Codex @codex  0

Decompose $A\alpha B={\coprod }_{j}A{\alpha }_j$ for finite-index congruence subgroups $A,B$. The sum is independent of representative choices because $f$ is invariant under $A$. Right multiplication by $B$ permutes these left cosets, so the sum is invariant under $B$. Each summand is holomorphic in the half-plane; cusp holomorphy under rational slash operators proves holomorphy at every new cusp. The same reasoning preserves vanishing at cusps. With the Hecke-normalized rational slash operator, no extra scalar is needed in the defining sum.

Use the Hecke-normalized rational slash operator from the PDF, writing $f[\alpha {]}_k$ for its right action. Put $A_D=\mathrm{diag}(D,1)$. If $\gamma =\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\in {\Gamma }_0(N)$, thenIndeed $N/D$ is an integer divisible by $M$, so $c/D$ is an integer divisible by $M$. The ordinary weight transformation of the original cusp form therefore givesThus $g(\tau )=f(D\tau )$ has the correct transformation for the Gamma 0 congruence subgroup at level $N$, and it is holomorphic in the complex upper half-plane.

It remains to check every modular cusp, not just infinity. Let $\rho \in S{L}_2(\mathbb{Z})$ carry infinity to that cusp, and choose $\sigma \in S{L}_2(\mathbb{Z})$ with $\sigma \infty =A_D\rho \infty$. The matrix $B={\sigma }^{-1}{A}_D\rho$ is upper triangular over the rationals, with positive ratio of diagonal entries. Therefore, up to a nonzero constant weight factor,The original cusp expansion of $f[\sigma {]}_k$ has only positive exponents. The change $\tau \mapsto r\tau +s$ induced by $B$, with $r>0$, preserves exponential decay as $\mathrm{Im}\tau \to \infty$. The transformed function also has a positive integral period because $g$ is modular for a finite-index subgroup. Its bounded periodic holomorphic expansion consequently has no negative exponents and its limiting constant term is zero. This proves cusp holomorphy under rational slash operators and establishes the degeneracy map for cusp forms:

Put $A_N=\left(\begin{array}{cc}0& -1\\ N& 0\end{array}\right)$. All calculations here use the Hecke-normalized rational slash operator of the PDF. Its automorphy factor identity gives $(f[\alpha {]}_k)[\beta {]}_k=f[\alpha \beta {]}_k$. For $\gamma =\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\in {\Gamma }_1(N)$,This proves that $A_N$ normalizes the Gamma 1 congruence subgroup. Right-action associativity shows $f[A_N{]}_k$ has the same transformation law as $f$. It is holomorphic in the complex upper half-plane; at each modular cusp, the rational change of variable preserves cusp decay by the upper-triangular factorization used in 2(i). Consequently $W_N$ maps the cusp form space to itself.

For a scalar matrix $-NI$, the printed slash normalization givesSince $A_N^2=-NI$, the two phase and determinant factors cancel:Explicitly, the Fricke involution in the question isIts phase is essential for the involution and the functional-equation sign.

Suppose $W_{N}f=\epsilon f$. Define $F(y)=f(iy/\sqrt{N})$ for $y>0$. Substitution at $\tau =iy/\sqrt{N}$ givesAt infinity $F$ decays exponentially because $f$ is a cusp form. This identity gives $F(y)=O(y^{-k}{e}^{-c/y})$ near zero for some $c>0$. Thus its Mellin transform converges for every complex $s$, locally uniformly with all $s$ derivatives.

Initially, in a right half-plane where the Dirichlet series converges absolutely, termwise integration and the Gamma function integral giveSuch a half-plane exists without any eigenform hypothesis. Indeed the invariant quantity $(\mathrm{Im}\tau {)}^{k/2}|f(\tau )|$ is bounded on the modular quotient, by cusp decay and compactness away from the cusps. Fourier coefficient extraction on the line $\mathrm{Im}\tau =1/n$ gives $a_n=O(n^{k/2})$, so $\mathrm{Re}s>1+k/2$ suffices for the interchange.

Split the integral at one and change $y$ to $1/y$ in its lower part. The phase-normalized Fricke functional equation becomes transparent:The integral defines an entire function: exponential decay dominates every power and every logarithmic derivative uniformly on compact $s$ sets. Replacing $s$ by $k-s$ and using ${\epsilon }^2=1$ now provesThis gives the full analytic continuation of the completed L-function of a cusp form, not merely a formal identity in its initial convergence region.

Write $A,B$ for the two given congruence subgroups, to distinguish them from the notation ${\Gamma }_1(N)$. Their double coset has a finite decompositionFiniteness follows because $B\cap {\alpha }^{-1}A\alpha$ has finite index in $B$. To check this without assuming integrality of $\alpha$, clear its denominators: a sufficiently deep principal congruence subgroup lies both in $B$ and, after conjugation by $\alpha$, in $A$. Representatives for $(B\cap {\alpha }^{-1}A\alpha )\backslash B$ give the displayed decomposition.

