No boundary symbol carries the full eigenpacket of a cusp form

Let $f$ be a normalised cuspidal eigenform of weight $k\ge2$ on $\Gamma_1(N)$ with nebentypus $\varepsilon$ and eigenvalues $a_\ell$ at every prime, and let $\Phi$ be a boundary datum whose boundary cochain $\partial\Phi$ satisfies the nebentype law for $\varepsilon$ and

Then $\partial\Phi=0$.

This is the exclusion of boundary (Eisenstein) eigensystems from a cuspidal eigenpacket. A nonzero boundary eigen-symbol has, at all but finitely many primes, eigenvalue $\psi_1(\ell)+\psi_2(\ell)\ell^{k-1}$ for a pair of Dirichlet characters, so $|a_\ell|\ge\ell^{k-1}-1$ at those primes. For $k\ge3$ this contradicts Hecke's bound $|a_\ell|\le C\ell^{k/2}$. For every $k\ge2$, including weight $2$, an elementary mean-square argument suffices: cuspidality gives $y^k|f(x+iy)|^2\le C_f$ on the upper half-plane, Parseval yields $\sum_{m\le X}|a_m|^2\ll_f X^k$, hence $\sum_m|a_m|^2/m^{2k-1}<\infty$, whereas the boundary spectrum gives $|a_\ell|^2/\ell^{2k-1}\ge1/(4\ell)$ at large primes, contradicting the divergence of $\sum_\ell1/\ell$. No Ramanujan–Petersson bound is needed.

Formalization Note The eigenvalues $a_\ell$ are those of the cusp form f.form, tied to its $q$-expansion by f.coeff_eq and f.eigen; the boundary spectrum is computed from the cusp components of $\Phi$.