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The modular discriminant has the convergent productis a weight-twelve level-one cusp form, and has order one at infinity. Applying the degeneracy map for cusp forms with $M=1,N=D=p$ gives $\boxed{\Delta (p\tau )\in S_{12}({\Gamma }_0(p))}$. The original form $\Delta$ is also a cusp form on this subgroup.

There are precisely two modular cusps for prime level, represented by infinity and zero. To see this directly, write a rational cusp as $a/c$ in lowest terms. If $p\mid c$, completing $(a,c{)}^T$ to a determinant-one matrix gives a member of ${\Gamma }_0(p)$ taking infinity there. If $p\nmid c$, the equation $Ac-aC=1$ has a solution with $p\mid C$: adjust a Bézout solution modulo $p$, using the invertibility of $c$. The matrix $\left(\begin{array}{cc}A& a\\ C& c\end{array}\right)$ then lies in ${\Gamma }_0(p)$ and takes zero there. The two classes are distinct because their denominator divisibility is preserved by the group.

At infinity the cusp width is one, so $q$ is the local parameter. The products start with $q$ and $q^p$, giving orders one and $p$. At zero use $S=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$. Since $S{T}^h{S}^{-1}=\left(\begin{array}{cc}1& 0\\ -h& 1\end{array}\right)$, the cusp width is $p$ and the parameter is $q_0=e^{2\pi i\tau /p}$. The weight-twelve inversion law of the modular discriminant givesHere determinant one makes the printed and ordinary slash operators for modular forms agree. In terms of $q_0$, the leading terms are $q_0^p$ and $p^{-12}{q}_0$. Hence the prime-level cusp orders of discriminant degeneracy forms areThe scalar $p^{-12}$ changes no order of vanishing.

Use the Hecke-normalized rational slash operator from the PDF, writing $f[\alpha {]}_k$ for its right action. Put $A_D=\mathrm{diag}(D,1)$. If $\gamma =\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\in {\Gamma }_0(N)$, thenIndeed $N/D$ is an integer divisible by $M$, so $c/D$ is an integer divisible by $M$. The ordinary weight transformation of the original cusp form therefore givesThus $g(\tau )=f(D\tau )$ has the correct transformation for the Gamma 0 congruence subgroup at level $N$, and it is holomorphic in the complex upper half-plane.

It remains to check every modular cusp, not just infinity. Let $\rho \in S{L}_2(\mathbb{Z})$ carry infinity to that cusp, and choose $\sigma \in S{L}_2(\mathbb{Z})$ with $\sigma \infty =A_D\rho \infty$. The matrix $B={\sigma }^{-1}{A}_D\rho$ is upper triangular over the rationals, with positive ratio of diagonal entries. Therefore, up to a nonzero constant weight factor,The original cusp expansion of $f[\sigma {]}_k$ has only positive exponents. The change $\tau \mapsto r\tau +s$ induced by $B$, with $r>0$, preserves exponential decay as $\mathrm{Im}\tau \to \infty$. The transformed function also has a positive integral period because $g$ is modular for a finite-index subgroup. Its bounded periodic holomorphic expansion consequently has no negative exponents and its limiting constant term is zero. This proves cusp holomorphy under rational slash operators and establishes the degeneracy map for cusp forms:

Put $A_N=\left(\begin{array}{cc}0& -1\\ N& 0\end{array}\right)$. All calculations here use the Hecke-normalized rational slash operator of the PDF. Its automorphy factor identity gives $(f[\alpha {]}_k)[\beta {]}_k=f[\alpha \beta {]}_k$. For $\gamma =\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\in {\Gamma }_1(N)$,This proves that $A_N$ normalizes the Gamma 1 congruence subgroup. Right-action associativity shows $f[A_N{]}_k$ has the same transformation law as $f$. It is holomorphic in the complex upper half-plane; at each modular cusp, the rational change of variable preserves cusp decay by the upper-triangular factorization used in 2(i). Consequently $W_N$ maps the cusp form space to itself.

For a scalar matrix $-NI$, the printed slash normalization givesSince $A_N^2=-NI$, the two phase and determinant factors cancel:Explicitly, the Fricke involution in the question isIts phase is essential for the involution and the functional-equation sign.

Suppose $W_{N}f=\epsilon f$. Define $F(y)=f(iy/\sqrt{N})$ for $y>0$. Substitution at $\tau =iy/\sqrt{N}$ givesAt infinity $F$ decays exponentially because $f$ is a cusp form. This identity gives $F(y)=O(y^{-k}{e}^{-c/y})$ near zero for some $c>0$. Thus its Mellin transform converges for every complex $s$, locally uniformly with all $s$ derivatives.

