Elliptic Functions: Genesis of Theta Functions

We have seen in the last post that the elliptic functions can be expressed in the form of infinite product and these products look actually like ratio of two infinite products. We wish to consider these products (which are more commonly known as theta functions) in more detail in this post.

Theta Functions $\theta_{3}$ and $\theta_{4}$

Let us consider the infinite product expansion of $\text{dn}(u, k)$ $$\text{dn}(u, k) = \sqrt{k^{\prime}}\prod_{n = 1}^{\infty}\dfrac{1 + 2q^{2n - 1}\cos\left(\dfrac{\pi u}{K}\right) + q^{4n - 2}}{1 - 2q^{2n - 1}\cos\left(\dfrac{\pi u}{K}\right) + q^{4n - 2}}$$ We put $\tau = iK' / K$ so that $q = \exp(\pi i \tau) = \exp(-\pi K' / K)$ and $z = \pi u / 2K$. The denominator of the above expansion can be taken as a function $$h(z) = G\prod_{n = 1}^{\infty}(1 - 2q^{2n - 1}\cos 2z + q^{4n - 2}) = G\prod_{n = 1}^{\infty}(1 - q^{2n - 1}e^{-2iz})(1 - q^{2n - 1}e^{2iz})$$ where $G$ is some constant independent of $z$ to be determined later.

It is easily seen that the function $h(z)$ has zeroes for $z = \pm (2n - 1) (\pi\tau / 2) + m\pi$ where $m, n$ are integers. We express the same by saying that the zeroes of $h(z)$ are $z = \pi\tau / 2\,(\text{mod}\, \pi, \pi\tau)$.

Then clearly $h(z)$ is periodic with period $\pi$. Also we have $$\begin{align} h(z + \pi\tau) &= G\prod_{n = 1}^{\infty}(1 - q^{-2}q^{2n - 1}e^{-2iz})(1 - q^{2}q^{2n - 1}e^{2iz})\notag\\ &=\frac{1 - q^{-1}e^{-2iz}}{1 - qe^{2iz}}\,h(z)\notag\\ &= -q^{-1}e^{-2iz}h(z)\notag \end{align}$$ Note that the function $h(z)$ we have defined by means of an infinite product is defined only when $|q| < 1$ and when that is the case the defining product is absolutely convergent. Since $h(z)$ is periodic with period $\pi$ we can express it as a Fourier series $$h(z) = \sum_{n = -\infty}^{\infty}a_{n}e^{2niz}$$ and since $h(z)$ is even we have $a_{-n} = a_{n}$. Since $h(z + \pi\tau) = -q^{-1}e^{-2iz}h(z)$ we have $$\begin{align} \sum_{n = -\infty}^{\infty}a_{n}e^{2ni(z + \pi\tau)} &= -q^{-1}\sum_{n = -\infty}^{\infty}a_{n}e^{2(n - 1)iz}\notag\\ \Rightarrow \sum_{n = -\infty}^{\infty}a_{n}q^{2n + 1}e^{2niz} &= -\sum_{n = -\infty}^{\infty}a_{n + 1}e^{2niz}\notag \end{align}$$ Since the Fourier series is unique we must have $a_{n + 1} = -q^{2n + 1}a_{n}$. By repeated application of this relation it is easy to see that $a_{n} = (-1)^{n}q^{n^{2}}a_{0}$.

Thus we have $$h(z) = G\prod_{n = 1}^{\infty}(1 - q^{2n - 1}e^{-2iz})(1 - q^{2n - 1}e^{2iz}) = a_{0}\sum_{n = -\infty}^{\infty}(-1)^{n}q^{n^{2}}e^{2niz}$$ To simplify things we choose the constant $G$ such that we have $a_{0} = 1$. This determines (although still indirectly) $G$ and we have the relation $$h(z) = G\prod_{n = 1}^{\infty}(1 - q^{2n - 1}e^{-2iz})(1 - q^{2n - 1}e^{2iz}) = \sum_{n = -\infty}^{\infty}(-1)^{n}q^{n^{2}}e^{2niz}$$ The above function $h(z)$ as we have defined is one of the four theta functions and historically it has been denoted by $\theta_{4}(z, q) = \theta_{4}(z \mid \tau)$

