Weierstrass p function

Let $\mathrm{\Lambda }\subseteq \mathbb{C}$ be a full complex lattice with oriented basis $({\omega }_1,{\omega }_2)$ (Complex lattice and quotient torus). The Weierstrass $\mathrm{\wp }$-function of $\mathrm{\Lambda }$ is the function

${\mathrm{\wp }}_{\mathrm{\Lambda }}(z):=\frac{1}{z^2}+{\sum }_{\omega \in \mathrm{\Lambda }\setminus \{0\}}(\frac{1}{(z-\omega {)}^2}-\frac{1}{{\omega }^2}),z\in \mathbb{C}\setminus \mathrm{\Lambda },$

where the sum over the lattice points different from $0$ is the unordered (finite-subset) sum: for the directed set of finite subsets $F\subseteq \mathrm{\Lambda }\setminus \{0\}$ ordered by inclusion, one forms the net of partial sums ${\sum }_{\omega \in F}((z-\omega {)}^{-2}-{\omega }^{-2})$, and ${\sum }_{\omega \in \mathrm{\Lambda }\setminus \{0\}}(\cdots )$ denotes its limit when the net converges and the limit does not depend on the directed set — equivalently, when the family is absolutely summable, i.e. when ${\sup }_F{\sum }_{\omega \in F}\mid (z-\omega {)}^{-2}-{\omega }^{-2}\mid <\mathrm{\infty }$ and the partial sums converge. The following theorem of this page proves that for every $z\in \mathbb{C}\setminus \mathrm{\Lambda }$ the family is absolutely summable, with normal (locally uniform, enumeration-free) convergence on $\mathbb{C}\setminus \mathrm{\Lambda }$; this is the sense in which ${\mathrm{\wp }}_{\mathrm{\Lambda }}$ is well defined beyond the displayed formula. In particular no ordering of $\mathrm{\Lambda }$ is used and the value does not depend on one.

The two correction terms. The summand $\frac{1}{(z-\omega {)}^2}-\frac{1}{{\omega }^2}=\frac{2z\omega -z^2}{(z-\omega {)}^2{\omega }^2}$ is holomorphic in $z$ on the disc $\mathrm{\mid }z\mathrm{\mid }<\mathrm{\mid }\omega \mathrm{\mid }$, so each summand is holomorphic near the origin; at $z=0$ its value is $(-\omega {)}^{-2}-{\omega }^{-2}=0$. The single uncorrected term $z^{-2}$ therefore supplies the entire principal part at the lattice point $0$, and the subtractions make the remaining series vanish at $0$: the constant term of the Laurent expansion of ${\mathrm{\wp }}_{\mathrm{\Lambda }}$ at $0$ is $0$. At a general lattice point $\lambda \in \mathrm{\Lambda }$ the same computation after the translation $z\mapsto z+\lambda$ shows that the principal part of ${\mathrm{\wp }}_{\mathrm{\Lambda }}$ at $\lambda$ is $(z-\lambda {)}^{-2}$.

Translation and parity. Reindexing the sum by $\omega \mapsto -\omega$ — a bijection of $\mathrm{\Lambda }\setminus \{0\}$ — replaces $(z-\omega {)}^{-2}-{\omega }^{-2}$ by $(z+\omega {)}^{-2}-{\omega }^{-2}$ and $z$ by $-z$, which is the same expression; consequently the absolutely convergent sum satisfies ${\mathrm{\wp }}_{\mathrm{\Lambda }}(-z)={\mathrm{\wp }}_{\mathrm{\Lambda }}(z)$ once its convergence is known, and ${\mathrm{\wp }}_{\mathrm{\Lambda }}$ is an even function. The sum depends only on the lattice $\mathrm{\Lambda }$, not on the oriented basis chosen to describe it, since the underlying index set $\mathrm{\Lambda }$ and every summand depend on $\mathrm{\Lambda }$ alone.

Finite-subset convergence and absolute summability. For a complex family $(a_i{)}_{i\in I}$, use the real and imaginary parts and modulus of Real and imaginary parts, complex conjugation, and modulus. Absolute summability implies convergence of its finite-subset net by Square-summable families on an arbitrary index set and the space ${\mathrm{\ell }}^2(I)$. For the reverse direction, suppose the finite-subset sums $s_F:={\sum }_{i\in F}{a}_i$ converge to $s\in \mathbb{C}$. Choose a finite $F_0$ such that $\mathrm{\mid }{s}_F-s\mathrm{\mid }<1$ whenever $F\supseteq F_0$; then $\mathrm{\mid }{s}_F\mathrm{\mid }\le \mathrm{\mid }s\mathrm{\mid }+1$ for every such $F$. For any finite set $P\subseteq I\setminus F_0$ on which $\mathrm{Re}(a_i)>0$, ${\sum }_{i\in P}\mathrm{Re}(a_i)=\mathrm{Re}(s_{F_0\cup P}-s_{F_0})\le \mathrm{\mid }s\mathrm{\mid }+1+\mathrm{\mid }{s}_{F_0}\mathrm{\mid }\mathrm{.}$ The same bound holds for finite sums of $-\mathrm{Re}(a_i)$ over negative terms, and likewise for the imaginary parts. Adding the finitely many terms in $F_0$ shows that the finite subsums of $\mathrm{\mid }\mathrm{Re}(a_i)\mathrm{\mid }$ and $\mathrm{\mid }\mathrm{Im}(a_i)\mathrm{\mid }$ are bounded. Since $\mathrm{\mid }{a}_i\mathrm{\mid }\le \mathrm{\mid }\mathrm{Re}(a_i)\mathrm{\mid }+\mathrm{\mid }\mathrm{Im}(a_i)\mathrm{\mid }$, the finite subsums of $\mathrm{\mid }{a}_i\mathrm{\mid }$ are bounded, so the family is absolutely summable by the definition in Square-summable families on an arbitrary index set and the space ${\mathrm{\ell }}^2(I)$. This argument uses no enumeration or choice principle.

The cubic tail. For $\mathrm{\mid }z\mathrm{\mid }\le R$ and $\mathrm{\mid }\omega \mathrm{\mid }\ge 2R$ the displayed numerator is bounded by $2R\mathrm{\mid }\omega \mathrm{\mid }+R^2$ and the denominator is bounded below by $\frac{1}{4}\mathrm{\mid }\omega {\mathrm{\mid }}^4$, so the summand is $O_R(\mathrm{\mid }\omega {\mathrm{\mid }}^{-3})$. The lattice alone therefore controls the size of the terms, and the convergence proof only has to count how many lattice vectors occur at each scale; that count and the resulting normal convergence are proved in the next items of this page.