaps, although different paths can lead to different branches.

Choose the principal complex logarithm and the slit domain $D=\mathbb{C}\setminus [1,\infty )$. On $D$, the functionis holomorphic, with the removable value $h(0)=1$. The domain $D$ is simply connected, so $h$ has a primitive and its integral from $0$ to $z$ is independent of the path in $D$. For $|z|<1$, the uniformly convergent power series on compact subdiscs givesTherefore $f$ is a single-valued analytic continuation of the dilogarithm to $\mathbb{C}\setminus [1,\infty )$. The apparent problem at $0$ is removable; $1$ is a genuine branch point, even though the function has a finite one-sided limit there, because its derivative has a logarithmic singularity. Other cuts from $1$ to infinity give corresponding branches.

Let $T$ be a closed triangle contained, together with a neighbourhood, in the domain of a holomorphic function $f$. We prove Cauchy theorem for a triangle without assuming continuity of $f^{\prime }$.

Join the edge midpoints to subdivide $T$ into four triangles, each with half its diameter and perimeter. Their positively oriented contour integrals sum to the integral around $T$, because the internal edges cancel. At least one subtriangle has integral of modulus at least one quarter of the original. Repeat that choice to obtain nested closed triangles $T_n$ satisfyingTheir intersection is a single point $z_{\ast }$, by compactness and shrinking diameter. complex differentiability at a point at $z_{\ast }$ givesThe first two terms have polynomial primitives, hence integrate to zero around each triangle. For sufficiently large $n$, $|r|<\epsilon$ on $T_n$, so the complex line integral estimate givesCombining the bounds gives $|{\int }_{\partial T}f(z)dz|\le \epsilon d_0{P}_0$. Since $\epsilon$ is arbitrary,The orientation can be reversed with no change to the conclusion.

Rouché's theorem. Suppose $f,g$ are holomorphic functions on a neighbourhood of the closure of a bounded domain enclosed by a simple, positively oriented, piecewise smooth curve. If $|g(z)|<|f(z)|$ on the boundary, then $f$ and $f+g$ have the same number of zeros inside, counted with multiplicity.

Apply Rouche theorem with $f(z)=z^4+3$ and $g(z)=e^{iz}$ on the first-quadrant quarter-disc of radius $R=\sqrt{2}$. On the two radial segments, $z^4$ is nonnegative real, so $|f|\ge 3$, whereas $|g|=e^{-\mathrm{Im}z}\le 1$. On the circular arc,At every arc point with positive imaginary part, $|g|<1$, giving the strict inequality; at the remaining endpoint $z=R$, $|f|=7>1=|g|$. Thus the strict boundary inequality holds everywhere, including the origin.

The four zeros of $f$ have modulus $3^{1/4}$ and arguments $\pi /4,3\pi /4,5\pi /4,7\pi /4$. Exactly one lies inside the chosen domain. Rouche theorem gives exactly one zero of $f+g$ there, counted with multiplicity, so that zero is simple. Finally any zero in the open first quadrant satisfiesThere can be no additional zero outside the quarter-disc. Hence

Non-characteristic. The first-order principal symbol is ${\nu }_t-i{\nu }_x$, which equals one on $(0,1)$.

Set $z=x+it$. The equation is equivalent to ${\partial }_{\bar{z}}\varphi =({\varphi }_x+i{\varphi }_t)/2=0$. Because $\varphi$ is continuously differentiable, the Cauchy-Riemann equations imply that it is a holomorphic function of $z$. Its convergent complex Taylor series restricts to a real power series along $t=0$, so $g$ is a real analytic function.

For smooth-data instability of the Cauchy-Riemann Cauchy problem, chooseThese are entire functions and satisfy the equation. Their initial derivatives obey ${\sup }_x|{\partial }_x^j{g}_n|\le n^j{e}^{-\sqrt{n}}$, so every finite sum tends to zero. ButThus analytic solutions can exist while continuous dependence on initial data fails for this smooth-data topology.

The Cauchy estimate follows directly from the differentiated Cauchy integral formula:The circle has length $2\pi r$, giving the bound for every integer $j\ge 0$.

Put $w=\varphi -\psi$ and integrate on the straight segment $z=\tau t/|t|$. For fixed $|x|<sR$, a disc centred at $x$ with radius approaching $R(\sigma (\tau )-s)$ stays inside $|\xi |\le \sigma (\tau )R$. Applying the first-derivative Cauchy estimate there givesIntegrating its modulus and taking the supremum over $|x|<sR$ is exactly the asserted integral estimate. The source cancels from $T\varphi -T\psi$.

