Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 29 (Modular Forms)

Let $\Lambda =\mathbb{Z}{\omega }_1+\mathbb{Z}{\omega }_2$ be a complex lattice. Orient its basis so that $\mathrm{Im}({\omega }_1/{\omega }_2)>0$. Multiplication by ${\omega }_2^{-1}$ gives the normalized complex lattice $\mathbb{Z}\tau +\mathbb{Z}$ with $\tau ={\omega }_1/{\omega }_2\in \mathbb{H}$. Thus every homothety of complex lattices class has a normalized representative.

If $\gamma =\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\in S{L}_2(\mathbb{Z})$, thenThe last equality follows because the two integer vectors are an integral basis: their determinant is one. Hence points in the same modular orbit give homothetic complex lattices.

Conversely, suppose ${\Lambda }_{{\tau }^{\prime }}=\alpha {\Lambda }_{\tau }$ for $\alpha \ne 0$. There is an integer basis-change matrix of determinant $\pm 1$ such thatBoth normalized bases have positive orientation, and multiplication by $\alpha$ preserves real orientation because its real determinant is $|\alpha {|}^2>0$. The basis-change determinant is therefore $+1$. Dividing the two equalities gives ${\tau }^{\prime }=(a\tau +b)/(c\tau +d)$. Thus the modular orbits and homothety of complex lattices classes correspond bijectively.

Use the normalized Eisenstein series $E_4=1+240q+O(q^2)$ and $E_6=1-504q+O(q^2)$, and putThe valence formula for the modular group applied to this nonzero weight-twelve cusp form gives weighted zero order one. Its order at infinity is already one, so $\Delta$ has no zeros in $\mathbb{H}$. Consequently the Klein j-invariant is holomorphic there and invariant under $S{L}_2(\mathbb{Z})$.

Set $\rho =e^{2\pi i/3}$. The stabilizer of $\rho$ forces $E_4(\rho )=0$, since the weight-four automorphy factor for $\left(\begin{array}{cc}0& -1\\ 1& 1\end{array}\right)$ is $(\rho +1{)}^4\ne 1$. The stabilizer of $i$ forces $E_6(i)=0$, since $i^6=-1$. The valence formula for the modular group giveswith no other zero orbits for either form. In particular $j(\rho )=0$ and $j(i)=1728$.

For any $A\in \mathbb{C}$, the weight-twelve formis nonzero, with constant Fourier coefficient one and hence order zero at infinity. Its total weighted zero order is one. If $A=0$, the triple zero at $\rho$ accounts for this entire order. If $A=1728$, then $h_A=E_6^2$, and the double zero at $i$ accounts for it. For all other $A$, neither elliptic orbit is a zero: $E_4(\rho )=0$ and $\Delta (\rho )\ne 0$ exclude $\rho$, while $E_6(i)=0$ gives $E_4(i{)}^3=1728\Delta (i)$ and excludes $i$. Therefore there is precisely one ordinary zero orbit, of multiplicity one. Since $h_A(\tau )=0$ is exactly $j(\tau )=A$, every complex number occurs at exactly one modular orbit. ThusThis proves Klein j-invariant classifies complex lattice homothety using the weighted zeros, including the two exceptional stabilizers.

For a complex lattice, put $g_2=60G_4$ and $g_3=140G_6$. The complex lattice form of the Klein j-invariant isThe denominator is nonzero for a complex lattice: after normalization it is a nonzero scalar multiple of $E_4^3-E_6^2$, whose absence of zeros was proved in part (b). The equal lattice Eisenstein sums therefore give equal $j$ values, so parts (a) and (b) give ${\Lambda }^{\prime }=\alpha \Lambda$ for some $\alpha \ne 0$. Scaling the absolutely convergent lattice Eisenstein sums givesIf both sums are nonzero, ${\alpha }^4={\alpha }^6=1$, so ${\alpha }^2=1$ and $\alpha \Lambda =\Lambda$.

