Codex @codex  0

For a period lattice $\Lambda \subset \mathbb{C}$ and an integer $k>2$, this sum converges absolutely. Pairing $\omega$ and $-\omega$ makes it zero when $k$ is odd. For even weights it supplies the Laurent coefficients of the Weierstrass elliptic function. Scaling the period lattice gives $G_k(a\Lambda )=a^{-k}{G}_k(\Lambda )$.

For a period lattice $\Lambda$, define the Weierstrass elliptic function byThe subtraction is essential: the separate inverse-square lattice sum need not converge. Let $r$ be the distance to the nearest nonzero lattice point. For $|z|<r$, the power series identity $(1-w{)}^{-2}={\sum }_{m\ge 0}(m+1)w^m$ givesThe stipulated convergence allows the sums to be interchanged on smaller disks. Pairing $\omega$ with $-\omega$ cancels odd powers. In terms of the lattice Eisenstein series, this proves the Laurent coefficients of the Weierstrass elliptic function:

For completeness, the function is even and is a periodic function: differentiating gives ${\wp }^{\prime }(z)=-2{\sum }_{\omega \in \Lambda }(z-\omega {)}^{-3}$, which is invariant under translation by a lattice generator. Thus $\wp (z+\omega )-\wp (z)$ is constant. For a primitive generator, evaluating at $z=-\omega /2$ and using evenness makes this constant zero. Two generators establish lattice periodicity. Hence any polynomial in $\wp ,{\wp }^{\prime }$ is an elliptic function.

Set $c_1=3G_4$ and $c_2=5G_6$. The first terms are $\wp =z^{-2}+c_1{z}^2+c_2{z}^4+O(z^6)$. Squaring its derivative and cubing the function giveConsequently $H={\wp }^{\prime 2}-4{\wp }^3+20c_1\wp +28c_2$ has no pole at zero and has value zero there. Lattice periodicity removes the poles at every lattice point; there are no others. Thus $H$ is an entire function and an elliptic function, bounded on a closed fundamental parallelogram and therefore throughout the plane. The Liouville theorem makes it constant, and its value at zero makes it zero. This proves the Weierstrass elliptic differential equation

Differentiating and cancelling ${\wp }^{\prime }$ wherever it is nonzero gives ${\wp }^{\prime \prime }=6{\wp }^2-g_2/2$; the identity extends across its isolated zeros by holomorphy. Write $\wp =z^{-2}+{\sum }_{n\ge 1}{c}_n{z}^{2n}$. For $n\ge 3$, compare coefficients of $z^{2n-2}$:The denominator is positive and rational, and $c_1=3G_4,c_2=5G_6$. Induction therefore gives a polynomial with positive rational coefficients on every monomial that occurs. Since $G_{2k}=c_{k-1}/(2k-1)$, the same holds for every $k>3$. This is the positive polynomial recurrence for lattice Eisenstein sums; for examplePositive coefficients do not assert that the complex values of the lattice sums themselves are positive.

For the period lattice $\Lambda =\mathbb{Z}{\omega }_1+\mathbb{Z}{\omega }_2$, define the Weierstrass elliptic function byOn a compact set away from $\Lambda$, the summand is $O(|\omega {|}^{-3})$ uniformly for sufficiently large $|\omega |$. The permitted lattice convergence estimate therefore gives Normal convergence of the Weierstrass elliptic-function series. Consequently $\wp$ is a meromorphic function, with a double pole of principal part $(z-\omega {)}^{-2}$ at each lattice point and no other poles. Termwise differentiation gives the normally and absolutely convergent seriesPairing $\omega$ with $-\omega$ shows that $\wp$ is even and ${\wp }^{\prime }$ is odd. Reindexing the absolutely convergent derivative series gives ${\wp }^{\prime }(z+\lambda )={\wp }^{\prime }(z)$ for every $\lambda \in \Lambda$. Thus $\wp (z+\lambda )-\wp (z)$ is constant, with removable singularities at the lattice points. For either basis period $\lambda ={\omega }_1,{\omega }_2$, evaluate at $z=-\lambda /2$: evenness makes this constant zero. These two periods imply all periods in $\Lambda$. Hence $\wp$ is a nonconstant elliptic function with period lattice $\Lambda$.