We thus have $$\begin{align}\theta_{4}(z, q) &= G\prod_{n = 1}^{\infty}(1 - 2q^{2n - 1}\cos 2z + q^{4n - 2})\notag\\ &= \sum_{n = -\infty}^{\infty}(-1)^{n}q^{n^{2}}e^{2niz}\notag\\ &= 1 + 2\sum_{n = 1}^{\infty}(-1)^{n}q^{n^{2}}\cos(2nz)\notag\end{align}$$ We can handle the numerator in the infinite product expansion of $\text{dn}(u, k)$ to get another theta function $\theta_{3}(z, q) = \theta_{3}(z \mid \tau)$ and get the following $$\theta_{3}(z, q) = G\prod_{n = 1}^{\infty}(1 + 2q^{2n - 1}\cos 2z + q^{4n - 2}) = \sum_{n = -\infty}^{\infty}q^{n^{2}}e^{2niz} = 1 + 2\sum_{n = 1}^{\infty}q^{n^{2}}\cos(2nz)$$ It is quite obvious that $\theta_{3}(z, q) = \theta_{4}(z \pm \pi / 2, q)$ and hence the zeroes of $\theta_{3}(z, q)$ are given by $z = (\pi / 2) + (\pi\tau) / 2\, (\text{mod}\, \pi, \pi\tau)$.

Like the case of $\theta_{4}(z, q)$ it is easy to see that $\theta_{3}(z + \pi\tau) = q^{-1}e^{-2iz}\theta_{3}(z, q)$ and therefore if we consider the function $F(z) = \theta_{3}(z, q) / \theta_{4}(z, q)$, we can easily see that $F(z + \pi) = F(z)$ and $F(z + \pi\tau) = -F(z)$ so that $F(z + 2\pi\tau) = F(z)$. It is now clear that the function $F(z)$ is doubly periodic with periods $\pi$ and $2\pi\tau$. This corresponds exactly with the periods $2K$ and $4iK'$ of $\text{dn}(u, k)$ if we notice that $z = \pi u / 2K$. Moreover the poles of the function $\theta_{4}(z, q)$ coincide with the those of $\text{dn}(u, k)$. It follows that the function $\text{dn}(u, k) / F(z)$ is doubly periodic with no poles and hence is a constant.

Thus we have $$\text{dn}(u, k) = D\,\frac{\theta_{3}(z, q)}{\theta_{4}(z, q)}$$ where $z = \pi u / 2K$ and $D$ is a constant independent of $u$. If we put $u = 0$ and $u = K$ we can evaluate $D = \sqrt{k'}$ and this justifies the product expansions obtained in last post. We also obtain the value of $k'$ in the form of theta functions as $$k' = \frac{\theta_{4}^{2}(0, q)}{\theta_{3}^{2}(0, q)}$$ It is customary to denote $\theta_{i}(0, q)$ by $\theta_{i}(q)$ or even by $\theta_{i}$ when there is no confusion. Thus we get $k' = \theta_{4}^{2} / \theta_{3}^{2}$

Theta Functions $\theta_{1}$ and $\theta_{2}$

We now focus on obtaining the product expansions of $\text{sn}(u, k)$ and $\text{cn}(u, k)$. It is clear that the poles of the $\text{sn}(u, k)$ match with those of the zeroes of $\theta_{4}(z, q)$ (noting that $z = \pi u / 2K$). Hence we expect that there could be a relation of the form $$\text{sn}(u, k) = \frac{g(z)}{\theta_{4}(z, q)}$$ Here we also expect $g(z)$ to be periodic with period $2\pi$ and the zeroes of $g(z)$ should match that of $\text{sn}(u, k)$. This means that the functions $g(z)$ should vanish at $z = 0 (\text{mod}\, \pi, \pi\tau)$. This latter property can be achieved by setting $$g(z) = \theta_{4}\left(z + \frac{1}{2}\pi\tau\right)$$ But this will not make $g(z)$ periodic with $2\pi$. Moreover we have another problem to solve, namely to make the ratio $g(z) / \theta_{4}(z, q)$ doubly periodic with the imaginary period $\pi\tau$. This would require us to have the periodicity factor $g(z + \pi\tau) / g(z)$ to match the periodicity factor $\theta_{4}(z + \pi\tau, q) / \theta_{4}(z, q)$.

