extending cusp forms - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T18:24:09Z http://mathoverflow.net/feeds/question/14959 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/14959/extending-cusp-forms extending cusp forms Neal Harris 2010-02-11T01:50:04Z 2010-02-17T14:20:25Z <p>Let $E/F$ be a quadratic extension of number fields, and let $V$ be an $n$-dimensional Hermitian space over $E$.</p> <p>Let $\tilde{G} := GU(V)$ and $G := U(V)$. Suppose that $(\pi, V_{\pi})$ is an irreducible cuspidal representation of $G.$</p> <p>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.</p> http://mathoverflow.net/questions/14959/extending-cusp-forms/15581#15581 Answer by Emerton for extending cusp forms Emerton 2010-02-17T14:20:25Z 2010-02-17T14:20:25Z <p>I believe the answer should be yes, by some version of the following sketch of an argument:</p> <p>(Note: by restriction of scalars, I regard all groups as being defined over $\mathbb Q$, and I write ${\mathbb A}$ for the adeles of $\mathbb Q$.)</p> <p>We are given $V_{\pi} \subset Cusp(G(F)\backslash G({\mathbb A}_F)).$</p> <p>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 $\pm 1$?)</p> <p>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 $\tilde{G}(\mathbb Q)\tilde{C}G(\mathbb A).$</p> <p>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 $\tilde{\chi}$).</p> <p>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 $\tilde{G}$.</p> <p>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.</p> <p>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 $V_{\pi}$. </p> <p>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!</p>