Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 24 / 5 / i

Here being modular of weight $k$ means obeying the weight transformation law, or equivalently being invariant under the slash operator for modular forms: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 $\mathrm{diag}(N,1)$ gives the integral determinant-one matrixSince the denominator in the latter action is $(c/N)(N\tau )+d=c\tau +d$, the full-group modularity of $f$ givesThus 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 $\mathrm{diag}(N,1)\gamma$ 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.