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)?


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 $f$.

  • $\begingroup$ Thank you. Your lemma 5 and its proof are simple and exactly what I was looking for. $\endgroup$ – davidlowryduda Nov 13 '19 at 4:21
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    $\begingroup$ For what it's worth, this lemma appears elsewhere: see, for example, Lemma 2.4 of "Low-Lying Zeros of Families of $L$-Functions" by Iwaniec, Luo, and Sarnak, or Lemma 3.13 of my paper "Density Theorems for Exceptional Eigenvalues for Congruence Subgroups" $\endgroup$ – Peter Humphries Nov 13 '19 at 11:36

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