# Finite dimensional automorphic representations of a definite quaternion with prime discriminant and Hecke action

Before stating the questions that I have, which are very specific and probably not so interesting to someone who has never thought about these things, I need to introduce some notation.

Let $p$ be any prime, and let $D$ be the quaternion algebra over $Q$ ramified precisely at $p$ and infinity. Choose a maximal order $R$ inside $D$. For any prime $\ell\neq p$, fix an isomorphism of $D_\ell:=D\otimes Q_\ell$ with the algebra $M_2(Q_\ell)$ in such a way that the maximal compact subring $R_\ell:=R\otimes Z_\ell$ corresponds to $M_2(Z_\ell)$.

If $N$ is an integer $>0$ not divisible by $p$, let $U_\ell(N)$ be the subgroup of $R_\ell^\star\simeq GL_2(Z_\ell)$ given by those matrices whose bottom row is congruent to $(0$ $1)$ modulo $N$ (equivalently, modulo the highest power of $\ell$ dividing $N$).

The ring $R_p:=R\otimes Z_p$ has a unique maximal, two-sided principal ideal $(\pi)$ generated by any uniformizer $\pi$, the residue field $R/(\pi)$ is a finite field with $p^2$ elements. We let $R_p^\star(1)$ denote the subgroup of $R_p^\star$ given by the units that are congruent to $1$ modulo $(\pi)$.

Let now $D^\star$ be the multiplicative group of $D$, viewed as an algebraic group over $Q$. For any integer $N>0$ not divisible by $p$, we are going to define an open subgroup $U(1,N)$ of the group $D^\star_A$ of points of $D^\star$ valued in $A$, the adele ring of $Q$. Namely $U(1,N)$ is the product of all the $U_\ell(N)$, for $\ell\neq p$; of $R_p^\star(1)$; and of the full (connected) component at infinity $D^\star_\infty$.

Consider the space $S(1,N)$ of complex valued functions on the double coset $D^\star\backslash D^\star_A/U(1,N)$, which is known to be finite. For any prime $\ell\neq p$ there is an Hecke operator $T_\ell$ acting on $S(1,N)$ that can be defined in terms of double cosets in the usual way.

(Let $\alpha_\ell$ be the matrix whose top row is $(\ell$ $0)$ and whose bottom is $(0$ $1)$; decompose $U_\ell(N)\alpha_\ell U_\ell(N)$ as a finite union of left cosets $\gamma_i U_\ell(N)$; for $f\in S(1,N)$ define $T_\ell(f)(x)=\sum f(x\gamma_i)$).

Let $V$ be the vector space of locally constant, complex valued functions on $D^\star_A$ that are left invariant by $D^\star$. Observe that $S(1,N)$ can be viewed as a finite dimensional subspace of $V$. Right translation defines an admissible representation of $D^\star_A$ on $V$ which is known to be completely decomposable into a discrete direct sum of irreducible admissible representations of $D^*_A$. If $f\in S(1,N)$, then denote by $V_f$ the smallest subspace of $V$ that is stable by $D^\star_A$.

Questions:

1) Let $f\in S(1,N)$, for some $N$. Is it true that the space $V_f$ is finite dimensional if and only if $f:D^\star_A\rightarrow C$ factors through the reduced norm map $Nr:D^\star_A\rightarrow A^\star$?

2) Does the subspace of $S(1,N)$ given by those functions that factor through the reduced Norm admit an Hecke stable complement?

3) Is the action of $T_\ell$ on $S(1,N)$, for $\ell\nmid pN$, semisimple?

4) Assuming that 2) holds, and letting $S_0(1,N)$ be such complement, how do we relate the C-subalgebra $T_0(N)$ of End($S_0(1,N)$) generated by the Hecke operators $T_\ell$, with $\ell\nmid pN$, to a C-algebra of Hecke operators acting on weight 2 cusps forms of a certain level? What I mean is: out of the J-L correspondence, can we read off an isomorphism between $T_0(N)$ and some Hecke algebra coming from classical modular forms?

Thanks.

[EDIT: In the 2nd and 3rd lines above "Questions:" I should have probably have said "discrete Hilbert direct sum", the "direct sum" being only dense in V]

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1) Yes, I think that's true. I guess it follows relatively easy from the statement that an automorphic representation of the algebraic group $D^\times$ is finite-dimensional iff it's 1-dimensional and factoring through the norm. This latter statement probably follows from strong approximation applied to (the adelic points of the algebraic group associated to) the norm 1 elements of $D^\times$, or you can apply Jacquet-Langlands and move to $GL(2)$ and basically use the standard classification of automorphic representations there.

2) Yes, I think it does. There's some analogue of the Petersson inner product here, right? It's just a combinatorial thing and you can use this. The space $S(1,n)$ is not a mysterious thing: multiplicity 1 etc is all true in this setting and the space is a sum of generalised eigenspaces just like classical spaces of cusp forms.

