- Show that if \(\tau\) is fixed with Im \(( \tau ) > 0\) , then the Jacobi theta function
\[ \Theta (z | \tau) = \sum_ {n = - \infty} ^ {\infty} e ^ {\pi i n ^ {2} \tau} e ^ {2 \pi i n z} \]
is of order 2 as a function of \(z .\) . Further properties of \(\Theta\) will be studied in Chapter 10.
[Hint: \(- n ^ { 2 } t + 2 n | z | \leq - n ^ { 2 } t / 2\) when \(t > 0\) and \(n \geq 4 | z | / t . ]\)
Solution
Write \[ \tau=u+it, \qquad t=\Im\tau>0. \] For \(z=x+iy\), the \(n\)th summand satisfies \[ \left|e^{\pi i n^2\tau}e^{2\pi inz}\right| =e^{-\pi t n^2-2\pi ny} \le e^{-\pi t n^2+2\pi|n||y|}. \] Using \[ 2|n||y|\le \frac t2 n^2+\frac{2}{t}y^2, \] we obtain \[ -\pi t n^2+2\pi|n||y| \le -\frac{\pi t}{2}n^2+\frac{2\pi}{t}y^2. \] Hence \[ |\Theta(z\mid\tau)| \le e^{2\pi y^2/t} \sum_{n\in\mathbb Z}e^{-\pi t n^2/2} \le C_t e^{(2\pi/t)|z|^2}. \] Thus the order is at most \(2\).
To prove that it is not smaller, first note the quasi-periodicity. Reindexing the absolutely convergent series gives \[ \Theta(z+\tau\mid\tau) =e^{-\pi i\tau-2\pi iz}\Theta(z\mid\tau), \] and therefore, for every integer \(m\ge1\), \[ \Theta(z+m\tau\mid\tau) =e^{-\pi i m^2\tau-2\pi i m z}\Theta(z\mid\tau). \tag{1} \] The theta function is not identically zero: on the real interval \([0,1]\) termwise integration gives \[ \int_0^1\Theta(x\mid\tau)\,dx=1, \] since all Fourier modes except \(n=0\) integrate to zero. Choose \(z_0\) with \(\Theta(z_0\mid\tau)\ne0\).
Taking absolute values in (1), \[ |\Theta(z_0+m\tau\mid\tau)| =|\Theta(z_0\mid\tau)| \exp\!\left(\pi t m^2+2\pi m\Im z_0\right). \] Meanwhile \[ |z_0+m\tau|=O(m). \] Thus along this sequence \[ \log|\Theta(z_0+m\tau\mid\tau)| =\pi t m^2+O(m), \] which cannot be bounded by \(O(m^\rho)\) for any \(\rho<2\). Therefore the order is at least \(2\).
Consequently \[ \boxed{\rho_\Theta=2}. \]