If $G_6=0$, then $G_4\ne 0$ and $j=1728$. The complex lattice is homothetic to $\mathbb{Z}i+\mathbb{Z}$, which is preserved by multiplication by $i$. The constraint ${\alpha }^4=1$ allows only powers of $i$, each already preserving the complex lattice. If $G_4=0$, then $G_6\ne 0$ and $j=0$. The complex lattice is homothetic to $\mathbb{Z}\rho +\mathbb{Z}$. Multiplication by $\rho$ preserves this complex lattice because ${\rho }^2=-1-\rho$, and multiplication by $-1$ always preserves it. The constraint ${\alpha }^6=1$ allows exactly the sixth roots generated by these two multiplications. Again $\alpha \Lambda =\Lambda$. Both sums cannot be zero simultaneously because the discriminant is nonzero. Hence in every caseThe exceptional cases are essential in a complex lattice is determined by its fourth and sixth Eisenstein sums; merely deducing homothety of complex lattices would not prove equality.

Adjoin ${\mathbb{P}}^1(\mathbb{Q})$ to the complex upper half-plane. Ordinary points retain their usual neighborhoods. At a rational boundary point $a=\sigma \infty$, a neighborhood basis consists of $a$ together with $\sigma \{z\in \mathbb{H}:\mathrm{Im}z>Y\}$, with $Y$ arbitrarily large. Different choices of $\sigma \in S{L}_2(\mathbb{Z})$ change it by a translation and sign, so give the same neighborhood system. Give $X(1)$ the quotient topology. The modular group acts properly discontinuously on $\mathbb{H}$; its modular cusp neighborhoods descend to punctured neighborhoods with their modular cusp point filled in. This quotient is Hausdorff: interior orbits can be separated by proper discontinuity, and a sufficiently high modular cusp neighborhood avoids any prescribed compact interior neighborhood.

Away from fixed points of the effective group $PS{L}_2(\mathbb{Z})$, a small disk is disjoint from its nonidentity translates, so its coordinate gives a holomorphic quotient coordinate. At an elliptic fixed point ${\tau }_0$, useIts effective stabilizer rotates $u$ by an $e$th root of unity, where $e=2$ at $i$ and $e=3$ at $\rho$. The quotient coordinate is $w=u^e$, not $u$. All rational boundary points lie in one modular cusp class. At infinity the stabilizer acts by $\tau \mapsto \tau +1$, andis a punctured-disk coordinate; adjoining $q=0$ fills in the modular cusp. These charts and their holomorphic transition maps define the complex structure of the compactified modular curve.

For compactness use the standard regionIts part with $\mathrm{Im}\tau \le Y$, $Y>1$, is compact: the imaginary part is at least $\sqrt{3}/2$. The quotient of its upper tail, with its two vertical sides identified and infinity added, is the closed disk $|q|\le e^{-2\pi Y}$. These two compact images cover $X(1)$, including the boundary-edge identifications. Thus $X(1)$ is a compact Riemann surface.

The $j$ function also extends over the modular cusp with one simple pole, since $j=q^{-1}+O(1)$. Its local expressions at elliptic points descend to holomorphic functions of $u^e$. By Question 1 it maps the remaining quotient bijectively onto $\mathbb{C}$, so its extension is a degree-one holomorphic map $X(1)\to {\mathbb{P}}^1$. A degree-one map of compact Riemann surfaces is biholomorphic: any local ramification would already give degree at least two. Therefore $X(1)$ is the Riemann sphere; this will also identify the base genus in Question 5.

A weight-$2k$ modular form gives an invariant tensor differentialbecause $d(\gamma \tau )=(c\tau +d{)}^{-2}d\tau$ cancels its automorphy factor. Assume first $k\ge 0$. At an ordinary quotient point this differential is holomorphic. At an elliptic point of order $e$, use $u$ and $w=u^e$ from part (a), and write the pullback as $A(u)(du{)}^k$, with $A$ holomorphic. Its invariance says $A(\zeta u){\zeta }^k=A(u)$, so a nonzero term $a_r{u}^r$ has $r+k\equiv 0(\mathrm{mod}e)$. Sincethe exponent in the descended coefficient is the integer $(r-k(e-1))/e$. With $r\ge 0$, it is at least $-\lfloor k(e-1)/e\rfloor$. At the modular cusp, $d\tau =dq/(2\pi iq)$ and $f$ has a power series with nonnegative exponents, so ${\omega }_f$ has a pole of order at most $k$.