To obtain a contraction mapping, use the weighted holomorphic norm on a shrinking time domain. Its finite-norm space consists of holomorphic functions on ${\mathcal{C}}_{\alpha }$ with zero initial value; the apparent quotient at $t=0$ is interpreted by a limit. Completeness follows because convergence in this norm implies locally uniform convergence of holomorphic functions, including near $t=0$, and the limiting pointwise bounds give convergence in the norm.

Here is the explicit contraction estimate on a shrinking holomorphic domain. Write $N=\Vert w{\Vert }_{\alpha }$, $r=|t|$, $A=\alpha (1-s)$, and chooseThen $\sigma -s=(A-\tau )/(2\alpha )$ and $\alpha (1-\sigma )-\tau =(A-\tau )/2$. The whole auxiliary polydisc lies inside ${\mathcal{C}}_{\alpha }$, and its supremum of $|w|$ is at most $2N\tau /(A-\tau )$. ConsequentlyThus $\Vert T\varphi -T\psi {\Vert }_{\alpha }\le (4\alpha /R)\Vert \varphi -\psi {\Vert }_{\alpha }$. Choose $0<\alpha <\min (\eta ,R/4)$. The source is bounded on the closed polydisc, say by $M_f$, and $\Vert T0{\Vert }_{\alpha }\le \alpha M_f$. Therefore $T$ maps the Banach space into itself and is a strict contraction mapping. The Banach fixed-point theorem gives a holomorphic fixed point with $\varphi (x,0)=0$. Differentiating the integral identity yields ${\varphi }_t-i{\varphi }_x=f$, establishing this case of the Cauchy-Kovalevskaya theorem.

Finally, near each real initial point, a real analytic function $g$ has a holomorphic extension $G$. Apply the same construction to $\varphi =G(x)+v$ with $v(x,0)=0$ and source $i{G}^{\prime }(x)$, restricting the discs if necessary. This gives a local real analytic function of $(x,t)$ with the prescribed initial value. Equivalently the solution is $\boxed{\varphi (x,t)=G(x+it)}$ wherever the extension is defined.

For the nonnegative Hodge Laplacian ${\Delta }_d^{\mathrm{H}}=d{d}^{\ast }+d^{\ast }d$, the Kähler Laplacian identity gives ${\Delta }_d^{\mathrm{H}}=2{\Delta }_{\bar{\partial }}$. If $u$ is a holomorphic function, then $\bar{\partial }u=0$, while ${\bar{\partial }}^{\ast }u=0$ by degree. ThereforeThe divergence-of-gradient convention printed in part (b) is ${\Delta }_d=-{\Delta }_d^{\mathrm{H}}$ on functions. The sign changes no kernel, so the requested harmonicity of holomorphic functions on a Kähler manifold isHolomorphic functions are harmonic without any compactness assumption. This is a local identity from the Kähler Laplacian identity, not an integration argument requiring a compact manifold.

The two opens are $U_1=\mathbb{C}\times {\mathbb{C}}^{\ast }$ and $U_2={\mathbb{C}}^{\ast }\times \mathbb{C}$; their intersection is $({\mathbb{C}}^{\ast }{)}^2$. There is just one degree-one term in the Čech cochain complex, and no degree-two term. The Čech coboundary sends $(g_1,g_2)$ to $g_2-g_1$, soThe denominator denotes the sum of the restricted function spaces.

To make the quotient explicit, expand a holomorphic function on the intersection in a normally convergent two-variable Laurent series,The terms with $m\ge 0$ extend to $U_1$. Of the remaining terms, those with $n\ge 0$ extend to $U_2$. Both subseries converge normally on their stated domains, by the coefficient bounds from the Cauchy integral formula. The unique remaining representative is the doubly negative partNo nonzero series of this form lies in the denominator, since a function on $U_1$ has no negative $z_1$ exponents and a function on $U_2$ has no negative $z_2$ exponents.

Writing $w_i=z_i^{-1}$, these representatives are exactly $w_1{w}_{2}F(w_1,w_2)$ with $F$ an entire holomorphic function on ${\mathbb{C}}^2$. To see that $F$ is entire, integrate for the coefficients on arbitrarily small product circles: $|a_{-i,-j}|\le M(r_1,r_2)r_1^i{r}_2^j$. Choosing $r_i{R}_i<1$ gives absolute convergence for $|w_i|\le R_i$, for every finite pair $R_i$. Conversely every such entire $F$ supplies a normally convergent representative on the intersection. Therefore the quotient consists of convergent doubly negative Laurent series, not merely finite Laurent polynomials:This gives the holomorphic first cohomology of punctured complex two-space; for instance $1/(z_1{z}_2)$ represents a nonzero class.

