I'm trying to understand the chapter on Elliptic functions in Stein's Complex Analysis. In particular, I am interested in the construction of Weierstrass's $\wp$ function.

Let $\Lambda = \{n + m\tau : n,m \in \mathbb{Z}\}$ be the usual lattice used when discussing Elliptic functions. In motivating the definition of the $\wp$ function, Stein mentions the (apparently formal) expression $$\sum_{\omega \in \Lambda} \frac{1}{(z+w)^2}$$ At a first glance this should satisfy the properties of an Elliptic function since (formally) it has poles at all lattice points and it is (formally) doubly periodic.

However, it mentions that it is not absolutely convergent. This made me realize I don't really understand what it means for a sum of this form to even be convergent in the first place.

Since we are summing over a countably infinite set (the lattice), what would the definition of absolute convergence be in this case? Moreover, why is this sum not absolutely convergent. Until now, I have only ever dealt with straightforward sums indexed by the integers so I am not sure what to make of this.


  • 1
    $\begingroup$ A simple way of saying it is that it means the sum is absolutely convergent if you index it by the integers in any possible way. $\endgroup$ – Qiaochu Yuan Sep 17 at 21:01
  • $\begingroup$ Actually the simplest way to define absolute convergence for any set of complex numbers is to impose the condition that the sum of absolute values of the elements in any finite subset is uniformly bounded by a constant $M$ and then the infimum of all such is the absolute sum. Note that this implies that the nonzero elements are countable since for any $n$, the set of elements with absolute value at least $\frac{1}{n}$ is finite $\endgroup$ – Conrad Sep 17 at 22:03

This is only a partial answer, but I think it is substantial enough to warrant posting:

I'm going to ignore the specific instance and say a bit about summing over general countable sets. First, let me give a rather abstract definition. If this is confusing, skip to the end for a more concrete one, but I do suggest trying to make sense of this one first - it will help set up the right intuition, in my opinion.

Suppose $\Xi$ is a countable set and $f:\Xi\rightarrow\mathbb{C}$ is some function (we could also replace $\mathbb{C}$ with something else without affecting the general picture). There is, I claim, a natural way to try to evaluate the expression $$\sum_{\xi\in\Xi}f(\xi),$$ namely by taking a limit of finite approximations. After all, for each finite $F\subseteq\Xi$ the expression $\sum_{\xi\in F}f(\xi)$ makes perfect sense, and for that matter that's basically how we thought about infinite sums in the "old" setting.

This suggests that we look instead at the expression $$\lim_{F\rightarrow\Xi, F\subseteq_{fin}\Xi}\sum_{\xi\in F}f(\xi).$$ But of course this just pushes us one step back: how do we take the relevant limit? There is, though, a natural way to try to do this: emulate the usual definition of limit at infinity, and say that we get "arbitrarily close" to the desired value as we look at "sufficiently big" finite subsets.

Formally, we say that $$\sum_{\xi\in\Xi}f(\xi)=L$$ iff for all $\epsilon>0$ there is some finite $X\subseteq \Xi$ such that for all finite $Y\subseteq \Xi$ with $X\subseteq Y$ we have $$\vert L-\sum_{\xi\in Y}f(\xi)\vert<\epsilon.$$

(At this point look back at the definition of $\lim_{n\rightarrow\infty}$.) When we say that $\sum_{\xi\in\Xi}f(\xi)$ converges absolutely, we mean that such an $L$ exists.

There is also a more concrete take on things. If $\Xi$ is a countable set, we'll say that an enumeration of $\Xi$ is just an explicit ordering $(\xi_i)_{i\in\mathbb{N}}$ of the elements of $\Xi$ - formally, an enumeration of $\Xi$ is a bijection $\beta:\mathbb{N}\rightarrow\Psi$.

Each enumeration $\beta=(\xi_i)_{i\in\mathbb{N}}$ gives rise to an associate "normal" infinite sum $$Sum(\beta):=\sum_{i\in\mathbb{N}}f(\xi_i)=\lim_{n\rightarrow\infty}\sum_{i=1}^nf(\xi_i).$$

When our countable set starts out with a "default" enumeration, this puts us in familiar territory: changing the enumeration is just rearranging the terms. At this point it's a good exercise to prove:

$\sum_{\xi\in\Xi}f(\xi)$ converges absolutely in the sense of the previous section iff for any two enumerations $\beta,\beta'$ of $\Xi$ we have $Sum(\beta)=Sum(\beta')$.

  • $\begingroup$ Incidentally, why couldn't we use an uncountable set in the first part? It clearly wouldn't work in the second part, since uncountable sets have no enumerations at all, but the first part seems unbothered by higher cardinalities. Well, we could, but things get boring - "most" of the set would have to be irrelevant! See here. $\endgroup$ – Noah Schweber Sep 17 at 20:34
  • $\begingroup$ Thanks, this write up certainly helped with the conceptual side of things. Admittedly, I'm still banging my head against the wall trying to understand why Stein's series is not absolutely convergent. I would have thought the quadratic $(z+w)^2$ term in the denominator would have been enough for absolute convergence... Well it is clear I have some gaps in my knowledge when it comes to manipulating series of complex numbers! $\endgroup$ – 1730 Sep 17 at 22:05

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