Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 27 / 4 / a / Solution

Use the normalized Fricke involution $W_N=f\mapsto f{|}_k\left(\begin{array}{cc}0& -1\\ N& 0\end{array}\right)$, and let $D_{\ell }=\left(\begin{array}{cc}\ell & 0\\ 0& 1\end{array}\right)$. Choose integers $m,n$ with $m\ell -Nn=1$. The matrix $\gamma =\left(\begin{array}{cc}1& n\\ N& m\ell \end{array}\right)$ lies in ${\Gamma }_1(N)$, and ${\gamma }^{\prime }=\left(\begin{array}{cc}\ell & n\\ N& m\end{array}\right)$ has determinant one and lies in ${\Gamma }_0(N)$, with $m\equiv {\ell }^{-1}(\mathrm{mod}N)$. The displayed identity in the PDF reads $D_{\ell }{\gamma }^{-1}=\mathrm{diag}(1,\ell ){\gamma }^{\prime }$. Taking double cosets therefore identifies the reversed correspondence with the diamond operator $⟨{\ell }^{-1}⟩$ times the original one. Conjugation by $W_N$ reverses the two diagonal entries, soThe same finite-correspondence change of variables in the Petersson inner product transfers a matrix to its inverse and yields $T_{\ell }^{\ast }=⟨{\ell }^{-1}⟩T_{\ell }$. The diamond operators commute with $T_{\ell }$, so $T_{\ell }{T}_{\ell }^{\ast }=T_{\ell }^{\ast }{T}_{\ell }$. Thus $T_{\ell }$ is a normal operator. Multiplying $W_N$ by a nonzero scalar does not change the conjugation formula.