Old subspace of cusp forms ($S_k({\Gamma }_1(N){)}^{\mathrm{old}}$)

The old subspace is spanned by $h(tz)$ for cusp forms $h$ of level $M<N$, where $M\mid N$ and $t\mid N/M$. It is stable under the Hecke operators, but need not have a basis of eigenvectors for every bad-prime operator. Its Petersson inner product orthogonal complement is the new subspace of cusp forms.

If $M\mid N$ and $D\mid N/M$, conjugating ${\Gamma }_0(N)$ by $\mathrm{diag}(D,1)$ gives matrices $\left(\begin{array}{cc}a& Db\\ c/D& d\end{array}\right)$ in ${\Gamma }_0(M)$. This proves the weight transformation of $f(D\tau )$. At any rational modular cusp, factor the rational change of variable into an integral cusp representative followed by an upper-triangular rational matrix of positive slope. The original cusp expansion then stays holomorphic and has zero constant term. Thus this map takes cusp forms of level $M$ to those of level $N$.

For prime $p$, the two cusps of ${\Gamma }_0(p)$ have widths one and $p$. At infinity the modular discriminant product gives the first pair of orders. At zero use $S\tau =-1/\tau$ and the parameter $q_0=e^{2\pi i\tau /p}$. The weight-twelve transforms are $\Delta {|}_{12}S=\Delta (\tau )$ and $\Delta (p\tau ){|}_{12}S=p^{-12}\Delta (\tau /p)$, giving orders $p$ and one in $q_0$.

If $p$ divides the primitive level $M$ of a trivial-character newform and $N=M{p}^r$, put $e_j=p^{jk/2}f(p^{j}z)$. Then $U_p{e}_0=a_p{e}_0$ and $U_p{e}_j=p^{k/2}{e}_{j-1}$ for $j\ge 1$. The normalized Fricke involution acts by $e_j\mapsto \epsilon e_{r-j}$, where $W_{M}f=\epsilon f$. These two operators act irreducibly: a nonzero invariant subspace contains an eigenvector of $U_p$; its eigenvalue is $a_p$ or zero. Either it contains $e_0$ immediately or applying the reversal and a power of $U_p$ produces $e_0$, using $a_p^2\ne p^k$. Reversal then produces $e_r$, and its successive images produce every basis vector.