Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 26 / 1

Write $\Gamma =S{L}_2(\mathbb{Z})$, $\mathbb{H}=\{\tau :\mathrm{Im}\tau >0\}$ and $q=e^{2\pi i\tau }$. The space $M_k$ consists of holomorphic functions on the complex upper half-plane satisfyingand holomorphic at the cusp: their Fourier expansion of a modular form is $f={\sum }_{n\ge 0}{a}_n{q}^n$. The subspace $S_k$ of cusp forms has $a_0=0$. Since all cusps are equivalent at level one, this condition treats every cusp. The element $-I$ forces odd-weight forms to vanish.

Use the permitted properties of the normalized Eisenstein series and modular discriminant:They have integer Fourier coefficients and weights $4,6,12$. For even $k\ge 0$, the assumed dimension formula gives $d=\lfloor k/12\rfloor +1$ except when $k\equiv 2(\mathrm{mod}12)$, when $d=\lfloor k/12\rfloor$. For $0\le j<d$, the remaining weight $k-12j$ is nonnegative and is not two. Choose $b_j\in \{0,1\}$ and $a_j\ge 0$ with $4a_j+6b_j=k-12j$. Every allowable remaining weight admits such a choice. The formsgive an integral triangular basis of level-one modular forms: their distinct first exponents prove independence, and their number $d$ proves that they span.

Now take $f$ with integral coefficients. Subtract its constant coefficient times $F_0$. The remainder still has integral coefficients and has no constant term. Subtract its coefficient at $q$ times $F_1$, then its current coefficient at $q^2$ times $F_2$, and continue. Every subtraction uses an integer. After $d$ steps the remainder's first $d$ coefficients vanish; triangularity and the basis property force that remainder to be zero. ConsequentlyThis proves integral polynomial generation without dividing by $1728$. The relation $E_4^3-E_6^2=1728\Delta$ explains why the additional integral generator is useful, although over $\mathbb{C}$ two generators suffice. For $k=0$ the basis is the constant one; when $M_k=0$, including weight two and odd weights, the claim is vacuous.

Start from the integral triangular basis of level-one modular forms. Proceed downwards in $j$. Set $g_{d-1}=F_{d-1}$; once $g_{j+1},\dots ,g_{d-1}$ are constructed, subtract from $F_j$ its coefficients at $q^{j+1},\dots ,q^{d-1}$ times these respective forms. The already constructed rows have a single nonzero coefficient among those initial positions, so this removes every unwanted initial coefficient without changing the coefficient one at $q^j$. All subtractions use integers. The resulting integral echelon basis of level-one modular forms therefore hasIts first $d$ coefficient vectors are the standard coordinate vectors, hence it is a basis. If another form had the same first $d$ coefficients as $g_j$, their difference would have those coefficients all zero; the triangular basis proves that difference vanishes. This proves uniqueness of every $g_j$, not merely uniqueness up to basis change.

For any $f\in M_k$, its expansion in this basis is $f={\sum }_{j=0}^{d-1}{a}_j(f)g_j$, since its first $d$ coefficients determine the basis coordinates. Comparing subsequent coefficients givesThe coefficients of $f$ need not be integral for this last identity. When needed at earlier indices, extend the notation by $c_n(j)={\delta }_{nj}$ for $0\le n<d$. This agrees with the actual coefficients of the basis forms and is the convention needed for the next part's all-index wording.

Use the standard normalization of the Hecke operator, for whichThe Hecke multiplication relations make these operators commute; a coefficient-based proof is included in 3(i). For a cusp form, $a_0(f)=0$, so the preceding coefficient relation saysfor every $f\in S_k$. Denote the bracketed operator by $D_n$. A vanishing first coefficient alone would not prove $D_n=0$. Instead, for every $m\ge 1$, commute it past $T_m$ and apply the same relation to the cusp form $T_{m}f$:Its constant coefficient is zero too. Thus every Fourier coefficient of $D_{n}f$ vanishes, and holomorphy gives $D_{n}f=0$. We have provedFor $n\ge d$ these are the integers constructed in (ii); for $1\le n<d$, the extension $c_n(j)={\delta }_{nj}$ makes the formula tautological. Without that implicit extension the symbols in the printed all-index assertion are undefined. The $j=0$ term is excluded because its coefficient functional is zero on cusp forms. If $S_k=0$, the operator statement is vacuous. This also exhibits the perfect integral Hecke pairing: the first $d-1$ coefficient functionals on the cusp space are dual to $T_1,\dots ,T_{d-1}$.