9
$\begingroup$

Let $G_2$ be the split exceptional group of type $G_2$ and $F$ be a p-adic field. Is it true that every irreducible smooth representation of $G_2(F)$ is self-contragredient? If the answer is Yes, can anybody give me a reference? If not, is there a way to describe the contragredient, for example, is there a MVW involution like the classical group case?

Edit:

I thank Paul Garrett and Jim Humphreys for their comments and Jeffrey Adams for his nice answer. According to Jeffrey Adams's answer, one expects that each L-packet of $G_2(F)$ is self-dual. On the other hand, according to the general philosophy of Gan–Gross–Prasad, in each L-packet, there should be at most one generic representation (GGP conjectured that in each generic local L-packet, there is at most one representation which has the given Bessel model or Fourier–Jacobi model for classical groups. I do not know if anybody conjectured this for exceptional groups. But I just think that we can expect this once we can define the corresponding model. In particular, one would expect the uniqueness of generic member in each L-packet. For the conjecture of uniqueness of generic element in each L-packet, there is probably early reference, but I learned it from GGP). Thus one would expect that:

Each generic smooth irreducible representation of $G_2(F)$ is self-dual.

Do we expect this or is this also false? If we do expect this, do we know anything related to this?

$\endgroup$
9
  • 3
    $\begingroup$ Surely not: (typical) non-unitarizable unramified principal series? Casselman's 1980 Compositio discussion of this applies to split $G_2$, among other things. $\endgroup$ May 5, 2016 at 19:35
  • $\begingroup$ "MVW involution"? I'm mildly surprised that I can't easily guess what this acronym is... For regular characters, the spherical Weyl group gives intertwinings among unramified principal series, in any case. $\endgroup$ May 5, 2016 at 22:28
  • $\begingroup$ @paulgarrett Thanks for your answer. MVW stands for Moeglin-Vigneras-Waldspurger. In their book "Correspondances de Howe sure un corps p-adique", they give a description of contragradient representations for unitary groups (which include symplectic, orthogonal,unitary...). For example, for symplectic groups, if $\pi$ is an irreducible smooth representation of $Sp_{2n}$ and $\delta$ is an element in $GSp_{2n}$ with similitude -1, then it is shown that $\tilde \pi$ is isomorphic to $\pi^\delta$. $\endgroup$
    – Q. Zhang
    May 6, 2016 at 0:43
  • $\begingroup$ @paulgarrett MVW also showed that irreducible smooth representations of $SO_{odd}$ are self-dual. I do not know what Casselman's 1980 Compositio say about $SO_{odd}$. Since $G_2$ can be embedded into $SO_7$, I thought the same would be true for $G_2$. $\endgroup$
    – Q. Zhang
    May 6, 2016 at 1:27
  • 1
    $\begingroup$ @JimHumphreys It's interesting to notice that the English translation of Bernstein-Zelevinski's classical paper " representations of GL(n,F), where $F$ is a non-archimedean local field" had the same mistake. See page 19 of that paper math1.tau.ac.il/~bernstei/Publication_list/publication_texts/… $\endgroup$
    – Q. Zhang
    Jul 2, 2016 at 1:58

2 Answers 2

17
$\begingroup$

Since $-1$ is in the Weyl group (over $F$, not just the algebraic closure) you might expect every irreducible representation to be self-dual. This is the case over $\mathbb R$. It is false over a $p$-adic field, but subtle, and it is not easy to construct an example.

There are non-self dual cuspidal unipotent represenations of $G_2(k)$ where $k$ is the (finite) residue field. By the standard pull back and induction procedure these give rise to non-self-dual supercuspidal representations of $G_2(F)$. The same thing works for all exceptional groups.

Dipendra Prasad has a discussion of some closely related matters in A 'relative' local Langlands Correspondence (arXiv:1512:04347).

In terms of involutions, $G_2(F)$ (for example) has an involution $\tau$ such that $\tau(g)$ is conjugate to $g^{-1}$ over the algebraic closure. This is the "Chevalley involution", and it is inner for $G_2(F)$. However $\tau(g)$ cannot always be $G_2(F)$-conjugate to $g^{-1}$ (exactly because then every irreducible representation would be self-dual, which is false.) See The Real Chevalley Involution (arXiv:1203:1901), page 4.

Finally, because of the involution just mentioned, one would expect that every L-packet for $G_2(F)$ is self-dual, but the duality operation could be nontrivial on the packet.

$\endgroup$
0
6
$\begingroup$

$\DeclareMathOperator\PGSp{PGSp}\DeclareMathOperator\PGL{PGL}$Generic representations of a $p$-adic $G_2$ are indeed self dual. It suffices to prove this for super-cuspidal representations. Observe that, for unitary representations, taking dual is the same as taking complex conjugate. Now all generic super-cuspidal lift one-to-one to generic representations of $\PGSp_6$ by the exceptional theta correspondence, see Savin–Weissman - Dichotomy for generic supercuspidal representations of $G_2$, Compositio Math (2011) (MSN). Since the exceptional theta correspondence commutes with complex conjugation, the statement follows from self-duality on $\PGSp_6$ side. In fact, since any representation of $G_2$ lifts either to $\PGSp_6$ or to a compact form of $\PGL_3$, one can completely classify representations of $G_2$ that are not self dual: the super-cuspidal representations that correspond to non-trivial representations of the compact form of $\PGL_3$.

$\endgroup$
2
  • $\begingroup$ Would it be possible briefly to compare your answer to @JeffreyAdams's? For example, how does the depth-$0$ representation of $\operatorname G_2(F)$ that he mentioned correspond to the compact form of $\operatorname{PGL}_3$? (Also, I took the liberty of correcting the spelling of Marty's name.) $\endgroup$
    – LSpice
    Sep 9, 2019 at 10:21
  • 3
    $\begingroup$ Those supircuspidal representations of $G_2$ correspond to unramified cubic characters of the compact form of $PGL_3$, see my paper "A class of supercuspidal representations of $G_2$" Canadian Math Bulletin (1999). $\endgroup$ Sep 9, 2019 at 13:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.