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For coefficients supported on $N$ consecutive integers,The star restricts to primitive Dirichlet characters. Their finite Fourier transform of a primitive Dirichlet character has normalization of absolute value $\sqrt{q}$. Orthogonality of Dirichlet characters therefore bounds the weighted contribution for one modulus by ${\sum }_{\begin{array}{c}a\mathrm{mod}q\\ (a,q)=1\end{array}}|{\sum }_n{a}_{n}e(an/q){|}^2$. Sum over $q$ and use the exponential-sum large sieve on the distinct reduced fractions of denominators at most $Q$.

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 $B=A^3$.

For a primitive Dirichlet character modulo $N>1$ of parity $a\in \{0,1\}$, put ${\Theta }_{\chi }(\tau )={\sum }_n{n}^a\chi (n)e^{\pi i{n}^2\tau }$, extending the character by zero on nonunits. The square root is analytic with positive value when $-i\tau$ is positive real. Apply Poisson summation separately in each residue class modulo $N$, and use the finite Fourier transform of a primitive Dirichlet character. The Fourier transforms of $e^{\pi i\tau x^2}$ and $x{e}^{\pi i\tau x^2}$ are $(-i\tau {)}^{-1/2}{e}^{-\pi i{t}^2/\tau }$ and $(t/\tau )(-i\tau {)}^{-1/2}{e}^{-\pi i{t}^2/\tau }$. These give the formula, including its odd-character phase.

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.

The least divisor of a character's modulus through whose unit-group reduction the Dirichlet character factors. The Chinese remainder theorem reduces its construction to choosing the least exponent at each prime. Surjectivity of reduction on units makes the descended primitive Dirichlet character unique.

For a primitive Dirichlet character whose conductor of a Dirichlet character is $q>1$ and whose character parity is $a$, Poisson summation gives ${\theta }_{\chi }(x)={\epsilon }_{\chi }{x}^{-a-1/2}{\theta }_{\overline{\chi }}(1/x)$, where ${\epsilon }_{\chi }=\tau (\chi )/(i^a\sqrt{q})$. Its rapid decay at both ends yields the entire Mellin representation of the completed Dirichlet L-function. The case of conductor of a Dirichlet character equal to one uses the ordinary Jacobi theta function and a subtracted constant term.

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.

The inducing primitive Dirichlet character has conductor of a Dirichlet character $q_0$. Increasing the modulus removes Euler factors only at new prime divisors. The correction creates zeros on the imaginary axis; raising the exponents of primes already dividing the conductor of a Dirichlet character changes no Euler factor. Primitive gamma factors continue to use $q_0$. For real-even characters this includes principal primitive conductor of a Dirichlet character one, with its Riemann zeta function pole treated separately.

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.

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 $|q|<1$.

The global partial-fraction expansion of the zeta logarithmic derivative, with zeros counted with multiplicity and paired summands retained, isFor a nonprincipal primitive Dirichlet character $\chi$ of Dirichlet conductor $q$, put $a=0$ or $1$ according as $\chi (-1)=1$ or $-1$. Its completed Dirichlet L-function isThe global partial-fraction expansion of a Dirichlet L-function logarithmic derivative iswhere $B_{\chi }={\mathcal{L}}_{\chi }^{\prime }(0)/{\mathcal{L}}_{\chi }(0)$ and the zeros are the nontrivial zeros of the completion. These expansions follow from genus-one Hadamard factorization; the combined terms converge locally normally. The primitive principal Dirichlet character has Dirichlet conductor one and gives the zeta formula. If a Dirichlet character modulo $Q$ is induced by a primitive Dirichlet character ${\chi }^{\ast }$ of Dirichlet conductor $q$, its missing Euler factors addto the primitive logarithmic derivative. This correction is $O(\log Q)$ on the real interval $1<\sigma \le 2$.

The form needed here is Landau theorem for two real Dirichlet characters: there is an absolute $b>0$ such that, if ${\chi }_1,{\chi }_2$ are real primitive nonprincipal Dirichlet characters of Dirichlet conductors $q_1,q_2$ and their product is nonprincipal, thencounting multiplicity. In particular, two distinct primitive real Dirichlet characters cannot each have a zero in that interval.

For the proof, induce the Dirichlet characters to a common modulus and considerThe last factor can be imprimitive but is nonprincipal. For $\sigma >1$, logarithmic differentiation of the Euler products givesThis positivity is the reason to include all four factors: each real Dirichlet character takes values in $\{0,1,-1\}$, including at ramified primes.