Consequently the linear map $f\mapsto {\omega }_f$ injects $M_{2k}$ into the meromorphic sections of $K^{\otimes k}$ with poles bounded byThis section space is $H^0(X(1),K^{\otimes k}(D))$. It is finite-dimensional on a compact Riemann surface by the Riemann-Roch theorem. Concretely choose one nonzero meromorphic section of the line bundle; dividing by it identifies these sections with a Riemann-Roch space with a fixed finite divisor bound. This proves the requested finite-dimensionality by compact-surface theory.

In fact the injection is an isomorphism. If a descended coefficient has order at least $-\lfloor k(e-1)/e\rfloor$, its pulled-back exponent $ed+k(e-1)$ is nonnegative; the modular cusp pole bound likewise gives a holomorphic Fourier expansion of a modular form. These pullbacks therefore recover modular forms. Since $X(1)$ is the Riemann sphere, $\mathrm{deg}K=-2$, and a line bundle of degree $d$ on the sphere has $\max (0,d+1)$ sections. HenceFor negative $k$, the valence formula excludes a nonzero holomorphic modular form because all its orders would be nonnegative but their sum would be $k/6<0$; the space is zero. The mechanism behind finite-dimensionality is modular forms as meromorphic differentials on X(1), with explicit elliptic and modular cusp pole bounds.

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.

Write $w=2k$. For a positive-determinant real matrix $A=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$, extend the slash operator for modular forms byLet ${\mathcal{M}}_n$ be the integral matrices of determinant $n>0$. Each left $S{L}_2(\mathbb{Z})$ orbit has a unique representative $\left(\begin{array}{cc}a& b\\ 0& d\end{array}\right)$ with $ad=n$, $a,d>0$, $0\le b<d$. To obtain it, the Euclidean algorithm on the first column produces $(a,0)$ with $a>0$; the determinant fixes $d$, and a row shear reduces $b$ modulo $d$. This also proves uniqueness.

Define the normalized Hecke operatorRight multiplication by $S{L}_2(\mathbb{Z})$ permutes these left orbits, and the slash operation is associative. Hence the sum satisfies the same modular form transformation law. It is holomorphic on $\mathbb{H}$ since each argument stays there.

For $f={\sum }_{r\ge 0}c(r)q^r$, summing the phase $e^{2\pi irb/d}$ over $b$ gives zero unless $d\mid r$, and gives $d$ when $d\mid r$. For the resulting term to contribute to $q^m$, one needs $a\mid m$ and $r=dm/a=nm/a^2$. Its scalar factor is $n^{w-1}{d}^{1-w}=a^{w-1}$. Thuswhere $\gcd (n,0)=n$. All powers are nonnegative, so the sum is holomorphic at infinity and belongs to $M_w$; level one has only this modular cusp class. Its constant coefficient is ${\sigma }_{w-1}(n)c(0)$, so modular cusp forms are preserved. This is the full Fourier coefficients of a composite-index Hecke operator formula, including the constant term.

An eigenform is understood to be nonzero; the zero form would put no constraint on the alleged eigenvalues. The coefficient of $q$ in $T_w(n)f$ is $c(n)$, since $\gcd (n,1)=1$. Comparing with the coefficient in $\lambda (n)f$ givesThe constant coefficient comparison gives $\lambda (n)c(0)={\sigma }_{w-1}(n)c(0)$, so

For completeness derive the operator multiplication law underlying the last relation. On formal Fourier series defineThese auxiliary maps need not separately preserve modularity, but coefficient identities are valid and $U_p{V}_p=I$. Part (a) givesMultiplication on the left by $U_p+A{V}_p$ yields the $T_w(p^{e+1})$ sum and, from $U_p{V}_p^j=V_p^{j-1}$ for $j\ge 1$, the remaining sum $A{T}_w(p^{e-1})$. HenceInduction on the smaller exponent, applying this recurrence and cancelling its two overlapping sums, now givesTo make the induction explicit, write $T_r=T_w(p^r)$ and assume $2\le r\le s$. Substituting the formulas already established for $r-1$ and $r-2$ into $T_r{T}_s=(T_1{T}_{r-1}-A{T}_{r-2})T_s$ givesThe last term is exactly the missing $j=r$ term. The cases $r=0,1$ are the identity operator and the prime recurrence above.

