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Given an $L$-homomorphism of Langlands dual groups $${}^LG \to {}^LG'$$

Langlands functoriality contectures predicts the existence of a tranfer map of automorphic representations $$Aut(G) \to Aut(G')$$

However, nothing in the functoriality results or conjectures seems to be concerned with cuspidality. Let us say I am interested in the set $Cusp(G)$ of cuspidal representations of $G$. The above map gives a a tranfer from it to $Aut(G')$.

My question is: what do we know about its image? Is it also cuspidal? Is it endowed at least with some extra properties?

Since those questions have in general negative answers I believe, I am more precisely interested in unitary groups: what about $G$ be a quasi-split unitary group in 2 or 3 variables, and $G'=GL_4$ or $GL_6$ on the quadratic extension defining the unitary group (from the base change transfer)?

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    $\begingroup$ I'm confused about what transfer you are interested in: if $G=U(n)$ defined with respect to a quadratic extension $E/F$, then the usual (stable or unstable) base change will be to $GL(n)/E$, not $GL(2n)/E$. However, you can then induce from $GL(n)/E$ to $GL(2n)/F$, but you're likely to lose even more cuspidality. $\endgroup$
    – Kimball
    Commented Jun 14, 2017 at 0:31

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You are quite correct that the Langlands transfer map does not preserve cuspidality in general. E.g. if you take a modular form of CM type, coming from a Groessencharacter $\psi$ of some imaginary quadratic field K that doesn't factor through the norm map to $\mathbf{Q}$, then this gives you a cuspidal automorphic representation $\pi$ of $GL_2 / \mathbf{Q}$, but the base-change of $\pi$ to $GL_2 / K$ will be Eisenstein.

This gives a flavour of what to expect: VERY roughly, the transfer of a cuspidal automorphic rep $\pi$ of $G$ to $G'$ will be cuspidal unless $\pi$ is itself a Langlands transfer from some other group $H$ via a homomorphism ${}^L H \to {}^L G$ whose image in ${}^L G'$ lands inside a Levi subgroup.

This naive picture is, of course, far from being the whole story -- for instance, there are "CAP representations" which are cuspidal despite being lifts from Levi subgroups.

A lot of these subtleties were first seen in the case of $G = GSp_4$, $G' = GL_4$; this is where CAP forms were first identified, for instance. There is a nice survey article by Arthur from 2005. At the end of the article, he states a classification of discrete-spectrum automorphic reps of $GSp_4$ into six types A-F, of which type A ("general type") has cuspidal transfer to $GL_4$, and the other five types are all functorial liftings from various subgroups of $GSp_4$, whose transfers to $GL_4$ are easily seen to be Eisenstein.

A lot of Arthur's work has been generalised to quasisplit unitary groups by Chung Pang Mok (Mem AMS, 2015) and that might go some way towards answering your more specific questions.

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  • $\begingroup$ I think your first statement is not quite correct: modular forms of CM type need not even be cuspidal. $\endgroup$
    – Peter Humphries
    Commented Nov 9, 2017 at 14:02
  • $\begingroup$ The simplest example is the following. Let $E$ be a quadratic extension of a number field $F$. Take a Hecke character $\psi$ of $E^{\times} \backslash \mathbb{A}_E^{\times}$ that factors through the norm map $N_{\mathbb{A}_E / \mathbb{A}_F}$. Then the automorphic induction $\pi(\psi)$ is an automorphic representation of $\mathrm{GL}_2(\mathbb{A}_F)$ that is noncuspidal. $\endgroup$
    – Peter Humphries
    Commented Nov 9, 2017 at 14:02
  • $\begingroup$ For example, take $\psi$ to be the trivial character, whose $L$-function is $\zeta_E(s)$, which factorises as $\zeta_F(s) L(s,\omega_{E/F})$, where $\omega_{E/F}$ is the quadratic Hecke character of $F^{\times} \backslash \mathbb{A}_F^{\times}$ corresponding to the extension $E/F$ via class field theory. $\endgroup$
    – Peter Humphries
    Commented Nov 9, 2017 at 14:03
  • $\begingroup$ Sure; I guess I don't usually think of Eisenstein series as being of CM type, but they sometimes are in weight 1. This is of course entirely consistent with my second paragraph, because these are exactly the forms for $GL1 / K$ which are functorial lifts from $GL1 / Q$. $\endgroup$
    – David Loeffler
    Commented Nov 9, 2017 at 18:42

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