The real partial-fraction identity for a primitive completion has no residual real constant. Indeed, after taking real parts in its Hadamard derivative, the constant is determined to be zero by the functional-equation symmetry of its modulus across the critical line; the zero kernels there cancel in reflected pairs. ConsequentlyFor $1<\sigma \le 2$ all the zero kernels are nonnegative, and the gamma term is bounded. The missing Euler factors have the previously bounded correction. Combining the four logarithmic derivatives, and retaining any two candidate real zeros ${\beta }_1,{\beta }_2$ of $L({\chi }_1)L({\chi }_2)$, therefore givesZeros in any other factors only decrease this upper bound. The sole pole term comes from zeta.

Put $H=\log (q_1{q}_2)$ and choose $a>0$ so small that $Ca<1/10$. If both $1-{\beta }_j<a/(4H)$, take $\sigma =1+a/H$. The upper bound is then at mosta contradiction. Taking $b=a/4$ proves the theorem, including the multiplicity assertion.

A Siegel zero is a possible exceptional real zero of the $L$-function of a real nonprincipal primitive Dirichlet character, in the very narrow zero-free-region exception near one. Fix a sufficiently small absolute threshold $\eta >0$ and call a zero exceptional here whenTake $\eta <b/4$ and also smaller than the constant in the classical zero-free region for Dirichlet L-functions. Such a zero, if present, is real and simple; its existence is not asserted. This choice fixes the constants consistently for the requested separation conclusions.

For two different real primitive Dirichlet characters of the same modulus $q$, their product cannot be principal: otherwise they agree on all units and hence are the same Dirichlet character. Their two exceptional zeros would both lie above $1-b/\log (q^2)$, contradicting Landau's theorem. Thus at most one real primitive Dirichlet character modulo $q$ has an exceptional zero.

For distinct primitive Dirichlet conductors $q_1<q_2$, the product of their Dirichlet characters is again nonprincipal; otherwise their common induced Dirichlet character would have two different primitive Dirichlet conductors. Suppose $q_2\le q_1^2$. ThenAgain both zeros violate the two-zero prohibition. Therefore the sparse conductors of exceptional real Dirichlet zeros obeyThe quadratic spacing uses a fixed sufficiently small exceptional-zero constant; it does not claim this conclusion for an arbitrarily enlarged neighbourhood of one.

A primitive Dirichlet character modulo $q$ is a Dirichlet character which is not induced from a character of a proper divisor of $q$. To determine the inducing primitive Dirichlet character, use the Chinese remainder theorem to decomposeOn each factor choose the least exponent $c_p\in \{0,\dots ,k\}$ through whose reduction the restricted character factors. Exponent zero means the trivial unit group modulo one. Put $q_0={\prod }_p{p}^{c_p}$ and define ${\chi }_0$ on the units modulo $q_0$ by these descended factors, extending by zero off the units. Every reduction of unit groups is surjective, so the descended character is unique. Its local exponents cannot be decreased, hence it is primitive. The original character is $\chi (n)={\chi }_0(n)$ when $(n,q)=1$, and zero otherwise.

Any other inducing modulus must have exponent at least $c_p$ at every prime, by restriction to the corresponding local factor. Therefore $q_0$ is the unique minimal modulus, the conductor of a Dirichlet character, and ${\chi }_0$ is the unique primitive Dirichlet character inducing $\chi$. The argument also explains why removing extra prime factors can change values at integers that were nonunits for $q$.

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.

The completed Dirichlet L-function isFor nonprincipal primitive Dirichlet characters, termwise Mellin transformation initially in $\Re s>1$ givesThe Dirichlet character theta function decays exponentially at infinity; its transformation makes it decay faster than any power at zero. Hence the integral is entire in $s$. For even parity, substitute $x=1/y$ and the theta transformation to obtainThe same calculation with the extra $x^{-1}$ power gives the odd functional equation with its corresponding root number.

The gamma function has no zeros and has simple poles at nonpositive integers. Thus the nontrivial zeros of $L$ and $\Lambda$ coincide with multiplicities. The trivial zeros of a Dirichlet L-function are $0,-2,-4,\dots$ for a nonprincipal even character, and $-1,-3,-5,\dots$ for an odd character. They cancel the gamma poles and are not zeros of $\Lambda$: the functional equation takes these points to the zero-free right-hand region, including the standard nonvanishing of nonprincipal Dirichlet L-functions at one at the even endpoint. The canceled zeros are simple.

The principal primitive Dirichlet character has conductor of a Dirichlet character equal to one and $L=\zeta$. In that case $\Lambda ={\pi }^{-s/2}\Gamma (s/2)\zeta (s)$ is meromorphic with poles at zero and one. Its canceled trivial zeros begin at $-2$, while $\zeta (0)=-1/2$ is not zero. Multiplication by $s(s-1)/2$ produces the entire Riemann xi function used below.

