In the previous post, we had some general discussion on the periodicity and came to the conclusion that an analytic function can have at most two independent periods and in that case the ratio of periods can not be real. It was also established that if $ \omega_{1}$ and $ \omega_{2}$ are two independent periods then any period $ \omega$ can be expressed as $ \omega = m\omega_{1} + n\omega_{2}$ where $ m, n$ are integers.

Let us then consider the lattice

$$\Lambda = \{ m\omega_{1} + n\omega_{2} \mid m, n \in \mathbb{Z}, \Im({\omega_{2} / \omega_{1}}) > 0\} = \langle\omega_{1}, \omega_{2}\rangle$$ where the notation of angle brackets on the right shows that the lattice is ultimately dependent on the basis.

Now it is possible that there could be more than basis of a given lattice. For example the lattice $ \langle 1, i\rangle$ also has the basis $ (1, 1 + i)$. We will now try to find if there is relation between any two bases of a given lattice. Thus let us suppose that for the above mentioned lattice $ \Lambda$ we also have another basis $ (\omega_{1}', \omega_{2}')$. Then we can express $ \omega_{1}', \omega_{2}'$ as a linear combination of $ \omega_{1}, \omega_{2}$ over the integers. Therefore we have \begin{align} \omega_{1}' &= a\omega_{1} + b\omega_{2}\notag\\ \omega_{2}' &= c\omega_{1} + d\omega_{2}\notag \end{align} for some integers $ a, b, c, d$ and by the same argument we have \begin{align} \omega_{1} &= a'\omega_{1}' + b'\omega_{2}'\notag\\ \omega_{2} &= c'\omega_{1}' + d'\omega_{2}'\notag \end{align} for some integers $ a', b', c', d'$.

The above equations can be written in the matrix form as \begin{align} \begin{bmatrix} \omega_{1}' \\ \omega_{2}' \end{bmatrix} &= \begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} \omega_{1} \\ \omega_{2} \end{bmatrix}\notag\\ \begin{bmatrix} \omega_{1} \\ \omega_{2} \end{bmatrix} &= \begin{bmatrix} a' & b' \\ c' & d' \end{bmatrix} \begin{bmatrix} \omega_{1}' \\ \omega_{2}' \end{bmatrix}\notag \end{align} $$\Rightarrow \begin{bmatrix} \omega_{1} \\ \omega_{2} \end{bmatrix} = \begin{bmatrix} a' & b' \\ c' & d' \end{bmatrix} \begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} \omega_{1} \\ \omega_{2} \end{bmatrix} = \begin{bmatrix} A & B \\ C & D \end{bmatrix} \begin{bmatrix} \omega_{1} \\ \omega_{2} \end{bmatrix} = \begin{bmatrix} A\omega_{1} + B\omega_{2} \\ C\omega_{1} + D\omega_{2} \end{bmatrix}$$ Since the ratio $ \omega_{2} / \omega_{1}$ is not real it follows that we have $ A = 1, B = 0, C = 0, D = 1$ and hence it follows that $$ \begin{bmatrix} a' & b' \\ c' & d' \end{bmatrix} \begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$ and hence $$\begin{vmatrix} a' & b' \\ c' & d' \end{vmatrix} \begin{vmatrix} a & b \\ c & d\end{vmatrix} = 1$$ Since $ a, b, c, d, a', b', c', d'$ are integers it follows that $$\begin{vmatrix} a & b \\ c & d\end{vmatrix} = \pm 1 \Rightarrow ad - bc = \pm 1$$ To determine the ambiguous sign we use the fact that the bases $ (\omega_{1}, \omega_{2})$ and $ (\omega_{1}', \omega_{2}')$ are both normalized and set $ \tau = \omega_{2} / \omega_{1}, \tau' = \omega_{2}' / \omega_{1}'$ and then we have \begin{align} \tau' &= \frac{\omega_{2}'}{\omega_{1}'} = \frac{c\omega_{1} + d\omega_{2}}{a\omega_{1} + b\omega_{2}} = \frac{c + d\tau}{a + b\tau} = \frac{(c + d\tau)(a + b\bar{\tau})}{|a + b\tau|^{2}}\notag\\ \Rightarrow \tau' &= \frac{ac + bd\tau\bar{\tau} + ad\tau + bc\bar{\tau}}{|a + b\tau|^{2}}\notag\\ \Rightarrow \Im(\tau') &= \frac{ad - bc}{|a + b\tau|^{2}}\Im(\tau)\notag \end{align} Since both $ \Im(\tau')$ and $ \Im(\tau)$ are positive it follows that $ ad - bc = 1$. Therefore it follows that if a lattice $ \Lambda$ has two bases $ (\omega_{1}, \omega_{2})$ and $ (\omega_{1}', \omega_{2}')$ then $$\begin{bmatrix} \omega_{1}' \\ \omega_{2}'\end{bmatrix} = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} \omega_{1} \\ \omega_{2} \end{bmatrix}$$ for some integers $ a, b, c, d$ such that $ ad - bc = 1$.

