Definition Weierstrass $\zeta$-function unclear

Question

The Weierstrass $\zeta$-function is defined as follows for a lattice $\Lambda$, where a lattice is a discrete subgroup of $\mathbb{C}$ containing an $\mathbb{R}$-basis for $\mathbb{C}$.

$$\zeta(z) = \frac{1}{z} + \sum_{\omega\in\Lambda\setminus\{0\}}\left(\frac{1}{z-\omega}+\frac{1}{\omega}+\frac{z}{\omega^2}\right)$$

Doesn't $\sum_{\omega\in\Lambda\setminus\{0\}}\frac{1}{\omega}$ equal 0, because if $x \in \Lambda$ then $-x\in\Lambda$? I understand that the term appears when differentiating the logarithm of the Weierstrass $\sigma$-function, but why is it written anywhere if it could just as well be left out?

I heavily recommend to change the notation $;\zeta;$ for the very well known and widely used $;\wp;$ — DonAntonio, Oct 13 '13 at 10:45
This is not the same function; in fact $\frac{\mathrm{d}}{\mathrm{d}z}\zeta(z) = -\wp(z)$. — G.L., Oct 13 '13 at 10:51
Related: https://math.stackexchange.com/q/1984530/532409 and https://math.stackexchange.com/q/2248789/532409 — Quillo, Nov 08 '22 at 19:20

score 4 · Accepted Answer · answered Oct 13 '13 at 09:35

You can'd do that rearrangement there because as written it is not absolutely convergent. On the other hand, with a small simplification,

$$\sum_{\omega\in\Lambda\setminus\{0\}}\left(\frac{z}{(z-\omega)\omega}+\frac{z}{\omega^2}\right)$$

is absolutely convergent, because $\omega$ is quadratic is both denominators.

On a related note, an addition of some constant is often required to construct holomorphic/meromorphic functions with prescribed zeroes and poles and residues. See Weierstrass theorem and Mittag-Leffler's theorem and their proofs on more along this line of thought.

Definition Weierstrass $\zeta$-function unclear

1 Answers1