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Let us consider the Theta function $$\theta(\tau) = \sum_{n \in \mathbb{Z}} e^{\pi i n^2 \tau} \text{ for } \mathrm{Im}(\tau)>0.$$ Then it is rather easy to see that $$\theta(\tau+2)=\theta(\tau)$$ and $$\theta(\tau)+\theta(\tau+1)=2 \cdot \theta(4 \tau).$$ Further, one can prove that $$\theta(-1/ \tau) = \sqrt{\frac{\tau}{i}} \cdot \theta(\tau).$$ All together, this yields $$\theta(1- \frac{1}{\tau}) = \theta(- \frac{1}{\tau}) + \theta(1- \frac{1}{\tau}) - \theta(- \frac{1}{\tau}) = 2 \cdot \theta(- \frac{4}{\tau}) - \theta(- \frac{1}{\tau}) = \sqrt{\frac{\tau}{i}} \cdot (\theta(\frac{\tau}{4}) - \theta(\tau)).$$ Here $\sqrt{}$ always denotes the principal branch of the square root.

Next, my textbook claims that $$f(\tau) := \theta^4(\tau)-\theta^4(\tau+1)+\tau^{-2} \cdot \theta^4(1- \frac{1}{\tau})$$ fulfils both $f(\tau+1) = - f(\tau)$ and $f(- \frac{1}{\tau}) = - \tau^{2} \cdot f(\tau)$ (and uses this to deduce that actually $f = 0$). It should be possible to derive these using the above $\theta$-identities, but I am completely stuck here. Any help is appreciated!

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Let us compute explicitly: $$ \begin{aligned} f(\tau)+f(\tau+1) &= \theta^4(\tau)-\theta^4(\tau+1) +\tau^{-2} \cdot \theta^4\left(1- \frac{1}{\tau}\right) \\ &\qquad\qquad + \theta^4(\tau+1)-\theta^4(\tau+2) +(\tau+1)^{-2} \cdot \theta^4\left(1- \frac{1}{\tau+1}\right) \\ &= \tau^{-2} \cdot \theta^4\left(1- \frac{1}{\tau}\right) +(\tau+1)^{-2} \cdot \theta^4\left(1- \frac{1}{\tau+1}\right) \\ &= \tau^{-2} \cdot\left[\ \theta^4\left(\frac{\tau-1}{\tau}\right) +\left(\frac\tau{\tau+1}\right)^2 \cdot \theta^4\left(\frac{\tau}{\tau+1}\right) \ \right] \\ &= \tau^{-2} \cdot\left[\ \theta^4\left(\frac{\tau-1}{\tau}-2\right) -\theta^4\left(-\frac 1{\qquad\frac{\tau}{\tau+1}\qquad}\right) \ \right] \\ &=0\ , \\[3mm] f\left(-\frac1\tau\right) &= \theta^4\left(-\frac1\tau\right) - \theta^4\left(-\frac1\tau+1\right) + \left(-\frac1\tau\right)^{-2}\theta^4\left(1-\frac1{-1/\tau}\right) \\ &= \theta^4\left(-\frac1\tau\right) - \theta^4\left(1-\frac1\tau\right) + \tau^2\theta^4(\tau+1) \\ &= -\tau^2\theta^4(\tau) +\tau^2\theta^4(\tau+1) - \tau^2\cdot\tau^{-2} \theta^4\left(1-\frac1\tau\right) \\ &= -\tau^2\;f(\tau)\ . \end{aligned} $$ (The power of $\tau$ differs in the last relation.)

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  • $\begingroup$ Yes, the power of $\tau$ in the last relation should be $2$. I edited it. Thank you! $\endgroup$
    – Algebrus
    Sep 3, 2019 at 12:10

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