Consider the group $GL(n,F_q)$ for finite field $F_q$, consider its irreducible representations over complex numbers.

Questions Is my understanding correct that the dimensions of all such irreps are polynomials in $q$ with integer coefficients having zeros only at roots of unity and zero ?

I understand that the answer should be contained in Green's 1955 paper THE CHARACTERS OF THE FINITE GENERAL LINEAR GROUPS, but as for me paper is difficult for extracting information. The case $q=2$ should be treated with care as demenstrated in comment from user148212.

Questions 2 If anwer is yes - is there any conceptual/nice reason for it ?

(R. Stanley comment below perfectly explains the part about roots).

Questions 3 If someone can give some nice formula for such dimensions that would be quite helpful.

Questions 4 For other finite groups of Lie type is there any similar phenomena ?

Questions 5 From the perspective of $F_1$ it would be nice to know is there always a manifold such that number of $F_q$ points is given by these polynomials ? (Flags are of that type).

Let me give some examples known to me supporting the positive answer to the question:

  • For $GL(2,F_q)$ dimensions are : $1$ (det-like irreps) , $q+1$ (principal series), $q-1$ (cuspidal), $q$ (Steinberg = irregular principal series). See e.g. MO273764, MO271389.

  • In general "regular princinpal series" - irreps induced from non-trivial characters of the Borel subgroup will have dimension $[n]_q!$. Just because $GL/Borel = Flag$ manifold has such number of points.

  • Cuspidal irreps: the degree of a cuspidal character of $GL(n, q)$ is $(q − 1)(q^2 − 1)· · ·(q^{n-1}-1)$ (see page 135 Corollary 5.4.5. of very nice thesis "Character Tables of the General Linear Group and Some of its Subroups" containing huge amount of concrete information).

  • For the so-called unipotent irreps there is q-analogue of "hook formula". The degrees of the unipotent characters are “polynomials in q”: $ q^{d(λ)} \frac{(q^n − 1)(q^{n−1} − 1)· · ·(q − 1)}{ \prod_{h(λ)}(q^h − 1) }$ with a certain d(λ) ∈ N, and where h(λ) runs through the hook lengths of λ. See nice survey by G. Hiss FINITE GROUPS OF LIE TYPE AND THEIR REPRESENTATIONS (top page 26, section 3.2.6).

  • From above source - section 3.2.7: The degrees of the unipotent characters of $GL(5,q)$ for table $(5)$ dim = $ 1 $, for table $(4, 1)$ dim = $q(q + 1)(q^2 + 1)$ for table $(3, 2)$ dim = $q^2(q^4 + q^3 + q^2 + q + 1)$ for table $(3, 1^2)$ dim = $q^3(q^2 + 1)(q^2 + q + 1)$ for table $ (2^2, 1)$ dim = $q^4(q^4 + q^3 + q^2 + q + 1)$ for table $(2, 1^3)$ dim = $q^6(q + 1)(q^2 + 1)$ for table $(1^5)$ dim = $ q^{10}$

  • characters for GL(3), GL(4) has been computed by R. Steinberg The representations of GL(3,q), GL(4,q), PGL(3,q), PGL(4,q) Canad. J. Math. 3(1951), 225-235. Which "This paper is part of a Ph.D. thesis written at the University of Toronto under the direction of Professor Richard Brauer". The degrees of the irreducible characters of GL3(q): $(q − 1)^2(q + 1)$, $(q − 1)(q^2 + q + 1)$, $ (q + 1)(q^2 + q + 1)$, $q^2 + q + 1$, $q(q^2 + q + 1)$, $q(q + 1)$, $q^3$, $1$. See e.g. G.Hiss quoted above section 3.3.6 page 28.

  • 18
    $\begingroup$ Note that if $\mathrm{GL}(n,F_q)$ has an irrep whose dimension is a polynomial in $q$, then the zeros of this polynomial are indeed roots of unity and zero. This is because the dimension of an irrep divides the order of the group, and if $f(q),g(q)\in \mathbb{Q}[q]$ have the property that $f(q)|g(q)$ for infinitely many integers $q$, then $f(q)|g(q)$ as polynomials. $\endgroup$ Aug 13, 2017 at 23:24
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    $\begingroup$ For a general finite gp of Lie type, the integer coefficient property may not be true; e.g. SL_2(F_q) has irreps of dim (q-1)/2 and (q+1)/2. $\endgroup$
    – user148212
    Aug 14, 2017 at 17:24
  • 1
    $\begingroup$ Maybe you need the assumption that q is big enough to avoid some "missing dim"; e.g. (q+1)(q^2+q+1)...(q^{n-1}+...+1) is the dim of some irreps only if q is big enough. $\endgroup$
    – user148212
    Aug 14, 2017 at 17:46
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    $\begingroup$ @user148212 thank you for your comments, however i am not sure i understand second one $\endgroup$ Aug 14, 2017 at 18:31
  • 1
    $\begingroup$ @AlexanderChervov This refers to the principal series one: E.g. in your GL_2 example, q+1 is not an irrep dim if q=2; in GL_2(F_2) there are 1 one dim cusp irrep, 1 trivial one dim irrep, and 1 two dim Steinberg irrep. $\endgroup$
    – user148212
    Aug 15, 2017 at 0:14

3 Answers 3


You are asking many questions here, about which much is written down in the rather large literature of the subject. Maybe I can point to some of the answers.

