For a lattice \(\Lambda\), define \[ G_n(\Lambda)=\sum_{\lambda\in\Lambda\setminus\{0\}}\lambda^{-2n}. \] Assume that uniqueness of modular forms of weight \(8\) gives a universal constant \(A\) such that \[ G_2(\Lambda)^2=A\,G_4(\Lambda) \] for every lattice \(\Lambda\). Determine \(A\).
Hint: use \[ \sum_{n\ne0}\frac1{n^8}=\frac{2\pi^8}{9450}=\frac{\pi^8}{4725}. \]
Solution
The universal constant is \[ \boxed{A=\frac73.} \]
1. Degenerate the lattice to a rank-one lattice.
For \(y>0\), let \[ \Lambda_y=\mathbb Z\oplus iy\mathbb Z. \] Write a lattice point as \[ m+iny, \qquad m,n\in\mathbb Z. \] For an exponent \(s>2\) and \(a\ge1\), \[ \sum_{m\in\mathbb Z}(m^2+a^2)^{-s/2} \le a^{-s}+2\int_0^\infty (x^2+a^2)^{-s/2}\,dx \le C_s a^{1-s} \] for a constant \(C_s\) depending only on \(s\). Hence the contribution to \[ \sum_{(m,n)\ne(0,0)}(m+iny)^{-s} \] from rows with \(n\ne0\) is bounded in absolute value by \[ C_s y^{1-s}\sum_{n\ne0}|n|^{1-s}, \] which tends to \(0\) as \(y\to\infty\).
Therefore, for \(s=4\) and \(s=8\), only the horizontal row survives in the limit.
2. The two relevant Eisenstein sums have cusp limits\[ G_2(\Lambda_y)\longrightarrow\frac{\pi^4}{45}, \qquad G_4(\Lambda_y)\longrightarrow\frac{\pi^8}{4725}. \]
By 1, \[ \lim_{y\to\infty}G_2(\Lambda_y) =\sum_{m\ne0}m^{-4} =2\zeta(4) =\frac{\pi^4}{45}. \] Likewise, using the value supplied in the hint, \[ \lim_{y\to\infty}G_4(\Lambda_y) =\sum_{m\ne0}m^{-8} =\frac{\pi^8}{4725}. \]
3. Take the limit in the universal identity.
For every \(y>0\), \[ G_2(\Lambda_y)^2=A\,G_4(\Lambda_y). \] Passing to the limit from 2 gives \[ \left(\frac{\pi^4}{45}\right)^2 =A\frac{\pi^8}{4725}. \] Cancelling \(\pi^8\) yields \[ A=\frac{4725}{45^2} =\frac{4725}{2025} =\boxed{\frac73}. \]