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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.

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  • $\begingroup$ In the setting of p-adic groups, the answer is "yes". (Taking generic to have its usual meaning of "admits a Whittaker model".) Assuming that the meaning of generic is similar in the finite group context, I would guess that answer is again yes. Could you remind me of the definition of generic in the finite group context? $\endgroup$
    – Emerton
    Mar 11, 2010 at 15:23

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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.

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    $\begingroup$ I agree with this answer. A few points: I think that the fact that the Steinberg representation is generic ("generic" = "having a Whittaker model") is proven in Carter's "Finite Groups of Lie type" (Chatper 8, maybe? That's what MathSciNet says - I don't have the book in front of me), where "generic" will be called "regular". This elaborates the Deligne-Lusztig reference given above. $\endgroup$
    – Marty
    Mar 12, 2010 at 1:38
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    $\begingroup$ Two textbook references: (1) the detailed treatment of Gelfand-Graev characters in Chapter 8 of R.W. Carter, Finite Groups of Lie Type (Chapter 6 is about the Steinberg character); (2) the concise treatment in Chapter 14 of Digne-Michel, Representations of Finite Groups of Lie Type, where 14.39 defines "regular" character as a constituent of Gelfand-Graev and gives the Steinberg character as an explicit example. The general theory requires extra care if the ambient algebraic group has a disconnected center. (For me the term "generic" isn't at all helpful in this setting.) $\endgroup$ Mar 12, 2010 at 11:57
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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 $1 - St$.

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    $\begingroup$ Hey Geordie, I think from the point of view of DL theory, "generic" is a terrible term, and "regular" is much better: the section of the map from the irreps of G to semisimple classes in the dual group given by the "regular' representations that Deligne-Lusztig produce gives a lovely analogous to the section you get from the adjoint quotient to the regular conjugacy classes. $\endgroup$ Mar 12, 2010 at 14:47
  • $\begingroup$ Kevin, yes, I agree! I was trying to guess what generic means (and even asked a few people here). By the way, do you know if the first bracketed statement in the second paragraph is true? Namely if one always gets the Steinberg when doing DL of the trivial modules from a torus? $\endgroup$ Mar 12, 2010 at 19:08
  • $\begingroup$ Geordie, the character St does occur (once) as a constituent of the DL character coming from the trivial character of any maximal torus. A quick reference is 7.6.6 in Carter's book, using the DL computation of inner products (as pointed out by Carter just after he defines "unipotent" characters in 12.1. I haven't traced this explicitly back to the DL paper or Lusztig's further development, but it's clearly a consequence of the earliest work on DL characters. $\endgroup$ Mar 12, 2010 at 21:19
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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".

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  • $\begingroup$ Yes, but these are "generic", with the standard interpretation of that adjective in the p-adic theory. $\endgroup$
    – Emerton
    Mar 11, 2010 at 16:22
  • $\begingroup$ There is one characteristic p situation in which representations like Steinberg (having maximum possible dimension) become generic in a geometric sense: consider all irreducible representations of the Lie algebra of a semisimple algebraic group, where those coming from the group ("restricted") such as the trivial representation are the least generic and where "most" representations have maximum dimension (p raised to the number of positive roots). But for finite or algebraic groups only the restricted ones are visible. $\endgroup$ Mar 11, 2010 at 17:19

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