For the comparison with sheaf cohomology, $U_1,U_2$ and their intersection are Stein manifolds. They can be realized as closed complex submanifolds of affine complex spaces by adding equations $z_i{w}_i=1$ for their nonzero coordinates. Cartan theorem B makes their higher cohomology with coefficients in the sheaf of holomorphic functions vanish. Thus the cover is acyclic for ${\mathcal{O}}_X$, and the acyclic cover theorem givesThe individual cover members are Stein; their union has the nonzero cohomology just computed.

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.

This is the Poincaré residue of a meromorphic top form with a simple pole. Write $z=z_1$ and locally $\alpha =(dz/z)\wedge \beta$, with $\beta$ a holomorphic $(n-1)$-form. The proposed residue is the pullback of $\beta$ to $Y$.

First it does not depend on the choice of $\beta$: if $dz\wedge (\beta -{\beta }^{\prime })=0$, then at each point the difference has a factor $dz$, and its pullback to $Y$ is zero. Now change the local defining coordinate to $t=az$, where $a$ is a nowhere-zero holomorphic function. Such a factor exists because both coordinates define the same smooth complex analytic hypersurface. If $\alpha =(dt/t)\wedge \eta$, thenAt points of $Y$, this is an equality in the ambient exterior-power fibre, and the last term is zero. It follows that $dz\wedge (\beta -\eta )=0$ there, so their pullbacks to $Y$ agree. This also covers changes of the tangential coordinates.

The restrictions are holomorphic top forms on $Y$, and their agreement on overlaps makes them a global section of the canonical bundle. Thus the residue is independent of the adapted coordinates:The ambient-fibre comparison is important: pulling an ambient $n$-form directly back to the $(n-1)$-dimensional hypersurface would give zero and would not prove the required independence.

On ${\mathbb{C}}^{n+1}\setminus \{0\}$, let $\Omega =d{z}_0\wedge \cdots \wedge d{z}_n$ and let $E={\sum }_i{z}_i{\partial }_{z_i}$ be the Euler vector field. The numerator is its interior product of a differential form, soBoth numerator and denominator have homogeneous scaling degree $n+1$. Thus this meromorphic $n$-form is invariant under constant nonzero scalings, and ${\iota }_E\alpha =0$ makes it horizontal for the quotient to Complex projective space.

To check descent using the hint, take a local holomorphic section $Z$ of the quotient and another section $Z^{\prime }=\lambda Z$, where $\lambda$ is a nowhere-zero holomorphic function. In the evaluationevery term involving $d\lambda$ has two collinear arguments and vanishes. The remaining term is ${\lambda }^{n+1}{Z}^{\ast }({\iota }_E\Omega )$, while $f(\lambda Z)={\lambda }^{n+1}f(Z)$. Hence $Z^{\prime \ast }\alpha =Z^{\ast }\alpha$. The forms therefore glue to a meromorphic section of the canonical bundle of Complex projective space.

On the affine chart $z_j\ne 0$, use the section $z_j=1$ and list the remaining coordinates as $t_1,\dots ,t_n$ in increasing original-index order. If $F$ is the resulting polynomial, thenAt a point of $Y$, some derivative $F_{t_r}$ is nonzero. Otherwise all the homogeneous derivatives except possibly $f_{z_j}$ would vanish; the Euler homogeneous function theorem gives ${\sum }_i{z}_i{f}_{z_i}=(n+1)f=0$ there, forcing $f_{z_j}=0$ too. This contradicts the permitted smoothness criterion. The holomorphic inverse function theorem now makes $F$ a transverse coordinate, so the pole is simple. On a neighbourhood with $F_{t_r}\ne 0$, rearranging the wedge product of differential forms givesBy part (a), its Poincaré residue is consequentlyThe remaining $t$ coordinates form a holomorphic coordinate system on $Y$, and $F_{t_r}$ is nonzero in this neighbourhood. The displayed top form is therefore nowhere zero. These neighbourhoods cover $Y$ and part (a) guarantees agreement on their overlaps. The residue is a nowhere-vanishing global holomorphic section:This is the explicit residue trivialization for a degree n+1 projective hypersurface, consistent with the Adjunction formula $K_Y\cong {\mathcal{O}}_Y((n+1)-n-1)$, but obtained here directly from the indicated form.

Let $M(x)={\sum }_{n\le x}\mu (n)$ be the Mertens function. Partial summation givesFor any $\Re s>0.9$, choose a sufficiently small $ϵ>0$ with $0.9+ϵ<\Re s$. The assumed bound makes the boundary term tend to zero and makes the integral absolutely convergent. Moreover, on each compact set of this half-plane, one can choose the same $ϵ$, so the integral has locally uniform convergence. Henceis a holomorphic function on $\Re s>0.9$, agreeing with the reciprocal Dirichlet series from part (b) on $\Re s>1$.

