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A sum of character values over a set. Arithmetic cancellation, rather than the termwise absolute-value bound, is the useful feature. A Gauss sum of a Dirichlet character is a weighted example, and the Pólya–Vinogradov inequality bounds interval sums.

Take the even primitive Dirichlet character ${\chi }_{12}$ whose values at $1,5,7,11$ modulo twelve are $1,-1,-1,1$. Its Gauss sum of a Dirichlet character is $\sqrt{12}$. The displayed series $A$ has $A(-1/\tau )=\sqrt{-i\tau }A(\tau )$ and $A(\tau +1)=e^{\pi i/12}A(\tau )$. Thus $A^{24}$ is the normalized weight-twelve modular discriminant. It has no interior zeros by the valence formula for the modular group. Its logarithmic derivative equals $E_2/24$, so integration gives the Dedekind eta function product. The odd character modulo four similarly produces a series $B$ with $B^8=A^{24}$ and leading term $q^{1/8}$; consequently $B=A^3$.

With $S_{\chi }(r)$ the finite Fourier transform in Gauss sum of a Dirichlet character, the constant coefficient of $G_k(\chi ,z)$ in $e^{2\pi iz/N}$ is $2L(\chi ,k)$, and its positive coefficients areFor a primitive Dirichlet character this simplifies to $2(-2\pi i{)}^{k}g(\chi )((k-1)!N^k{)}^{-1}{\sum }_{r\mid n}\overline{\chi }(r)r^{k-1}$. Without primitivity the simplified formula is generally false.

Use the determinant-normalized slash operator and $W_M=\left(\begin{array}{cc}0& -1\\ M& 0\end{array}\right)$. For units $u,v$ with $Nuv\equiv -1(\mathrm{mod}D)$, the matrix identity $U_u{W}_{N{D}^2}=D{\gamma }_u{W}_N{U}_v$ has ${\gamma }_u=\left(\begin{array}{cc}(1+Nuv)/D& u\\ Nv& D\end{array}\right)\in {\Gamma }_0(N)$. A positive scalar matrix acts trivially. The nebentypus supplies $\psi (D)$; reindexing gives $\overline{\chi }(u)=\chi (-N)\chi (v)$. Thus the factor is initially $\psi (D)\chi (-N)\tau (\chi )/\tau (\overline{\chi })$. The primitive Gauss sum of a Dirichlet character identity $\tau (\chi )\tau (\overline{\chi })=\chi (-1)D$ gives the displayed equivalent factor. No assumption that $f$ itself is a Fricke eigenform is needed.

Write $U_u=\left(\begin{array}{cc}1& u/D\\ 0& 1\end{array}\right)$ and $\tau (\chi )={\sum }_{u\mathrm{mod}D}\chi (u)e^{2\pi iu/D}$. The finite Fourier transform of a primitive Dirichlet character givesFor $(n,D)=1$ this follows by multiplying residues by $n$. For a nonunit $n$, primitivity supplies a unit $v\equiv 1(\mathrm{mod}D/p)$ with nontrivial character value, for a prime $p\mid (n,D)$; multiplication by $v$ leaves the exponential unchanged and forces the sum to vanish. Orthogonality of the finite exponentials gives ${\sum }_{n\mathrm{mod}D}|{\sum }_u\chi (u)e^{2\pi inu/D}{|}^2=D{\sum }_u|\chi (u){|}^2=D\phi (D)$. The primitive transform identity makes the left side $\phi (D)|\tau (\chi ){|}^2$. Thus the primitive Gauss sum of a Dirichlet character is nonzero, with $|\tau (\chi ){|}^2=D$. Consequently the twist is the finite rational-translation sumThis representation will prove both the level statement and the Fricke formula.

Take $\gamma =\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\in {\Gamma }_0(N{D}^2)$, and choose $v\equiv d^{2}u(\mathrm{mod}D)$. Since $ad\equiv 1(\mathrm{mod}D)$, this means $av\equiv ud(\mathrm{mod}D)$. Direct multiplication givesEvery entry is integral, its determinant is one, and its lower-left entry is divisible by $N$. Thus ${\gamma }_u\in {\Gamma }_0(N)$. Its lower-right entry is congruent to $d$ modulo $N$, so $f|{\gamma }_u=\psi (d)f$ by its nebentypus character. Using the right-action property of the slash operator,The map $u\mapsto v=d^{2}u$ permutes the residue classes. Reindexing gives $\overline{\chi }(u)=\chi (d{)}^2\overline{\chi }(v)$, soThis proves the required modular weight and character on ${\Gamma }_0(N{D}^2)$.

