I made a previous post which was unclear and mistaken in fundamental aspects, so that it was actually more worthy making this new post than actually editing the previous one.

I'm studying the construction of the well known Gelfand-Graev character of a finite group of Lie type $G$. In an attempt to understand the general case, I'm trying to adapt the general construction to the particular case when $G=GL(n,q)$.

Here is the the main ideia of the construction in this case: you pick a linear complex character from a certain family of linear complex characters of the unitriangular group $U$ and induce it to the whole $G$. This induced character is well defined in the sense that it does not depend on the choice of character from the family mentioned above, and is called the Gelfand-Graev character of $G$.

The characters of this family are called non-degenerate. From what I've seen, the "non-degeneracy" concept is somewhat recurrent in representation theory.

Now for the question: I'm having trouble in identifing these "non-degenerate" characters of $U$. What are they?

To reiterate the question in a more specific way, I make the following observation:

One knows that $U/[U,U]$ has, in a sense, all the information about the complex linear characters $\chi$ of $U$ (note that $ker\chi$ must contain $[U,U]$ since $\chi(U) $ is a finite subgroup of $\mathbb{C}^*$, hence cyclic and so abelian). Furthermore, we have a bijection between the irreducible characters of $U/[U,U]$ (which are linear since $U/[U,U]$ is abelian) and the linear characters of $U$. Since $U/[U,U]$ is finite abelian, it as the same number of irreducible characters as conjugacy classes, which is order of $U/[U,U]$. So, one can say that the elements of $U/[U,U]$ are in bijection with the linear complex characters of $U$ Finally, it can be shown that $U/[U,U] \simeq \underbrace{\mathbb{F}_q^+\times...\times\mathbb{F}_q^+}_{n-1}$ (see section 8.1 p. 252 of R.W. Carter, Finite Groups of Lie Type: Conjugacy Classes and Complex Characters).

Hence, the elements of $\underbrace{\mathbb{F}_q^+\times...\times\mathbb{F}_q^+}_{n-1}$ are in bijection with the linear complex characters of $U$

So, reiterating the question, hoping that it makes sense: to which elements of $\underbrace{\mathbb{F}_q^+\times...\times\mathbb{F}_q^+}_{n-1}$ are the so called non-degenerate characters associated with?

As to why I hope the question makes sense: It is known that the subgroup $T$ of diagonal matrices of $G$ acts by conjugation on $U$ such that $[U,U]$ is invariant. Then $T$ acts on $U/[U,U]$, which is to say $T$ acts on the set of linear complex characters of $U$. Via the above isomorphism, one can look to the corresponding action of $T$ on $\underbrace{\mathbb{F}_q^+\times...\times\mathbb{F}_q^+}_{n-1}$, making things somewhat more tractable. According to the construction in Carter's book mentioned above (p. 253 proposition 8.1.2), it is the well behaved action of $T$ over the non-degenerate characters of $U$ that allows the Gelfand-Graev character to be well defined, which is what I'm trying to understand at this point.

Thank you for your patience.

NOTE: Since I have not defined what actually are these non-degenerate characters, one can check the general definition in Carter's book mentioned above, on page 253. That is what I'm trying to understand and adapt to the case of $GL(n,q)$.

  • $\begingroup$ By the way, "$\chi(U)$ is a finite subgroup of $\mathbb C^\times$, hence cyclic and so Abelian" is true but, I think, overkill: in any case (even if we look at analogues over infinite fields), $\chi(U)$ is a subgroup of an Abelian group, hence Abelian. $\endgroup$
    – LSpice
    Jan 13, 2015 at 20:00

2 Answers 2


Also, in the action of the diagonal subgroup $T$ on $U$, certainly $[U,U]$ is preserved, so $T$ acts on $U/[U,U]$. Dually, $T$ acts on characters of $U$, by $(t\cdot \chi)(u)=\chi(t^{-1}ut)$. Either way, there are finitely-many orbits. The smallest orbit among characters is the trivial character, the largest is the collection of "generic" characters.

  • 1
    $\begingroup$ I particularly like this point of view because it explains the motivation for this occurrence of the term 'generic' (which otherwise can be badly abused in harmonic analysis). $\endgroup$
    – LSpice
    Jan 13, 2015 at 20:29
  • 1
    $\begingroup$ I forgot to emphasize that the transitivity of the action of $T$ on all "generic" characters implies for elementary reasons that all the induced representations are isomorphic. $\endgroup$ Jan 13, 2015 at 21:00

The ones for which no $\mathbb F_q$-entry is 0. Probably a more useful way to think of it is that a character is non-degenerate if it is non-trivial on each simple root subgroup.


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