The identity theorem applied to the holomorphic functions $(s-1)\zeta (s)F(s)$ and $s-1$ extends their equality throughout $\Re s>0.9$; the assumed simple pole makes the first function holomorphic at $1$. Thus $\zeta (s)F(s)=1$ away from $1$, proving a zero-free half-plane:

To obtain the strongest symmetric conclusion, use the following basic facts explicitly: the functional equation of the Riemann zeta function gives the entire function $\xi (s)=\frac{1}{2}s(s-1){\pi }^{-s/2}\Gamma (s/2)\zeta (s)$ with $\xi (s)=\xi (1-s)$, and the trivial zeros of the Riemann zeta function are the negative even integers. Every Nontrivial zero of the Riemann zeta function is therefore accompanied by its reflection $1-s$. A Nontrivial zero of the Riemann zeta function with real part below $0.1$ would reflect to one above $0.9$, which is impossible. Consequently all nontrivial zeros lie in the closed stripThe $ϵ$ in the hypothesis prevents this argument from excluding either boundary line; the trivial zeros of the Riemann zeta function remain present.

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.

A weight-$k$ modular form for $\Gamma =S{L}_2(\mathbb{Z})$ is a holomorphic function $f$ on the complex upper half-plane satisfyingand holomorphic at a cusp. Since the full modular group has one cusp class, this last condition means that the period-one function has a convergent expansion $f(\tau )={\sum }_{n\ge 0}{a}_n{q}^n$, $q=e^{2\pi i\tau }$, near infinity. A cusp form has $a_0=0$. The element $-I$ makes a nonzero weight odd form impossible.

For weight zero, $f$ descends to a holomorphic function on the quotient of the complex upper half-plane by the modular group. At an elliptic fixed point, invariance under its finite stabilizer makes the local Taylor series a series in the quotient coordinate. At the cusp, the $q$ expansion extends it over $q=0$. The standard fundamental domain of the modular group, with its boundary identified and its cusp added, is a compact Riemann surface, the compactified modular curve $X(1)$. The maximum modulus principle now provesHolomorphy at the cusp is essential to this conclusion.

For positive even $k\ge 4$, define the unnormalized Eisenstein seriesThe real-linear map $(m,n)\mapsto m\tau +n$ is invertible, and on a compact subset of the complex upper half-plane its norm is uniformly comparable to $\sqrt{m^2+n^2}$. Thus the sum converges absolutely and locally uniformly for $k>2$, proving holomorphy. For $\gamma =\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$,The map $(m,n)\mapsto (ma+nc,mb+nd)$ is a bijection on the nonzero integer pairs, giving $G_k(\gamma \tau )=(c\tau +d{)}^k{G}_k(\tau )$.

For the expansion, the cotangent partial-fraction Fourier kernel follows by differentiating the partial-fraction expansion of $\pi \cot \pi z$ and its geometric-series expansion $-\pi i(1+2{\sum }_{r\ge 1}{e}^{2\pi irz})$, valid for $\mathrm{Im}z>0$. It gives, for even $k\ge 2$,The $m=0$ part of $G_k$ is $2\zeta (k)$, where $\zeta$ is the Riemann zeta function; positive and negative $m$ contribute equally. Apply the kernel at $z=m\tau$ for $m>0$, and regroup the absolutely convergent sum according to $N=mr$. With the divisor sum ${\sigma }_{k-1}(N)={\sum }_{r\mid N}{r}^{k-1}$, this provesThe expansion has no negative powers and converges for $|q|<1$, so it also proves holomorphy at a cusp and completes the proof of modularity.

The supplied zeta values yield the Fourier expansion of a normalized Eisenstein seriesBoth $E_4^3$ and $E_6^2$ are modular forms of weight twelve. Their constant terms cancel, so their difference divided by $1728$ is a cusp form. Integrality requires an argument: dividing an integral series by $1728$ alone would not suffice. Expanding directly gives the integrality identity for the modular discriminantFor every integer $d$, $5d^3+7d^5$ is divisible by twelve. Modulo three, $d^3\equiv d^5\equiv d$; modulo four, an even $d$ makes both powers divisible by four, while an odd $d$ has $d^3\equiv d^5\equiv d(\mathrm{mod}4)$. Thus each coefficient $(5{\sigma }_3(n)+7{\sigma }_5(n))/12$ is integral. The other three series visibly have integral coefficients, provingThis proves integrality without assuming a product expansion for the modular discriminant.