The periodicity factor for $g(z)$ can be found by noting that $$\begin{align} g(z + \pi\tau) &= \theta_{4}\left(z + \frac{1}{2}\pi\tau + \pi\tau\right)\notag\\ &= -q^{-1}e^{-2i(z + \pi\tau / 2)}\theta_{4}\left(z + \frac{1}{2}\pi\tau\right)\notag\\ &= -q^{-1}e^{-2iz}e^{-\pi i\tau}g(z)\notag\end{align}$$ It is now easy to see that we can get our job done if we change the definition of $g(z)$ to $$g(z) = Ae^{iz}\theta_{4}\left(z + \frac{1}{2}\pi\tau\right)$$ where $A$ is a constant to be determined later. The above function $g(z)$ is better known historically as $\theta_{1}(z, q)$. Thus we have $$\theta_{1}(z, q) = Ae^{iz}\theta_{4}\left(z + \frac{1}{2}\pi\tau\right)$$ It is then clear that $\theta_{1}(z, q)$ is periodic with period $2\pi$ and the periodicity factor $\theta_{1}(z + \pi\tau, q) / \theta_{1}(z, q) = -q^{-1}e^{-2iz}$. Therefore the function $F(z) = \theta_{1}(z, q) / \theta_{4}(z, q)$ is doubly periodic with periods $2\pi, \pi\tau$ and has poles at $z = 0(\text{mod}\,\pi,\pi\tau)$. It thus follows that the functions $\text{sn}(u, k) / F(z)$ is doubly periodic entire function and thus is a constant. We have now established that
$$\text{sn}(u, k) = S\frac{\theta_{1}(z, q)}{\theta_{4}(z, q)}$$ for some constant $S$ depending upon $A$ and $k$.

Similarly we can handle the elliptic function $\text{cn}(u, k)$. Clearly we can take $\text{cn}(u, k) = g(z) / \theta_{4}(z, q)$ where $g(z)$ is having a period $2\pi$ and the periodicity factor should be such that the function $g(z) / \theta_{4}(z, q)$ is periodic with period $2\pi\tau$. Hence we must have the periodicity factor of $g(z)$ the same as that of $\theta_{4}(z, q)$ but opposite in sign. Both these goals can be achieved by setting $g(z) = \theta_{1}(z + \pi / 2, q)$.

Thus we come to our last theta function $\theta_{2}$ defined by $$\theta_{2}(z, q) = \theta_{1}\left(z + \frac{\pi}{2}, q\right)$$ And like before, we have the relation $$\text{cn}(u, k) = C\frac{\theta_{2}(z, q)}{\theta_{4}(z, q)}$$ where $C$ is a constant dependent on $A$ and $k$.

