Steinberg Representations of Finite Groups of Lie Type - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T06:25:14Z http://mathoverflow.net/feeds/question/17821 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/17821/steinberg-representations-of-finite-groups-of-lie-type Steinberg Representations of Finite Groups of Lie Type Jeff Breeding 2010-03-11T04:26:28Z 2010-10-30T01:37:06Z <p>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? </p> <p>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. </p> http://mathoverflow.net/questions/17821/steinberg-representations-of-finite-groups-of-lie-type/17865#17865 Answer by Geordie Williamson for Steinberg Representations of Finite Groups of Lie Type Geordie Williamson 2010-03-11T14:11:11Z 2010-03-11T14:55:22Z <p>I don't think so. Let \$T\$ be an \$F\$-stable torus. A character of \$T^F\$ is in <em>general position</em> if its stabiliser under \$N_G(T)/T\$ is trivial. I assume by <em>generic</em> 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".)</p> <p>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 <em>any</em> \$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\$.</p> http://mathoverflow.net/questions/17821/steinberg-representations-of-finite-groups-of-lie-type/17876#17876 Answer by Jim Humphreys for Steinberg Representations of Finite Groups of Lie Type Jim Humphreys 2010-03-11T15:52:47Z 2010-03-11T15:52:47Z <p>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". </p> http://mathoverflow.net/questions/17821/steinberg-representations-of-finite-groups-of-lie-type/17934#17934 Answer by Kevin McGerty for Steinberg Representations of Finite Groups of Lie Type Kevin McGerty 2010-03-12T01:11:54Z 2010-10-30T01:37:06Z <p>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".</p> <p>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.</p>