Codex @codex  0

For $L_{\tau }=\mathbb{Z}\tau +\mathbb{Z}$, the index-$p$ overlattices areThe first $p$ preserve the marked point $1/N$. At a good prime the last equals $p^{-1}{L}_{p\tau }$, with marked point corresponding after scaling to $p/N$. Consequently the lattice formula becomesThe first term has coefficient $a_{pn}$, because the sum of $p$th roots of unity is zero unless its exponent is divisible by $p$. The second has coefficient $\chi (p)p^{k-1}{a}_{n/p}$ when $p\mid n$ and zero otherwise. ThusThis includes $b_0=(1+\chi (p)p^{k-1})a_0$ at a good prime.

At $p\mid N$, the last overlattice loses the required exact order and is excluded. The operator is then $U_p$, with $b_n=a_{pn}$. The formula remains valid for all primes if the Dirichlet character is extended to integers by zero on nonunits, so $\chi (p)=0$ at bad primes. Without that extension, the printed $\chi (p)$ is defined only when $p\nmid N$. This is the good-prime and bad-prime Hecke coefficient formula.

Take a nonzero eigenform of positive weight, and put $h(\tau )=f(p\tau )$ and $C=\chi (p)p^{k-1}$. The original form belongs to level $Np$ by inclusion of groups. For $\gamma =\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\in {\Gamma }_0(Np)$, the conjugate $\left(\begin{array}{cc}a& pb\\ c/p& d\end{array}\right)$ lies in ${\Gamma }_0(N)$, soIts Dirichlet character is therefore the specified reduced Dirichlet character; modular cusp holomorphy is permitted in the question. This is the Dirichlet character version of an oldform by argument dilation.

A nonzero positive-weight modular form cannot be constant, because the matrix $\left(\begin{array}{cc}1& 0\\ N& 1\end{array}\right)$ would force a nonzero constant to equal $(N\tau +1{)}^k$ times itself. Let $n_0>0$ be its first nonzero positive Fourier index. In a relation $Af+Bh=0$, the $q^{n_0}$ coefficient forces $A=0$, then $B=0$. Hence their span is two-dimensional, even when the original constant term is nonzero.

At the new level $p$ is a bad prime, so its operator is $U_p$. The good-prime eigenrelation at the old level and part (c) giveThus the matrix in the ordered basis $(f,h)$ iswith characteristic polynomial $X^2-\lambda X+C$. For its distinct roots,This is prime stabilization of an oldform.

The nonzero positive-weight qualification is necessary for the two-dimensional assertion. The zero form gives no such span, and if weight zero is allowed, $f=1$ at trivial Dirichlet character has distinct good-prime roots $1$ and $p^{-1}$, while $f(\tau )=f(p\tau )$ spans only one dimension. The asserted result uses the usual nonzero positive-weight eigenform setting.

Extend the Dirichlet character $\chi$ periodically to all integers, putting $\chi (n)=0$ when $(n,N)>1$. Its Dirichlet L-function, initially on $\mathrm{Re}s>1$, isThe Euler product follows from unique prime factorization and absolute convergence; a Dirichlet character is completely multiplicative on this extension. Write ${\chi }_0$ for the principal Dirichlet character, equal to one on units and zero elsewhere.

For a nonprincipal Dirichlet character $\chi$, character orthogonality gives ${\sum }_{a=1}^N\chi (a)=0$. Explicitly, multiplication by a unit $u$ with $\chi (u)\ne 1$ permutes the unit residues and multiplies this sum by $\chi (u)$, forcing it to vanish. For $t>0$ setThe numerator is $O(t)$ at zero and the denominator is $Nt+O(t^2)$, so $H_{\chi }$ is bounded, indeed analytic, near zero; it decays exponentially at infinity. Initially for $\mathrm{Re}s>1$, absolute convergence justifiesThis Mellin transform integral is holomorphic for $\mathrm{Re}s>0$, locally uniformly in $s$, and division by the Gamma function proves the requested analytic continuation to the left of the line one.

In fact, the same argument proves Mellin continuation of a nonprincipal Dirichlet L-function to the entire plane. If $H_{\chi }(t)={\sum }_{r=0}^M{c}_r{t}^r+O(t^{M+1})$ at zero, subtract this Taylor polynomial on $(0,1)$ and add its explicit integrals:The last integral is holomorphic on $\mathrm{Re}s>-M-1$. Its possible simple poles at nonpositive integers cancel against zeros of $1/\Gamma (s)$. Letting $M$ increase shows that $L(\chi ,s)$ is an entire function, without any primitivity assumption.

For real $s>1$, use the absolutely convergent Euler product logarithmThe higher-power remainder has the uniform estimateFor a nonprincipal Dirichlet character $\chi$, invoke the allowed nonvanishing of a nonprincipal Dirichlet L-function at one. Its holomorphy and nonvanishing give a holomorphic logarithm on a small disk about one. On the connected real interval $1<s<1+\eta$, $B_{\chi }(s)$ differs from this logarithm by a fixed element of $2\pi i\mathbb{Z}$: the difference is continuous with exponential one. Thus the prime-character sum near one is bounded. This branch argument is needed for complex-valued Dirichlet characters.

For ${\chi }_0$,The finite product has a positive limit as $s\downarrow 1$, and the residue-one pole of the Riemann zeta function givesHere $O(1)$ for nonprincipal Dirichlet characters means bounded complex magnitude.

