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For a level-one cusp form and any Dirichlet character modulo $N>1$, this translation sum is a cusp form on ${\Gamma }_1(N)\cap {\Gamma }_0(N^2)$. Conjugating that subgroup by $\left(\begin{array}{cc}1& j/N\\ 0& 1\end{array}\right)$ yields integral determinant-one matrices; cusp holomorphy under rational slash operators supplies all cusp conditions. Its exact Fourier coefficients at positive indices are $a_n(f){\sum }_{j\in (\mathbb{Z}/N\mathbb{Z}{)}^{\times }}\chi (j{)}^{-1}{e}^{2\pi inj/N}$. For a primitive Dirichlet character they equal $g(\overline{\chi })\chi (n)a_n(f)$, by the finite Fourier transform of a primitive Dirichlet character. For imprimitive characters the sum can be nonzero at nonunits, so the simplified twist formula need not hold.

For a primitive Dirichlet character, $S_{\chi }(r)=\overline{\chi }(r)g(\chi )$ for every integer $r$, with zero on nonunits. For a unit $r$ this follows by substitution. If a prime $p$ divides both $r$ and $N$, primitivity supplies a unit $u\equiv 1(\mathrm{mod}N/p)$ with $\chi (u)\ne 1$; substitution by $u$ fixes the additive exponential and forces $S_{\chi }(r)=0$. For an imprimitive Dirichlet character this vanishing can fail.

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.

For a nonprincipal Dirichlet character modulo $N$, setThe numerator vanishes at $t=0$, so $H_{\chi }$ is analytic there and exponentially decreasing as $t\to \infty$. Initially for $\mathrm{Re}s>1$, $\Gamma (s)L(\chi ,s)={\int }_0^{\infty }{H}_{\chi }(t)t^{s-1}dt$. This integral already extends holomorphically to $\mathrm{Re}s>0$. Subtracting finite Taylor polynomials near zero, as in meromorphic continuation of a Mellin transform from an asymptotic expansion, continues it with possible simple poles only at nonpositive integers. Zeros of the reciprocal Gamma function cancel these, proving that $L(\chi ,s)$ is an entire function. This does not require a primitive Dirichlet character.

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.

The principal Dirichlet character modulo $N$ is one on integers coprime to $N$ and zero on all other integers. Its Dirichlet L-function is $L({\chi }_0,s)=\zeta (s){\prod }_{p\mid N}(1-p^{-s})$, so it has a simple pole at one. For $N>1$ it is induced from the character of modulus one and is not a primitive Dirichlet character.