Modular Function of Weight k > 2: The Eisenstein Series

Let $k > 2$ be an integer, and consider the following function defined on the upper half plane of the complex numbers: $$G_k(τ) = \sum_{(c,d)∈ℤ^2-(0,0)} \dfrac{1}{(cτ+d)^k}$$ We shall prove that the series above converges absolutely for all $τ$ in the upper half plane, and as such the summands may be permuted freely. Then, we shall prove that the series converges uniformly on compact subsets, and thus defines a holomorphic function on the upper half plane. Finally, we shall prove that $G_k(τ)$ is weakly modular of weight $k$, and also $G_k(τ)$ is bounded as $\text{Im}(τ)→∞$, and so defines a modular function of weight $k$.

Proving that the series absolutely converges for any given $τ$ is more or less the same as showing the the same for the Weierstrass series with respect to the lattice $[τ,1]$; we outline the idea. One may contain the lattice points in disjoint balls of uniform size $δ > 0$. Since each annulus of radii $R-1, R+2$ has area $O(R)$, the convergence of $\sum \dfrac{1}{|cτ+d|^k}$ follows by comparison with the convergent series $\sum_{R=1} \dfrac{1}{R^{k-1}}$.

For positive real numbers $A,B$, let $Ω = \{τ~|~|\text{Re}(τ)| ≤ A, |\text{Im}(τ)| ≥ B\}$. We show $G_k(τ)$ converges uniformly on $Ω$, and hence on compact subsets in general, thereby proving $G_k(τ)$ defines a holomorphic function. We do this by showing that $\sum_{(c,d)∈ℤ^2-(0,0)} σ_{c,d}$ converges on $Ω$, where $$σ_{c,d} = \sup_{τ∈Ω} \dfrac{1}{|cτ+d|^k}$$ Note that this supremum is approached by bringing $cτ$ as close to $-d$ as possible; when $|d|$ is large enough, say $|d| > N$, this supremum is in fact achieved by letting $τ$ be one of $\pm A + Bi$. Thus, we know that $\sum σ_{c,d}$ converges when only large enough $d$ are summed, by considering $G_k(τ)$ converges absolutely for each $τ$. As for $\sum_c σ_{c,d}$ for each fixed $d$ with $|d| ≤ N$, we see $σ_{c,d} ≤ \dfrac{1}{|cB|^k} = \dfrac{1}{|B|^k} \dfrac{1}{|c|^k}$ generally when $c ≠ 0$, so the aforementioned sum converges. We conclude $G_k(τ)$ converges uniformly on $Ω$ and is holomorphic on the upper half plane. This calculation also shows that $G_k(τ)$ is bounded on $Ω$; the fact that $G_k(τ)$ is bounded as $\text{Im}(τ)→∞$ will therefore follow once we show it is weakly modular, and thereby $ℤ$-periodic.

Consider the action of a special fractional linear transformation on the lattice $L=[τ,1]$; for generic integers $α,β$ we see $$ατ+β↦α \dfrac{aτ+b}{cτ+d} + β = \dfrac{(aα+cβ)τ + (bα + dβ)}{cτ+d}$$ Therefore, noting that $\begin{bmatrix}a & b \\ c & d \end{bmatrix}^T$ permutes $ℤ^2$, we see the special linear transformation induces a bijection $L→\dfrac{L}{cτ+d}$. We conclude that $G_k(\dfrac{aτ+b}{cτ+d}) = (cτ+d)^k G_k(τ)$, so $G_k(τ)$ is weakly modular, hence fully modular by the above comment.$~\square$

Note: There is an error in the above proof, namely "when $|d|$ is large enough, say $|d| > N$, this supremum is in fact achieved by letting $τ$ be one of $\pm A + Bi$." One can instead show that there exists a constant $C > 0$ such that $|γτ+δ| > C \cdot \sup \{|γ|,|δ|\}$ for all $τ∈Ω$ and real $γ, δ$ not both zero, implying $$\sum σ_{c,d} ≤ \dfrac{1}{C} \sum \dfrac{1}{(\sup \{|c|,|d|\})^k}$$ is finite.