Elliptic Functions: Theta Function Identities

In this post we will prove some theta function identities. We will try to pass from an existing identity between the elliptic functions to a corresponding identity between theta functions. Sometimes we will also establish identities which follow quite obviously from the series or product expansions of the theta functions. Most of the times we will also make use of the Liouville's theorem that any doubly periodic entire function is a constant.

Squares of Theta Functions

We can start with the standard relation $$\text{cn}^{2}(u, k) + \text{sn}^{2}(u, k) = 1$$ to get the following relation between theta functions \begin{align} \frac{\theta_{4}^{2}}{\theta_{2}^{2}}\frac{\theta_{2}^{2}(z, q)}{\theta_{4}^{2}(z, q)} &+ \frac{\theta_{3}^{2}}{\theta_{2}^{2}}\frac{\theta_{1}^{2}(z, q)}{\theta_{4}^{2}(z, q)} = 1\notag\\ \Rightarrow \theta_{2}^{2}\theta_{4}^{2}(z, q) &= \theta_{4}^{2}\theta_{2}^{2}(z, q) + \theta_{3}^{2}\theta_{1}^{2}(z, q)\notag \end{align} and thus we get $$\theta_{4}^{2}\theta_{2}^{2}(z, q) = \theta_{2}^{2}\theta_{4}^{2}(z, q) - \theta_{3}^{2}\theta_{1}^{2}(z, q)$$ Similarly from the equation $ \text{dn}^{2}(u, k) + k^{2}\,\text{sn}^{2}(u, k) = 1$ we get $$\theta_{4}^{2}\theta_{3}^{2}(z, q) = \theta_{3}^{2}\theta_{4}^{2}(z, q) - \theta_{2}^{2}\theta_{1}^{2}(z, q)$$ Replacing $ z$ by $ z + \pi / 2$ in the above relations we get \begin{align} \theta_{4}^{2}\theta_{1}^{2}(z, q) &= \theta_{2}^{2}\theta_{3}^{2}(z, q) - \theta_{3}^{2}\theta_{2}^{2}(z, q)\notag\\ \theta_{4}^{2}\theta_{4}^{2}(z, q) &= \theta_{3}^{2}\theta_{3}^{2}(z, q) - \theta_{2}^{2}\theta_{2}^{2}(z, q)\notag \end{align} Therefore we can express the square of a theta functions in terms of the squares of two other theta functions. To this list we can add the following identities which are easily seen as consequence of the series expansions of theta functions: \begin{align} \theta_{3}(z, q) &= \theta_{3}(2z, q^{4}) + \theta_{2}(2z, q^{4})\notag\\ \theta_{4}(z, q) &= \theta_{3}(2z, q^{4}) - \theta_{2}(2z, q^{4})\notag \end{align}