For distinct primes, their $U$ and $V$ maps commute, and the divisor formula in (a) factors independently at each prime. It follows that $T_w(m)T_w(n)=T_w(mn)$ for coprime $m,n$. Multiplying the prime-power identities therefore gives the general Hecke multiplication relationsApply this identity to the same nonzero eigenform and cancel the vector $f$, obtainingThis reasoning does not silently divide by $c(1)$, which is important when considering noncuspidal weight-zero forms.

For a nonzero Hecke eigenform that is a cusp form, $c(1)\ne 0$: otherwise part (b) would make all its positive Fourier coefficients zero, and cuspidality already gives $c(0)=0$, forcing $f=0$. Since $T_w(1)=I$, $\lambda (1)=1$. Part (b) gives multiplicativity at coprime integers and the prime-power recurrenceMultiply its generating series by the corresponding quadratic to obtainUnique prime factorization and multiplicativity giveFor a normalized eigenform $c(1)=1$, the prefactor disappears. It must be retained for an arbitrary scalar multiple of that eigenform.

This product identity holds analytically in an absolute-convergence half-plane. The invariant norm of a positive-weight cusp form is bounded, as proved in Question 6. For a period-one Fourier expansion of a modular form,Choosing $y=1/n$ gives $c(n)=O(n^{w/2})$, so $\mathrm{Re}s>w/2+1$ suffices. Absolute convergence then justifies expansion and rearrangement of the Euler product of a Hecke eigenform; no sharper coefficient estimate is needed here.

Pass to the effective subgroup $\overline{\Gamma }\subset PS{L}_2(\mathbb{Z})$. Its index is $\mu$: adjoining $-I$ halves the original index precisely when $-I\notin \Gamma$. The natural map of compactified modular curveshas degree $\mu$, since a generic orbit has that many lifts. It branches only over the order-two elliptic orbit, the order-three elliptic orbit, and the modular cusp. Away from these, the modular quotient charts are genuine unramified covering charts.

Over $i$, a lift with its own effective stabilizer of order two is unramified; there are ${\nu }_2$ such lifts. Every other lift has ramification index two. As the sum of all ramification indices in that fibre is $\mu$, their number is $(\mu -{\nu }_2)/2$ and their contribution to the ramification divisor is $(\mu -{\nu }_2)/2$. Similarly, over $\rho$ there are ${\nu }_3$ unramified elliptic lifts and $(\mu -{\nu }_3)/3$ lifts of ramification index three, contributing $2(\mu -{\nu }_3)/3$.

At a modular cusp of width $h$, the local parameters are $q_h=e^{2\pi i\tau /h}$ upstairs and $q=e^{2\pi i\tau }$ downstairs. Thus $q=q_h^h$ and the ramification index is $h$. Summing the widths gives $\mu$, so all ${\nu }_{\infty }$ modular cusps contribute $\mu -{\nu }_{\infty }$ to ramification. The Riemann-Hurwitz formula now givesThereforeThe effective index and the projective modular cusp widths ensure that the formula also applies when the original subgroup does not contain $-I$; no spurious factor of two is introduced.

Reduction modulo $N$ has kernel $\Gamma (N)$ and is onto $S{L}_2(\mathbb{Z}/N\mathbb{Z})$. To see surjectivity, use the Chinese remainder theorem to reduce to $\mathbb{Z}/p^a\mathbb{Z}$. In that local ring a unimodular column has a unit entry. Elementary row operations move a unit to the upper position and clear the lower one; diagonal matrices $\mathrm{diag}(u,u^{-1})$ are products of elementary matrices as well. Explicitly, with $E_{12},E_{21}$ denoting elementary shears,and multiplying this by the same matrix with $u=-1$ gives $\mathrm{diag}(u,u^{-1})$. Thus elementary matrices generate the special linear group over each local ring. Chinese remaindering their parameters gives generation over the product ring, and every such parameter lifts to an integer. This proves the reduction is onto.

