Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 29 / 3

The infinite product converges locally uniformly in $\mathbb{H}$, because ${\sum }_{n\ge 1}|q{|}^n<\infty$ uniformly on compact subsets. None of its factors vanishes there, and the convergent logarithm of its tail shows that the product is nonzero. Logarithmic differentiation, justified by locally uniform convergence, givesThe middle equality groups the absolutely convergent terms $n{q}^{nr}$ by $m=nr$. Thus $F^{\prime }/F=2\pi i{E}_2$.

Let $H(\tau )={\tau }^{-12}F(-1/\tau )/F(\tau )$. It is nonzero and holomorphic on the simply connected upper half-plane. Using the printed transformation $E_2(-1/\tau )={\tau }^2{E}_2(\tau )+6\tau /(\pi i)$, its logarithmic derivative isSo $H$ is constant. At $\tau =i$ we have $-1/i=i$ and $i^{-12}=1$, giving $H(i)=1$. Therefore $F(-1/\tau )={\tau }^{12}F(\tau )$. Also $F(\tau +1)=F(\tau )$ directly from $q=e^{2\pi i\tau }$. Since these two transformations generate $S{L}_2(\mathbb{Z})$, $F$ has the weight-twelve modular form transformation law.

At the unique modular cusp its expansion is $F=q+O(q^2)$, so it extends holomorphically with zero value. HenceThis is the modular discriminant. The transformed argument contributes $6\tau /(\pi i)$; the converted TeX's numerator $67$ is a transcription error, not the printed relation.

Use $\Theta =(2\pi i{)}^{-1}d/d\tau$. For $\gamma =\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$, differentiating the weight-$k$ law givesThe transformations of $E_2$ under the generators $S,T$, and the chain rule for their compositions, giveEquivalently, one can check just $S,T$ in the next cancellation, since they generate the group. Substituting these formulas intocancels the extra term and gives $(D_{k}f)(\gamma \tau )=(c\tau +d{)}^{k+2}{D}_{k}f(\tau )$. Both constituents are holomorphic on $\mathbb{H}$ and have Fourier expansions of a modular form with nonnegative exponents, so $D_{k}f$ is holomorphic at the modular cusp. ThusThis operator is the Serre derivative.

If $f={\sum }_{n\ge 0}{a}_n{q}^n$, then its derivative has constant coefficient zero, while $E_2$ has constant coefficient one. ThereforeFor $k\ne 0$ this vanishes exactly when $a_0(f)=0$, proving the claimed cusp form equivalence in the intended positive-weight setting. At weight zero the printed equivalence is false: $f=1$ is a noncuspidal weight-zero modular form, but $D_{0}f=0$ belongs to $S_2$. Indeed all weight-zero forms are constants by compactness of $X(1)$. Negative-weight level-one modular forms are zero by the valence formula for the modular group, so there is no further nonzero exception. The transformation and holomorphy conclusion for $D_{k}f$ remains true at weight zero.

From part (a), $\Theta F=E_{2}F$. Its product has integer coefficients and begins with $q$, so write $F={\sum }_{n\ge 1}\tau (n)q^n$, with $\tau (1)=1$. Comparing the coefficient of $q^n$ giveswhich rearranges toThis is a recurrence for the Ramanujan tau function; for example it gives $\tau (2)=-24$ and $\tau (3)=252$.

For the congruence, use the normalized Eisenstein expansionsThe modular form $E_4^3-E_6^2$ is a cusp form of weight twelve, and its leading coefficient is $720-(-1008)=1728$. Since $S_{12}$ is one-dimensional and $F$ starts with $q$,Also $D_6{E}_6$ is a weight-eight form, with constant coefficient $-1/2$. Since $M_8$ is one-dimensional and $E_4^2$ has constant coefficient one, $D_6{E}_6=-E_4^2/2$, orThese identities have rational coefficients whose only displayed denominators are prime to five, so coefficientwise reduction modulo five is legitimate. We have $E_4\equiv 1$. Furthermore $d^5\equiv d(\mathrm{mod}5)$ gives ${\sigma }_5(n)\equiv {\sigma }_1(n)$, while $-504\equiv -24\equiv 1$, so $E_6\equiv E_2$. ThereforeThe equality in the middle is in ${\mathbb{F}}_5$, where $-1/3=1/2$. Comparing coefficients in $\Theta E_6=-504{\sum }_{n\ge 1}n{\sigma }_5(n)q^n$ givesThis proves the Ramanujan tau congruence modulo five from the given small-dimensional modular-form spaces, without assuming that congruence as a theorem.