Let be $f:\mathbb C\rightarrow\overline {\mathbb C}$ a meromorphic function. The set of periods $\Omega_f$ is a discrete (additive) group and we have one of these possibilities:

i) $\Omega_f= \{0\}$

ii) $\Omega_f= \mathbb Z\omega$

iii) $\Omega_f= \mathbb Z\omega_1+\mathbb Z\omega_2$

In the case iii) we say that $f$ is an elliptic function and we now that the foundamental regions are compact subset of $\mathbb C$ (for example foundamental parallelograms). Elliptic fuctions, respect a fixed group of periodicities $\Omega_f=\Lambda$, form a field $E(\Lambda)$ and one can prove that $E(\Lambda)=\mathbb C(\wp,\wp')$.

Now I have not found in literature similar results for the case ii) of meromorphic simply periodic functions. In this case foudamental regions are not compact sets and, fixed the group $\Omega_f$, simply periodic functions respect $\Omega_f$ form a field. I ask if this field is in fact $\mathbb C\big(e^\frac{2\pi iz}{\omega}\big)$; moreover what is the relation beethween the meromorphic functions:

$$\varepsilon_k=\sum_{n\in\mathbb Z}(z-n)^{-k}$$

and the field of simply periodic funcion, where $\Omega_f=\mathbb Z$?

  • 2
    $\begingroup$ Assume for simplicity that $\Lambda_f=\mathbb{Z}$. Note that elliptic functions will automatically also be periodic (i.e. your case ii) above), but they will give rise to periodic functions with infinitely many poles in the strip $\mathbb{C}/\mathbb{Z}$. However I think it should be easy to prove that any function belonging to the field $\mathbb{C}(e^{2\pi i z})$ will only have finitely many poles in this strip, since $e^{2\pi i z}$ will be injective when restricted to this strip. $\endgroup$ – newguy Aug 21 '12 at 12:16
  • $\begingroup$ So $\mathbb C (e^{\frac {2\pi iz}{\omega}})$ is only contained in the field of simply periodic functions..... $\endgroup$ – Dubious Aug 21 '12 at 12:57

This is not really an answer, but too long for a comment. In any case it is also fairly trivial, but I hope you still find some use in it.

I think that it there is unlikely to be a "nice" classification of such functions comparable to (iii). This is mainly because your region of interest (the fundamental strip) is not compact, so things can go "too crazy".

If you want a more concrete reason, note that if $f:\mathbb{C} \to \mathbb{C}$ is any holomorphic function, then $f\left(e^{2\pi i z}\right)$ is still holomorphic, but singly periodic! So surely there are very many such functions.

There are ways of getting out nicer answers, essentially by introducing artificial compactness conditions. For example, you might consider only functions such that $\lim_{Re(z) \to \pm \infty} f(z)$ exist (possibly $\infty$). The field of such functions is precisely $\mathbb{C}\left(e^{2\pi i z}\right)$. Indeed, let $X_0$ be the topological space obtained by gluing the left and right side of your fundamental strip, and let $X$ be its end-compactification (i.e. the sphere obtained by adding two "infinitely far" points to the "infinitely long cylinder" $X_0$). The conditions then precisely say that $f$ is meromorphic on $\mathbb{C}$ and extends to a continuous function from $X$ to the Riemann sphere $P^1$. The function $e(z) = e^{2\pi i z}$ extends to a continuous bijection between $X$ and $P^1$. If $f$ is a continuous function on $X$ holomorphic on $\mathbb{C}$ (i.e. $X_0$), then $f \circ e^{-1}$ is a continuous map from $P^1$ to itself, holomorphic away from the north and south pole. Such a function is meromorphic by the theorem on removable singularities, and hence a rational function (by the classification of meromorphic functions on $P^1$ - a compact riemann surface, whence the nice answer!). Hence $f$ is a rational function in $e(z)$, as was to be shown.

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  • $\begingroup$ your comment is very usefull. You said: "if $f:\rightarrow\mathbb C→\mathbb C $ is any holomorphic function, then $f(e^{2πiz})$ is still holomorphic, but singly periodic". But this situation is true also for elliptic functions infact $f(\wp(z))$ is still an elliptic function. $\endgroup$ – Dubious Sep 21 '12 at 7:33
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    $\begingroup$ But $\wp$ has poles. Hence $f(\wp(z))$ will be meromorphic only if $f$ is "meromorphic at infity", i.e. a rational function. All other meromorphic functions have essential singularities at infinity, so $f(\wp(z)$ will have essential singularities. $\endgroup$ – Tom Bachmann Sep 21 '12 at 16:56

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