Addition Formulas

We can next focus on the addition formulae and set \begin{align} s_{1} &= \text{sn}(u, k), c_{1} = \text{cn}(u, k), d_{1} = \text{dn}(u, k)\notag\\ s_{2} &= \text{sn}(v, k), c_{2} = \text{cn}(v, k), d_{2} = \text{dn}(v, k)\notag\\ \Delta &= 1 - k^{2}s_{1}^{2}s_{2}^{2}\notag \end{align} then we have $$\text{cn}(u + v, k) = \frac{c_{1}c_{2} - s_{1}s_{2}d_{1}d_{2}}{\Delta},\,\, \text{cn}(u - v, k) = \frac{c_{1}c_{2} + s_{1}s_{2}d_{1}d_{2}}{\Delta}$$ so that \begin{align} \text{cn}(u + v, k)\,\text{cn}(u - v, k) &= \frac{c_{1}^{2}c_{2}^{2} - s_{1}^{2}s_{2}^{2}d_{1}^{2}d_{2}^{2}}{\Delta^{2}}\notag\\ &= \frac{c_{1}^{2}(\Delta - s_{2}^{2}d_{1}^{2}) - s_{2}^{2}d_{1}^{2}(\Delta - c_{1}^{2})}{\Delta^{2}}\notag\\ &= \frac{c_{1}^{2} - s_{2}^{2}d_{1}^{2}}{\Delta}\notag\\ \Rightarrow \text{cn}(u + v, k)\,\text{cn}(u - v, k) &= \frac{\text{cn}^{2}(u, k) - \text{sn}^{2}(v, k)\,\text{dn}^{2}(u, k)}{1 - k^{2}\,\text{sn}^{2}(u, k)\,\text{sn}^{2}(v, k)}\notag \end{align} Putting $ x = \pi u / 2K, y = \pi v / 2K$ we get $$\frac{\theta_{4}}{\theta_{2}}\frac{\theta_{2}(x + y, q)}{\theta_{4}(x + y, q)}\frac{\theta_{4}}{\theta_{2}}\frac{\theta_{2}(x - y, q)}{\theta_{4}(x - y, q)} = \dfrac{\dfrac{\theta_{4}^{2}}{\theta_{2}^{2}}\dfrac{\theta_{2}^{2}(x, q)}{\theta_{4}^{2}(x, q)} - \dfrac{\theta_{3}^{2}}{\theta_{2}^{2}}\dfrac{\theta_{1}^{2}(y, q)}{\theta_{4}^{2}(y, q)}\dfrac{\theta_{4}^{2}}{\theta_{3}^{2}}\dfrac{\theta_{3}^{2}(x, q)}{\theta_{4}^{2}(x, q)}}{1 - \dfrac{\theta_{2}^{4}}{\theta_{3}^{4}}\dfrac{\theta_{3}^{2}}{\theta_{2}^{2}}\dfrac{\theta_{1}^{2}(x, q)}{\theta_{4}^{2}(x, q)}\dfrac{\theta_{3}^{2}}{\theta_{2}^{2}}\dfrac{\theta_{1}^{2}(y, q)}{\theta_{4}^{2}(y, q)}}$$ or $$\frac{\theta_{2}(x + y, q)\theta_{2}(x - y, q)}{\theta_{4}(x + y, q)\theta_{4}(x - y, q)} = \dfrac{\theta_{2}^{2}(x, q)\theta_{4}^{2}(y, q) - \theta_{3}^{2}(x, q)\theta_{1}^{2}(y, q)}{\theta_{4}^{2}(x, q)\theta_{4}^{2}(y, q) - \theta_{1}^{2}(x, q)\theta_{1}^{2}(y, q)}$$ Dropping the parameter $ q$ to simplify the notation, we get $$\frac{\theta_{2}(x + y)\theta_{2}(x - y)}{\theta_{4}(x + y)\theta_{4}(x - y)} = \frac{\theta_{2}^{2}(x)\theta_{4}^{2}(y) - \theta_{3}^{2}(x)\theta_{1}^{2}(y)}{\theta_{4}^{2}(x)\theta_{4}^{2}(y) - \theta_{1}^{2}(x)\theta_{1}^{2}(y)}$$ or $$\frac{\theta_{4}^{2}(x)\theta_{4}^{2}(y) - \theta_{1}^{2}(x)\theta_{1}^{2}(y)}{\theta_{4}(x + y)\theta_{4}(x - y)} = \frac{\theta_{2}^{2}(x)\theta_{4}^{2}(y) - \theta_{3}^{2}(x)\theta_{1}^{2}(y)}{\theta_{2}(x + y)\theta_{2}(x - y)}$$ Now we can keep $ y$ as constant and view the above two expressions as representing a function $ f(x)$. It is then easy to notice that $ f(x)$ is periodic with periods $ \pi, \pi\tau$ and moreover it has singularities only when $ \theta_{4}(x + y) = 0$ or $ \theta_{4}(x - y) = 0$ or $ \theta_{2}(x + y) = 0$ or $ \theta_{2}(x - y) = 0$. Now both the $ \theta_{4}$ cannot vanish simultaneously, neither can both the $ \theta_{2}$. It thus follow that the functions $ f(x)$ can have at most one simple pole in a period parallelogram and this is not possible unless $ f(x)$ is a constant. Hence we have $ f(x) = f(0) = \theta_{4}^{2}$.

It now follows that \begin{align} \theta_{4}^{2}\theta_{4}(x + y)\theta_{4}(x - y) &= \theta_{4}^{2}(x)\theta_{4}^{2}(y) - \theta_{1}^{2}(x)\theta_{1}^{2}(y)\notag\\ \theta_{4}^{2}\theta_{2}(x + y)\theta_{2}(x - y) &= \theta_{2}^{2}(x)\theta_{4}^{2}(y) - \theta_{3}^{2}(x)\theta_{1}^{2}(y)\notag \end{align} Since we can express the squares of a theta function in terms of squares of other theta functions, these relations can be used to generation many similar results. Also we can replace both $ x, y$ by $ x + \pi / 2, y + \pi / 2$ to get further results. Using these techniques, for example, we can get \begin{align} \theta_{2}^{2}\theta_{2}(x + y)\theta_{2}(x - y) &= \theta_{4}^{-2}\theta_{2}^{2}\{\theta_{2}^{2}(x)\theta_{4}^{2}(y) - \theta_{3}^{2}(x)\theta_{1}^{2}(y)\}\notag\\ &= \theta_{4}^{-2}\{\theta_{2}^{2}\theta_{2}^{2}(x)\theta_{4}^{2}(y) - \theta_{2}^{2}\theta_{3}^{2}(x)\theta_{1}^{2}(y)\}\notag\\ &= \theta_{4}^{-2}\{\{\theta_{3}^{2}\theta_{3}^{2}(x) - \theta_{4}^{2}\theta_{4}^{2}(x)\}\theta_{4}^{2}(y) - \theta_{3}^{2}(x)\{\theta_{3}^{2}\theta_{4}^{2}(y) - \theta_{4}^{2}\theta_{3}^{2}(y)\}\}\notag \end{align} i.e. $$\theta_{2}^{2}\theta_{2}(x + y)\theta_{2}(x - y) = \theta_{3}^{2}(x)\theta_{3}^{2}(y) - \theta_{4}^{2}(x)\theta_{4}^{2}(y)$$

