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A modular form is a holomorphic function on the complex upper half-plane that transforms with a fixed weight under a congruence subgroup and is holomorphic at every cusp.

A modular form on an arbitrary finite-index subgroup $\Gamma \le S{L}_2(\mathbb{Z})$ is a holomorphic function on the complex upper half-plane, invariant under the weight-$k$ slash operator for modular forms, and holomorphic at a cusp at every cusp of a modular group for $\Gamma$. If $\sigma \in S{L}_2(\mathbb{Z})$ maps infinity to a cusp, choose a genuine translation period $h>0$ with $\sigma T^h{\sigma }^{-1}\in \Gamma$. Then $f{|}_k\sigma$ has a convergent series in $e^{2\pi iz/h}$ with nonnegative exponents. A cusp form has zero constant term at every cusp. This definition allows noncongruence subgroups and avoids sign ambiguities in odd weights when a smaller width is defined only modulo the center.

A weight-$k$ modular form for $\Gamma =S{L}_2(\mathbb{Z})$ is a holomorphic function $f$ on the complex upper half-plane satisfyingand holomorphic at a cusp. Since the full modular group has one cusp class, this last condition means that the period-one function has a convergent expansion $f(\tau )={\sum }_{n\ge 0}{a}_n{q}^n$, $q=e^{2\pi i\tau }$, near infinity. A cusp form has $a_0=0$. The element $-I$ makes a nonzero weight odd form impossible.

For weight zero, $f$ descends to a holomorphic function on the quotient of the complex upper half-plane by the modular group. At an elliptic fixed point, invariance under its finite stabilizer makes the local Taylor series a series in the quotient coordinate. At the cusp, the $q$ expansion extends it over $q=0$. The standard fundamental domain of the modular group, with its boundary identified and its cusp added, is a compact Riemann surface, the compactified modular curve $X(1)$. The maximum modulus principle now provesHolomorphy at the cusp is essential to this conclusion.

For positive even $k\ge 4$, define the unnormalized Eisenstein seriesThe real-linear map $(m,n)\mapsto m\tau +n$ is invertible, and on a compact subset of the complex upper half-plane its norm is uniformly comparable to $\sqrt{m^2+n^2}$. Thus the sum converges absolutely and locally uniformly for $k>2$, proving holomorphy. For $\gamma =\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$,The map $(m,n)\mapsto (ma+nc,mb+nd)$ is a bijection on the nonzero integer pairs, giving $G_k(\gamma \tau )=(c\tau +d{)}^k{G}_k(\tau )$.

For the expansion, the cotangent partial-fraction Fourier kernel follows by differentiating the partial-fraction expansion of $\pi \cot \pi z$ and its geometric-series expansion $-\pi i(1+2{\sum }_{r\ge 1}{e}^{2\pi irz})$, valid for $\mathrm{Im}z>0$. It gives, for even $k\ge 2$,The $m=0$ part of $G_k$ is $2\zeta (k)$, where $\zeta$ is the Riemann zeta function; positive and negative $m$ contribute equally. Apply the kernel at $z=m\tau$ for $m>0$, and regroup the absolutely convergent sum according to $N=mr$. With the divisor sum ${\sigma }_{k-1}(N)={\sum }_{r\mid N}{r}^{k-1}$, this provesThe expansion has no negative powers and converges for $|q|<1$, so it also proves holomorphy at a cusp and completes the proof of modularity.

The supplied zeta values yield the Fourier expansion of a normalized Eisenstein seriesBoth $E_4^3$ and $E_6^2$ are modular forms of weight twelve. Their constant terms cancel, so their difference divided by $1728$ is a cusp form. Integrality requires an argument: dividing an integral series by $1728$ alone would not suffice. Expanding directly gives the integrality identity for the modular discriminantFor every integer $d$, $5d^3+7d^5$ is divisible by twelve. Modulo three, $d^3\equiv d^5\equiv d$; modulo four, an even $d$ makes both powers divisible by four, while an odd $d$ has $d^3\equiv d^5\equiv d(\mathrm{mod}4)$. Thus each coefficient $(5{\sigma }_3(n)+7{\sigma }_5(n))/12$ is integral. The other three series visibly have integral coefficients, provingThis proves integrality without assuming a product expansion for the modular discriminant.

