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The required anomaly cancellation depends on the exact Eisenstein series of weight two lawHere is a way to establish it without assuming that $E_2$ is a modular form. The lattice definition of the completed nonholomorphic Eisenstein series in Question 6 is modular invariant. Its Poisson summation calculation, proved there, gives the finite Laurent coefficient at $s=1$ asThe residue is constant, so $C$ is also invariant. This calculation uses only convergent Gaussian integrals and the analytic infinite product, not modularity of that product. Using the Wirtinger derivative ${\partial }_{\tau }$, with ${\partial }_{\tau }y=1/(2i)$, differentiation of the locally uniformly convergent series givesDifferentiate $C(\gamma \tau )=C(\tau )$ to see that $E_2^{\ast }=E_2-3/(\pi y)$ transforms with weight two. Substituting $\Im (\gamma \tau )=y/|j{|}^2$ yieldssince $\overline{j}-j=-2icy$. This is the weight-two transformation from an invariant Eisenstein limit.

Combine this law with the derivative transformation of a weak modular form proved in part (i). The anomalous terms in $Df-(k/12)E_{2}f$ cancel exactly, leaving the weight-$k+2$ transformation law. Both functions are holomorphic on the complex upper half-plane, and their Fourier expansions contain no negative powers. Hence the Serre derivative satisfiesIts constant coefficient is $-k{a}_0(f)/12$; it need not be a cusp form when $f$ is not cuspidal.

Write $F(\tau )={\sum }_{M\mid N}c(M)E_2(M\tau )$ and $C={\sum }_{M\mid N}c(M)/M$. For $\gamma =\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\in {\Gamma }_0(N)$, each divisor $M$ of $N$ also divides the lower-left entry $c$. ThusThe full transformation of the Eisenstein series of weight two consequently givesSumming with the prescribed coefficients provesIf $C=0$, this is exactly the desired weight-two law. Conversely, test it at $\gamma =\left(\begin{array}{cc}1& 0\\ N& 1\end{array}\right)\in {\Gamma }_0(N)$. Its anomaly factor $6N(N\tau +1)/(\pi i)$ is nonzero throughout the upper half-plane; if $F$ is modular it forces $C=0$. This proves cancellation of weight-two Eisenstein anomalies:Equivalently the inverse-height terms cancel in $\sum c(M)E_2^{\ast }(M\tau )$, leaving precisely the holomorphic function $F$. Eliminating $c(1)$ shows that all such combinations are spanned by $E_2(\tau )-M{E}_2(M\tau )$ for $M>1$ dividing $N$.

Logarithmically differentiate the convergent product of the modular discriminant. On compact subsets of $\mathbb{H}$, absolute convergence permits termwise differentiation and rearrangement of the resulting divisor sums:The division is valid because $\Delta$ has no interior zeros. Its weight-twelve inversion law is $\Delta (-1/\tau )={\tau }^{12}\Delta (\tau )$. Taking a logarithmic derivative, including the derivative $1/{\tau }^2$ of $-1/\tau$, yieldsThus the Eisenstein series of weight two has the anomalous transformation

Let $y=\mathrm{Im}\tau$. Under inversion, $y$ becomes $y/|\tau {|}^2$. Using ${\tau }^2-|\tau {|}^2=2iy\tau$, the anomalous term cancels:The Fourier expansion gives $E_2(\tau +1)=E_2(\tau )$, and translation leaves $y$ unchanged. Since translation and inversion generate the projective modular group and weight two is unchanged by $-I$, the almost holomorphic weight-two Eisenstein series satisfiesIt is modular in the transformation sense, but is not a holomorphic modular form. Substituting $\mathrm{Im}(\gamma \tau )=y/|c\tau +d{|}^2$ into this law also gives the useful full-group anomaly

The displayed definition uses iterated Eisenstein summation in weight two: the sum in $n$ is evaluated before the sum in $m$, with the single $(0,0)$ term omitted. This order is essential. The two-dimensional lattice series does not have absolute convergence, so arbitrary rearrangement would not be justified.

