Is "self-dual" equivalent to "dihedral" for Maass forms on $\mathrm{GL}(2)$?

Suppose $$f$$ is a Maass cusp form on $$\Gamma_0(D)$$. The associated symmetric square $$L$$-function $$L(\mathrm{sym}^2 f, s)$$ has a pole at $$s = 1$$ if and only if $$f = \overline{f}$$ (if $$f$$ is self-dual).

One way for $$f$$ to be self-dual is if $$f$$ comes from a Hecke character $$\eta$$ (sometimes called a grossencharacter) on a real quadratic field and $$L(s, f) = L(s, \eta)$$. This is a construction of Maass forms that was described by Maass. I know it better from section 1.9 of Bump's Automorphic Forms and Representations. I don't use this terminology often, but I think this is equivalent to $$f$$ being a CM Maass form, or to $$f$$ corresponding to a dihedral Galois representation.

Is this the only way to get a self-dual Maass form?

I believe this is true, but I don't know if this is folklore or conjectural or known.

• If $\pi$ is an automorhpic representation for $\mathrm{GL}(2)$, I would say that $\pi$ is self-dual if and only if $\pi^{\vee} \simeq \pi$. While this is implied by $L(\mathrm{Sym}^2 \pi,s)$ having a pole at $s = 1$, it is not equivalent. If $\pi$ is associated to an even Galois representation with image $\mathrm{SL}_2(\mathbf{F}_3) = \widetilde{A}_4$ then $\pi$ is self-dual but the symmetric square is still cuspidal. Perhaps my next comment will clarify what is going on. Commented 2 days ago
• Write $\pi^{\vee} = \pi \chi$ and assume $L(\mathrm{Sym}^2 \pi,s)$ has a pole at $s = 1$. Consider $$L(\pi \times \pi,s) = L(\mathrm{Sym}^2 \pi,s) L(\wedge^2 \pi,s) = L(\mathrm{Sym}^2 \pi,s) L(\chi,s)$$ By Rankin-Selberg, your assumptions imply both that $\chi$ is non-trivial and $$\pi \simeq \pi^{\vee} \simeq \pi \otimes \chi.$$ Since $\chi \ne 1$, it follows that $\chi$ is a character of a quadratic extension $F$ and $\pi$ is induced from a character on $\mathrm{GL}(1)/F$. This characterization of automorphic induction is probably due to Hecke and Maass in this case. Commented 2 days ago
Let $$f$$ be a Maass form on the upper half-plane with nebentypus $$\chi$$. It is known that \begin{align*} \text{f is self-dual}&\qquad\Longleftrightarrow\qquad\text{L(f\times f,s) has a pole at s=1};\\ \text{f is dihedral}&\qquad\Longleftrightarrow\qquad\text{L(\mathrm{sym}^2f,s) has a pole at s=1}. \end{align*} On the other hand, $$L(f\times f,s)=L(\mathrm{sym}^2f,s)L(\chi,s),$$ and the $$L$$-functions on the right-hand side do not vanish at $$s=1$$, so we conclude that $$\text{f is self-dual}\qquad\Longleftrightarrow\qquad\text{f is dihedral}\quad\text{or}\quad \text{\chi is trivial}.$$
• I guess the broader point here is the difference between "orthogonally-self-dual" (pole of $L(\mathrm{Sym}^2)$) versus "sympletically-self-dual" (pole of $L(\wedge^2)$). Everything in dimension $2$ is "symplectic up to twist" whereas "orthogonal up to twist" implies induced. @WillSawin I think the (removed) sentence was intended to be about the distinction between orthogonal and generalized orthogonal --- you can be induced and not self-dual! Commented 2 days ago
• @user491858 Indeed, dimension $2$ is special. The OP was partially right in that the only self-dual Maass forms (on the upper half-plane) with nontrivial central character are the dihedral ones. I updated my post accordingly, inspired by your valuable comments. Regarding my removed sentence: my momentary confusion came from the wrong idea that a Hecke character can be reconstructed from its $L$-function, which is of course false (e.g. a Galois twist does not change the $L$-function). In fact I went down this path earlier at MathOverflow (at which time Peter Scholze corrected me). Commented yesterday