Steinberg Representations of Finite Groups of Lie Type - MathOverflow

Steinberg Representations of Finite Groups of Lie Type

2010-03-11T04:26:28Z

Let G be a finite group of Lie type. Assume G is also of universal type. Is the Steinberg representation of G generic, i.e., does the Steinberg representation admit a Whittaker model?

A Whittaker model for a representation of G is defined in a similar fashion as in the case of GL(2, F) in Bump's "Automorphic Forms and Representations." I am interested in the genericity of the Steinberg representation of a group of matrices over a finite field.

Answer by Geordie Williamson for Steinberg Representations of Finite Groups of Lie Type

2010-03-11T14:11:11Z

I don't think so. Let $T$ be an $F$-stable torus. A character of $T^F$ is in general position if its stabiliser under $N_G(T)/T$ is trivial. I assume by generic you mean "obtained by Deligne-Lusztig induction from a character in general position". (These are exactly the characters which appear in MacDonald's conjecture, and are therefore "generic".)

In this setting the Steinberg character is the opposite of generic. It appears, for example, when one induces the trivial character from a split torus (and I think it occurs in the Deligne-Lusztig induction from any $F$-stable torus, but am not sure). For example, in $SL_2$ the (Harish-Chandra = Deligne-Lusztig) induction of the trivial character yields $1 + St$ and Deligne-Ludztig induction of the trivial character from the non-split torus yields $1 - St$.

Answer by Jim Humphreys for Steinberg Representations of Finite Groups of Lie Type

2010-03-11T15:52:47Z

Yes, what does "generic" mean for a finite group? Geordie is correct that the Steinberg representation is far from being a typical Deligne-Lusztig character. In fact, its unique features make it "special" for both ordinary and modular representation theory of finite groups of Lie type. I surveyed a lot of this in Bull. Amer. Math. Soc. 16 (1987), openly accessible at AMS e-math. Even for p-adic groups, it seems the correct analogue of the Steinberg representation is the "special representation".

Answer by Kevin McGerty for Steinberg Representations of Finite Groups of Lie Type

2010-03-12T01:11:54Z

I think that for finite groups of Lie type, the analogue of "having a Whittaker model" is that the representation occurs in a Gelfand-Graev representation: these are the representations obtained by inducing a "regular" character from the unipotent subgroup of a rational Borel. Such representations are multiplicity free and so constitute a "model" (in the sense I think people say "Whittaker model"). Now when the center of $G$ is connected, all regular characters are conjugate under the action of the maximal torus of the Borel, so the Gelfand-Graev representation is unique (otherwise there is a family of such representations). In their famous paper, Deligne and Lusztig decompose the Gelfand-Graev representation in this case and show that there is exactly one constituent in each "geometric conjugacy class" of irreducible representations (which can be thought of as a semisimple conjugacy class in the dual group). The Steinberg representation is then the representative in the conjugacy class of the identity element -- that is the representative among the "unipotent representations".

To focus more on the actual question (!) the character of the Steinberg representation is explicitly known, and it is easy to check from this that its restriction to $U$ is the regular representation, so it certainly occurs in the Gelfand-Graev representation.