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For a nonzero elliptic function, the weighted sum of its zero positions minus its pole positions belongs to the period lattice. Integrating $z{f}^{\prime }/f$ around a fundamental parallelogram proves this: translating opposite edges gives a lattice-linear combination of integrals of $f^{\prime }/f$, each an integer multiple of $2\pi i$. The unweighted integral proves that the zero and pole multiplicities have equal total.

On $\mathbb{C}/(\mathbb{Z}+\mathbb{Z}i)$, a Hermitian form linear in its first argument has integral imaginary part on the period lattice exactly when its real scalar coefficient is integral. Positivity is equivalent to $t>0$. The associated holomorphic line bundle has degree $t$ because its First Chern class is represented by $(it/2)dz\wedge d\overline{z}$. In particular the oriented Chern pairing is minus the imaginary part under this convention.

Suppose ${\tau }_1=(a{\tau }_2+b)/(c{\tau }_2+d)$, with $ad-bc=1$, and put $\alpha =(c{\tau }_2+d{)}^{-1}$. The identitiesshow that multiplication by $\alpha$ takes one period lattice isomorphically to the other: its basis matrix has determinant $ad-bc=1$. If $N\mid c$, the image of $1/N$ is congruent to $a/N$ modulo the target period lattice. Moreover $ad\equiv 1(\mathrm{mod}N)$, so $a$ is a unit modulo $N$. The image therefore generates the same cyclic subgroup.

Conversely, a holomorphic group isomorphism of complex tori lifts to $z\mapsto \alpha z$, with $\alpha {\Lambda }_{{\tau }_2}={\Lambda }_{{\tau }_1}$. Indeed its lift fixing zero is additive: its failure to be additive is a continuous lattice-valued function, hence zero. An additive holomorphic map is complex linear. WriteThe corresponding integral basis matrix has determinant $1$: complex multiplication preserves the real orientation, and both bases $(1,{\tau }_j)$ are positively oriented. Solving gives ${\tau }_1=(a{\tau }_2+b)/(c{\tau }_2+d)$. Preservation of the cyclic subgroup means $\alpha /N\equiv u/N(\mathrm{mod}{\Lambda }_{{\tau }_1})$ for some unit $u$ modulo $N$. Comparing the two real basis coefficients gives $N\mid c$. Thus the matrix lies in the Gamma 0 congruence subgroup, and the classification isThis is the noncompact modular curve classification; no cusp points have been added.

Use the convention that a Hermitian form is complex linear in its first argument. In the integral Hermitian forms on the square complex torus calculation, every Hermitian form on $\mathbb{C}$ has the form $H_t(z,w)=tz\overline{w}$, where $t=H_t(1,1)$ is real. For $\lambda =m+ni$ and $\mu =r+si$ in the period lattice,Consequently integrality holds precisely when $t\in \mathbb{Z}$: necessity follows from $E_t(1,i)=-t$, and sufficiency follows from the displayed formula. Addition of Hermitian forms corresponds to addition of $t$. Positivity is equivalent to $t>0$, since $H_t(z,z)=t|z{|}^2$. The answer isWith the conjugate-linear-first convention the formula is $t\overline{z}w$ and the sign of $E_t$ reverses. The positive integer parameter is unchanged.

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.

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.