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.

Galois Correspondence of Deck Transformations (5.2)

Let $p : Y → X$ be the universal covering, with $\text{Deck}(Y/X)$ its set of deck transformations. If $G ≤ \text{Deck}(Y/X)$ is a subgroup, $π : Y → Z$ is the appropriate quotient of $Y$ by equivalence modulo $G$, and $q : Z → X$ factors $p$ through the quotient, show that $q$ is a covering map that is Galois precisely when $G$ is a normal subgroup of $\text{Deck}(Y/X)$, in which case $$\text{Deck}(Z/X) ≅ \text{Deck}(Y/X)/G$$
Proof: We show $q$ is a covering map. Throughout let bars denote passage into the quotient. Let $x∈X$ be given. Since $p$ is a covering map, let $U$ be a neighborhood of $x$ with $$p^{-1}(U) = ∪_{j∈J} V_j$$ for disjoint open subsets $V_j$ of $Y$ with each $p|V_j → U$ a homeomomorphism. If $J$ is empty, then we are done. Otherwise, let $V$ be some given $V_j$, and define $V_g = g(V)$ for $g∈\text{Deck}(Y/X)$. By virtue of the universality of $p$, we have $$p^{-1}(U) = ∪_{g∈\text{Deck}(Y/X)} V_g$$ for disjoint open subsets $V_g$ of $Y$ with each $p|V_g → U$ a homeomomorphism. If $\{Gh_i\}_{i∈I}$ is a choice of representatives for the right cosets of $G$, and $W_i = \overline{∪_{g∈Gh_i}V_g}$, we claim $$q^{-1}(U) = ∪_{i∈I} W_i$$ for disjoint open subsets $W_i$ of $W$ with each $q|W_i → U$ a homeomomorphism. Indeed, each $W_i$ is open as it is the image of a $π$-saturated open set in $Y$, and are disjoint since the $V_g$ are disjoint. As well, it is clear that each $π|h_i(V)→W_i$ is a homeomorphism with inverse $φ_i$, hence each $q|W_i→U$ is a homeomorphism as $q = p∘φ_i$ on $W_i$. Thus $q$ is a covering map.

Suppose $G$ is a normal subgroup of deck transformations. Then for $\overline{y_0},\overline{y_1}∈Z$ lying in a common fiber over $q$, we see $y_0,y_1∈Y$ lie in a common fiber over $p$, so let $σ∈\text{Deck}(Y/X)$ be such that $σ(y_0) = y_1$. We define $\overline{σ}(\overline{y}) = \overline{σ(y)}$. This is seen to be independent of the choice of representative in $y$, since if $g∈G$, then $$\overline{σ(g(y))} = \overline{σ(g(σ^{-1}(σ(y))))} = \overline{g'(σ(y))} = \overline{σ(y)}$$ for some other $g'∈G$ since $G$ is normal. $\overline{σ}$ is seen to be a continuous fiber-preserving endomorphism of $Z$ with inverse $\overline{σ^{-1}}$, hence $\overline{σ}∈\text{Deck}(Z/X)$. Therefore $q : Z → X$ is Galois, and since $\overline{σ}\overline{ψ} = \overline{σψ}$ with $\overline{σ} = 1$ iff $σ∈G$, we have $$\text{Deck}(Z/X) ≅ \text{Deck}(Y/X)/G$$ Conversely, suppose $q : Z → X$ is Galois, with $σ∈\text{Deck}(Y/X)$, $g∈G$ given. We will show $h = σgσ^{-1}∈G$. Choose some $y_0∈Y$, and let $σ(y_0)=y_1$. Let $ψ∈\text{Deck}(Z/X)$ be such that $ψ(\overline{y_0}) = \overline{y_1}$. Then both $π∘σ$ and $ψ∘π$ are continuous fiber-preserving maps from $Y$ to $Z$ (relative to $p$ and $q$) sending $y_0$ to $\overline{y_1}$, thus they are equal on all of $Y$. Thus, $$\overline{σ(y)} = ψ(\overline{y})$$ Substituting $y$ with $g(y)$, we have $$\overline{σ(g(y))} = \overline{h(σ(y))} = ψ(\overline{g(y)}) = ψ(\overline{y}) = \overline{σ(y)}$$ so that, since $σ$ is a bijection and $y$ is arbitrary, $$\overline{y} = \overline{h(y)}$$ for all $y∈Y$. But this is to say $h$ preserves the fibers of $π$; necessarily $h∈G$, which concludes the proof that $G$ is normal.$~\square$