In the 1955 paper of J.A. Green on finite general linear groups, the results were heavily combinatorial and left open quite a few questions about special linear and other groups of Lie type. Green's own students Srinivasan and Lehrer made significant progress on Sp$_4$ and on special linear groups, over finite fields. But here the results get far more complicated, with for example many fractions occurring in degree polynomials. This is true also for the finite groups of type G$_2$, treated by Chang and Ree, even though the ambient algebraic group in this case has a (trivial) connected center. This connected center case was first worked out most thoroughly later on, but by now the efforts of Lusztig (and others) have made it possible in principle to compute all degrees of irreducible representations for finite groups of Lie type.

An exposition by T.A. Springer of some of this including Green's work may be most useful to you. This arose from the special IAS year 1968-69, published here as Part D. Note in particular the last part of I, in which degrees of irreducible representations of finite general linear groups are expressed explicitly as polynomials in $q$. Ian Macdonald also participated in that special year (and later wrote up a version of Green's work in his book on symmetric functions); he formulated some conjectures about what should occur for other groups of Lie type, which were then essentially proved by Deligne and Lusztig in their fundamental 1976 paper. The key point in all types is the determination of the elusive "cuspidal" or "discrete series" characters. (A helpful textbook source is the 1985 treatise by Roger Carter here; but afterwards Lusztig made many further advances.)

[ADDED] It's not clear how far one can conceptualize the polynomial result, but at least for the induced representations which give irreducibles it's obvious how this property follows recursively. The cuspidal representations seem more mysterious, along with many others which are unipotent (among those which arise from decomposing induced representations non-trivially), I don't know how to account conceptually for the degrees to be given by polynomials in $q$. Of course, these degrees do have to divide the group orders, which themselves are polynomials in $q$.

  • $\begingroup$ Thank you for the answer (and my is UPvote). About ' many questions ' my basic question is the first one - the one in the title. $\endgroup$ Aug 14, 2017 at 17:51
  • $\begingroup$ Concerning degrees, this has not been the sole concern: the natural problem is to determine all character values (i.e., the character table). This gets understood so far by a complicated recursive method, which makes it seem at the moment impossible to write degrees in all ranks as polynomials in $q$ with integral (or rational) coefficients except for finite general linear groups. I'm not sure how far one can expect to write down closed degree formulas for all finite groups of Lie type, but Frank Luebeck and others have computed a lot of special cases. $\endgroup$ Aug 14, 2017 at 18:20
  • $\begingroup$ Yes that is true , however what I observe in literature that even GL case except GL2 is not exposed in some good pedagogical manner... So MO seems best way to ask for help... There are really lots of expositions of GL2.... $\endgroup$ Aug 14, 2017 at 18:41
  • $\begingroup$ The nice 2011 Springer text by Cedric Bonnafe does provide a good modern treatment of complex representations of the special linear group SL$_2$, while my older exposition (limited to groups over the prime field but freely available online) is more elementary even though much less up-to-date: maa.org/sites/default/files/pdf/upload_library/22/Ford/… $\endgroup$ Aug 14, 2017 at 23:05
  • $\begingroup$ Your paper is "on my table". It is wonderful ! $\endgroup$ Aug 16, 2017 at 6:33

As Jim has noted, for $\mathrm{GL}_n$ the formulas were given long ago by Green, Springer, and Macdonald and it was shown that they are polynomials in q with integer coefficients. Since the question asked for an explicit formula, I will now produce their formula.

First, let us discuss the parameterisation of the irreducible representations. Let $\Phi$ denote the orbits of the Frobenius automorphism $x\mapsto x^q$ on $\overline{\mathbb{F}_q}^\times$. For an element $\gamma\in \overline{\mathbb{F}_q}^\times$, we let $\{\gamma\}$ denote the corresponding orbit, and let $d(\gamma)$ denote the size of this orbit. Let $\mathcal{P}$ denote the set of partitions.

The irreducible complex representations of $G_n:=\mathrm{GL}_n(\mathbb{F}_q)$ are parameterised by the set of maps $$\Lambda: \Phi\rightarrow \mathcal{P}$$ such that $\Lambda$ is constant on the orbits and the sum of all the partitions in the image of $\Lambda$ equals $n$. We let $d(\Lambda)$ denote the dimension of the corresponding irreducible representation. Then showing $d(\lambda)\in \mathbb{Z}[q]$ is more or less equivalent to showing that $|G_n|/d(\lambda)\in \mathbb{Z}[q]$. (One can use Gauss's Lemma, primitive polynomials, etc.).