Jacobi's Imaginary Transformation

The Jacobi's imaginary transformation expressed the elliptic functions of argument $ iu$ in terms of functions of argument $ u$, but in doing so changes the modulus from $ k$ to its complementary modulus $ k^{\prime}$. Thus for example we have $$\text{dn}(iu, k) = \frac{\text{dn}(u, k')}{\text{cn}(u, k')}$$ The transformation from $ k \to k'$ leads to $ K \to K'$ and changes the parameter $ \tau = iK' / K$ to $ \tau' = iK / K'$ so that $ \tau\tau' = -1$. We then have \begin{align} \text{dn}(iu, k) &= \sqrt{k^{\prime}}\,\frac{\theta_{3}(\pi iu / 2K \mid \tau)}{\theta_{4}(\pi iu / 2K \mid \tau)}\notag\\ \text{dn}(u, k') &= \sqrt{k}\,\frac{\theta_{3}(\pi u / 2K' \mid \tau')}{\theta_{4}(\pi u / 2K' \mid \tau')}\notag\\ \text{cn}(u, k') &= \frac{\sqrt{k}}{\sqrt{k'}}\frac{\theta_{2}(\pi u / 2K' \mid \tau')}{\theta_{4}(\pi u / 2K' \mid \tau')}\notag \end{align} and if we set $ z = \pi i u / 2K$ so that $ \pi u / 2K' = (\pi i u / 2K)(-iK / K') = -\tau' z$ then we get $$\frac{\theta_{3}(z \mid \tau)}{\theta_{4}(z \mid \tau)} = \frac{\theta_{3}(-\tau' z \mid \tau')}{\theta_{2}(-\tau' z \mid \tau')} = \frac{\theta_{3}(\tau' z \mid \tau')}{\theta_{2}(\tau' z \mid \tau')}$$ so that $$\frac{\theta_{3}(z \mid \tau)}{\theta_{3}(\tau' z \mid \tau')} = \frac{\theta_{4}(z \mid \tau)}{\theta_{2}(\tau' z \mid \tau')} = f(z)\text{ (say)}$$ Since the zeroes of the theta functions $ \theta_{2}, \theta_{3}, \theta_{4}$ are pairwise disjoint, therefore the function $ f(z)$ is an entire function with no zeroes. Also f(z) is clearly an even function. From these properties of $ f(z)$ it is clear that f(z) is not doubly periodic. We need to find some other function with these similar properties and also ensure that this new function has same periodicity factors as that of $ f(z)$. An example of such a function is $ F(z) = e^{az^{2}}$.