Cuspidality must also be checked. For any modular cusp matrix $\sigma \in S{L}_2(\mathbb{Z})$ and any rational translation $U_u$, choose $\rho \in S{L}_2(\mathbb{Z})$ with $\rho \infty =U_u\sigma \infty$. Then ${\rho }^{-1}{U}_u\sigma$ is upper triangular with positive dilation. Since $f|\rho$ decays exponentially at infinity, so does $f|U_u\sigma$. The finite sum defining $f_{\chi }$ therefore tends to zero at every modular cusp; its transformation law supplies a modular cusp period, so its holomorphic Fourier expansion has zero constant term there. Hence the primitive twist at coprime level is

Now put $W_M=\left(\begin{array}{cc}0& -1\\ M& 0\end{array}\right)$ and $g=f{|}_k{W}_N$, always using the determinant factor specified in the question. For each unit $u$ choose a unit $v$ with $Nuv\equiv -1(\mathrm{mod}D)$, possible because $(N,D)=1$. The crucial integral matrix factorization isThe right side multiplies out to $\left(\begin{array}{cc}uND& -1\\ N{D}^2& 0\end{array}\right)$, so the identity is exact. Its determinant-one matrix has lower-right entry $D$, and therefore $f|{\gamma }_u=\psi (D)f$. A positive scalar matrix $DI$ acts trivially under the determinant-normalized slash operator, since its determinant contributes $D^k$ and its denominator contributes $D^{-k}$. ThusThe inverse-residue substitution is $u\equiv -(Nv{)}^{-1}(\mathrm{mod}D)$, giving $\overline{\chi }(u)=\chi (-N)\chi (v)$. The sum over $v$ is $\tau (\chi )g_{\overline{\chi }}$. Hence the Fricke transform of a primitive coprime twist isThe second version uses $\tau (\chi )\tau (\overline{\chi })=\chi (-1)D$, which follows from $\overline{\tau (\chi )}=\chi (-1)\tau (\overline{\chi })$ and $|\tau (\chi ){|}^2=D$. Here $g_{\overline{\chi }}=\sum b(n)\overline{\chi }(n)q^n$ if $g=\sum b(n)q^n$; it is not the complex conjugate function $\overline{g}$. The coprimality of $N,D$ is used both in the matrix factorization and in evaluating $\psi (D)$. For $D=1$ the twist is trivial and the formula simply reads $f|W_N=g$.

Extend $\chi$ by zero to integers not coprime to $N$, and write $g(\chi )={\sum }_{a\mathrm{mod}N}\chi (a)e^{2\pi ia/N}$ for its Gauss sum of a Dirichlet character. We use the analytic branch of $\sqrt{-i\tau }$ that is positive when $\tau$ is positive imaginary. With the convention $\hat{F}(t)={\int }_{\mathbb{R}}F(x)e^{-2\pi ixt}dx$, the Gaussian Fourier transform and its differentiated version areThe first formula follows by completing the square, first on the imaginary axis and then by analytic continuation; the second follows by differentiating the Fourier transform in $t$. All the functions are Schwartz functions on the real axis.

For completeness, the finite Fourier transform of a primitive Dirichlet character satisfiesFor a unit $r$ this is substitution by its inverse. If $p\mid (r,N)$, primitivity supplies a unit $u\equiv 1(\mathrm{mod}N/p)$ with $\chi (u)\ne 1$: otherwise the character would factor through modulus $N/p$. Multiplication by $u$ fixes the additive exponential but multiplies the sum by $\chi (u)$, so the sum vanishes. This proves the identity also on nonunits. Finite Parseval identity now gives $|g(\chi ){|}^2=N$, since exactly $\phi (N)$ frequencies have nonzero character values. In particular this Gauss sum of a Dirichlet character is nonzero.

Apply Poisson summation to each residue class: ${\sum }_{j}F(a+Nj)=N^{-1}{\sum }_r{e}^{2\pi ira/N}\hat{F}(r/N)$. Multiply by $\chi (a)$ and sum over $a$. The formulas above give the two character theta transformationsandIn the odd case the prefactor originally reads $g(\chi )/(N^2\tau \sqrt{-i\tau })$; using $\tau =i(-i\tau )$ explains the factor $i$. Opposite parity makes the corresponding series identically zero by pairing $n,-n$. This is why the weight-three-halves series is needed for odd Dirichlet characters.

