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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.

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    $\begingroup$ 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. $\endgroup$
    – user491858
    Commented Jun 11 at 9:53
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    $\begingroup$ 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. $\endgroup$
    – user491858
    Commented Jun 11 at 9:54
  • $\begingroup$ @user491858 Thank you. I think I should learn more about automorphic induction too. $\endgroup$
    – davidlowryduda
    Commented Jun 11 at 18:13

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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 $$\text{$f$ is self-dual}\qquad\Longleftrightarrow\qquad\text{$f$ is dihedral}\quad\text{or}\quad \text{$\chi$ is trivial}.$$

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    $\begingroup$ 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! $\endgroup$
    – user491858
    Commented Jun 11 at 10:08
  • $\begingroup$ @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). $\endgroup$
    – GH from MO
    Commented Jun 11 at 12:18
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    $\begingroup$ @GHfromMO Thank you! Reading and thinking about this answer led me to Venkatesh's Beyond Endoscopy and special forms on GL(2), and from there to some work of Labesse and Langlands. This is very informative and I have to read them more closely. Do you have any other references that you would suggest for me to learn about this better? $\endgroup$
    – davidlowryduda
    Commented Jun 11 at 16:50
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    $\begingroup$ @davidlowryduda I am not the best person to ask, because I am not too familiar with Langlands functoriality. I think reading some original papers on functoriality (Gelbart-Jacquet, Ramakrishnan, Kim-Shahidi, Cogdell-PiatetskiShapiro, etc.) or reading Shahidi's book (bookstore.ams.org/view?ProductCode=COLL/58) might be beneficial. In fact asking some of these experts (by email) is perhaps the best advice I can give. $\endgroup$
    – GH from MO
    Commented Jun 12 at 17:15

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