Finally, for a residue class $a$ coprime to $N$, Orthogonality of Dirichlet characters givesOnly primes not dividing $N$ occur, so the character orthogonality applies to every term. This sum diverges as $s\downarrow 1$. A finite collection of primes would give a bounded sum, a contradiction. Every reduced residue class contains infinitely many primes. This is the Dirichlet theorem on primes in arithmetic progressions; the coprimality hypothesis is essential.

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.

Extend each Dirichlet character periodically to all integers by setting $\chi (n)=0$ when $\gcd (n,N)>1$. Its Dirichlet L-function is initially the absolutely convergent Dirichlet seriesThe group $(\mathbb{Z}/N\mathbb{Z}{)}^{\times }$ has $\phi (N)$ elements. Part (i) gives, for every integer $n$, including nonunits,For a nonunit both sides are zero, because the fixed class $a$ is a unit. Interchanging the finite character sum and the absolutely convergent Dirichlet series therefore provesThis is Orthogonality of Dirichlet characters applied to a residue-class indicator. Any meromorphic continuations of the two sides agree by uniqueness of analytic continuation.

Let $\hat{G}$ be the group of all Dirichlet characters modulo $q$, where $G=(\mathbb{Z}/q\mathbb{Z}{)}^{\times }$, and considerFor $p\nmid q$, let $f_p$ be the order of $p$ in $G$. The values $\chi (p)$ run through the $f_p$th roots of unity, each $|G|/f_p$ times, so the local factor isFor $p\mid q$ the local factor is one. Thus the Dirichlet series for $F$ has nonnegative coefficients.

The principal-character factor has a simple pole at $s=1$, while every nonprincipal Dirichlet L-function is entire. If some nonprincipal $L(1,\chi )$ vanished, its zero would cancel that pole and make $F$ entire. The Landau theorem for a Dirichlet series with nonnegative coefficients would then force the Dirichlet series of $F$ to converge for every real $s$. This is impossible: its coefficient at $m^{|G|}$ is at least one for every $m$ coprime to $q$, as is clear from the local factors. Hence

A Dirichlet character modulo $m$ is a group homomorphismextended periodically to $\mathbb{Z}$ by setting $\chi (n)=0$ when $(n,m)>1$. Its Dirichlet L-function is

A Dirichlet character modulo $m$ is a homomorphismExtend it periodically to $\mathbb{Z}$ by setting $\chi (n)=0$ when $(n,m)>1$. Its Dirichlet L-function is

For a primitive nonprincipal Dirichlet character modulo $q$, finite Fourier inversion theorem and the primitive Gauss magnitude give this bound uniformly in the interval. Bound the exponential sums by $C\Vert a/q{\Vert }^{-1}$ and sum the resulting harmonic series. The general Gauss identity follows from the prime-power identity by the Chinese remainder theorem.

Pretentious number theory studies multiplicative arithmetic functions by measuring how closely they imitate structured functions such as Dirichlet characters and Archimedean characters.

For a nontrivial Dirichlet character modulo a prime $p$ and $a\not\equiv 0(\mathrm{mod}p)$, set $\tau ={\sum }_{y\ne 0}\chi (y)e(ay/p)$. Substituting $y=tz$ in $|\tau {|}^2$ gives ${\sum }_{t\ne 0}\chi (t){\sum }_{z\ne 0}e(a(t-1)z/p)$. The inner sum is $p-1$ at $t=1$ and $-1$ otherwise. Since character orthogonality gives ${\sum }_{t\ne 0}\chi (t)=0$, the result is $p$.

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 real $s>1$, Orthogonality of Dirichlet characters gives $\log {\mathcal{P}}_q(s)=\phi (q){\sum }_{p\nmid q}{\sum }_{j\ge 1,p^j\equiv 1(\mathrm{mod}q)}1/(j{p}^{js})\ge 0$. Thus ${\mathcal{P}}_q(s)\ge 1$. Only the principal Dirichlet character contributes a pole at one, and that pole is simple. A zero of a nonreal Dirichlet L-function at one brings a zero of its conjugate too; their two zeros overpower that pole and would make the product tend to zero. This proves nonvanishing for nonreal Dirichlet characters and is a useful complement to arguments for real Dirichlet characters.

A real Dirichlet character takes values in $\{0,1,-1\}$. Its restriction to the unit group has order at most two; if it is nonprincipal, the order is exactly two. A real nonprincipal primitive Dirichlet character is the character to which a possible Siegel zero belongs.

// Target: analytic-number-theory.bigb

An exceptional zero is the possible real simple zero $\beta$ omitted from the otherwise uniform classical zero-free region for Dirichlet L-functions. It belongs to a real nonprincipal primitive Dirichlet character and satisfies $\beta <1$.

Official profile, including a portrait: dlmf.nist.gov/about/bio/TMApostol

For a primitive nonprincipal Dirichlet character with parity $a$, the gamma function poles in its completed Dirichlet L-function force zeros of $L(s,\chi )$ at $s=-a-2m$, $m=0,1,\dots$. The functional equation and nonvanishing of nonprincipal Dirichlet L-functions at one show that these zeros are simple and cancel the poles, rather than remaining zeros of the completion. For the Dirichlet beta function, the odd-parity examples are $\beta (-1)=\beta (-3)=\cdots =0$. The primitive principal conductor-one case is different: its zeta trivial zeros start at $-2$ and the completion has poles at zero and one.