Simple explicit example of local Jacquet-Langlands theorem for inner forms of GL(n), and consequences - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T03:22:07Z http://mathoverflow.net/feeds/question/27446 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/27446/simple-explicit-example-of-local-jacquet-langlands-theorem-for-inner-forms-of-gl Simple explicit example of local Jacquet-Langlands theorem for inner forms of GL(n), and consequences Kevin Buzzard 2010-06-08T08:25:22Z 2010-06-08T13:28:09Z <p>This one will be very easy for the experts.</p> <p>Let $F$ be a nonarch local field, let $n\geq1$ be an integer, choose <code>$0\leq d&lt;n$</code> 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 $G(F)$.</p> <p>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 $\pi$.</p> <p>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!</p> <p>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 $gcd(d_1,n)=gcd(d_2,n)$? </p> <p>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?</p> <p>I don't know where in the literature to look for such statements :-/ so I ask here.</p> http://mathoverflow.net/questions/27446/simple-explicit-example-of-local-jacquet-langlands-theorem-for-inner-forms-of-gl/27463#27463 Answer by Paul Broussous for Simple explicit example of local Jacquet-Langlands theorem for inner forms of GL(n), and consequences Paul Broussous 2010-06-08T13:28:09Z 2010-06-08T13:28:09Z <p>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:</p> <p>"Jacquet-Langlands et unitarisabilité", Journal de l'Institut de Mathématiques de Jussieu, 2007, </p> <p>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. </p>