Clearly under the action of $ S$ we have $ \tau' = \tau + 1$ and under the action of $ T$ we have $ \tau' = -1 / \tau$. Therefore we can make a special choice of the ratio $ \tau = \omega_{2} / \omega_{1}$ for a given lattice. Applying $ S$ or $ S^{-1}$ a number of times, we can always get the real part of $ \tau$ in the range $ -1 / 2 < \Re(\tau) \leq 1 / 2$. Similarly we can assume that $ |\tau| \geq 1$ otherwise we can apply $ T$ to make $ |\tau| \geq 1$.

Thus we can consider the region of the complex plane $$\mathcal{F} = \{ z \in \mathbb{C} \mid \Im(z) > 0, |z| \geq 1, -\frac{1}{2} < \Re(z) \leq \frac{1}{2}\,\, \text{and}\,\, \Re(z) \geq 0\,\, \text{if}\,\, |z| = 1\}$$ This region is called the

By periodicity the values of the function $ f(z)$ are same in any period parallelogram and therefore its values are completely determined by the values it takes in any given period parallelogram. Moreover it should also be noted that it is really not necessary to keep the periods as the vertices of the parallelogram. Any four points of type $$ c, c + \omega_{1}, c + \omega_{1} + \omega_{2}, c + \omega_{2}$$ will make a parallelogram and the values of the function in this parallelogram will determines its values in the entire complex plane.

In general since the elliptic functions have poles at isolated points in the complex plane, it is possible to find a period parallelogram such that the the function has no poles on the the boundary of the parallelogram. Thus let $ P_{a}$ be the parallelogram with vertices $$ a, a + \omega_{1}, a + \omega_{1} + \omega_{2}, a + \omega_{2}$$ such that there are no poles of the elliptic function $ f(z)$ (here by $ f(z)$ we will mean $ \text{sn}(z), \text{cn}(z)$ etc, although the results which we establish here are valid for any doubly periodic function) on the boundary $ \partial P_{a}$ of the parallelogram $ P_{a}$.

We can now see that the doubly periodic function must possess poles in the parallelogram $ P_{a}$. For if it were not so then $ f(z)$ would be bounded in $ P_{a}$ and hence bounded in the whole complex plane (by periodicity the values in $ P_{a}$ are repeated all over the complex plane). By Liouville's theorem this is not possible unless $ f(z)$ is constant. Hence it follows that:

It is this theorem which Liouville had actually proved but historically the statement "

Another result which is a direct consequence of the double periodicity is the following:

To establish this we only need to observe that the sum of residues of the functions $ f(z)$ in a period parallelogram $ P_{a}$ is given by the integral $$\frac{1}{2\pi i}\int_{\partial P_{a}}f(z)\,dz$$ By double periodicity the integrals over the opposites sides of the parallelogram cancel each other and thereby the integral over the whole parallelogram is zero.

The above result implies that we can not have a doubly periodic function with arbitrary number and nature of poles. For example a doubly periodic function can't have only one simple pole in a period parallelogram. In general a doubly periodic function is completely determined by its periods and location and nature of its poles in the period parallelogram.

### Lattices

Such a set of complex numbers of the form $ m\omega_{1} + n\omega_{2}$ with $ m, n$ being integers is called a*lattice*or more properly a*complex lattice*. Also we say that the complex numbers $ \omega_{1}, \omega_{2}$ form a*basis*of the lattice. It is quite obvious that if $ \omega_{1}, \omega_{2}$ is a basis, then so is $ \omega_{2}, \omega_{1}$. We choose a particular order so that the imaginary part of $ \omega_{2} / \omega_{1}$ is positive. Such a basis is then called a*normalized*basis. Henceforth we shall only consider normalized bases.Let us then consider the lattice

$$\Lambda = \{ m\omega_{1} + n\omega_{2} \mid m, n \in \mathbb{Z}, \Im({\omega_{2} / \omega_{1}}) > 0\} = \langle\omega_{1}, \omega_{2}\rangle$$ where the notation of angle brackets on the right shows that the lattice is ultimately dependent on the basis.

