- Let
\[ E _ {4} (\tau) = \sum_ {(n, m) \neq (0, 0)} \frac {1}{(n + m \tau) ^ {4}} \]
be the Eisenstein series of order 4.
Show that \(E _ { 4 } ( \tau ) \to \pi ^ { 4 } / 4 5\) as \(\operatorname{Im} ( \tau ) \to \infty\)
More precisely,
\[ \left| E _ {4} (\tau) - \frac {\pi^ {4}}{4 5} \right| \leq c e ^ {- 2 \pi t} \quad \text { if } \tau = x + i t \text { and } t \geq 1. \]
- Deduce that
\[ \left| E _ {4} (\tau) - \tau^ {- 4} \frac {\pi^ {4}}{4 5} \right| \leq c t ^ {- 4} e ^ {- 2 \pi / t} \quad \text { if } \tau = i t \text { and } 0 < t \leq 1. \]
Solution
Separate the terms with \(m=0\): \[ E_4(\tau)=2\zeta(4)+\sum_{m\ne0}\sum_{n\in\mathbb Z}(n+m\tau)^{-4}. \] For \(\Im w>0\), Exercise 7 of Chapter 4 gives \[ \sum_{n\in\mathbb Z}(n+w)^{-4} =\frac{8\pi^4}{3}\sum_{r=1}^\infty r^3e^{2\pi irw}. \tag{1} \] The contributions of \(m\) and \(-m\) are equal (replace \(n\) by \(-n\)). Hence, for \(\tau=x+it\), \[ E_4(\tau)=\frac{\pi^4}{45} +\frac{16\pi^4}{3}\sum_{m=1}^\infty\sum_{r=1}^\infty r^3e^{2\pi irm\tau}, \tag{2} \] because \(2\zeta(4)=\pi^4/45\).
If \(t\ge1\), taking absolute values in (2) gives \[ \left|E_4(\tau)-\frac{\pi^4}{45}\right| \le \frac{16\pi^4}{3}\sum_{m,r\ge1}r^3e^{-2\pi rmt}. \] Since \(e^{-2\pi rmt}\le e^{-2\pi t}e^{-2\pi(rm-1)}\) for \(t\ge1\), the remaining double sum is finite and independent of \(t\). Thus \[ \left|E_4(\tau)-\frac{\pi^4}{45}\right|\le Ce^{-2\pi t}. \] Part (a) follows immediately.
For part (c), reindexing the lattice gives the modular transformation \[ E_4(-1/\tau) =\sum_{(n,m)\ne(0,0)}(n-m/\tau)^{-4} =\tau^4E_4(\tau). \tag{3} \] Take \(\tau=it\), \(0<t\le1\). Since \(-1/(it)=i/t\), equation (3) yields \[ E_4(it)=t^{-4}E_4(i/t). \] Applying part (b) at height \(1/t\ge1\), \[ \left|E_4(it)-t^{-4}\frac{\pi^4}{45}\right| =t^{-4}\left|E_4(i/t)-\frac{\pi^4}{45}\right| \le Ct^{-4}e^{-2\pi/t}, \] as required.