= Solution
Here being modular of weight $k$ means obeying the weight transformation law, or equivalently being invariant under the <slash operator for modular forms>:
$$
f\!\left(\frac{a\tau+b}{c\tau+d}\right)=(c\tau+d)^kf(\tau)\qquad\begin{pmatrix}a&b\\c&d\end{pmatrix}\in\Gamma.
$$
This transformation property alone does not impose holomorphy; a holomorphic <modular form> additionally satisfies interior and cusp holomorphy. That distinction is needed for the nonholomorphic completion in the next part.
Put $F(\tau)=f(N\tau)$. For $\gamma\in\Gamma_0(N)$, conjugation by $\operatorname{diag}(N,1)$ gives the integral determinant-one <matrix>
$$
\gamma_N=\begin{pmatrix}a&Nb\\c/N&d\end{pmatrix}\in SL_2(\mathbb Z),\qquad N\gamma\tau=\gamma_N(N\tau).
$$
Since the denominator in the latter action is $(c/N)(N\tau)+d=c\tau+d$, the full-group modularity of $f$ gives
$$
\boxed{F(\gamma\tau)=(c\tau+d)^kF(\tau)\qquad\gamma\in\Gamma_0(N).}
$$
Thus argument dilation lowers the group to the <Gamma 0 congruence subgroup> while retaining the weight. If $f$ is a holomorphic <modular form>, cusp holomorphy is preserved too: an integral <matrix> $\operatorname{diag}(N,1)\gamma$ of determinant $N$ factors as an integral determinant-one <matrix> times an upper triangular positive-determinant <matrix>. The corresponding slash translate is a constant times $f((a\tau+b)/d)$ with positive $a/d$, whose expansion at infinity has no negative exponents. This checks every rational cusp; for a <cusp form> its constant term remains zero. It is the level-one instance of the <degeneracy map for cusp forms>.
Back to article page
