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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.

For a rational matrix of positive determinant, this slash normalization differs from the determinant-normalized slash operator by $(\det \alpha {)}^{k/2-1}$. Both are right actions, by the automorphy factor identity. For a positive scalar $t$, the displayed action of $tI$ is multiplication by $t^{k-2}$. On determinant-one matrices it agrees with the ordinary slash operator for modular forms. This normalization makes a good-prime Hecke operator a plain sum over its double-coset representatives.

Use the determinant-normalized slash operator and let ${\mathcal{M}}_n$ be all integral matrices of positive determinant $n$. Their left $S{L}_2(\mathbb{Z})$ cosets have the unique determinant-n matrix representatives for Hecke operatorsInteger row operations reduce the first column to $(a,0{)}^T$, and a row shear then reduces $b$ modulo $d$. DefineChanging representatives does not affect the sum, while right multiplication by the modular group permutes the cosets. Hence $T_{n}f$ has the same modular transformation law. Rational positive-determinant slash transforms are holomorphic on $\mathbb{H}$ and preserve modular cusp vanishing: choose an integral matrix carrying infinity to their rational image modular cusp, after which the remaining matrix is upper triangular. Its positive dilation sends the original decaying modular cusp expansion to another decaying expansion. Thus these are well-defined Hecke operators on $S_k$.

For the Fourier expansion, insert $f(z)={\sum }_{m\ge 1}a(m)e^{2\pi imz}$. The sum over $b$ vanishes unless $d\mid m$, in which case it is $d$. Writing $m=d\ell$, the resulting exponential is $q^{a\ell }$, and its factor is $n^{k-1}{d}^{1-k}=a^{k-1}$. ThereforeIn particular, the first coefficient of $T_{n}f$ is $a(n;f)$.

Assume $f\ne 0$ is a simultaneous Hecke eigenform. The first-coefficient identity gives $a(n)=\lambda (n)a(1)$. If $a(1)=0$, all coefficients vanish, contradicting $f\ne 0$. After dividing by $a(1)$, we may therefore suppose $a(1)=1$, and then $\lambda (n)=a(n)$.

We derive the relations needed for the Euler product. For coprime $m,n$, applying the coefficient formula twice splits the divisors into their disjoint prime supports and yields $T_m{T}_n=T_{mn}$. For one prime define formal Fourier operators $U_p(\sum a(r)q^r)=\sum a(pr)q^r$ and $V_{p}f(z)=f(pz)$. The coefficient formula givesMultiplying this finite sum by $T_p=U_p+p^{k-1}{V}_p$ shows thatThese auxiliary operators need not separately preserve level one; the identities concern their action on Fourier series. Applied to the eigenform, they giveThe prime-power generating function consequently satisfiesMultiplicativity and unique prime factorization now prove the Euler product of a Hecke eigenform:For the original unnormalized form, the right side is multiplied by $a(1)$ and $a(p)$ in the factors is replaced by $\lambda (p)=a(p)/a(1)$. The zero form has $L=0$ and is excluded from the nonzero-eigenform product assertion.

These manipulations initially hold in an absolutely convergent right half-plane. For example, $y^{k/2}|f(z)|$ is bounded by modular invariance and modular cusp decay. The coefficient integral at $y=1/n$ gives $|a(n)|\le C{n}^{k/2}$, so $\mathrm{Re}s>k/2+1$ is sufficient. No analytic continuation is needed to justify the initial product.

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 $f|W_N=g$.

Write $\Gamma =S{L}_2(\mathbb{Z})$ and use the determinant-normalized slash operatorThe automorphy factor identity proves $(f{|}_k\alpha ){|}_k\beta =f{|}_k(\alpha \beta )$, and positive scalar matrices act trivially. Let ${\mathcal{M}}_n$ be all integral matrices of determinant $n>0$. It is stable under left and right multiplication by $\Gamma$. Its left orbits have unique representativesIndeed Bézout's identity reduces the first column to $(a,0)$, the determinant fixes $d=n/a$, and a row shear reduces $b$ modulo $d$. These are the determinant-n matrix representatives for Hecke operators. For composite $n$, ${\mathcal{M}}_n$ can contain more than one double coset; keeping every integral matrix, including nonprimitive ones, is essential.

Define the Hecke operator byChanging a representative on the left leaves its summand unchanged. Right multiplication by $\Gamma$ permutes left orbits of ${\mathcal{M}}_n$, so the result satisfies the weight-$k$ transformation law. Each summand is holomorphic. Its expansion at infinity is bounded, and vanishes there when $f$ is a cusp form; the group has just one cusp class. Therefore $T_n$ preserves both $M_k$ and $S_k$, and $T_1$ is the identity.

For a Fourier expansion $f={\sum }_{r\ge 0}{a}_r{q}^r$, the sum over $b$ vanishes unless $d\mid r$. Substitute $r=dt$ and then collect the coefficient of $q^m$ to obtain the Fourier coefficients of a composite-index Hecke operator:Here $\gcd (0,n)=n$, so the constant coefficient is ${\sigma }_{k-1}(n)a_0$; at $m=1$ it is $a_n$. This proves preservation of integral coefficients. At a prime,with $a_{m/p}=0$ if $p\nmid m$.

We can prove the entire Hecke multiplication relations directly, not just quote them. On formal power series put $U_p(\sum a_m{q}^m)=\sum a_{pm}{q}^m$ and $V_{p}f(q)=f(q^p)$. Although these individual operators need not preserve level one, $U_p{V}_p=1$ and the coefficient formula givesMultiply on the left by $T_p=U_p+p^{k-1}{V}_p$. In the first part the $j=0$ term is $U_p^{r+1}$; every $j\ge 1$ reduces by $U_p{V}_p=1$ to $p^{j(k-1)}{V}_p^{j-1}{U}_p^{r-j}$. The second part, together with the $j=0$ term, is $T_{p^{r+1}}$. The remaining sum is $p^{k-1}{T}_{p^{r-1}}$. ThusFor coprime indices, the coefficient divisors split independently in the displayed formula, proving $T_m{T}_n=T_{mn}$ when $(m,n)=1$. Induction with the prime-power recurrence gives $T_{p^r}{T}_{p^t}={\sum }_{j=0}^{\min (r,t)}{p}^{j(k-1)}{T}_{p^{r+t-2j}}$. Combining the primes provesIn particular the modular Hecke algebra is commutative and is generated by the prime operators.

For cusp forms the Petersson inner product isCusp decay makes it finite and positive definite. The Hecke operators are self-adjoint for the Petersson inner product. To see the change-of-variables mechanism, fix a double coset $\Gamma \alpha \Gamma \subset {\mathcal{M}}_n$ and put $H=\Gamma \cap {\alpha }^{-1}\Gamma \alpha$. Representatives $\gamma$ of $H\backslash \Gamma$ give left representatives $\alpha \gamma$. Unfolding the sum in $⟨{\sum }_{\gamma }f{|}_k\alpha \gamma ,g⟩$ changes the integral over $\Gamma \backslash \mathbb{H}$ to the integral over $H\backslash \mathbb{H}$ of $(f{|}_k\alpha )\overline{g}$. The substitution $w=\alpha \tau$ preserves $dxdy/y^2$ and has $\Im w=(\det \alpha )y/|c\tau +d{|}^2$; these factors move the slash from $f$ to $g{|}_k{\alpha }^{-1}$. Refold over $\alpha H{\alpha }^{-1}=\Gamma \cap \alpha \Gamma {\alpha }^{-1}$ to get the reverse double coset. Since scalar matrices act trivially, ${\alpha }^{-1}$ has the same slash action as its adjugate ${\alpha }^{\ast }=n{\alpha }^{-1}$, an integral matrix of determinant $n$. The adjugate involution permutes all the double cosets in ${\mathcal{M}}_n$. The common real factor $n^{k/2-1}$ is unchanged, giving $⟨T_{n}f,g⟩=⟨f,T_{n}g⟩$.

Since $S_k$ is finite dimensional, commuting self-adjoint linear operators have a common orthonormal basis of Hecke eigenforms: diagonalize one operator, restrict the others to its eigenspaces, and repeat until these spaces cannot split further. If $f$ is a nonzero common Hecke eigenform with $T_{n}f={\lambda }_{n}f$, the coefficient at $q$ gives $a_n={\lambda }_n{a}_1$. Were $a_1=0$, every positive coefficient would vanish, contrary to $f\ne 0$ being cuspidal. Normalize $a_1=1$; then ${\lambda }_n=a_n$, and the Hecke multiplication relations yieldThe eigenvalues are real by self-adjointness. Their multiplicativity and prime-power recurrence give the Euler product of a Hecke eigenform, initially in an absolutely convergent half-plane:The Mellin transform in Question 5 supplies its analytic continuation and functional equation.

Finally, for positive even $k\ge 4$, $M_k=\mathbb{C}{E}_k\oplus S_k$, by subtracting the constant coefficient times the normalized Eisenstein series. The divisor identityfollows by multiplying the geometric sums at each prime; it and the constant-coefficient formula prove $T_n{E}_k={\sigma }_{k-1}(n)E_k$. Thus $M_k$ has an Eisenstein series eigenline and a basis of cuspidal Hecke eigenforms. Odd and negative weights have no forms, $M_2=0$, and on $M_0=\mathbb{C}$ the operator is the scalar ${\sigma }_{-1}(n)$. These statements fix the normalization and describe the exceptional small weights as well.

Fix an even weight $k$ and write $\Gamma =S{L}_2(\mathbb{Z})$. Extend the slash operator for modular forms to positive-determinant rational matrices by the determinant-normalized slash operatorThe automorphy factor identity makes this a right action. Let ${\mathcal{D}}_n$ be the integral matrices of determinant $n>0$. Their left $\Gamma$-cosets have unique determinant-n matrix representatives for Hecke operatorsAn integral row operation obtained from Bézout's identity makes the first column $(a,0{)}^T$, and a row shear reduces the other top entry modulo $d$, proving this description. The usual Hecke operator isThis normalization has $T_1=1$. Right multiplication by $\Gamma$ permutes the left cosets in ${\mathcal{D}}_n$, so $T_{n}f$ has the same modular transformation law as $f$. Each summand is holomorphic in $\mathbb{H}$. Holomorphy at the cusp and preservation of cusp forms follow directly from the coefficient formula below; the full modular group has only one cusp orbit. Thus the operators act on both $M_k$ and $S_k$.

If $f={\sum }_{r\ge 0}{a}_r{q}^r$, summing over $b$ kills terms unless $d\mid r$. Writing $r=d\ell$ gives output exponent $a\ell$, so a term of exponent $m$ requires $a\mid m$ and has original index $r=dm/a=nm/a^2$. Its multiplier is $n^{k-1}{d}^{1-k}=a^{k-1}$. The Fourier coefficients of a composite-index Hecke operator are consequentlyFor $m=0$, this means $a_0(T_{n}f)={\sigma }_{k-1}(n)a_0(f)$. In particular $a_1(T_{n}f)=a_n(f)$. For a prime number $p$,with a nonexistent fractional-index coefficient interpreted as zero.

The Hecke multiplication relations follow from these coefficients. For coprime indices, the two divisor choices combine uniquely and give $T_m{T}_n=T_{mn}$. For prime powers put $A=p^{k-1}$, and, on formal Fourier series, let $U_p(\sum a_m{q}^m)=\sum a_{pm}{q}^m$ and $V_p(\sum a_m{q}^m)=\sum a_m{q}^{pm}$. These individual maps need not preserve level one, but $U_p{V}_p=1$, and the coefficient formula givesMultiplying on the left by $T_p=U_p+A{V}_p$ and cancelling $U_p{V}_p$ yieldsInduction gives $T_{p^u}{T}_{p^v}={\sum }_{j=0}^{\min (u,v)}{p}^{j(k-1)}{T}_{p^{u+v-2j}}$; combining distinct prime factors provesThe modular Hecke algebra is therefore commutative and generated by its prime-index operators. The identity $U_p{V}_p=1$, rather than the generally false reversed identity, is what proves the prime-power recurrence.

On $S_k$ use the positive definite Petersson inner productExponential cusp decay makes this integral finite. The Hecke operators are self-adjoint for the Petersson inner product. Here is the correspondence argument. For a positive-determinant matrix $\alpha$, set $H=\Gamma \cap {\alpha }^{-1}\Gamma \alpha$. The finite sum over the double coset $\Gamma \alpha \Gamma$ unfolds its pairing intoChanging variables $z=\alpha \tau$ preserves hyperbolic area, and $\mathrm{Im}(\alpha \tau )=\det (\alpha )y/|c\tau +d{|}^2$ shows that the determinant-normalized slash factors transfer $\alpha$ from $f$ to ${\alpha }^{-1}$ on $g$. The new quotient is by $\alpha H{\alpha }^{-1}=\Gamma \cap \alpha \Gamma {\alpha }^{-1}$; folding it back gives the operator for the reversed double coset. For $\alpha \in {\mathcal{D}}_n$, its adjugate $n{\alpha }^{-1}$ again lies in ${\mathcal{D}}_n$. Positive scalar matrices act trivially in the determinant normalization, so ${\alpha }^{-1}$ and its adjugate give the same action. The adjugate involution preserves the entire determinant-$n$ correspondence, and the external normalization $n^{k/2-1}$ is real. Thus

Finite-dimensional spectral theory now gives a simultaneous orthonormal eigenbasis on $S_k$, since the operators commute and are self-adjoint. For a common Hecke eigenform $f$ with eigenvalues ${\lambda }_n$, the first-coefficient formula gives $a_n(f)={\lambda }_n{a}_1(f)$. If $a_1(f)=0$, every coefficient vanishes and $f=0$, a contradiction. Rescale each eigenform to have $a_1=1$. The resulting basis is orthogonal, andAll these eigenvalues and coefficients are real. A fixed simultaneous eigenspace is one-dimensional, because its normalized Fourier coefficients are uniquely determined by the eigenvalues.

Applying the multiplication relations to an eigenform givesThe recurrence yields ${\sum }_{r\ge 0}{a}_{p^r}{X}^r=(1-a_{p}X+p^{k-1}{X}^2{)}^{-1}$. Multiplicativity then gives the Euler product of a Hecke eigenforminitially in a right half-plane, for example $\mathrm{Re}s>k/2+1$ using the Fourier coefficient bound for a cusp form. Modular inversion and the Mellin transform of a cusp-form L-function continue $\Lambda (f,s)=(2\pi {)}^{-s}\Gamma (s)L(f,s)$ to an entire function and give $\Lambda (f,s)=i^k\Lambda (f,k-s)$.

There is also an integral structure. If $d=\dim S_k$, choose $F_j={\Delta }^j{R}_{k-12j}$, $1\le j\le d$, with $R_w$ a monomial in $E_4,E_6$ of weight $w$ and constant term one. These forms have integral Fourier coefficients and start with $q^j$; their first $d$ coefficients form a unitriangular matrix. They are a basis, and every cusp form with integral coefficients is an integral combination of them, by successive elimination of its first $d$ coefficients. The coefficient formula shows that $T_n$ preserves this lattice. Its characteristic polynomial is consequently monic with integer coefficients, proving that the Hecke eigenvalues are algebraic integers. Together with self-adjointness, this gives real algebraic integers without requiring any unproved bound for their sizes.

For even $k\ge 4$, the full space decomposes as $M_k=\mathbb{C}{E}_k\oplus S_k$. The normalized Eisenstein series is a simultaneous eigenform withIts coefficients are $-2k{\sigma }_{k-1}(m)/B_k$, and the elementary divisor identity ${\sigma }_{k-1}(m){\sigma }_{k-1}(n)={\sum }_{d\mid \gcd (m,n)}{d}^{k-1}{\sigma }_{k-1}(mn/d^2)$ verifies the eigenvalue formula, including the constant coefficient. Thus the cusp eigenbasis plus $E_k$ diagonalizes the operators on $M_k$. Weight two and odd weights have no forms; in weight zero the constant function has eigenvalue ${\sigma }_{-1}(n)$. As a concrete cuspidal example, $S_{12}=\mathbb{C}\Delta$, so $T_n\Delta =\tau (n)\Delta$ for its Fourier coefficients, and the recurrence gives $\tau (p^2)=\tau (p{)}^2-p^{11}$.

Use the determinant-normalized slash operator and the usual Hecke operator normalization. A normalized eigenform is a nonzero cusp form $f={\sum }_{r\ge 1}{a}_r{q}^r$ with $a_1=1$ and $T_{n}f=\lambda (n)f$ for every $n$. We first establish the operator facts needed here, rather than assume the Hecke multiplication relations.

Integer matrices of determinant $n$ have the determinant-n matrix representatives for Hecke operators ${\gamma }_{a,b,d}=\left(\begin{array}{cc}a& b\\ 0& d\end{array}\right)$, with $ad=n$ and $0\le b<d$. Integer row operations give these representatives: the Euclidean algorithm reduces the first column to $(a,0)$, and a shear reduces $b$ modulo $d$. Right multiplication by the modular group permutes the left cosets, so $T_{n}f=n^{k/2-1}{\sum }_{ad=n}{\sum }_{b\mathrm{mod}d}f{|}_k{\gamma }_{a,b,d}$ is modular. Rational changes of coordinate preserve cusp vanishing, so it remains a cusp form. Substituting its Fourier expansion of a modular form and summing $e^{2\pi irb/d}$ givesIn particular $a_1(T_{n}f)=a_n(f)$, and the normalization gives $\lambda (n)=a_n(f)$ for every $n$.

For coprime $u,v$, applying this coefficient formula twice amounts to choosing independently a divisor of $\gcd (m,u)$ and one of $\gcd (m,v)$. Their product runs exactly once through the divisors of $\gcd (m,uv)$, with the same factor $d^{k-1}$ and coefficient index. Thus $T_u{T}_v=T_{uv}$. For a prime $p$, put $P=p^{k-1}$ and $\nu =v_p(m)$. The coefficient of $q^m$ in $T_{p^e}f$ isThe coefficient in $T_p{T}_{p^e}f$ is $A_e(pm)+P{A}_e(m/p)$, the second summand being zero if $p\nmid m$. Expanding the two finite sums and matching their endpoints gives $A_{e+1}(m)+P{A}_{e-1}(m)$ for $e\ge 1$. Equality of all coefficients proves $T_p{T}_{p^e}=T_{p^{e+1}}+p^{k-1}{T}_{p^{e-1}}$. Applying these identities to the nonzero eigenvector $f$ yields

The commuting normal operators $T_{\ell }$, together with the finite-order diamond operators, admit simultaneous diagonalization for the Petersson inner product. Their restrictions to the new subspace of cusp forms have one-dimensional common eigenspaces by the newform multiplicity-one theorem. This additional theorem is the essential newform input: normality alone would allow higher multiplicities. Since every $U_p$ commutes with the good-prime operators and preserves the new space, it preserves each such line and is scalar on it. Normalize its nonzero first Fourier coefficient to one to obtain a newform. This gives the requested simultaneous eigenbasis of the new space.

By definition, every old form is a linear combination of lower-level oldforms by argument dilation. Repeatedly split each lower-level cusp space into its old space and its Petersson inner product orthogonal new space; induction on the level shows that the specified dilations of newforms span the full space. To prove independence, separate distinct primitive eigenforms by their good-prime eigenvalues, using the newform multiplicity-one theorem. For a fixed normalized form, its dilations $f(tz)$ are independent: order the distinct positive integers $t$ increasingly, and the coefficient of the least occurring $q^t$ kills its coefficient in any vanishing linear combination. Repeat for the next least $t$. Thus these dilations are indeed a basis.

For trivial character, part (a) shows that $W_N$ commutes with all the good-prime operators, so it preserves the one-dimensional primitive line. Also $W_N^2=(-1{)}^k$ for the determinant-normalized slash operator; a nonzero trivial-character form has even $k$. ThereforeThe power of $N$ in an unnormalized Fricke eigenvalue depends on the convention. The plain slash action $(cz+d{)}^{-k}f(\gamma z)$ gives $\epsilon N^{-k/2}$, while the Hecke-style action $(\det \gamma {)}^{k-1}(cz+d{)}^{-k}f(\gamma z)$ gives $\epsilon N^{k/2-1}$. The PDF prints $\pm N^{k-1}$; that value requires the deliberate rescaling $w_N=N^{k-1}{W}_N$ and does not follow under either usual convention. Under that rescaling its displayed eigenvalue is $\epsilon N^{k-1}$; all the preceding eigenspace and conjugation statements are unchanged.

With the determinant-normalized slash operator, let $W_N=\left(\begin{array}{cc}0& -1\\ N& 0\end{array}\right)$ and $Jf=i^k(f{|}_k{W}_N)=g$. Conjugation by $W_N$ preserves ${\Gamma }_1(N)$, and rational slash operators preserve modular cusp holomorphy and vanishing. Hence $g$ is also a cusp form at that level. Direct substitution gives $J^{2}f=f$; the factor $i^k$ cancels the central weight sign.

Put $F(y)=f(iy/\sqrt{N})$ and $G(y)=g(iy/\sqrt{N})$. The defining formula gives the exact relationsIn the initial half-plane of absolute convergence, termwise integration of the Fourier series and the gamma function yield the Mellin transform of a cusp-form L-functionThe scaling of $y$ accounts for $N^{s/2}$ in the completion. At infinity $F$ and $G$ decay exponentially. At zero the boxed relation expresses $F$ as a power times an exponentially decaying function of $1/y$. Thus this integral converges locally uniformly for every complex $s$, including after differentiation in $s$, and defines an entire function.

Splitting at one and changing $y$ to $1/y$ in the lower integral givesApplying the same formula to $g$ interchanges $F,G$, because $J^2=1$. It proves the phase-normalized Fricke functional equationThe entire function here is the completion, despite the apparent poles of the gamma factor in its initial product formula.

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.