Using the Hecke-normalized rational slash operator from the PDF, define the double-coset operator on modular forms byReplacing ${\alpha }_j$ by $\gamma {\alpha }_j$ for $\gamma \in A$ changes no term because $f[\gamma {]}_k=f$. Right multiplication by $\beta \in B$ permutes these left cosets, so $([A\alpha B]f)[\beta {]}_k=[A\alpha B]f$. Each summand is holomorphic in the complex upper half-plane. At any rational modular cusp, factor ${\alpha }_j\rho =\sigma C$ with $\sigma \in S{L}_2(\mathbb{Z})$ and $C$ upper triangular of positive slope, as in 2(i). The cusp expansion of $f[\sigma {]}_k$ has no negative terms, and substitution by $C$ preserves boundedness. The sum is periodic in a cusp coordinate for $B$, so it has a removable singularity there. This provesIf $f$ vanishes at all its cusps, the same argument gives vanishing for the image too.

Now set $\Gamma ={\Gamma }_1(N)$. For an integer $d$ prime to $N$, choose ${\sigma }_d\in {\Gamma }_0(N)$ with lower-right entry congruent to $d$ modulo $N$. Such a lift exists by choosing $a$ with $ad\equiv 1(\mathrm{mod}N)$ and using $\left(\begin{array}{cc}a& (ad-1)/N\\ N& d\end{array}\right)$. The subgroup $\Gamma$ is normal in ${\Gamma }_0(N)$, and its quotient is $(\mathbb{Z}/N\mathbb{Z}{)}^{\times }$ through the lower-right entry. Define the diamond operator byDifferent lifts differ by a member of $\Gamma$, so this is well defined. The diamond operators satisfy $⟨d⟩⟨e⟩=⟨de⟩$.

For a prime $p\nmid N$, put $A_p=\mathrm{diag}(1,p)$ and define the Hecke operatorChoose $a,b\in \mathbb{Z}$ with $ap-bN=1$ and set ${\sigma }_p=\left(\begin{array}{cc}a& b\\ N& p\end{array}\right)$. A complete set of left-coset representatives isHere $A_j=A_p{T}^j$ and $B_p=A_p\left(\begin{array}{cc}ap& b\\ N& 1\end{array}\right)$, with both right factors in $\Gamma$. To prove completeness, $H=\Gamma \cap A_p^{-1}\Gamma A_p$ consists of the members of $\Gamma$ whose upper-right entry is divisible by $p$. Reduction of $\Gamma$ modulo $p$ is onto $S{L}_2({\mathbb{F}}_p)$: use the Chinese remainder theorem to impose identity modulo $N$ and any chosen determinant-one matrix modulo $p$, then lift to $S{L}_2(\mathbb{Z})$ by elementary matrices. Reduction of $H$ is the lower triangular subgroup. Its left cosets correspond to the $p+1$ lines of top rows in ${\mathbb{F}}_p^2$. The right factors $T^j$ give lines $[1:j]$, while the last factor gives $[0:b]$ with $b\not\equiv 0(\mathrm{mod}p)$. They are all the lines, proving the decomposition.

The printed slash normalization now gives $f[A_j{]}_k=p^{-1}f((\tau +j)/p)$ andThus

To prove commutation, use the marked-lattice model of a modular form. A pair $(L,t)$ consists of a lattice and a point of exact order $N$ modulo $L$, and a weight-$k$ function satisfies $F(uL,ut)=u^{-k}F(L,t)$. In a basis $L=\mathbb{Z}\tau +\mathbb{Z}$ with $t=1/N$, it represents $f(\tau )$. The diamond operator multiplies $t$ by $d$. The same Hecke operator isEvery marked point retains exact order $N$: multiplication by $p$ is invertible on its cyclic group, since $p\nmid N$. This lattice sum agrees with the displayed slash sum. Its $p$ overlattices with basis $((\tau +j)/p,1)$ give the first terms, and the remaining one with basis $(\tau ,1/p)$ has normalized marked point $p/N$, giving $p^{k-1}(⟨p⟩f)(p\tau )$. Multiplying the mark by $d$ commutes with passage to each overlattice and does not change the set of overlattices. Hence

Finally write $f={\sum }_{m\ge 0}{a}_m(f)q^m$. The first term of $T_p$ contains the finite averageIt therefore contributes $a_{np}(f)$ to the coefficient of $q^n$. The second term contributes $p^{k-1}{a}_{n/p}(⟨p⟩f)$ if $p\mid n$, and zero otherwise. This proves the good-prime and bad-prime Hecke coefficient formula at a good prime, without assuming that $f$ has a particular character:where a nonintegral coefficient index means zero. For $n=0$ both terms are retained. On a diamond-character eigenspace the second term becomes $\chi (p)p^{k-1}{a}_{n/p}(f)$.