# extending cusp forms

Let $E/F$ be a quadratic extension of number fields, and let $V$ be an $n$-dimensional Hermitian space over $E$.

Let $\tilde{G} := GU(V)$ and $G := U(V)$. Suppose that $(\pi, V_{\pi})$ is an irreducible cuspidal representation of $G.$

Is there an irreducible cuspidal representation $(\tilde{\pi}, V_{\tilde{\pi}})$ of $\tilde{G}$ such that $V_\pi \subset V_{\tilde{\pi}}|_{G}$? Note that here, the restriction is that of cusp forms, not of the representation itself.

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what does "the restriction is that of cusp forms" mean? – Max Flander Feb 11 '10 at 3:04
$V_{\tilde{\pi}}$ is a space of functions on $\tilde{G}\supset G$. I want to restrict those functions to $G$, and see if I recover all of $V_\pi$ by doing this. This is in contrast to restricting the map $\tilde{\pi} : \tilde{G}\to GL(V_{\tilde{\pi}})$ to $G$, which is what people might normally think of when they hear 'restriction' in the context of a representation. – Neal Harris Feb 11 '10 at 3:39
Not an answer but just a reformulation. The $L$-group of $G-tilde$ maps onto the $L$-group of $G$, with kernel $C^*$, and so there should be some functoriality map from auto reps on G-tilde to auto reps on G. A question close to yours is "is this map surjective"? I know nothing about the analytic theory of auto forms but on the Galois side this question conjecturally says something like "does a rep of the Langlands group of F to LG always lift to one to LG-tilde?". The obstruction to lifting is in H^2(Langlands,C^*). Now $H^2(Galois,C^*)=0 so that's a good start... – Kevin Buzzard Feb 11 '10 at 7:41 So I guess implicit in Kevin's reformulation is that the functoriality map from auto reps of$\tilde{G}$to auto reps of$G$is something like apply restriction of cusp forms and take an irreducible auto subrepresentation... is it known that there's a unique such subrepresentation? – jnewton Feb 12 '10 at 11:24 ## 1 Answer I believe the answer should be yes, by some version of the following sketch of an argument: (Note: by restriction of scalars, I regard all groups as being defined over$\mathbb Q$, and I write${\mathbb A}$for the adeles of$\mathbb Q$.) We are given$V_{\pi} \subset Cusp(G(F)\backslash G({\mathbb A}_F)).$Let$\tilde{C}$denote the maximal$\mathbb Q$-split torus in the centre of$\tilde{G}$(this is just a copy of$\mathbb G_m$), and write$C = \tilde{C}\cap G$. (I guess this is just$\pm 1$?) Now$C(\mathbb A)$acts on$V_{\pi}$through some character$\chi$of$(\mathbb A)/C(\mathbb Q)$. Choose an extension$\tilde{\chi}$of$\chi$to a character of$\tilde{C}(\mathbb A)/\tilde{C}(\mathbb Q)$, and regard$V_{\pi}$as a representation of$\tilde{C} G$by have$\tilde{C}$act through$\tilde{\chi}$. Since$\tilde{C} G$is normal and Zariksi open in$\tilde{G}$, we should be able to further extend the$\tilde{C} G(\mathbb A)$-action on$V_{\pi}$to an action of$\tilde{G}(\mathbb Q)\tilde{C}G(\mathbb A).$Now if we consider$Ind_{\tilde{G}(\mathbb Q)\tilde{C} G(\mathbb A)}^{\tilde{G}(\mathbb A)} V_{\pi},$we should be able to find a cupsidal representation$V_{\tilde{\pi}}$of the form you want (with$\tilde{C}(A)$acting via$\tilde{\chi}$). The intuition is that automorphic forms on$G(\mathbb A)$are$Ind_{G(\mathbb Q)}^{G(\mathbb A)} 1,$and similarly for$\tilde{G}$. We will consider variants of this formula that takes into account central characters, and think about how to compare them for$G$and$\tilde{G}$. Inside the automorphic forms, we have those where$C(\mathbb A)$acts by$\chi$; this we can write as$Ind_{G(\mathbb Q)C(\mathbb A)}^{G(\mathbb A)} \chi$, and then rewrite as$Ind_{\tilde{G}(\mathbb Q)\tilde{C}(\mathbb A)}^{\tilde{G}(\mathbb Q)\tilde{C}G(\mathbb A)} \tilde{\chi}.$This is where$V_{\pi}$lives, once we extend it to a repreresentation of$\tilde{G}(\mathbb Q)\tilde{C}G(\mathbb A)$as above. Now the automorphic forms on$\tilde{G}(\mathbb A)$with central character$\tilde{\chi}$are$Ind_{\tilde{G}(\mathbb Q)\tilde{C}(\mathbb A)}^{\tilde{G}(\mathbb A)} \tilde{\chi},$which we can rewrite as$Ind_{\tilde{G}(\mathbb Q) \tilde{C}G(\mathbb A)}^{\tilde{G}(\mathbb A)} Ind_{\tilde{G}(\mathbb Q)\tilde{C}(\mathbb A)}^{\tilde{G}(\mathbb Q)\tilde{C}G(\mathbb A)} \tilde{\chi}.$This thus contains$Ind_{\tilde{G}(\mathbb Q)\tilde{C}G(\mathbb A)}^{\tilde{G}(\mathbb A)}V_{\pi}$inside it, and so an irreducible constituent of the latter should be a$V_{\tilde{\pi}}$whose restriction (as a space of functions) to$G(\mathbb A)$contains$V_{\pi}$. What I have just discussed is the analogue for$G$and$\tilde{G}$of the relation between automorphic forms on$SL_2$and$GL_2$discussed e.g. in Langlands--Labesse. Hopefully I haven't blundered! - Thanks Matthew! It looks as though this argument should work for any$\tilde{G}$and$G$such that$\tilde{G}_{der}\subset G\subset\tilde{G}\$. – Neal Harris Feb 18 '10 at 1:43
Yes, I think so. – Emerton Feb 18 '10 at 2:20