The standard primitive-character multiplicative large sieve inequality iswhere the star restricts to primitive Dirichlet characters. This is the form used in analytic arguments for Linnik's theorem. The prime-power Gauss identities extend to arbitrary primitive conductors by the Chinese remainder theorem. Thus the primitive Dirichlet character sum is, up to a factor of modulus $q^{-1/2}$, the character-weighted sum of the additive values $A(a/q)={\sum }_n{a}_{n}e(an/q)$ over units $a$. Orthogonality of Dirichlet characters, extending the primitive-character summation to all characters, givesThe additive sieve on the $Q^{-2}$-spaced Farey points proves the displayed bound.

For both prime-interval applications, use the following large sieve upper bound for sifted intervals. Suppose $S$ is in an interval of length $H$ and avoids one residue modulo every prime $p\le Q$ not dividing a fixed $q$. ThenTo prove it, choose the forbidden Chinese remainder theorem residue $r_d$ for each squarefree $d$. The Ramanujan sum $c_d(n-r_d)$ equals $\mu (d)$ on $S$, since $n-r_d$ is a unit modulo $d$. Therefore Cauchy-Schwarz inequality givesIndeed the linear combination with coefficients $e(-a{r}_d/d)$ has value $\mu (d)|S|$, and these coefficients have squared norm $\phi (d)$. Sum over the allowed squarefree $d$, apply the additive large sieve, and cancel $|S|$; the empty set is immediate.

Finally ${\mathcal{L}}_1(Q)\ge c\log (2Q)$. Squarefree integers have a positive elementary lower density: the nonsquarefree integers up to $X$ are covered by multiples of $k^2$, and ${\sum }_{k\ge 2}{k}^{-2}\le 3/4$. Partial summation turns this density into the harmonic lower bound. Splitting each squarefree $d$ into its factors supported on primes dividing $q$ and its coprime part givesConsequently ${\mathcal{L}}_q(Q)\ge c\phi (q)\log (2Q)/q$, uniformly in $q$ and $Q$.

Let $1(n)=1$ and $ϵ(n)=1_{n=1}$, the identity for Dirichlet convolution. The elementary identities needed areFor the first, prime factorization gives ${\sum }_{d\mid n}\mu (d)={\prod }_{p\mid n}(1-1)$ when $n>1$, and one when $n=1$. For the second, if $n=\prod p^{a_p}$, then ${\sum }_{d\mid n}\Lambda (d)={\sum }_p{a}_p\log p=\log n$, the Von Mangoldt divisor identity. Convolving the second identity with $\mu$ proves the third.

For $U,V\ge 1$, let the subscripts $\le U$, $>U$, $\le V$, $>V$ denote truncations. The Vaughan identity isIts pointwise form isTo prove it, use ${\mu }_{>U}\ast 1=ϵ-{\mu }_{\le U}\ast 1$ and $1\ast {\Lambda }_{>V}=\log -1\ast {\Lambda }_{\le V}$. This givesand rearrangement proves the formula, as in the Vaughan identity proof.

The Bombieri–Vinogradov theorem states that for every $A>0$ there is $B>0$ such thatHere $\psi (y;q,a)={\sum }_{\begin{array}{c}n\le y\\ n\equiv a(\mathrm{mod}q)\end{array}}\Lambda (n)$ is the Chebyshev function in an arithmetic progression, and $\varphi$ is the Euler totient function. In particular, partial summation givesafter choosing the logarithmic saving in the weighted theorem sufficiently large and absorbing the prime power terms. This is the offset logarithmic integral function.

For the proof strategy, Orthogonality of Dirichlet characters converts errors in arithmetic progressions into character sums of Dirichlet characters weighted by $\Lambda$. Reduction to primitive Dirichlet characters is followed by the character large sieve, which bounds their mean square by $O((N+Q^2)\sum |a_n{|}^2)$ with the usual $q/\varphi (q)$ weights. The Vaughan identity separates the weighted sums into short terms, Type I sums and Type II sums. Direct inner-sum estimates handle the Type I sums; the Cauchy-Schwarz inequality and the large sieve control the Type II sums. The Siegel–Walfisz theorem supplies arbitrarily strong logarithmic savings for small moduli, where an average estimate alone would not suffice. A dyadic decomposition, suitable $U,V$ and a sufficiently large $B$ absorb the divisor and logarithmic losses. This explains why the range is essentially $Q^2\le x$, with logarithmic room to spare.

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: