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Choose an oriented lattice basis with $t={\omega }_2/N$ modulo the lattice. Changing such a basis is exactly the Gamma 1 congruence subgroup action, so the weight law makes the displayed function independent of the choice. It satisfies $F(uL,ut)=u^{-k}F(L,t)$. Holomorphy on the half-plane and at all modular cusp degenerations identifies the functions coming from modular 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.

The Gamma 1 congruence subgroup isThe space $M_k({\Gamma }_1(N))$ consists of the holomorphic functions $f:\mathfrak{h}\to \mathbb{C}$ satisfyingfor every $\gamma \in {\Gamma }_1(N)$ and which are holomorphic at a cusp for every cusp. Its subspace $S_k({\Gamma }_1(N))$ consists of the forms that vanish at every cusp.

The Gamma 1 congruence subgroup is

The cusps are the ${\Gamma }_1(p)$-orbits of primitive columns $(a,c{)}^T$, with simultaneous negation representing the same point of ${\mathbb{P}}^1(\mathbb{Q})$. For the Gamma 1 congruence subgroup, the standard primitive-vector classification separates the orbits according to $d=\gcd (c,p)$. For each divisor $d\mid p$, the two surviving unit coordinates giveorbits when $p>2$. Hence an odd prime gives

For $p=2$, simultaneous negation is already trivial modulo $2$, so the division by two does not apply. In that case ${\Gamma }_1(2)={\Gamma }_0(2)$ has two cusps, represented by infinity and zero. Thus the Number of cusps of Gamma 1 of prime level is

For $p\ge 5$, the Gamma 1 congruence subgroup has $(p-1)/2$ modular cusps of cusp width one and $(p-1)/2$ of cusp width $p$. Primitive columns modulo $p$, taken up to sign, are acted on by $(a,c)\mapsto (a+bc,c)$. The classes with $c=0$ have cusp width one, the others cusp width $p$. There are no effective elliptic stabilizers. The genus formula for a modular curve gives genus $(p-5)(p-7)/24$.