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Take $F$ a number field, $\pi$ a cuspidal automorphic representation of $GL(3, \mathbb{A}_F).$ Suppose $\pi \cong \pi\otimes \chi.$ Comparing central characters we see that $\chi$ must be cubic.

Now suppose $\chi$ is a cubic character of $\mathbb{A}_F^\times$. Is it known how to construct $\pi$ such that $\pi \cong \pi\otimes \chi$ (or whether it is possible to do so)?

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Let $E / F$ be the cyclic cubic extension corresponding to $\chi$ by class field theory. Let $\sigma$ be a generator of $\operatorname{Gal}(E / F)$, and let $\psi$ be a character of $E^\times \backslash \mathbb{A}_E^\times$ such that $\psi \ne \psi^\sigma$.

Then we can consider the automorphic induction of $\psi$ to an automorphic representation $\pi$ o $GL(3, \mathbb{A}_F)$. This exists, by Example 1.1 of Clozel's 1986 ICM survey; it is cuspidal, because of our assumption on $\psi$ (Theorem 2.4(iv) of op.cit.); and it satisfies $\pi \cong \pi \otimes \chi$.

(If I understand Clozel's article correctly, these are actually the only examples of cuspidal $\pi$ such that $\pi \cong \pi \otimes \chi$.)

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    $\begingroup$ Yes, this is the converse to automorphic induction. For $\mathrm{GL}_2(\mathbb{A}_F)$, this is a result of Labesse and Langlands; see my answer here: mathoverflow.net/questions/97006/… $\endgroup$
    – Peter Humphries
    Commented Oct 24, 2016 at 22:01

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