Theta transformation formula

Remarks

Part 1 holds term by term, because $e^{\pi i n^2} = (-1)^n = e^{2\pi i n/2}$. Proof idea for part 2: for fixed $\tau$, apply the Poisson summation formula to the Gaussian $x \mapsto \exp(\pi i x^2\tau + 2\pi i x z)$, whose Fourier transform is again a Gaussian, with $-1/\tau$ in place of $\tau$. Both sides are holomorphic in $\tau$, so it is enough to check $\tau = it$ with $t > 0$ and continue analytically. Example: at $z = 0$ and $\tau = it$ the formula reads $\sum_{n} e^{-\pi n^2/t} = \sqrt{t}\sum_n e^{-\pi n^2 t}$, the classical identity behind the functional equation of the Riemann zeta function. The hypothesis $\tau \in \mathbb{H}$ is needed: for real $\tau$ the series does not converge.

Going deeper

For a student

The maps $\tau \mapsto \tau + 1$ and $\tau \mapsto -1/\tau$ generate the modular group, so the formula describes how $\vartheta$ behaves under it. The theta constant $\vartheta(0, \tau)$ is invariant under $\tau \mapsto \tau + 2$ and picks up the factor $\sqrt{-i\tau}$ under $\tau \mapsto -1/\tau$. So it is a modular form of weight $1/2$ for the subgroup generated by these two maps. Raising to the fourth power and using part 2 gives the transformation law used in counting sums of four squares. In the integral $\int_0^\infty (\vartheta(0, it) - 1)\,t^{s/2 - 1}\,dt$, the formula of the example yields Riemann's proof of the functional equation of $\zeta(s)$.