For noninteger $w$, the cosecant partial-fraction identity isFor completeness, apply the residue theorem to $\pi \cot (\pi \zeta )/(\zeta -w{)}^2$ on squares with large half-integer sides. The cotangent is bounded on the contours and the integral is $O(R^{-1})$. Its residues at the integers are $(n-w{)}^{-2}$ and its residue at $w$ is $-{\pi }^2{\csc }^2(\pi w)$, proving the formula. If $\mathrm{Im}w>0$, the geometric-series expression $\cot (\pi w)=-i(1+2{\sum }_{r\ge 1}{e}^{2\pi irw})$, differentiated termwise, gives the cotangent partial-fraction Fourier kernel

Put $q=e^{2\pi iz}$ with $z$ in the complex upper half-plane. For positive $m$, this gives $-4{\pi }^2{\sum }_{r\ge 1}r{q}^{mr}$. Negative $m$ gives the same value, by replacing $n$ with $-n$ in its inner sum. The $m=0$ row is ${\sum }_{n\ne 0}{n}^{-2}={\pi }^2/3$, by the Basel problem. The resulting series in $m,r$ does have absolute convergence, locally uniformly in $z$, so collecting the coefficient at $q^{\ell }$ is legitimate:The coefficient is the sum-of-divisors function, since $r$ runs over the positive divisors of $\ell$. ThusThere is no conflict with vanishing of weight-two level-one modular forms: the Eisenstein series of weight two has an anomalous transformation term, so it is not a weight-two modular form.

Write $D=(2\pi i{)}^{-1}d/dz=qd/dq$ and use the Serre derivative $D_{k}f=Df-(k/12)E_{2}f$. Translation invariance of $f$ and the Eisenstein series of weight two gives $D_{k}f(z+1)=D_{k}f(z)$.

Differentiate the weight-$k$ transformation $f(-1/z)=z^{k}f(z)$. The chain rule yieldsThe assumed transformation of the Eisenstein series of weight two givesSubtracting cancels the extra term. Hence $D_{k}f(-1/z)=z^{k+2}{D}_{k}f(z)$. Since $S$ and $T=\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)$ generate the modular group, these two transformations prove the weight-$k+2$ law for all its elements. Holomorphy on the complex upper half-plane follows from the formula.

For the Fourier expansion of a modular form $f={\sum }_{n\ge 0}{a}_n{q}^n$, differentiation gives $Df={\sum }_{n\ge 1}n{a}_n{q}^n$. The product of the convergent series for $E_2$ and $f$ likewise has no negative powers. Thus $D_{k}f$ is holomorphic at a cusp at infinity, and therefore at every cusp of the modular group. Its constant coefficient is $-k{a}_0/12$. Since $k>0$, it vanishes exactly when $a_0=0$. Consequently

PutThe infinite product converges locally uniformly and never vanishes on the complex upper half-plane. With $D=qd/dq=(2\pi i{)}^{-1}d/d\tau$, logarithmic differentiation gives

The product is unchanged by $T:\tau \mapsto \tau +1$. To study $S:\tau \mapsto -1/\tau$, defineUsing the transformation law for the Eisenstein series of weight two,Thus $R$ is constant. At the fixed point $\tau =i$, one has $i^{12}=1$, so $R(i)=1$. HenceThe transformations under $S$ and $T$, which generate the modular group, show that $P$ is a weight-twelve modular form. Its Fourier expansion begins $q+O(q^2)$, so it is a cusp form. The normalized element of $S_{12}(\Gamma (1))$ is unique by part (a), and therefore

The Serre derivative sends a level-one weight-$k$ modular form to weight $k+2$. Differentiating $f(-1/z)=z^{k}f(z)$ produces the extra term $k{z}^{k+1}f/(2\pi i)$; the anomalous transformation of the Eisenstein series of weight two cancels it. Translation invariance and the Fourier expansion of a modular form verify the remaining holomorphy conditions. For $k>0$, its constant coefficient is $-k{a}_0(f)/12$, so it is a cusp form exactly when $f$ is.