Hi,


Let $D$ be a quaternion algebra over $\mathbf Q$ such that $D\otimes\mathbf R = M\_2(\mathbf R)$.


Let $\pi = \pi\_\infty \otimes\_p \pi\_p$ be an irreducible automorphic representation of $D^\times$.


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?


Thanks



strong approximation and one-dimensional automorphic representations
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 $G$.


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).


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 $f$).


Our assumption on $\pi$ shows that furthermore $g\_{\infty}f = f$ for all $g\_{\infty} \in G(\mathbb R)$.


Thus if $\gamma \in G(\mathbb Q)$, $g \in G(\mathbb A)$,
$g\_{\infty} \in G(\mathbb R)$, and $u \in U$, then
$$f(\gamma g ug\_{\infty}) = (g\_{\infty} f)(\gamma g u) = f(\gamma g u)
= f(g).$$


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 $G(\mathbb A)$.


Since $f$ was an arbitrary element of $\pi,$ we see that $\pi$ is trivial,
i.e. that all $\pi\_v$ are one-dimesional.




---


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.


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 *are* 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.)


