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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
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up vote 3 down vote accepted

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!

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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
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