3) Yes. Use either Jacquet-Langlands and classical results, or mimic the standard proof for $GL(2)$ using the pairing on $S(1,N)$.

4) This question has a bit more meat to it. You seem to be asking what the Jacquet-Langlands correspondence is explicitly. I know something this but it's a bit messy. Let me write down some stuff; the answer isn't quite as "clean" as you'd like it to be though, perhaps.

In $S_0(N)$ you're seeing automorphic representations which have a fixed vector under $R_p^*(1)$. These come in two flavours. The first are those which actually have a fixed vector under all of $R_p^*$. These contribute one dimension, locally, to $S_1(N)$ and under J-L correspond to forms of level $Np$ which are new at $p$ with trivial character. The second sort are those without an $R_p^*$-fixed vector. These are 2-dimensional (I mean their $\pi_p$ is 2-dimensional) [Edit: as Tomasso pointed out, this isn't true. There are some more 1-dimensional ones factoring through the local norm map and coming from tamely ramified but not unramified quasicharacters; let's ignore them in what follows.] and all of $\pi_p$ is fixed by $R_p^*(1)$, so here you're getting 2-dimensional spaces in $S_0(N)$ on which Hecke is acting as scalars (so a super-naive version of multiplicity 1 is failing; what's really going on is that the classical theory of oldforms and newforms works in a completely different way to the behaviour of $S_0(N)$ at $p$). Applying JL, if memory serves, you see forms which are supercuspidal at $p$; their level will be something like $Np^2$, their character will be something like the norm of the character of $\mathbf{F}_{p^2}$ associated to the form on $S_0(N)$. So the answer to your question is "yes, you can write down a classical space of forms that corresponds to $S_0(N)$" but it's not pleasant to write it down---you have to understand the possibilities of what forms of level $Np^2$ look like, and only choose certain of them. It's much easier to understand from the point of view of representation theory, either of $GL(2,\mathbf{Q}_p)$ or the local Weil group.

All this is very well-understood by lots of people here, so feel free to correct me or ask more.

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Kevin, thanks a lot for your answer. I strongly suspected these things to be true, but I never found complete proofs in the literature. I will look more carefully. Concerning your answer to 4), I think there might also be infinite dimensional automorphic representations $\Pi$ arising from S(1,N) whose local component $\pi_p$ at $p$ is one-dimensional and the action of $R^*_p/R^*_p(1)$ is through a non trivial character of order dividing p-1 (a character that factors through the norm to ${\bf F}_p^*$). These guys also should correspond to newforms on the $GL_2$-side with $p^2$ in the level. –  Tommaso Centeleghe Sep 26 '10 at 18:07
Can you tell me more about the representations of the Weil group of ${\bf Q}_p$ that these $\pi_p$ correspond to? Are they easy to described starting from the finite dimensional representation of $D_p^*$ given by the local component of $\Pi$? –  Tommaso Centeleghe Sep 26 '10 at 18:08
Finally, a consequence of 1) is that, I believe, the subspace of S(1,N) giving rise to finite dimensional representations of the adele group of D^* does not grow with N, it is constant. The corresponding systems of Hecke eigenvalues are of the form $(1+\ell)\psi(\ell)$, where $\psi$ is a Dirichlet character of conductor dividing $p$. I found a bit surprising that Dirichlet characters mod $N$ do not enter the picture. And this was actually the reason why I started to doubt that 1) was correct... Thanks again. –  Tommaso Centeleghe Sep 26 '10 at 18:08
Tommaso: if $\pi_p$ is 1-dimensional then the whole of $D_p^\times$ must act via a character. The quotient of $D_p^\times$ by the normal subgroup $R_p^*(1)$ is not abelian though (conjugation by a uniformiser acts via raising to the $p$th power, which is never trivial) so I can't see how $\pi_p$ can be 1-dimensional in your situation with $R_p^\times$ acting via a non-trivial character. Let me know if I made a slip. The Weil group reps are almost "guessable"---the 1-d ones are twists of Steinberg and the 2-d ones are induced from a character on the unramified quadratic extension. –  Kevin Buzzard Sep 26 '10 at 19:09
There are references for this stuff but I don't know them; I just picked it up by osmosis. Sorry I can't be more helpful. Finally, I don't really understand your comments about things not growing with $N$: don't we pick up more and more Dirichlet characters. In fact we are at odds about Dirichlet characters here! Dirichlet chars give rise to characters of $\mathbf{A}^\times$ and hence to auto reps of $D^\times$ that factor through the norm, right? So I am a bit confused by your comments about growth/Dirichlet chars. –  Kevin Buzzard Sep 26 '10 at 19:12