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Let $f(z)$ denote a weight $0$ Hecke-Maass form of level $N$ and let $\theta(z)$ denote the Jacobi theta function. Then $y^{1/4} f(z) \overline{\theta(z)}$ transforms as an automorphic form of weight $-\frac{1}{2}$ and level $N'$, with $N' \mid 4N$. Let $g(z)$ be a Maass form of weight $-\frac{1}{2}$ and level $N'$, with the same multiplier system as $y^{1/4} f(z)\overline{\theta(z)}$.

Question: Under what circumstances can we prove that the Petersson inner product $\langle y^{1/4} f \overline{\theta}, g \rangle$ vanishes?

Heuristics: The fucntion $y^{1/4} \theta(z)$ is, up to a constant, the residue at $w=\frac{3}{4}$ of a weight $\frac{1}{2}$ non-holomorphic Eisenstein series $E^{1/2}(z,w)$ on $\Gamma_0(4)$. Thus \begin{align*} \langle y^{\frac{1}{4}} f \overline{\theta}, g \rangle & = c \cdot \mathrm{Res}_{w=\frac{3}{4}} \langle y^{\frac{1}{4}} f \overline{E^{1/2}(z,\overline{w})}, g \rangle \\ & = c \cdot \mathrm{Res}_{w=\frac{3}{4}} \int_0^\infty \int_0^1 y^{w+\frac{1}{4}} f(z) \overline{g(z)} \frac{dxdy}{y^2} \\ & = c \cdot \mathrm{Res}_{w=3/4} \sum_{n \neq 0} \frac{a_f(n) \overline{a_g(n)}}{(4\pi \vert n \vert)^{w-\frac{3}{4}}} \int_0^\infty y^{w-\frac{3}{4}} W_{0 ,it_f}(y) W_{\frac{-n}{4 \vert n \vert}, it_g}(y) \frac{dy}{y}, \end{align*} in which $a_f(n)$ and $a_g(n)$ are the Fourier-Whittaker coefficients of $f$ and $g$.

One may verify directly that the integral is analytic at $w=\frac{3}{4}$. Thus a necessary condition for $\langle y^{1/4} f \overline{\theta}, f \rangle \neq 0$ is that the Dirichlet series \begin{align*} L(s, f \times \overline{g}) := \sum_{n \neq 0} \frac{a_f(n) \overline{a_g(n)}}{\vert n \vert^{s-1}} \end{align*} must have a pole at $s=1$. (The on-average estimates $a_f(n), a_g(n) \approx \vert n \vert^{-1/2}$ imply that $L(s,f \times \overline{g})$ is analytic in $\Re s > 1$.)

Written in this way, it seems very likely that $\langle y^{1/4} f \overline{\theta}, g \rangle$ would vanish in most, if not all, circumstances. If $f$ and $g$ were both of integral weight, this conclusion would follow from Selberg's orthogonality conjecture. Of course, $g$ is half-integral weight, so the situation is less clear.

Note: In my particular case of interest, $f(z)$ is a dihedral Maass form associated to $\mathbb{Q}(\sqrt{2})$. Please feel free to assume this if it adds useful structure to the problem.

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    $\begingroup$ In general, it is nonvanishing. Let $G$ be the Shimura lift of $g$ to a Maass form of weight $0$. Then $|\langle y^{1/4} f\overline{\theta},g\rangle|^2$ is equal, up to local constants and normalising factors, to $\Lambda(1/2,\operatorname{ad}f \otimes g)$. See Theorem 4.5 of the following paper: doi.org/10.1093/imrn/rnt179 $\endgroup$
    – Peter Humphries
    Commented Apr 25 at 14:39
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    $\begingroup$ Also note that if $f$ is dihedral, then $\operatorname{ad} f$ factorises further, as in my answer here: mathoverflow.net/a/285653/3803. $\endgroup$
    – Peter Humphries
    Commented Apr 25 at 14:48
  • $\begingroup$ Thanks for the reference! Presumably this should be $\Lambda(1/2, \mathrm{ad} f \otimes G)$? $\endgroup$
    – awwalker
    Commented Apr 25 at 14:56
  • $\begingroup$ To clarify the definitions: Does $g$ being a Maass form of weight $-1/2$ mean that $g$ is the complex conjugate of a Maass form of weight $1/2$? $\endgroup$
    – Joseph
    Commented Apr 25 at 15:09
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    $\begingroup$ If $f$ is the automorphic induction of a Hecke Grossencharakter $\psi$ of $\mathbb{Q}(\sqrt{D})$ with $D$ a positive fundamental discriminant, it should be possible to use the theta correspondence to directly show that $\langle y^{1/4}f\overline{\theta},g\rangle$ is equal to the product of two period integrals: one period integral of $G \psi^2$ over the closed geodesics of discriminant $D$, and one period integral given by the $D$-th Fourier coefficient of $g$. But I don't think this kind of identity appears anywhere in the literature... $\endgroup$
    – Peter Humphries
    Commented Apr 25 at 15:58

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