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

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.

The commuting normal operators $T_{\ell }$, together with the finite-order diamond operators, admit simultaneous diagonalization for the Petersson inner product. Their restrictions to the new subspace of cusp forms have one-dimensional common eigenspaces by the newform multiplicity-one theorem. This additional theorem is the essential newform input: normality alone would allow higher multiplicities. Since every $U_p$ commutes with the good-prime operators and preserves the new space, it preserves each such line and is scalar on it. Normalize its nonzero first Fourier coefficient to one to obtain a newform. This gives the requested simultaneous eigenbasis of the new space.

By definition, every old form is a linear combination of lower-level oldforms by argument dilation. Repeatedly split each lower-level cusp space into its old space and its Petersson inner product orthogonal new space; induction on the level shows that the specified dilations of newforms span the full space. To prove independence, separate distinct primitive eigenforms by their good-prime eigenvalues, using the newform multiplicity-one theorem. For a fixed normalized form, its dilations $f(tz)$ are independent: order the distinct positive integers $t$ increasingly, and the coefficient of the least occurring $q^t$ kills its coefficient in any vanishing linear combination. Repeat for the next least $t$. Thus these dilations are indeed a basis.

For trivial character, part (a) shows that $W_N$ commutes with all the good-prime operators, so it preserves the one-dimensional primitive line. Also $W_N^2=(-1{)}^k$ for the determinant-normalized slash operator; a nonzero trivial-character form has even $k$. ThereforeThe power of $N$ in an unnormalized Fricke eigenvalue depends on the convention. The plain slash action $(cz+d{)}^{-k}f(\gamma z)$ gives $\epsilon N^{-k/2}$, while the Hecke-style action $(\det \gamma {)}^{k-1}(cz+d{)}^{-k}f(\gamma z)$ gives $\epsilon N^{k/2-1}$. The PDF prints $\pm N^{k-1}$; that value requires the deliberate rescaling $w_N=N^{k-1}{W}_N$ and does not follow under either usual convention. Under that rescaling its displayed eigenvalue is $\epsilon N^{k-1}$; all the preceding eigenspace and conjugation statements are unchanged.

Let $f_j(z)=f(p^{j}z)$ for $0\le j\le r$, and write $a=a_p(f)$ and $W_{M}f=\epsilon f$. The Hecke operator $U_p$ acts on the Fourier expansion by $\sum a_n{q}^n\mapsto \sum a_{pn}{q}^n$. Since $p\mid M$, the primitive relation is $a_{pn}=a_p{a}_n$. Hence $U_p{f}_0=a{f}_0$ and $U_p{f}_j=f_{j-1}$ for $j\ge 1$. Substituting the argument directly in the Fricke involution givesThus, in the basis $e_j=p^{jk/2}{f}_j$, put $b=p^{k/2}$ and let $R$ be the reversal matrix, $R{e}_j=e_{r-j}$. The matrices areIf the exam's rescaling $w_N=N^{k-1}{W}_N$ is used, multiply the second matrix by $N^{k-1}$.

Here is an explicit irreducibility argument for the prime-power old block at a bad prime. Any nonzero common invariant complex subspace $V$ contains an eigenvector of the restricted $U_p$. Its eigenvalue is either $a$ or zero. If $a\ne 0$, the $a$-eigenspace is $\mathbb{C}{e}_0$; containing it gives $e_r$ by reversal, then every $e_j$ by successive powers of $U_p$. The kernel of $U_p$ is $\mathbb{C}(e_1-(b/a)e_0)$. If $V$ instead contains this vector, reversal and $U_p^r$ giveThis is nonzero: the bad-prime eigenvalue of a trivial-character newform satisfies $a^2=p^{k-2}$ when $p\parallel M$, and $a=0$ when $p^2\mid M$, whereas $b^2=p^k$. If $a=0$, the kernel is already $\mathbb{C}{e}_0$, and the same reversal-and-shift argument works. Therefore the block has no nonzero proper subspace invariant under $U_p$ and the Fricke operator, and hence none invariant under the full ring in the question.