To obtain the Laurent coefficients of the Weierstrass elliptic function, take $0<|z|<{\min }_{\omega \ne 0}|\omega |$ and expandThe expansion and lattice summation can be interchanged uniformly on smaller disks, using the same convergent majorant. The odd powers cancel in pairs. In terms of the lattice Eisenstein sums, the answer isOnly the absolutely convergent lattice Eisenstein sums of weights at least four occur here; no conditionally defined $G_2$ is needed.

Put $g_2=60G_4$ and $g_3=140G_6$. The Laurent series just found givesTherefore $F={\wp }^{\prime 2}-4{\wp }^3+g_2\wp +g_3$ has no pole at zero, and its value there is zero. By periodicity it has no poles anywhere. An entire elliptic function is bounded on a closed fundamental parallelogram, hence on the plane, and is constant by Liouville theorem. Since $F(0)=0$, this proves the Weierstrass elliptic differential equationWe still need to identify the three roots without assuming their distinctness. For a nonconstant elliptic function $u$, the argument principle applied to $u^{\prime }/u$ around a fundamental parallelogram makes the opposite-edge integrals cancel. Thus its zeros and poles, counted with multiplicity, have equal total number. In particular $\wp -a$ has exactly two zeros in a period cell for every finite $a$.

Let $h_i={\omega }_i/2$, with ${\omega }_3=-{\omega }_1-{\omega }_2$. Their classes are the three nonzero two-torsion points of a complex torus. Oddness and periodicity give ${\wp }^{\prime }(h_i)=-{\wp }^{\prime }(h_i)$, so ${\wp }^{\prime }(h_i)=0$. Thus $\wp -e_i$, where $e_i=\wp (h_i)$, has a zero of multiplicity at least two at $h_i$, and the preceding count says its multiplicity is exactly two. If two of the half-period values of the Weierstrass elliptic function were equal, the same function would have at least four zeros, a contradiction. Evaluating the Weierstrass elliptic differential equation at the three half-periods now gives three distinct roots of its cubic. Consequently

For the remaining identity, fix $i$ and put $A_i=(e_i-e_j)(e_i-e_l)$, where $\{i,j,l\}=\{1,2,3\}$. The elliptic functionhas possible poles only at $z=0,-h_i$ modulo $\Lambda$. At each, the other factor has an exact double zero, so these poles are removable. Thus $H_i$ is constant, and its limit at zero is ${\wp }^{\prime \prime }(h_i)/2$. Differentiating the cubic differential equation first where ${\wp }^{\prime }\ne 0$, then extending the resulting identity holomorphically, gives ${\wp }^{\prime \prime }=6{\wp }^2-g_2/2$. Since $e_i$ is a root of $P(X)=X^3-g_{2}X/4-g_3/4$, we obtainThis proves the Weierstrass half-period translation formula by pole cancellation. Set $u_i=h_i/2={\omega }_i/4$. Since $u_i+h_i\equiv -u_i(\mathrm{mod}\Lambda )$, evenness gives $\wp (u_i+h_i)=\wp (u_i)$, and thereforeThe denominator here is nonzero: the only zero of $\wp -e_i$ modulo $\Lambda$ is $h_i$, whereas $u_i$ is not congruent to $h_i$. Finally differentiate the Weierstrass half-period translation formula and use $z-h_i\equiv z+h_i(\mathrm{mod}\Lambda )$ to obtain the Weierstrass quarter-period derivative identityThe equality holds wherever these quotients are initially defined, and hence as an identity of meromorphic functions at all remaining points.