The periodicity factors of $ f(z)$ are given by \begin{align} f(z + \pi) &= e^{i\tau'(2z + \pi)}f(z)\notag\\ f(z + \pi\tau) &= e^{-i(2z + \pi\tau)}f(z)\notag \end{align} whereas we have $$F(z + h) = e^{ah(2z + h)}F(z)$$ The periodicity of $ f(z)$ and $ F(z)$ match if $ h = \pi$ and $ ah = i\tau^{\prime}$ so that $ a = i\tau' / \pi = 1 / (\pi i \tau)$. Note that with this choice of parameter $ a$, both the periodicity factors match and hence the function $ f(z) / F(z)$ becomes a doubly periodic entire function and thereby reduces to a constant. Thus we have $$\frac{f(z)}{F(z)} = \frac{f(0)}{F(0)} = \frac{\theta_{3}(0 \mid \tau)}{\theta_{3}(0 \mid \tau')} = \frac{\sqrt{2K / \pi}}{\sqrt{2K' / \pi}} = \left(\frac{K'}{K}\right)^{-1/2} = (-i\tau)^{-1/2}$$ Putting the pieces together we get $$\frac{\theta_{3}(z \mid \tau)}{\theta_{3}(\tau' z \mid \tau')} = \frac{\theta_{4}(z \mid \tau)}{\theta_{2}(\tau' z \mid \tau')} = (-i\tau)^{-1/2}\exp\left(\frac{z^{2}}{\pi i \tau}\right)$$ and we arrive at the theta function transformation formulas: \begin{align} \theta_{3}(z \mid \tau) &= (-i\tau)^{-1/2}\exp\left(\frac{z^{2}}{\pi i \tau}\right)\theta_{3}(\tau' z \mid \tau')\notag\\ \theta_{4}(z \mid \tau) &= (-i\tau)^{-1/2}\exp\left(\frac{z^{2}}{\pi i \tau}\right)\theta_{2}(\tau' z \mid \tau')\notag \end{align} Replacing $ z$ by $ z + \pi\tau / 2$ we get two another transformation formulas \begin{align} \theta_{2}(z \mid \tau) &= (-i\tau)^{-1/2}\exp\left(\frac{z^{2}}{\pi i \tau}\right)\theta_{4}(\tau' z \mid \tau')\notag\\ \theta_{1}(z \mid \tau) &= -i(-i\tau)^{-1/2}\exp\left(\frac{z^{2}}{\pi i \tau}\right)\theta_{1}(\tau' z \mid \tau')\notag \end{align}

Landen's Transformation

The Landen's transformation is given by $$\text{sn}((1 + k')u, \lambda) = \frac{(1 + k')\,\text{sn}(u, k)\,\text{cn}(u, k)}{\text{dn}(u, k)}$$ where $ \lambda = (1 - k') / (1 + k')$. Then we have following relation between the complete elliptic integrals: $$\frac{\Lambda'}{\Lambda} = \frac{2K'}{K}$$ Thus it follows that we have the relation $ \tau_{1} = 2\tau$ where $ \tau_{1}$ corresponds to $ \lambda$ in the same way as $ \tau$ corresponds to $ k$. Also note that if $ z = \pi u / 2K$ then $$(1 + k')u \cdot \frac{\pi}{2\Lambda} = \frac{2u}{1 + \lambda}\frac{\pi}{2\Lambda} = \frac{\pi u}{(1 + \lambda)\Lambda} = \frac{\pi u}{K} = 2z$$ Hence if we switch to theta functions, the Landen's transformation becomes $$\frac{1}{\sqrt{\lambda}}\frac{\theta_{1}(2z \mid 2\tau)}{\theta_{4}(2z \mid 2\tau)} = (1 + k')\frac{1}{\sqrt{k}}\frac{\theta_{1}(z \mid \tau)}{\theta_{4}(z \mid \tau)}\frac{\sqrt{k'}}{\sqrt{k}}\frac{\theta_{2}(z \mid \tau)}{\theta_{4}(z \mid \tau)}\frac{1}{\sqrt{k'}}\frac{\theta_{4}(z \mid \tau)}{\theta_{3}(z \mid \tau)}$$ Now $ \lambda = (1 - k'^{2}) / (1 + k')^{2} = k^{2} / (1 + k')^{2}$ so we have $ \sqrt{\lambda} = k / (1 + k')$ and then $$\frac{\theta_{1}(2z \mid 2\tau)}{\theta_{4}(2z \mid 2\tau)} = \frac{\theta_{1}(z \mid \tau)\theta_{2}(z \mid \tau)}{\theta_{3}(z \mid \tau)\theta_{4}(z \mid \tau)}$$ or $$\frac{\theta_{3}(z \mid \tau)\theta_{4}(z \mid \tau)}{\theta_{4}(2z \mid 2\tau)} = \frac{\theta_{1}(z \mid \tau)\theta_{2}(z \mid \tau)}{\theta_{1}(2z \mid 2\tau)}$$ Let $ h(z)$ be the common value of these two expressions. Since the zeroes of $ \theta_{1}$ and $ \theta_{4}$ are disjoint we can see that the function $ h(z)$ is an entire function. It is easily seen that the function $ h(z)$ is also doubly periodic with periods $ \pi, \pi\tau$ and hence reduces to a constant namely $ h(0)$.

Thus we have the identity $$\frac{\theta_{3}(z \mid \tau)\theta_{4}(z \mid \tau)}{\theta_{4}(2z \mid 2\tau)} = \frac{\theta_{1}(z \mid \tau)\theta_{2}(z \mid \tau)}{\theta_{1}(2z \mid 2\tau)} = \frac{\theta_{3}(0 \mid \tau)\theta_{4}(0 \mid \tau)}{\theta_{4}(0 \mid 2\tau)}$$
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