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

First establish rational conjugation of finite-index modular subgroups without assuming that $\Gamma$ is a congruence subgroup. Multiply $\gamma$ by a positive integer to obtain an integral matrix $A$, and let $D=\det A>0$. Conjugation is unchanged by this scalar. If $h=I+DB\in \Gamma (D)$, thenThus the principal congruence subgroup $\Gamma (D)$ is contained in $H=S{L}_2(\mathbb{Z})\cap {\gamma }^{-1}S{L}_2(\mathbb{Z})\gamma$. It has finite index because reduction modulo $D$ has finite image. Inside $H$, pullback under conjugation of $\Gamma$ has relative index at most $[S{L}_2(\mathbb{Z}):\Gamma ]$. ConsequentlyThis argument does not assert that an arbitrary finite-index subgroup contains a principal congruence subgroup.

Use the determinant-normalized slash operatorThe positive real power of the determinant is used; on $S{L}_2(\mathbb{Z})$ this reduces to the usual slash operator for modular forms. The automorphy factor identity gives the right-action rule $(f{|}_k\gamma ){|}_k\eta =f{|}_k(\gamma \eta )$.

A modular form on a finite-index subgroup of integer weight $k$ is a holomorphic function on the complex upper half-plane, invariant under this weight-$k$ action of $\Gamma$, and holomorphic at a cusp at each of its cusps. A cusp of a modular group is a $\Gamma$ orbit in ${\mathbb{P}}^1(\mathbb{Q})$. If $\sigma \in S{L}_2(\mathbb{Z})$ carries infinity to its representative, choose a positive integer $h$ with $\sigma T^h{\sigma }^{-1}\in \Gamma$. Such $h$ exists by finite index. Then $f{|}_k\sigma$ is periodic and has a convergent expansion in $q_h=e^{2\pi iz/h}$ near zero; holomorphy means no negative exponents, and being a cusp form means zero constant term. Using an actual translation period avoids possible signs if a smaller width of a cusp is defined only modulo the center, particularly in odd weights.

For cusp holomorphy under rational slash operators, choose $\sigma \in S{L}_2(\mathbb{Z})$ with $\sigma \infty =\gamma \infty$, possible by completing a primitive integer pair to a determinant-one matrix. Then ${\sigma }^{-1}\gamma =\left(\begin{array}{cc}a& b\\ 0& d\end{array}\right)$, with $a/d>0$. Up to a nonzero constant factor,The imaginary part of the argument tends to infinity with that of $z$, so this remains bounded by the cusp expansion of $f{|}_k\sigma$. It tends to zero if $f$ is a cusp form. Moreover $f{|}_k\gamma$ is invariant under ${\Gamma }^{\prime }$: for $\eta \in {\Gamma }^{\prime }$, $\gamma \eta {\gamma }^{-1}\in \Gamma$ and the right-action rule applies. Finite index gives a translation period for ${\Gamma }^{\prime }$, so boundedness is a removable singularity at zero in that periodic parameter. This proves holomorphy at infinity. For every other cusp, apply the same argument to the rational matrix $\gamma \rho$, with $\rho \in S{L}_2(\mathbb{Z})$. Thus all cusp conditions hold, and

For the character twist by rational translations of a cusp form, put $B_j=\left(\begin{array}{cc}1& j/N\\ 0& 1\end{array}\right)$ and $K={\Gamma }_1(N)\cap {\Gamma }_0(N^2)$. For $\eta =\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\in K$, direct conjugation givesIndeed $N^2\mid c$ and $a\equiv d\equiv 1(\mathrm{mod}N)$. Every $f{|}_k{B}_j=f(z+j/N)$ is therefore $K$-invariant and vanishes at all its cusps by the preceding rational-translate argument. Their finite weighted sum is a cusp form, for every Dirichlet character:

There is, however, a missing primitivity hypothesis in the printed final expansion claim. The exact Fourier expansion of a modular form is alwaysValues of a Dirichlet character on units have modulus one, so $\chi (j{)}^{-1}=\overline{\chi (j)}$. For unit $n$, substitution gives $S_{\overline{\chi }}(n)=\chi (n)g(\overline{\chi })$, where $g$ is the Gauss sum of a Dirichlet character. For nonunit $n$, this vanishing formula requires a primitive Dirichlet character.

Here is its proof in that case. Choose a prime $p\mid (n,N)$. Primitivity supplies a unit $u\equiv 1(\mathrm{mod}N/p)$ with $\chi (u)\ne 1$: otherwise the character would factor through the surjective reduction to units modulo $N/p$. Surjectivity follows by lifting a unit and, if needed, adjusting the lift to avoid the additional prime $p$, using the Chinese remainder theorem. Multiplication by $u$ fixes $e^{2\pi inj/N}$ because $p\mid n$, but multiplies the character factor by a nontrivial constant. Hence $S_{\overline{\chi }}(n)=0$. The finite Fourier transform of a primitive Dirichlet character now gives the corrected formulaThe constant is nonzero: finite exponential orthogonality gives ${\sum }_{n\mathrm{mod}N}|S_{\overline{\chi }}(n){|}^2=N\phi (N)$, whereas the proved formula makes this $\phi (N)|g(\overline{\chi }){|}^2$. Thus $|g(\overline{\chi })|=\sqrt{N}$.