It now remains to evaluate the constants $A, C, S$. To begin with we have $$\begin{align} \theta_{1}(z, q) &= Ae^{iz}\theta_{4}\left(z + \frac{1}{2}\pi\tau\right) = Ae^{iz}\sum_{n = -\infty}^{\infty}(-1)^{n}q^{n^{2}}e^{2ni(z + \pi\tau / 2)}\notag\\ &= Ae^{iz}\sum_{n = -\infty}^{\infty}(-1)^{n}q^{n^{2}}q^{n}e^{2niz} = Aq^{-1 / 4}\sum_{n = -\infty}^{\infty}(-1)^{n}q^{(n + 1/2)^{2}}e^{(2n + 1)iz}\notag\\ &= 2Aiq^{-1 / 4}\sum_{n = 0}^{\infty}(-1)^{n}q^{(n + 1/2)^{2}}\sin(2n + 1)z\notag \end{align}$$ A natural choice for the constant $A$ is $A = -iq^{1/4}$ and then $$\theta_{1}(z, q) = 2\sum_{n = 0}^{\infty}(-1)^{n}q^{(n + 1/2)^{2}}\sin(2n + 1)z$$ Also it is worth noticing the infinite product expansion of $\theta_{1}(z, q)$. We have $$\begin{align} \theta_{1}(z, q) &= -iq^{1/4}\theta_{4}\left(z + \frac{1}{2}\pi\tau\right)e^{iz}\notag\\ &= -iq^{1/4}G\prod_{n = 1}^{\infty}(1 - q^{2n - 2}e^{-2iz})(1 - q^{2n}e^{2iz})e^{iz}\notag\\ &= 2Gq^{1/4}\sin z\prod_{n = 1}^{\infty}(1 - q^{2n}e^{-2iz})(1 - q^{2n}e^{2iz})\notag \end{align}$$ Thus we finally have $$\begin{align} \theta_{1}(z, q) &= 2\sum_{n = 0}^{\infty}(-1)^{n}q^{(n + 1/2)^{2}}\sin(2n + 1)z\notag\\ &= 2Gq^{1/4}\sin z\prod_{n = 1}^{\infty}(1 - 2q^{2n}\cos 2z + q^{4n})\notag\ \end{align}$$ and since $\theta_{2}(z, q) = \theta_{1}(z + \pi / 2, q)$ it follows that $$\begin{align} \theta_{2}(z, q) &= 2\sum_{n = 0}^{\infty}q^{(n + 1/2)^{2}}\cos(2n + 1)z\notag\\ &= 2Gq^{1/4}\cos z\prod_{n = 1}^{\infty}(1 + 2q^{2n}\cos 2z + q^{4n})\notag \end{align}$$ Again note that since $$\theta_{1}(z, q) = -iq^{1/4}e^{iz}\theta_{4}\left(z + \frac{1}{2}\pi\tau\right)\tag{A}$$ we have $$\theta_{4}\left(z + \frac{1}{2}\pi\tau, q\right) = iq^{-1/4}e^{-iz}\theta_{1}(z, q)\tag{B}$$ Replacing $z$ with $z + \pi\tau / 2$ in $(A)$ we get $$\begin{align} \theta_{1}\left(z + \frac{1}{2}\pi\tau, q\right) &= -iq^{3/4}e^{iz}\theta_{4}(z + \pi\tau)\notag\\ \Rightarrow \theta_{1}\left(z + \frac{1}{2}\pi\tau, q\right) &= iq^{-1/4}e^{-iz}\theta_{4}(z, q)\tag{C} \end{align}$$ Dividing $(C)$ by $(B)$ we get $$\dfrac{\theta_{1}\left(z + \dfrac{1}{2}\pi\tau, q\right)}{\theta_{4}\left(z + \dfrac{1}{2}\pi\tau, q\right)} = \dfrac{\theta_{4}(z, q)}{\theta_{1}(z, q)}\tag{D}$$ This is an identity which will be used in the evaluation of the constant $S$ and $C$. To evaluate them we just need to put $u = K$ in the formula for $\text{sn}(u, k)$ to get $$1 = S\dfrac{\theta_{1}\left(\dfrac{\pi}{2}, q\right)}{\theta_{4}\left(\dfrac{\pi}{2}, q\right)} = S\frac{\theta_{2}(0, q)}{\theta_{3}(0, q)}\Rightarrow S = \frac{\theta_{3}(0, q)}{\theta_{2}(0, q)} = \frac{\theta_{3}}{\theta_{2}}$$ Again putting $u = K + iK'$ in the formula for $\text{sn}(u, k)$ we get $$\frac{1}{k} = S\dfrac{\theta_{1}\left(\dfrac{\pi}{2} + \dfrac{\pi\tau}{2}, q\right)}{\theta_{4}\left(\dfrac{\pi}{2} + \dfrac{\pi\tau}{2}, q\right)} = S\dfrac{\theta_{4}\left(\dfrac{\pi}{2}, q\right)}{\theta_{1}\left(\dfrac{\pi}{2}, q\right)} = S \frac{\theta_{3}(0, q)}{\theta_{2}(0, q)} = S^{2}$$ and thus we get $$S = k^{-1/2},\,\,\,\, k = \frac{\theta_{2}^{2}}{\theta_{3}^{2}}$$ To evaluate $C$ we put $u = 0$ in the formula for $\text{cn}(u, k)$ to get $$1 = C\frac{\theta_{2}(0, q)}{\theta_{4}(0, q)} \Rightarrow C = \frac{\theta_{4}}{\theta_{2}} = \frac{\theta_{4} / \theta_{3}}{\theta_{2} / \theta_{3}} = \frac{\sqrt{k'}}{\sqrt{k}}$$