Now it is possible that there could be more than basis of a given lattice. For example the lattice $ \langle 1, i\rangle$ also has the basis $ (1, 1 + i)$. We will now try to find if there is relation between any two bases of a given lattice. Thus let us suppose that for the above mentioned lattice $ \Lambda$ we also have another basis $ (\omega_{1}', \omega_{2}')$. Then we can express $ \omega_{1}', \omega_{2}'$ as a linear combination of $ \omega_{1}, \omega_{2}$ over the integers. Therefore we have \begin{align} \omega_{1}' &= a\omega_{1} + b\omega_{2}\notag\\ \omega_{2}' &= c\omega_{1} + d\omega_{2}\notag \end{align} for some integers $ a, b, c, d$ and by the same argument we have \begin{align} \omega_{1} &= a'\omega_{1}' + b'\omega_{2}'\notag\\ \omega_{2} &= c'\omega_{1}' + d'\omega_{2}'\notag \end{align} for some integers $ a', b', c', d'$.

The above equations can be written in the matrix form as \begin{align} \begin{bmatrix} \omega_{1}' \\ \omega_{2}' \end{bmatrix} &= \begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} \omega_{1} \\ \omega_{2} \end{bmatrix}\notag\\ \begin{bmatrix} \omega_{1} \\ \omega_{2} \end{bmatrix} &= \begin{bmatrix} a' & b' \\ c' & d' \end{bmatrix} \begin{bmatrix} \omega_{1}' \\ \omega_{2}' \end{bmatrix}\notag \end{align} $$\Rightarrow \begin{bmatrix} \omega_{1} \\ \omega_{2} \end{bmatrix} = \begin{bmatrix} a' & b' \\ c' & d' \end{bmatrix} \begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} \omega_{1} \\ \omega_{2} \end{bmatrix} = \begin{bmatrix} A & B \\ C & D \end{bmatrix} \begin{bmatrix} \omega_{1} \\ \omega_{2} \end{bmatrix} = \begin{bmatrix} A\omega_{1} + B\omega_{2} \\ C\omega_{1} + D\omega_{2} \end{bmatrix}$$ Since the ratio $ \omega_{2} / \omega_{1}$ is not real it follows that we have $ A = 1, B = 0, C = 0, D = 1$ and hence it follows that $$ \begin{bmatrix} a' & b' \\ c' & d' \end{bmatrix} \begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$ and hence $$\begin{vmatrix} a' & b' \\ c' & d' \end{vmatrix} \begin{vmatrix} a & b \\ c & d\end{vmatrix} = 1$$ Since $ a, b, c, d, a', b', c', d'$ are integers it follows that $$\begin{vmatrix} a & b \\ c & d\end{vmatrix} = \pm 1 \Rightarrow ad - bc = \pm 1$$ To determine the ambiguous sign we use the fact that the bases $ (\omega_{1}, \omega_{2})$ and $ (\omega_{1}', \omega_{2}')$ are both normalized and set $ \tau = \omega_{2} / \omega_{1}, \tau' = \omega_{2}' / \omega_{1}'$ and then we have \begin{align} \tau' &= \frac{\omega_{2}'}{\omega_{1}'} = \frac{c\omega_{1} + d\omega_{2}}{a\omega_{1} + b\omega_{2}} = \frac{c + d\tau}{a + b\tau} = \frac{(c + d\tau)(a + b\bar{\tau})}{|a + b\tau|^{2}}\notag\\ \Rightarrow \tau' &= \frac{ac + bd\tau\bar{\tau} + ad\tau + bc\bar{\tau}}{|a + b\tau|^{2}}\notag\\ \Rightarrow \Im(\tau') &= \frac{ad - bc}{|a + b\tau|^{2}}\Im(\tau)\notag \end{align} Since both $ \Im(\tau')$ and $ \Im(\tau)$ are positive it follows that $ ad - bc = 1$. Therefore it follows that if a lattice $ \Lambda$ has two bases $ (\omega_{1}, \omega_{2})$ and $ (\omega_{1}', \omega_{2}')$ then $$\begin{bmatrix} \omega_{1}' \\ \omega_{2}'\end{bmatrix} = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} \omega_{1} \\ \omega_{2} \end{bmatrix}$$ for some integers $ a, b, c, d$ such that $ ad - bc = 1$.

