How to prove that the Peterson inner product of $E_k$ with a cusp eigenform is $0$?

Question

How to prove that the Peterson inner product of $E_k$ and $f$ is $0$ where $f \in S_k$ is an eigenform for all of the level $1$ Hecke operators?

Answer 1

The Hecke algebra is self-adjoint with respect to the Petersson inner product. For a modular form $g$ denote $a_n(g)$ its $n$-th Fourier coefficient. Since $f$ and $E_k$ are eigenforms, for any $n>0$ we have

$$a_n(E_k) \left<E_k, f\right> = \left<T_n E_k, f\right> = \left<E_k, T_n f\right> = a_n(f) \left<E_k, f\right>$$

so if we had $\left <E_k, f\right> \neq 0$, we would have $a_n(E_k) = a_n(f)$ for all $n>0$. This would imply that $E_k-f$ is the constant $1$, which is absurd because constants are of weight $0$ but here $k>0$.