We have seen in the previous post that the elliptic functions have at least two distinct periods. In the current post we shall discuss in details the ramifications of the double periodicity and will get a flavor of some of the methods of analytic function theory. An outline of the topics we would specifically discuss is provided below:

It is also obvious that if $ \omega$ is any period then so is $ n\omega$ where $ n$ is an integer. Similarly if there are two or more periods $ \omega_{1}, \omega_{2}, \ldots, \omega_{k}$ then for any integers $ n_{1}, n_{2}, \ldots, n_{k}$, the number $ \omega = n_{1}\omega_{1} + n_{2}\omega_{2} + \cdots + n_{k}\omega_{k}$ is also a period. We will now analyze the structure of set of non-trivial periods of a function in more detail.

Let the set of non-trivial periods of $ f(z)$ be denoted by $ M(f)$. Thus $$M(f) = \{ \omega \mid \omega \neq 0, f(z + \omega) = f(z)\}$$ The first thing to notice here is that we can't have periods with arbitrarily small modulus, i.e. the lower bound for the set $ \{|\omega| : \omega \neq 0, f(z + \omega) = f(z)\}$ is definitely positive. If it were not so we could find a sequence of non-trivial periods $ \omega_{n}$ such that $ \omega_{n} \to 0$ as $ n \to \infty$. This means that $$\lim_{n \to \infty}\frac{f(z + \omega_{n}) - f(z)}{\omega_{n}} = 0$$ and this means that $ f'(z) = 0$ at all points in the domain of definition and therefore $ f(z)$ is constant.

Next it is not possible that the set of non-trivial periods has a limit point i.e. the set of non-trivial periods is discrete. This is quite obvious if we notice that in case there were a limit point say $ \omega_{0}$, then we could find a sequence $ \{\omega_{n}\}$ of non-trivial distinct periods such that $ \omega_{n} \to \omega_{0}$ as $ n \to \infty$. And then we would have $$ f(z + \omega_{0}) = \lim_{n \to \infty}f(z + \omega_{n}) = \lim_{n \to \infty} f(z) = f(z)$$ (assuming continuity of $ f(z)$ at $ z + \omega_{0}$) and thereby $ \omega_{0}$ would also be a non-trivial period. This would finally lead to the fact the sequence of non-trivial periods $ \omega_{0} - \omega_{n} \to 0$ as $ n \to \infty$ contrary to what we have proved above.

Thus

Since the set of non-trivial periods is discrete we can find a non-trivial period which has the smallest modulus and thus we have the following result:

Let us now pick a period $ \omega_{1}$ in $ M(f)$ which has the least modulus. It is then clear that $ n\omega_{1}$ is a period where $ n$ is integer. It should now be observed that if $ p$ is real and $ p\omega_{1}$ is a period then $ p$ must be an integer. For if it is not the case then we can write $ p = I + g$ where $ I$ is the integral part of $ p$ and $ g$ with $ 0 < g < 1$ is the fractional part of $ p$. Now $ p\omega_{1}$ and $ I\omega_{1}$ are periods and therefore $ p\omega_{1} - I\omega_{1} = g\omega_{1}$ is a period and clearly $ |g\omega_{1}| < |\omega_{1}|$ contrary to the fact that $ \omega_{1}$ is a period with least modulus.

It may happen that all the periods turn out to be of the form $ n\omega_{1}$ with $ n$ being an integer. If this is the case we say that the functions $ f(z)$ is

We can now show that any arbitrary period $ \omega$ can be expressed

In the above case we say that the function $ f(z)$ is

Our interest will mainly be in doubly periodic functions (because we have seen earlier that elliptic functions are doubly periodic). In the next post we will see that the periods of a doubly periodic function form a lattice and we will try to discuss some general properties of period lattices.

- Periodicity in general - a function can have at most two most periods and in case it has two periods the ratio between periods can not be real.
- Lattices and period parallelogram - position of zeroes and poles of elliptic functions.
- Liouville's theorem on elliptic functions

### Periodicity in General

We will assume in the following that the functions involved are analytic in their domain of definition. We can then consider some general notions about periodicity. A function $ f(z)$ is said to be*periodic*if there is a complex number $ \omega$ such that $ f(z + \omega) = f(z)$ for all points $ z$ in the domain of definition of $ f(z)$. In such cases we say that $ \omega$ is a*period*of $ f(z)$. It is quite obvious that any function is periodic with the*trivial*period $ \omega = 0$ and hence we will be interested in non-trivial periods $ \omega \neq 0$. Another trivial case is when the function is a constant in which case any complex number is a period.It is also obvious that if $ \omega$ is any period then so is $ n\omega$ where $ n$ is an integer. Similarly if there are two or more periods $ \omega_{1}, \omega_{2}, \ldots, \omega_{k}$ then for any integers $ n_{1}, n_{2}, \ldots, n_{k}$, the number $ \omega = n_{1}\omega_{1} + n_{2}\omega_{2} + \cdots + n_{k}\omega_{k}$ is also a period. We will now analyze the structure of set of non-trivial periods of a function in more detail.

Let the set of non-trivial periods of $ f(z)$ be denoted by $ M(f)$. Thus $$M(f) = \{ \omega \mid \omega \neq 0, f(z + \omega) = f(z)\}$$ The first thing to notice here is that we can't have periods with arbitrarily small modulus, i.e. the lower bound for the set $ \{|\omega| : \omega \neq 0, f(z + \omega) = f(z)\}$ is definitely positive. If it were not so we could find a sequence of non-trivial periods $ \omega_{n}$ such that $ \omega_{n} \to 0$ as $ n \to \infty$. This means that $$\lim_{n \to \infty}\frac{f(z + \omega_{n}) - f(z)}{\omega_{n}} = 0$$ and this means that $ f'(z) = 0$ at all points in the domain of definition and therefore $ f(z)$ is constant.