Now we state the formula for $|G_n|/d(\lambda)$ and explain why it is a polynomial in $\mathbb{Z}[q]$. For each partition $\lambda$, define the normalised Hook polynomial of $\lambda$ by $$ H_\lambda(q):=q^{-\frac{1}{2}<\lambda, \lambda>} \prod(q^h-1). $$ Here, the product is taken over the boxes in the Young diagram of $\lambda$ and $h$ is the hook length of the box. Moreover, $<\lambda, \lambda>:=\sum_{i} (\lambda'_i)^2$, where $\lambda'$ is the dual partition. Next, we defined the normalised hook polynomial of $\Lambda$. Let $$ m_{\lambda, d}:=\#\{\{\gamma\} \, | \, d(\gamma)=d, \Lambda(\gamma)=\lambda\} $$ and set $$H_\Lambda(q):= \prod_{d,\lambda} H_\lambda(q^d)^{m_{d,\lambda}}. $$

Finally, we have the formula $$|G_n|/d(\Lambda')=(-1)^n q^{\frac{1}{2}n^2} H_{\Lambda},$$ where $\Lambda'$ is obtained from $\Lambda$ by dualizing all the partitions; see also this paper, equation 3.1.5.

The only thing remaining is to see why this expression, which is seemingly a polynomial in $q^{\frac{1}{2}}$, is actually a polynomial in $q$. This boils down to showing that $$ n^2- \sum_{d,\lambda} m_{\lambda,d} d\sum_{i}(\lambda_i')^2 $$ is even. But note that since a partition and its dual have the same size, we have $$ n=|\Lambda|=\sum_{d,\Lambda'} m_{d,\lambda} d|\lambda| = \sum_{d,\lambda} m_{d,\lambda} d (\sum_i \lambda_i'). $$ Thus, modulo $2$, we have $$ n^2- \sum_{d,\lambda} m_{\lambda,d} d\sum_{i}(\lambda_i')^2 = \sum_{d,\lambda} \left((m_{d,\lambda} d)^2 (\sum_i (\lambda_i')^2) - (m_{\lambda,d} d)\sum_{i}(\lambda_i')^2\right)=0, $$ where the last equality follows from the fact that $x^2-x$ is always even.

  • $\begingroup$ A bit of a correction: Functions $\Phi \to \mathcal{P}$ canonically index conjugacy classes in $GL_n(\mathbb{F}_q)$. In order to canonically parametrize irreps, one should replace $\Phi$ with $\Gamma$, which is an inductive limit of $\mathrm{Hom}(\mathbb{F}_{q^n}^{\times}, \mathbb{C}^{\times})$ along norm maps; see Section 3.1 of arxiv.org/abs/math/0612668 . $\endgroup$ Oct 18, 2021 at 13:07
  • $\begingroup$ Are the two not canonically isomorphic? For instance, for S_n we know that the set of conjugacy classes is in canonical bijection with the set of irreducible characters (because both are in canonical bijections with partitions)...I had always assumed something similar holds for GL_n(F_q). Are you saying this is not the case? $\endgroup$
    – Dr. Evil
    Oct 18, 2021 at 21:22
  • $\begingroup$ I don't believe so. For $GL_1$, there is no canonical bijection between $\mathbb{F}_q^{\times}$ and $\mathrm{Hom}(\mathbb{F}_q^{\times}, \mathbb{C}^{\times})$, so I don't see why it would be better for $n$ larger, and all of the sources I know make this distinction. PS: Thanks for the useful answer! $\endgroup$ Oct 18, 2021 at 21:29
  • $\begingroup$ Right...Thanks for the clarification. $\endgroup$
    – Dr. Evil
    Oct 18, 2021 at 22:27

This is just an ``asymptotic'' answer which add some supports.

Let $T$ be a maximal torus of $\mathrm{GL}_n(\mathbb{F}_q)$ (I am following the terminology of Carter's book), let $\theta\in \mathrm{Hom}(T,\overline{\mathbb{Q}}_{\ell})$, and let $R_{T}^{\theta}$ be the corresponding Deligne--Lusztig representation. Suppose $T$ corresponds to a partition $P$ of $n$.

It is known that, if $\theta$ is in general position (i.e. not fixed by any non-trivial elements in the Weyl group relative to $T$), then $R_{T}^{\theta}$ is up to $\pm1$ irreducible. Also, every irreducible representation appears in some $R_{T}^{\theta}$. Moreover, for fixed $n$ and $P$, when $q$ tends to $+\infty$, the ratio of the number of the general position $\theta$'s by the number of all the $\theta$'s tends to $1$. Finally, the dimensions of the $R_{T}^{\theta}$'s are integer coefficient polynomials in $q$.

(All these should be in Carter's book mentioned in Professor Humphrey's answer.)

The above properties imply that:

There is a finite set of polynomials $D(n)\subseteq \mathbb{Z}[x]$ such that, when $q$ tends to $+\infty$, the ratio of the number of irreducible representations of $\mathrm{GL}_n(\mathbb{F}_q)$ whose dimension is the evaluation of an element of $D(n)$ at $q$ by the number of all irreducible representations tends to $1$.


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