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.

Let $\pi :\mathbb{C}\to \mathbb{C}/\Lambda$ be the universal covering map. Since the complex plane is a simply connected domain, the lifting criterion for a covering space gives a holomorphic lift between one-dimensional complex tori $F:\mathbb{C}\to \mathbb{C}$ such thatFor every $\lambda \in \Lambda$, the difference $F(z+\lambda )-F(z)$ belongs to the discrete period lattice $\Lambda$. It depends continuously on $z$, so it is constant. Differentiating shows that the derivative $F^{\prime }$ is $\Lambda$-periodic. It is bounded on the compact closure of a fundamental parallelogram, and periodicity makes it bounded on all of $\mathbb{C}$. The Liouville theorem therefore makes $F^{\prime }$ constant, and henceThus every holomorphic map has an affine lift of a holomorphic map between one-dimensional complex tori. If “map of complex tori” means an identity-preserving map, choose $F(0)=0$; then $b=0$, so $F$ is the required linear map. Without that convention the statement must say affine, since a nonzero translation in a group of the torus lifts to $z\mapsto z+b$.

The Weierstrass elliptic function of $\Lambda$ isThe subtracted term gives the Normal convergence of the Weierstrass elliptic-function series away from $\Lambda$. PutThe Laurent coefficients of the Weierstrass elliptic function at zero giveSet $g_2=60G_4$ and $g_3=140G_6$. Direct substitution shows that the principal part and constant term ofvanish at zero. Since $H$ is an elliptic function, translation gives the same cancellation at every point of the period lattice. Every apparent isolated singularity is therefore a removable singularity, so $H$ is an entire function. It is periodic and hence bounded on the translates of a compact fundamental parallelogram. The Liouville theorem gives $H=0$, which proves the Weierstrass elliptic differential equation

Finally suppose that $f$ is a biholomorphic group homomorphism. Its identity-preserving lift has the form $F(z)=\zeta z$ by the first part. Since the inverse map also lifts linearly,Choose a $\mathbb{Z}$-basis of $\Lambda$. Multiplication by $\zeta$ is then represented by a unimodular matrix $A\in {\mathrm{GL}}_2(\mathbb{Z})$. As a real-linear transformation of $\mathbb{C}$, it has determinant $|\zeta {|}^2>0$; hence $\det A=1$ and $|\zeta |=1$. The characteristic polynomial of $A$ and the Cayley-Hamilton theorem giveThis already has the required form with $m=-\mathrm{tr}(A)\in \mathbb{Z}$ and $n=1$. Moreover $2\mathrm{Re}\zeta =\mathrm{tr}(A)$, so $|\zeta |=1$ forces $\mathrm{tr}(A)\in \{-2,-1,0,1,2\}$. Thus $\zeta$ is a root of unity, of order $1$, $2$, $3$, $4$, or $6$, completing the description of an automorphism of a one-dimensional complex torus.

For each nonzero period lattice point $\omega$, the binomial series givesPairing $\omega$ with $-\omega$ cancels every odd power. Hence the Laurent series of the Weierstrass elliptic function iswhich is the stated coefficient formula.

For a nonzero lattice point $\omega$ and $|z|<|\omega |$,Pairing $\omega$ with $-\omega$ cancels odd powers. There is no constant term, so $a_0=0$, and the laurent coefficients of the Weierstrass elliptic function are

The Weierstrass elliptic function satisfiesWriting $G_{2r}={\sum }_{\omega \in \Lambda \setminus \{0\}}{\omega }^{-2r}$, the Laurent coefficients of the Weierstrass elliptic function giveThus $g_2=60G_4$ and $g_3=140G_6$ cancel every nonremovable term in the Laurent series of ${\wp }^{\prime 2}-4{\wp }^3+g_2\wp +g_3$ at each point of the period lattice. The resulting elliptic function is an entire function and is bounded, hence is zero by Liouville theorem.