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

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)?

## 1 Answer

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 $$f$$.

• Thank you. Your lemma 5 and its proof are simple and exactly what I was looking for. – davidlowryduda Nov 13 '19 at 4:21
• 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" – Peter Humphries Nov 13 '19 at 11:36