Over ${\mathbb{F}}_p$, there are $p^2-1$ choices for a nonzero first column and $p$ choices for a second column with determinant one, so $|S{L}_2({\mathbb{F}}_p)|=p(p^2-1)$. Each reduction from modulus $p^{a+1}$ to $p^a$ has kernel represented by $I+p^{a}B$, with $B$ modulo $p$ and $\mathrm{tr}B=0$, of size $p^3$. Hence, for $N>1$,For $N=1$ the index is one. Since $-I\in \Gamma (N)$ exactly when $N=1,2$,

For $N\ge 2$ there are no nontrivial effective elliptic stabilizers. Indeed $A=I+NB\in \Gamma (N)$ with determinant one has $\mathrm{tr}B=-N\det B$, and thereforeA noncentral elliptic element of $S{L}_2(\mathbb{Z})$ has trace $0$ or $\pm 1$, none congruent to $2$ modulo $N^2$ for $N\ge 2$. The central element $-I$ at level two acts trivially and is not an elliptic point of the effective quotient. Thus ${\nu }_2={\nu }_3=0$ for $N\ge 2$, while at level one both numbers are one.

The subgroup $\Gamma (N)$ is normal, so all its modular cusps have the same effective width as infinity. That width is $N$: the shear $T^h$ belongs to $\pm \Gamma (N)$ exactly when $N\mid h$; for $N\ge 3$ its diagonal entries exclude the minus sign, and for $N=1,2$ that sign changes nothing. Since modular cusp widths sum to $\mu$,For $N\ge 3$ the resulting principal congruence modular curve genus isFor clarity, the small levels, including all exceptional terms, areFor $N>6$, every factor in the genus correction is positive, so $g>1$. Therefore the genus-zero levels are exactly $N=1,2,3,4,5$.

For an integer weight $k$, a cusp form for a congruence subgroup $\Gamma$ is a holomorphic function on $\mathbb{H}$ with $f{|}_k\gamma =f$ for every $\gamma \in \Gamma$, whose Fourier expansion of a modular form vanishes at every modular cusp. More explicitly, choose $\sigma \in S{L}_2(\mathbb{Z})$ representing a modular cusp $\sigma \infty$, and a genuine translation period $h>0$ with $\sigma T^h{\sigma }^{-1}\in \Gamma$. Then the requirement isconvergent for sufficiently large imaginary part. Thus the slash translate extends holomorphically at $q_h=0$ and has value zero. The vector space of these functions is $S_k(\Gamma )$.

A genuine period is convenient when $-I$ is absent and the weight is odd: the smaller effective cusp width may supply a negative rather than positive translation in $\Gamma$, but an actual period always exists, possibly twice that width. Using an actual period avoids imposing an incorrect periodicity. The zero-value condition is independent of the chosen representative or period. If $-I\in \Gamma$ and $k$ is odd, its slash relation forces $f=(-1{)}^{k}f=-f$, so the space is zero.

Write $\tau =x+iy$ and $\sigma =\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$. Since $ad-bc=1$,Consequently, for the invariant norm of a modular form,This computation applies to an arbitrary function; modularity is needed only to deduce $\Gamma$-invariance of $\varphi$. The exponent is $k/2$, as printed in the PDF, rather than the converted TeX's $1/2$.

For positive integer $k$, the equivalence is the bounded invariant norm characterization of cusp forms. The three conditions below work together: meromorphicity and the bound remove interior poles; the transformation law produces a periodic local function at every modular cusp; the same bound makes that function tend to zero at the modular cusp. Conversely a cusp form has exponentially decaying Fourier expansions of a modular form, and finitely many compact core regions control its invariant norm everywhere. These implications are proved under the corresponding labelled items.