### The Unimodular Group and its Fundamental Region

The matrices of the form $ \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ where $ a, b, c, d$ are integers with $ ad - bc = 1$ form a group under matrix multiplication and this group is called the*unimodular group*and denoted by $ SL(2, \mathbb{Z})$. The two particular elements of this group $ S = \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}$ and $ T = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}$ are worth noticing in the sense that they generate the entire unimodular group. In other words any member of this group can be expressed as a product of a finite number of $ S$ and $ T$.Clearly under the action of $ S$ we have $ \tau' = \tau + 1$ and under the action of $ T$ we have $ \tau' = -1 / \tau$. Therefore we can make a special choice of the ratio $ \tau = \omega_{2} / \omega_{1}$ for a given lattice. Applying $ S$ or $ S^{-1}$ a number of times, we can always get the real part of $ \tau$ in the range $ -1 / 2 < \Re(\tau) \leq 1 / 2$. Similarly we can assume that $ |\tau| \geq 1$ otherwise we can apply $ T$ to make $ |\tau| \geq 1$.

Thus we can consider the region of the complex plane $$\mathcal{F} = \{ z \in \mathbb{C} \mid \Im(z) > 0, |z| \geq 1, -\frac{1}{2} < \Re(z) \leq \frac{1}{2}\,\, \text{and}\,\, \Re(z) \geq 0\,\, \text{if}\,\, |z| = 1\}$$ This region is called the

*fundamental region*for the unimodular group $ SL(2, \mathbb{Z})$. Given any lattice $ \Lambda$ we can find a basis $ (\omega_{1}, \omega_{2})$ such that the ratio $ \tau = \omega_{2} / \omega_{1}$ lies in the fundamental region. Moreover this $ \tau$ will be unique in $ \mathcal{F}$ for a given lattice.### Period Parallelogram and Liouville's Theorem

Let $ f(z)$ be a doubly periodic function with two independent periods $ \omega_{1}$ and $ \omega_{2}$. Consider the points $ 0, \omega_{1}, \omega_{1} + \omega_{2}, \omega_{2}$ in that particular order. They form a parallelogram in the complex plane. In general the points $$ m\omega_{1} + n\omega_{2}, (m + 1)\omega_{1} + n\omega_{2}, (m + 1)\omega_{1} + (n + 1)\omega_{2}, m\omega_{1} + (n + 1)\omega_{2}$$ also form a parallelogram and set of such parallelograms for all integers $ m, n$ actually forms a tiling of the complex plane.By periodicity the values of the function $ f(z)$ are same in any period parallelogram and therefore its values are completely determined by the values it takes in any given period parallelogram. Moreover it should also be noted that it is really not necessary to keep the periods as the vertices of the parallelogram. Any four points of type $$ c, c + \omega_{1}, c + \omega_{1} + \omega_{2}, c + \omega_{2}$$ will make a parallelogram and the values of the function in this parallelogram will determines its values in the entire complex plane.

In general since the elliptic functions have poles at isolated points in the complex plane, it is possible to find a period parallelogram such that the the function has no poles on the the boundary of the parallelogram. Thus let $ P_{a}$ be the parallelogram with vertices $$ a, a + \omega_{1}, a + \omega_{1} + \omega_{2}, a + \omega_{2}$$ such that there are no poles of the elliptic function $ f(z)$ (here by $ f(z)$ we will mean $ \text{sn}(z), \text{cn}(z)$ etc, although the results which we establish here are valid for any doubly periodic function) on the boundary $ \partial P_{a}$ of the parallelogram $ P_{a}$.

We can now see that the doubly periodic function must possess poles in the parallelogram $ P_{a}$. For if it were not so then $ f(z)$ would be bounded in $ P_{a}$ and hence bounded in the whole complex plane (by periodicity the values in $ P_{a}$ are repeated all over the complex plane). By Liouville's theorem this is not possible unless $ f(z)$ is constant. Hence it follows that:

**Liouville's Theorem:***There does not exist a non constant doubly periodic entire function.*It is this theorem which Liouville had actually proved but historically the statement "

*an entire function cannot be bounded unless it is a constant*" is referred to as the Liouville's theorem.Another result which is a direct consequence of the double periodicity is the following:

*The sum of residues of a doubly periodic function in a period parallelogram is zero.*To establish this we only need to observe that the sum of residues of the functions $ f(z)$ in a period parallelogram $ P_{a}$ is given by the integral $$\frac{1}{2\pi i}\int_{\partial P_{a}}f(z)\,dz$$ By double periodicity the integrals over the opposites sides of the parallelogram cancel each other and thereby the integral over the whole parallelogram is zero.

The above result implies that we can not have a doubly periodic function with arbitrary number and nature of poles. For example a doubly periodic function can't have only one simple pole in a period parallelogram. In general a doubly periodic function is completely determined by its periods and location and nature of its poles in the period parallelogram.

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