Summary

To summarize we have the following results $$\begin{align} \theta_{1}(z, q) &= 2\sum_{n = 0}^{\infty}(-1)^{n}q^{(n + 1/2)^{2}}\sin(2n + 1)z\notag\\ &= 2Gq^{1/4}\sin z\prod_{n = 1}^{\infty}(1 - 2q^{2n}\cos 2z + q^{4n})\notag\\ \theta_{2}(z, q) &= 2\sum_{n = 0}^{\infty}q^{(n + 1/2)^{2}}\cos(2n + 1)z\notag\\ &= 2Gq^{1/4}\cos z\prod_{n = 1}^{\infty}(1 + 2q^{2n}\cos 2z + q^{4n})\notag\\ \theta_{3}(z, q) &= G\prod_{n = 1}^{\infty}(1 + 2q^{2n - 1}\cos 2z + q^{4n - 2})\notag\\ &= \sum_{n = -\infty}^{\infty}q^{n^{2}}e^{2niz} = 1 + 2\sum_{n = 1}^{\infty}q^{n^{2}}\cos(2nz)\notag\\ \theta_{4}(z, q) &= G\prod_{n = 1}^{\infty}(1 - 2q^{2n - 1}\cos 2z + q^{4n - 2})\notag\\ &= \sum_{n = -\infty}^{\infty}(-1)^{n}q^{n^{2}}e^{2niz} = 1 + 2\sum_{n = 1}^{\infty}(-1)^{n}q^{n^{2}}\cos(2nz)\notag\\ \text{sn}(u, k) &= \frac{\theta_{3}}{\theta_{2}}\frac{\theta_{1}(z, q)}{\theta_{4}(z, q)} = \dfrac{2q^{1/4}}{\sqrt{k}}\sin\left(\dfrac{\pi u}{2K}\right)\prod_{n = 1}^{\infty}\dfrac{1 - 2q^{2n}\cos\left(\dfrac{\pi u}{K}\right) + q^{4n}}{1 - 2q^{2n - 1}\cos\left(\dfrac{\pi u}{K}\right) + q^{4n - 2}}\notag\\ \text{cn}(u, k) &= \frac{\theta_{4}}{\theta_{2}}\frac{\theta_{2}(z, q)}{\theta_{4}(z, q)}=\dfrac{2q^{1/4}\sqrt{k'}}{\sqrt{k}}\cos\left(\dfrac{\pi u}{2K}\right)\prod_{n = 1}^{\infty}\dfrac{1 + 2q^{2n}\cos\left(\dfrac{\pi u}{K}\right) + q^{4n}}{1 - 2q^{2n - 1}\cos\left(\dfrac{\pi u}{K}\right) + q^{4n - 2}}\notag\\ \text{dn}(u, k) &= \frac{\theta_{4}}{\theta_{3}}\frac{\theta_{3}(z, q)}{\theta_{4}(z, q)}=\sqrt{k'}\prod_{n = 1}^{\infty}\dfrac{1 + 2q^{2n - 1}\cos\left(\dfrac{\pi u}{K}\right) + q^{4n - 2}}{1 - 2q^{2n - 1}\cos\left(\dfrac{\pi u}{K}\right) + q^{4n - 2}}\notag \end{align}$$ where $$\begin{align} z &= \frac{\pi u}{2K},\,\, q = \exp\left(-\frac{\pi K'}{K}\right)\notag\\ k &= \frac{\theta_{2}^{2}}{\theta_{3}^{2}} = 4\sqrt{q}\prod_{n = 1}^{\infty}\left(\frac{1 + q^{2n}}{1 + q^{2n - 1}}\right)^{4}\notag\\ k' &= \frac{\theta_{4}^{2}}{\theta_{3}^{2}} = \prod_{n = 1}^{\infty}\left(\frac{1 - q^{2n - 1}}{1 + q^{2n - 1}}\right)^{4}\notag \end{align}$$ and since we have $k^{2} + k'^{2} = 1$ we get the fundamental identity $$\theta_{2}^{4} + \theta_{4}^{4} = \theta_{3}^{4}$$ To complete the story we have yet to evaluate the constant $G$ and this will be done in the next post.

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