To deduce the products from these transformations, use two specific primitive Dirichlet characters. The even character ${\chi }_{12}$ takes values $1,-1,-1,1$ on $1,5,7,11$ modulo twelve. It is primitive: its values do not factor through six or four, hence through no proper divisor of twelve. Its Gauss sum of a Dirichlet character is $2\sqrt{3}=\sqrt{12}$. PutThe even transformation gives $A(-1/\tau )=\sqrt{-i\tau }A(\tau )$; also $A(\tau +1)=e^{\pi i/12}A(\tau )$, since each nonzero term has $n^2\equiv 1(\mathrm{mod}24)$. Thus $A^{24}$ transforms as a weight-twelve modular form under the generators $S,T$ of the modular group. It is holomorphic and starts with $q$, so it is a normalized cusp form. The proof in Question 1 shows $S_{12}=\mathbb{C}{\Delta }_0$, giving $A^{24}={\Delta }_0$. In particular $A$ has no interior zeros, by the valence formula for the modular group.

The Serre derivative argument in Question 2 gives $D{\Delta }_0=E_2{\Delta }_0$. Hence $DA/A=E_2/24$. The analytic product $P(\tau )=q^{1/24}{\prod }_{n\ge 1}(1-q^n)$ is nonzero and has the same logarithmic derivative, since $DP/P=1/24-\sum {\sigma }_1(n)q^n$. Therefore $A/P$ is constant on the connected complex upper half-plane; its limit at infinity is one. This proves $A=P$, without assuming either of the product identities being deduced. Pair the terms $n,-n$ in $A$, choosing representatives $n=6r+1$. Since ${\chi }_{12}(6r+1)=(-1{)}^r$, we getThis is the Euler pentagonal identity, derived by the character theta proof of the eta product.

For the second product use the odd primitive character ${\chi }_4(1)=1$, ${\chi }_4(3)=-1$, with Gauss sum of a Dirichlet character $2i$. SetThe odd transformation gives $B(-1/\tau )=(-i\tau {)}^{3/2}B(\tau )$ and the series gives $B(\tau +1)=e^{\pi i/4}B(\tau )$. Thus $B^8$ is another normalized weight-twelve cusp form, and $B^8={\Delta }_0=A^{24}$. Because $A$ is nonzero, $B/A^3$ has eighth power one, so it is constant; its leading coefficient fixes it to one. Pairing positive and negative odd integers now givesSince $B=A^3=q^{1/8}{\prod }_{n\ge 1}(1-q^n{)}^3$, the Jacobi cubic product identity follows:All pairings and differentiations take place in locally uniformly convergent theta functions or products on $|q|<1$.

For a primitive Dirichlet character the previous formula gives $|\hat{\chi }(a)|=|\chi (a)||\hat{\chi }(1)|$. The Plancherel theorem for the unitary finite transform yieldsThus $|\hat{\chi }(1)|=1$, the normalized Gauss sum of a Dirichlet character magnitude.

For an imprimitive character modulo $p^k$ with $k\ge 2$, its values on residues are periodic modulo $p^{k-1}$: descent preserves the values on units, and divisibility by $p$ is unchanged by that shift. Splitting the sum into these residue classes gives a factor ${\sum }_{j=0}^{p-1}e(-j/p)=0$. Hence $\hat{\chi }(1)=0$. If $k=1$, the only imprimitive character is principal, and its Gauss sum is ${\sum }_{n=1}^{p-1}e(-n/p)=-1$, giving $|\hat{\chi }(1)|=p^{-1/2}$. All cases satisfy the required bound.

An even Dirichlet character satisfies $\chi (-1)=1$, while an odd Dirichlet character satisfies $\chi (-1)=-1$. Write $a=0$ or $1$ for its character parity and, for $x>0$, define the Dirichlet character theta functionFor a primitive Dirichlet character whose conductor of a Dirichlet character is $q>1$, the term at zero is zero. Put $\tau (\chi )={\sum }_{r\mathrm{mod}q}\chi (r)e(r/q)$, using the positive exponential, and ${\epsilon }_{\chi }=\tau (\chi )/(i^a\sqrt{q})$. The primitive Gauss sum of a Dirichlet character has magnitude $\sqrt{q}$, so $|{\epsilon }_{\chi }|=1$. The theta transformation isThus the powers are $x^{-1/2}$ in the even case and $x^{-3/2}$ in the odd case; the odd root number contains $1/i$. The conjugate character is necessary for a nonreal character. These formulas also follow by applying Poisson summation to the Gaussian function on each residue class, and to its derivative for odd parity. For the primitive principal Dirichlet character whose conductor of a Dirichlet character is one, use the ordinary Jacobi theta function with constant term one; its transformation has root number one.

