Can the Petersson inner product $\langle f(z), f(2z) \rangle$ be zero?

Question

Suppose $f$ is a weight $k$ cuspidal Hecke eigenform on $\Gamma_0(N)$. Then $f(2z)$ is a weight $k$ cuspform on $\Gamma_0(2N)$.

Is it possible that $f(z)$ and $f(2z)$ can be orthogonal (regarded as forms on $\Gamma_0(2N)$)? That is, can the Petersson inner product $\langle f(z), f(2z) \rangle = 0$, where the product is taken over $\Gamma_0(2N) \backslash\mathcal{H}$?

More generally, can $\langle f(z), f(nz) \rangle = 0$ (where the product is regarded over the appropriate quotient of the upper half plane)?

Answer 1

Yes, the Petersson inner product can be zero. In my paper "Explicit bounds for sums of squares (see Lemma 5) I show that if $f$ is a newform of level $N$ and $p$ is a prime that does not divide $N$, then $$ \langle f(z), f(pz) \rangle = \frac{a(p)}{p^{k-1} (p+1)} \langle f(z), f(z) \rangle. $$ So if you can find a weight $k$ level $N$ newform with odd $N$ for which $a(2)$ vanishes, then this gives you an example. The newform $f(z) = q - 2q^{3} + \cdots$ of weight $2$, level $19$ and trivial character is an example of such an $f$.