For a concrete counterexample to the printed unrestricted claim, take $N=2$, the principal Dirichlet character, and $f=\Delta$. The translation sum is $\Delta (z+1/2)={\sum }_n(-1{)}^n\tau (n)q^n$, whose $q^2$ coefficient is $-24$. Any constant multiple of the proposed odd-index-only series has $q^2$ coefficient zero. Thus the general modularity conclusion is proved, while the claimed simplification is false without the stated extra hypothesis.

Use the convention that a weak modular form is a holomorphic function $f$ on the complex upper half-plane satisfyingwith no condition at infinity. In particular $f(z+1)=f(z)$, so it has a Laurent series $f(z)={\sum }_{n\in \mathbb{Z}}{a}_n{q}^n$ on $0<|q|<1$, where $q=e^{2\pi iz}$. It is a modular form when it is holomorphic at a cusp, here the unique level-one cusp at infinity, equivalently $a_n=0$ for $n<0$. It is a cusp form when additionally $a_0=0$. If one uses the narrower weakly holomorphic convention, one also requires finitely many negative terms; the derivative argument below holds in that subspace as well. The matrix $-I$ forces odd-weight forms to be zero.

Put $(a{)}_r=a(a+1)\cdots (a+r-1)$, with $(a{)}_0=1$, andWe prove the derivative transformation of a weak modular form in the uniform formFor $\ell =0$ this is the weight-$k$ transformation under $S$. Differentiating the formula and multiplying by $z^2$, since $d(-1/z)/dz=z^{-2}$, gives the recurrencewhere out-of-range coefficients are zero. The displayed explicit coefficients satisfy it. Indeed, for interior $j$, extract the common polynomial $(k+j{)}_{\ell -j}$; the remaining identity isThis follows from Pascal's identity and $j\left(\genfrac{}{}{0ex}{}{\ell }{j}\right)=(\ell -j+1)\left(\genfrac{}{}{0ex}{}{\ell }{j-1}\right)$. The endpoint coefficients satisfy the recurrence directly. This polynomial proof also applies when a factor is zero, avoiding division by that factor. As $C_{\ell ,\ell }=1$, the result isThe letter $l$ in the printed exponent is the same derivative order $\ell$.

For integer $k<0$, take $\ell =1-k$. If $j<\ell$, the product runs from $k+j\le 0$ through $k+\ell -1=0$, and therefore vanishes. Consequently $f^{(1-k)}$ has weight $k+2(1-k)=2-k$ under $S$. It is also invariant under $T$, by differentiating $f(z+1)=f(z)$, and $S,T$ generate the modular group. Its holomorphy is preserved by differentiation. With $D=(2\pi i{)}^{-1}d/dz=qd/dq$, termwise differentiation of the locally convergent Laurent series gives the Bol identity for modular forms:For negative indices the exponent $1-k$ is a positive integer, and the constant term is killed, so every term is unambiguous.

The Gamma 1 congruence subgroup isThe space $M_k({\Gamma }_1(N))$ consists of the holomorphic functions $f:\mathfrak{h}\to \mathbb{C}$ satisfyingfor every $\gamma \in {\Gamma }_1(N)$ and which are holomorphic at a cusp for every cusp. Its subspace $S_k({\Gamma }_1(N))$ consists of the forms that vanish at every cusp.

For an integer $k$ and a congruence subgroup $\Gamma \le S{L}_2(\mathbb{Z})$, the space $M_k(\Gamma )$ consists of holomorphic functions $f:\mathfrak{h}\to \mathbb{C}$ satisfyingfor every $\gamma =\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\in \Gamma$, and which are holomorphic at every cusp. This is the space of modular forms of weight $k$ and level $\Gamma$.

A modular function of weight $k$ and level $\Gamma (1)$ is a meromorphic function $f:\mathfrak{h}\to \mathbb{C}$ satisfyingfor every $\gamma =\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\in \Gamma (1)$, and having a meromorphic Fourier expansion at the cusp infinity. Equivalently, $f{|}_k\gamma =f$ for the slash operator for modular forms.

A modular form is a modular function that is holomorphic on $\mathfrak{h}$ and holomorphic at infinity. Its Fourier expansion therefore has the formSince $-I\in \Gamma (1)$, a nonzero level-one form must have even weight.