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This one will be very easy for the experts.

Let $F$ be a nonarch local field, let $n\geq1$ be an integer, choose $0\leq d<n$ and let $D$ be the central simple algebra over $F$ with invariant $d/n$ in the Brauer group [EDIT: and with $F$-dimension $n^2$, so e.g. if $gcd(d,n)>1$ then $D$ will not be a division algebra but rather a matrix algebra over the division algebra with invariant $d/n$]. Let $G$ be the algebraic group $D^\times$ and let $\pi_0$ denote the trivial 1-dimensional representation of $G(F)$.

The local Jacquet-Langlands theorem guarantees us the existence of a smooth irreducible representation $\pi=JL(\pi_0)$ of the group $GL(n,F)$ canonically associated to $\pi_0$. This construction gives a completely canonical map from the set $\{0,1/n,2/n,\ldots,(n-1)/n\}$ to the set of smooth irreducible representations of $GL(n,F)$, sending $d/n$ to $\pi$.

If you put a gun to my head and asked me to guess what $\pi=\pi(d/n)$ was in this situation, I would probably go for the following construction: set $h=gcd(d,n)$, Let $P$ be the standard parabolic in $GL(n)$ whose Levi is $GL(h)^{n/h}$, and (non-normally) induce the trivial 1-dimensional representation of $P$ up to $G$; such a representation will, I suspect, have a canonical "biggest" irreducible subquotient, corresponding on the Galois side to an $n$-dimensional representation of the Weil-Deligne group of $F$ which is a direct sum of $h$ representations of degree $n/h$ each of which is Steinberg (in the sense that $N$ is maximally unipotent). I only envisage this because I can't imagine any other such map which agrees with what I know in the $GL(2)$ case!

Here is a consequence of my guess: if $d$ is coprime to $n$ then the trivial 1-dimensional representation of $D^\times$ corresponds to the Steinberg representation of $GL(n)$, whatever $d$ is. This makes me wonder whether the following is true: say $d_1$ and $d_2$ are both coprime to $n$ and let $G_i$ be the group of units of the central simple algebra over $F$ with invariant $d_i/n$. Are the smooth irreducible representations of $G_i$ canonically in bijection with one another? Does this remain true if I relax the condition that the $d_i$ are coprime to $n$ but instead only demand that $gcd(d_1,n)=gcd(d_2,n)$?

More generally is it true that if $gcd(d_1,n)$ divides $gcd(d_2,n)$ then there's a canonical injection from the irreps of $G_1$ to the irreps of $G_2$, which is a bijection iff the gcd's coincide?

I don't know where in the literature to look for such statements :-/ so I ask here.

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A toy example which supports your conjectures: we can take $d_1 = -d_2$ modulo n. Then $G_1 = G_2^{\rm op}$, and composing with the inverse map will give you a bijection between representations of $G_1$ and of $G_2$. But I'm sure you already thought of that. –  David Loeffler Jun 8 '10 at 10:21
    
Hi David---no, I didn't think of that and it's a neat observation :-) Thanks! –  Kevin Buzzard Jun 8 '10 at 12:43
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Sorry Kevin, I read your question too quickly. For non essentially square integrable representations you have to give a meaning to the Jacquet-Langlands correspondence. This is done in Badulescu's article:

"Jacquet-Langlands et unitarisabilité", Journal de l'Institut de Mathématiques de Jussieu, 2007,

where the correspondence is generalized to the Grothendieck group of finite length representations. Moreover Badulescu proves that the extended correspondence is compatible with the Zelevinsky-Aubert duality and one knows that the dual of the trivial representation is a (generalized ???) Steinberg representation. So you're right : one should obtain a (generalized) Steinberg representation.

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So thanks for the references. I am minded to accept one of your answers, but I don't know which one to accept. Perhaps you should delete one of them and move the comments to the other one? If you don't then I'll just accept the first one. –  Kevin Buzzard Jun 8 '10 at 15:02
    
grin Paul---I was suggesting that you copied the useful comments from one answer into the other! You can edit your own answers! If you have two things to say, the best thing is to say them both in one answer, not to make two answers. –  Kevin Buzzard Jun 8 '10 at 19:07
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