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

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.