Next it is not possible that the set of non-trivial periods has a limit point i.e. the set of non-trivial periods is discrete. This is quite obvious if we notice that in case there were a limit point say $ \omega_{0}$, then we could find a sequence $ \{\omega_{n}\}$ of non-trivial distinct periods such that $ \omega_{n} \to \omega_{0}$ as $ n \to \infty$. And then we would have $$ f(z + \omega_{0}) = \lim_{n \to \infty}f(z + \omega_{n}) = \lim_{n \to \infty} f(z) = f(z)$$ (assuming continuity of $ f(z)$ at $ z + \omega_{0}$) and thereby $ \omega_{0}$ would also be a non-trivial period. This would finally lead to the fact the sequence of non-trivial periods $ \omega_{0} - \omega_{n} \to 0$ as $ n \to \infty$ contrary to what we have proved above.

Thus

*the set of periods $ M(f)$ does not have any limit points in the finite portion of the complex plane.*Since the set of non-trivial periods is discrete we can find a non-trivial period which has the smallest modulus and thus we have the following result:

*If $ f(z)$ is not constant and periodic then there is a period $ \omega \neq 0$ such that $ |\omega| \leq |\omega'|$ for all periods $ \omega' \neq 0$.*Let us now pick a period $ \omega_{1}$ in $ M(f)$ which has the least modulus. It is then clear that $ n\omega_{1}$ is a period where $ n$ is integer. It should now be observed that if $ p$ is real and $ p\omega_{1}$ is a period then $ p$ must be an integer. For if it is not the case then we can write $ p = I + g$ where $ I$ is the integral part of $ p$ and $ g$ with $ 0 < g < 1$ is the fractional part of $ p$. Now $ p\omega_{1}$ and $ I\omega_{1}$ are periods and therefore $ p\omega_{1} - I\omega_{1} = g\omega_{1}$ is a period and clearly $ |g\omega_{1}| < |\omega_{1}|$ contrary to the fact that $ \omega_{1}$ is a period with least modulus.

It may happen that all the periods turn out to be of the form $ n\omega_{1}$ with $ n$ being an integer. If this is the case we say that the functions $ f(z)$ is

*singly periodic*with period $ \omega_{1}$. However if this is not the case then we have other periods which are not of the form $ n\omega_{1}$ ($ n$ being integer) and by the argument in the previous paragraph they are not in the form $ p\omega_{1}$ ($ p$ being real). Out of all these periods there will again be a period $ \omega_{2}$ with smallest modulus and then the ratio $ \omega_{2} / \omega_{1}$ is not real.We can now show that any arbitrary period $ \omega$ can be expressed

*uniquely*in the form $ \omega = m\omega_{1} + n\omega_{2}$ where $ m, n$ are integers. Since the ratio $ \omega_{2} / \omega_{1}$ is not real we can express $ \omega$ as $ \omega = p_{1}\omega_{1} + p_{2}\omega_{2}$ where $ p_{1}, p_{2}$ are real. We now choose integers $ k_{1}, k_{2}$ such that $ |p_{1} - k_{1}| \leq 1/2$ and $ |p_{2} - k_{2}| \leq 1/2$ so that $ \omega' = \omega - k_{1}\omega_{1} - k_{2}\omega_{2}$ is a period and \begin{align} |\omega'| &= |\omega - k_{1}\omega_{1} - k_{2}\omega_{2}| = |(p_{1} - k_{1})\omega_{1} + (p_{2} - k_{2})\omega_{2}|\notag\\ &< |p_{1} - k_{1}||\omega_{1}| + |p_{2} - k_{2}||\omega_{2}|\notag\\ &\leq \frac{1}{2}|\omega_{1}| + \frac{1}{2}|\omega_{2}| \leq |\omega_{2}|\notag \end{align} and therefore the period $ \omega'$ must be of the form $ n\omega_{1}$ where $ n$ is an integer. It now follows that $ \omega = (n + k_{1})\omega_{1} + k_{2}\omega_{2}$ and clearly this expression has to be unique otherwise subtracting two such expressions we shall get $ a_{1}\omega_{1} + a_{2}\omega_{2} = 0$ where $ a_{1}, a_{2}$ are integers. This would contradict the fact that the ratio $ \omega_{2} / \omega_{1}$ is not real.In the above case we say that the function $ f(z)$ is

*doubly periodic*with periods $ \omega_{1}$ and $ \omega_{2}$. To summarize:*An analytic function $ f(z)$ is either**constant or**not periodic or**singly periodic with a period $ \omega$ and each period $ \omega'$ is of the form $ \omega' = n\omega$ where $ n$ is an integer or**doubly periodic with two periods $ \omega_{1}$ and $ \omega_{2}$ such that the ratio $ \omega_{2} / \omega_{1}$ is not real and every period $ \omega$ can be expressed uniquely as $ \omega = m\omega_{1} + n\omega_{2}$ where $ m, n$ are integers.*

Our interest will mainly be in doubly periodic functions (because we have seen earlier that elliptic functions are doubly periodic). In the next post we will see that the periods of a doubly periodic function form a lattice and we will try to discuss some general properties of period lattices.

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