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.

The original PDF has the summation condition $\gcd (n,N)=1$; the TeX's $\gcd (n,n)=1$ is a transcription error. There is also an actual missing hypothesis in the PDF's coefficient formula: that simplified formula requires a primitive Dirichlet character. We first derive a formula valid for every character, and then show both the primitive specialization and a counterexample to the unrestricted version.

For $k>2$, the character-twisted Eisenstein series converges absolutely and locally uniformly on the complex upper half-plane. On a compact subset, $|mz+n|$ is bounded below by a positive constant times $(m^2+n^2{)}^{1/2}$, and the corresponding two-dimensional lattice sum converges. Changing $n$ to $n+mN$ provesThe condition $\chi (-1)=(-1{)}^k$ makes the terms for $(m,n)$ and $(-m,-n)$ equal. The terms with $m=0$ contribute $2L(\chi ,k)$, and all other terms are twice the sum over $m\ge 1$.

For $\mathrm{Im}w>0$, the cotangent identity andgive, after $k-1$ differentiations,Differentiation is justified by locally uniform convergence. This is the cotangent partial-fraction Fourier kernel.

Write $n=a+Nt$, with $a$ running through the unit classes modulo $N$, and define the finite Fourier transformHere $g(\chi )$ is the Gauss sum of a Dirichlet character.
Applying the kernel with $w=(mz+a)/N$ yieldsThe double series converges absolutely: $|S_{\chi }(r)|\le \phi (N)$ and the exponential decay controls ${\sum }_{m,r}{r}^{k-1}{e}^{-2\pi mr\mathrm{Im}z/N}$. Grouping the terms with $mr=n$ proves the general Fourier expansion of a character-twisted Eisenstein series:

If $r$ is a unit modulo $N$, substitution $a\mapsto r^{-1}a$ gives $S_{\chi }(r)=\chi (r{)}^{-1}g(\chi )$. Suppose now that $\chi$ is primitive and $r$ is a nonunit. Choose a prime $p\mid \gcd (r,N)$. Reduction of units modulo $N$ onto units modulo $N/p$ is surjective. Primitivity means that $\chi$ is nontrivial on its kernel, so there is a unit $u\equiv 1(\mathrm{mod}N/p)$ with $\chi (u)\ne 1$. Since $ru\equiv r(\mathrm{mod}N)$, substitution by $u$ forces $S_{\chi }(r)=\chi (u{)}^{-1}{S}_{\chi }(r)$ and thus $S_{\chi }(r)=0$. This proves the finite Fourier transform of a primitive Dirichlet character identity, and consequentlyThe inverse-character notation in the question is understood to mean $\overline{\chi }(r)$, extended by zero on nonunits; literal inversion of $\chi (r)=0$ would be undefined.

For a counterexample without primitivity, take $N=2$, the principal character, and $k=4$. Then $S_{\chi }(r)=(-1{)}^r$, so $g(\chi )=-1$ but $S_{\chi }(2)=1$. At $n=2$, the general formula gives $c_2=7A_4/8$, whereas the printed simplified formula, interpreted as zero on nonunits, gives $-A_4/8$. Hence the printed coefficient formula is false for arbitrary imprimitive characters; the general boxed formula above supplies the correction.

A Dirichlet character modulo $N$ is primitive if its character on $(\mathbb{Z}/N\mathbb{Z}{)}^{\times }$ does not factor through reduction to the unit group for any proper divisor of $N$. Its least inducing modulus is called its conductor. Primitivity is essential in the usual formula expressing its finite Fourier transform as a conjugate character times a Gauss sum of a Dirichlet character.

For a primitive Dirichlet character $\chi$ modulo $D$, its Gauss sum of a Dirichlet character gives $f_{\chi }=\tau (\overline{\chi }{)}^{-1}{\sum }_u\overline{\chi }(u)f|U_u$, with $U_u=\left(\begin{array}{cc}1& u/D\\ 0& 1\end{array}\right)$. If $\gamma =\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\in {\Gamma }_0(N{D}^2)$, take $v\equiv d^{2}u(\mathrm{mod}D)$. Then $U_u\gamma U_v^{-1}\in {\Gamma }_0(N)$ and its lower-right entry is congruent to $d$ modulo $N$. Reindexing the sum yields the character $\psi (d)\chi (d{)}^2$. Rational translations preserve vanishing at modular cusps, proving the stated cuspidality as well as the transformation law.