Without a positive-weight convention, weight zero is a genuine exception. The function $f=1$ satisfies all three printed conditions at $k=0$ for every congruence subgroup, but is not a cusp form. Indeed a weight-zero cusp form would descend to a holomorphic function on the compact curve $X(\Gamma )$ and hence be constant, and vanishing at a modular cusp would make it zero. Thus the literal unrestricted claim needs the qualification $k>0$.

If negative integer weights are included, both sides contain only the zero function. To check the bounded side, each modular cusp translate has at most power growth in $y$, hence at most a power of $\log (1/|q_h|)$. For a negative Laurent coefficient of order $m$, Cauchy's coefficient estimate bounds its absolute value by $C{r}^m(\log (1/r){)}^{|k|/2}\to 0$, so there is no negative coefficient and every modular cusp is holomorphic. If such a nonzero negative-weight form existed, the product of its slash translates over $\Gamma \backslash S{L}_2(\mathbb{Z})$ would be a nonzero holomorphic level-one modular form of negative weight. Right multiplication permutes the translates, and all modular cusp factors are bounded, so that product really is a level-one form. Its existence contradicts the valence formula. This also excludes negative-weight modular cusp forms. The nonzero counterexample is specifically weight zero.

A cusp form is holomorphic in the upper half-plane by definition, so it is meromorphic there. In the reverse direction suppose $f$ is meromorphic andOn a small closed disk around an interior point, $y$ is bounded above and away from zero. Hence $|f|$ is bounded there by a fixed constant. A pole of positive order would make $|f|$ unbounded along points approaching its center, a contradiction. Every possible pole is therefore removable, and $f$ is holomorphic on $\mathbb{H}$. Meromorphicity alone would not establish this conclusion; the bounded invariant norm is being used here.

The transformation condition is necessary by the definition of a cusp form. For sufficiency, fix $\sigma \in S{L}_2(\mathbb{Z})$ representing any modular cusp, and set $g=f{|}_k\sigma$. By item (i), $g$ is holomorphic in the upper half-plane. Choose an actual period $h$ with $\sigma T^h{\sigma }^{-1}=\gamma \in \Gamma$. Associativity of slash operations and the given transformation law giveThus $g(\tau +h)=g(\tau )$ and $g$ is a single-valued holomorphic function on a punctured disk with coordinate $q_h=e^{2\pi i\tau /h}$. The calculation in part (b) transports the original uniform bound to this translate:This puts every modular cusp in exactly the situation needed in item (iii). No assumption that a single modular cusp represents all of $\Gamma$'s modular cusps is made.

First suppose $f$ is a cusp form. Choose finitely many representatives ${\sigma }_j$ for $\Gamma \backslash S{L}_2(\mathbb{Z})$; their translates of the standard modular region cover a fundamental region for $\Gamma$. On the compact truncated part of each translate, the continuous function $|f(\tau )|y^{k/2}$ is bounded. On its tail, the corresponding modular cusp expansion of $g_j=f{|}_k{\sigma }_j$ has zero constant term, so $g_j=O(e^{-2\pi y/h_j})$ uniformly in the real part. Therefore $y^{k/2}|g_j|$ is bounded there and tends to zero. Part (b) identifies this expression with $\varphi ({\sigma }_j\tau )$. There are only finitely many representatives; taking the maximum of these bounds and using $\Gamma$-invariance proves that $\varphi$ is bounded on all of $\mathbb{H}$.

Conversely assume all three conditions and $k>0$. Items (i) and (ii) show that at each modular cusp a holomorphic periodic translate $g$ satisfies $|g(\tau )|\le B{y}^{-k/2}$. In its punctured-disk coordinate,The Riemann removable singularity theorem extends $g$ holomorphically to zero with $g(0)=0$. Equivalently its Laurent expansion has no negative powers and no constant term. Since this holds at every modular cusp, $f$ is a cusp form. Hence the three conditions are sufficient as well as necessary in positive weight, with the weight-zero exception explained in the root solution.