I've been studying the Gelfand-Graev character's general construction for a finite group of Lie type. I wish to discuss its particularization in a seminar for the general linear group over a finite field, but I'm lacking some motivational ideas for introducing the concept in a simple fashion. In I.M Gelfand and M.I Graev's 1962 motivational paper "Construction of irreducible representations of simple algebraic groups over a finite field", Gelfand refers back to the general construction of irreducible representations of Dickson-Chevalley groups, which I would prefer to avoid since discussing it would easily go beyond the scope of the seminar. Also, I can't seem to relate the ideas discussed in Gelfand's paper with the construction presented in R. W Carter's extensive book Finite Groups of Lie Type: Conjugacy Classes and Complex Characters in chapter 8.1.

I think that an answer to the following question will help me (I will be using the same notation as in Carter's book that I mentioned before):

Let $G$ be a reductive group with connected center, $F$ a Frobenius morphism of $G$, $B$ an $F$-stable Borel subgroup, and $U$ its unipotent radical. How are the ideas from Gelfand's paper related to the fact that one is interested in inducing a linear nondegenerate character from $U^F$ to $G^F$? (I've asked about the nature of these nondegenerate characters in another post for the case where $U^F$ is the group of unitriangular matrices over a finite field).

Thank you in advance.


1 Answer 1


Motivation or intuition is usually difficult to recover from older published work, but I think it helps a lot here to put the work of I.M. Gelfand and his collaborators in perspective. Finite groups of Lie type were never their main focus, and in any case by the 1970s the ideas of Deligne and Lusztig (developed much further by Lusztig and others) largely took over that subject. Gelfand looked much more broadly at representations of Lie groups and at linear algebraic groups over many kinds of local fields, along with finite fields.

For Gelfand's own work it's useful to look at the English translations of various papers in volume 2 of the Collected Papers (Springer, 1988), especially those gathered in Part IV: Models of representations; representations of groups over various fields. This includes the short 1962 papers with Graev which introduced Gelfand-Graev characters. (Note that the Russian papers are now freely available online, but usually it's harder to get access to the translation journals -- and some of the translations are less reliable than others.)

It's worth quoting the opening lines of a 1974 note by Bernstein-Gelfand-Gelfand A new model for representations of finite semisimple algebraic groups (in a sometimes non-idiomatic English translation):

"An important problem in representation theory is to construct the so-called model, i.e. representations of the given group $G$ that contain almost every irreducible representation of $G$ exactly once."

This theme was pursued for various classes of Lie groups (compact and noncompact) and finite groups of Lie type, where the technical methods naturally vary a lot. But roughly speaking the emphasis is placed on working with induced representations from a maximal unipotent subgroup. In the finite case, this presumably led to experimentation with inducing the simplest types of characters (those of degree 1) from a Sylow $p$-subgroup such as the upper triangular unipotent subgroup in a finite general linear group. Since the subgroup is relatively small in this case, it's of course not to be expected that such an induced character is irreducible. Indeed, if one induces the trivial character of a Borel subgroup such as the full triangular subgroup of a general linear group, the decomposition of the induced character is complicated to work out. But if the ambient algebraic group has a connected center, the more subtle idea of Gelfand-Graev is remarkably successful: inducing a "regular" (meaning most non-trivial) character of the unipotent group yields an induced character whose constituents all have multiplicity 1 and exhaust the "regular" characters of $G$. Connectedness of the center is essential for getting a unique character of $G$ regardless of which regular character of the unipotent group is used in the construction.

(Gelfand-Graev could only prove the multiplicity 1 property in a special case, but Steinberg then provided a general proof given in Carter's book. It's worth looking at Steinberg's discussion toward the end of his last section 14; his 1967-68 Yale lectures are still available online here.)

This notion of "regular" character was later shown by Lusztig to fit well with the Deligne-Lusztig theory, in terms of a more precise analogue of the notion of "regular" element in an algebraic group. See also Chapter 14 in the concise textbook by Digne and Michel Representations of Finite Groups of Lie Type (Cambridge, 1991).

  • 1
    $\begingroup$ Unfortunately, the time-dependent 'still' in "still available online here" is now false. A quick Google turned up a version on Peter Trapa's home page: Lectures on Chevalley groups. It seems also to be on Semantic Scholar, although I'm never sure how reliable those links are; and it's finally been published (I've never been clear why it wasn't in the collected works). $\endgroup$
    – LSpice
    Commented Nov 4, 2019 at 21:40
  • $\begingroup$ @LSpice: I've always been somewhat naive about this issue, but in the case of Steinberg's Yale lectures there have always been efforts to replace the mimeographed notes by typeset and edited ones. Eventually this was done by AMS after Steinberg's death. What I don't know is how the ownership changed then. $\endgroup$ Commented Nov 6, 2019 at 0:12
  • 1
    $\begingroup$ P.S. This is one of the many questions I should have asked him when we spent a month at INI in Cambridge (UK), sharing an office. Probably it was in my first visit to IAS in 1968-69 that he posted a note from a publisher who declined to publish his lectures on "Chevrolet" groups. $\endgroup$ Commented Nov 6, 2019 at 0:13

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.