Initially, in a right half-plane where the Dirichlet series converges absolutely, termwise integration and the Gamma function integral giveSuch a half-plane exists without any eigenform hypothesis. Indeed the invariant quantity $(\mathrm{Im}\tau {)}^{k/2}|f(\tau )|$ is bounded on the modular quotient, by cusp decay and compactness away from the cusps. Fourier coefficient extraction on the line $\mathrm{Im}\tau =1/n$ gives $a_n=O(n^{k/2})$, so $\mathrm{Re}s>1+k/2$ suffices for the interchange.

Split the integral at one and change $y$ to $1/y$ in its lower part. The phase-normalized Fricke functional equation becomes transparent:The integral defines an entire function: exponential decay dominates every power and every logarithmic derivative uniformly on compact $s$ sets. Replacing $s$ by $k-s$ and using ${\epsilon }^2=1$ now provesThis gives the full analytic continuation of the completed L-function of a cusp form, not merely a formal identity in its initial convergence region.

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.

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.

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.

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.

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.

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.

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.

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.

For the full modular group $\Gamma =S{L}_2(\mathbb{Z})$ and integer weight $k$, a modular form is a holomorphic function on the upper half-plane satisfyingand holomorphic at the cusp at infinity. Explicitly, periodicity under $\tau \mapsto \tau +1$ gives a Fourier expansion of a modular form in $q=e^{2\pi i\tau }$, and holomorphy at the cusp means $f={\sum }_{n\ge 0}{a}_n{q}^n$. A cusp form has $a_0=0$. All cusps are equivalent for this group. The matrix $-I$ forces $f=(-1{)}^{k}f$, so every odd-weight modular form is zero.

There is a corresponding qualification to the displayed Eisenstein series formula: the Bernoulli-normalized formula is for even integers $k\ge 4$. For odd integers $k>2$, pairing $(m,n)$ with $(-m,-n)$ gives $G_k=0$, while $B_k=0$, so the printed expression divided by $B_k$ is undefined. We prove the asserted nonzero formula in its even-weight range.

On each compact set in the upper half-plane, the real-linear map $(m,n)\mapsto m\tau +n$ has a uniformly bounded inverse. Hence $|m\tau +n|\ge C(m^2+n^2{)}^{1/2}$ there. For $k>2$, lattice convergence gives locally uniform absolute convergence of $G_k$, and therefore holomorphy. If $\gamma =\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\in \Gamma$, thenThe map $(m,n)\mapsto (ma+nc,mb+nd)$ permutes the nonzero integer pairs. Reindexing proves the weight-$k$ transformation law.

To compute the Fourier expansion of a normalized Eisenstein series, start from the cotangent partial-fraction expansionFor $\mathrm{Im}z>0$, the cotangent Fourier expansion is $\pi \cot (\pi z)=-\pi i-2\pi i{\sum }_{r\ge 1}{e}^{2\pi irz}$. Differentiating both normally convergent expressions $k-1$ times givesFor even $k$, the terms with $m=0$ sum to $2\zeta (k)$, and the terms with $m<0$ duplicate those with $m>0$. Substituting $z=m\tau$ in the last identity and collecting $N=mr$ givesThe rearrangement is absolutely convergent for every $|q|<1$, since the divisor function grows at most polynomially.

For completeness, the normalization follows directly from the definition of the Bernoulli numbers. Near zero, the cotangent partial-fraction expansion givesOn the other hand $\pi \cot (\pi z)=\pi i+2\pi i/(e^{2\pi iz}-1)$. Substitute $t=2\pi iz$ in the generating function of the Bernoulli numbers. The term $B_1=-1/2$ cancels $\pi i$, and the higher odd terms vanish because $t/(e^t-1)+t/2$ is even. Comparison of Laurent coefficients proves the Euler evaluation of even zeta valuesThusThis Fourier series also proves holomorphy at infinity, completing the modular form verification.

For the Ramanujan congruence modulo 691, the standard facts we use are that the modular discriminant is the weight-twelve cusp form $\Delta =q{\prod }_{n\ge 1}(1-q^n{)}^{24}={\sum }_{n\ge 1}\tau (n)q^n$, and that $S_{12}=\mathbb{C}\Delta$. The latter also follows from the valence formula for the modular group: a nonzero weight-twelve modular form has total weighted zero order one, so a cusp form whose coefficient of $q$ vanishes would have cusp order at least two and must be zero. The product makes the Ramanujan tau function integer-valued.

Write $E_k=G_k/(2\zeta (k))$. Using $B_4=-1/30$ and the supplied $B_{12}=-691/2730$, we haveBoth $E_{12}$ and $E_4^3$ have weight twelve and constant term one. Their difference is a cusp form, so is a multiple of $\Delta$. Its coefficient of $q$ is $65520/691-720=-432000/691$, which determines that multiple:The coefficients of $E_4^3$ are integers. Comparing the positive-index Fourier coefficients and reducing modulo $691$ therefore gives $65520{\sigma }_{11}(n)\equiv -432000\tau (n)$. Both numerical constants are congruent to $566$ modulo $691$, and $\gcd (566,691)=1$. Cancellation proves