strong approximation and one-dimensional automorphic representations - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T09:22:53Z http://mathoverflow.net/feeds/question/67207 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/67207/strong-approximation-and-one-dimensional-automorphic-representations strong approximation and one-dimensional automorphic representations unknown 2011-06-08T05:09:13Z 2011-06-08T08:15:55Z <p>Hi,</p> <p>Let $D$ be a quaternion algebra over $\mathbf Q$ such that $D\otimes\mathbf R = M_2(\mathbf R)$.</p> <p>Let $\pi = \pi_\infty \otimes_p \pi_p$ be an irreducible automorphic representation of $D^\times$.</p> <p>Supposedly if $\pi_\infty$ is one-dimensional, then every $\pi_p$ is one-dimensional, and it should follow from strong approximation, but how to prove it?</p> <p>Thanks</p> http://mathoverflow.net/questions/67207/strong-approximation-and-one-dimensional-automorphic-representations/67221#67221 Answer by Emerton for strong approximation and one-dimensional automorphic representations Emerton 2011-06-08T08:15:55Z 2011-06-08T08:15:55Z <p>Let $G$ be the group of norm one elements in $D^{\times}$. An easy argument shows that it suffices to prove the claim for $G$ in place of $D^{\times}$. In other words, I will let $\pi$ be an automorphic rep. of $G$.</p> <p>Now suppose that $\pi_{\infty}$ is one-dimensional. This means that in fact $\pi_{\infty}$ is trivial (because $G(\mathbb Q_v) = SL_2(\mathbb R)$ has no non-trivial characters).</p> <p>Now we use the fact that the automorphic representation lives inside the space of automorphic forms on $G(\mathbb Q)\backslash G(\mathbb A).$ Thus $\pi$ is a space of functions $f(g)$ on $G(\mathbb A)$ such that $f(\gamma g u) = f(g)$ for any $\gamma \in G(\mathbb Q)$ and $u \in U$, a compact open subgroup of $G(\mathbb A^{\infty})$ (with $U$ depending on $f$).</p> <p>Our assumption on $\pi$ shows that furthermore $g_{\infty}f = f$ for all $g_{\infty} \in G(\mathbb R)$.</p> <p>Thus if $\gamma \in G(\mathbb Q)$, $g \in G(\mathbb A)$, $g_{\infty} \in G(\mathbb R)$, and $u \in U$, then $$f(\gamma g ug_{\infty}) = (g_{\infty} f)(\gamma g u) = f(\gamma g u) = f(g).$$</p> <p>Now strong approximation says that the double coset space $G(\mathbb Q)\backslash G(\mathbb A)/ U G(\mathbb R)$ is a point, and combined with the above calculation, this shows that $f$ is constant, and thus generates the trivial representation under the action of $G(\mathbb A)$.</p> <p>Since $f$ was an arbitrary element of $\pi,$ we see that $\pi$ is trivial, i.e. that all $\pi_v$ are one-dimesional.</p> <hr> <p>Note that it is important here that $\pi$ was a subspace of automorphic forms, and not just a subquotient. This is automatic when $D$ is a (non-trivial) quaternion algebra, because then $G$ is anisotropic, hence all automorphic forms are automatically $L^2$, and so the space of automorphic forms is semi-simple.</p> <p>On the other hand, if we consider $GL_2$, then while the above proof goes through for cuspidal representations (which are necessarily subreps., and not just subquotients, of automorphic forms), one can find (non-cuspidal) automorphic reps. of $GL_2$ which <em>are</em> trivial at $\infty$ but infinite-dimensional at the other places. (Of course, these are then subquotients of automorphic forms which can't be split off as subrepresentations.)</p>