Let $q$ be a prime power and $U_n(\mathbb{F}_q)$ be the group of unitriangular $n\times n$-matrices. I've read and heard in several places (see e.g. this mathoverflow question) that classifying irreducible complex characters of $U_n(\mathbb{F}_q)$ is a wild problem.

This seems to be in contrast to the fact that $U_n(\mathbb{F}_q)$ is a finite group, and hence there are only finitely many irreducible represenations, i.e. classifying them for a fixed $n$ and $q$ is a finite problem. So my first question is:

$1.$ In what sense is classifying irreducible characters for $U_n(\mathbb{F}_q)$ a wild problem?

The only reference I found so far is the following paper by Poljak not accessible to me:

[Poljak:On the Characters of Irreducible Complex Representations of the Group of Triangular Matrices over a Prime Finite Field. (Ukrainian) Dopovīdī Akad. Nauk Ukraïn. RSR, 1966, 434-436.

So my other questions are:

$2.$ What is the rough idea to prove this? (A bit more precise than "the problem is translated to the problem of classifying indecomposable representations of a wild quiver).

$3.$ Are there other instances of classifying irreducible representations being a wild problem in this particular sense?

  • $\begingroup$ I tried this once for the more general situation, where $F_q$ is being replaced by $\mathbb{Z}_p$ or the ring of integers in a local field. I tried it with Clifford theory. I thought this should be the way to go because you have a sequence of semidirect products of abelian groups. $\endgroup$
    – Marc Palm
    Jan 30, 2014 at 10:23
  • $\begingroup$ Should the translation in the parenthetical remark in question 2 go in the opposite direction? $\endgroup$
    – S. Carnahan
    Jan 30, 2014 at 10:26
  • $\begingroup$ @S.Carnahan What he appearently showed is that "a nice description of the conjugacy classes would lead to a nice description of wild quivers" [Arias-Castro, Diaconis, Stanley: A super-class walk on upper-triangular matrices] $\endgroup$ Jan 30, 2014 at 10:35
  • 1
    $\begingroup$ Have you looked at "Classes of conjugate elements of the unitriangular group" by Gudinov, Kapitonova, Polyak, Rud'ko and Tsitkin? The English translation is in vol. 26 of the Springer journal Cybernetics, available online. $\endgroup$ Jan 30, 2014 at 11:54
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    $\begingroup$ Perhaps the wildness is in families, e.g. it is something about the behavior of the problem as $n$ and $q$ vary? $\endgroup$ Jan 31, 2014 at 1:34

3 Answers 3

  1. I believe that one sense in which classifying irreducible characters of $U_n(\mathbb{F}_q)$ is wild is the one given in Geoff Robinson's comment, namely that a parametrisation of the conjugacy classes in $U_n(\mathbb{F}_q)$ (at least for all $n$ and $q$) involves parametrising similarity classes of pairs of matrices over finite fields. The latter is a classical 'wild' problem and I think it is equivalent to (or contained in) a host of other intractible classification problems. Roughly speaking, a classification problem is tame if the indecomposable reps are described by a finite number of discrete parameters and a single continuous one (see Mariano Suárez-Alvarez's answer here and Qiaochu Yuan's answer here). By a theorem of Drozd a finite dimensional algebra is either of tame or wild representation type. The (English version of the) relevant paper by Drozd is "Tame and wild matrix problems", Lect. Notes Math. 832 (1980) 242-258. The terminology of wild and tame was apparently introduced by Donovan and Freislich in "Some evidence for an extension of the Brauer–Thrall conjecture", Sonderforschungsbereich Theor. Math. 40 (1972) 24-26.

  2. The paper mentioned by Jeremy Rickard above claims to prove this. I have not read this paper, however.

  3. Wild classification problems seem to be somewhat ubiquitous in representation theory. The prime example is perhaps the finite dimensional modules of a free $k$-algebra on two generators, for $k$ a field. Moreover, according to Nagornyj, "Complex representations of the general linear group of degree three modulo a power of a prime", Zap. Naucn. Sem. Leningrad. Otdel. Mat. Inst. Steklov. (LOMI), 75:143–150, 197–198, 1978, the problem of parametrising the conjugacy classes in $\mathrm{GL}_{4n}(\mathbb{Z}/p^2)$ contains the matrix pair problem in $\mathrm{M}_{4n}(\mathbb{Z}/p)$. Another example is the classification of all (not necessarily finite-dimensional) representations of the Lie algebra $\mathrm{sl}_2(\mathbb{C})$, as shown in an amazing answer by Torsten Ekedahl to this question.

  • 1
    $\begingroup$ "Wild classification problems seem to be somewhat ubiquitous in representation theory". But the question remains: what does "wild" means in this context ? For each of those problems termed "wild", the question to be asked is : is there a positive result saying that the problem cannot be solved in certain terms (like the equation of the fifth degree cannon be solved in radicals), or is the "wildness" just another name for our current incompetence (as the equation of the third degree was certainly "wild" for ancient Greeks) ? $\endgroup$
    – Joël
    Feb 2, 2014 at 0:33
  • $\begingroup$ I've added some more details about this. $\endgroup$ Feb 2, 2014 at 8:30

Well, to understand how this problem is wild it may be useful to contrast it with the situation of finite reductive groups where we do have a classification statement. The first part of this post considers the reductive case and then after the cut I give some information about $\mathrm{U}_n(q)$. Hopefully some of this is clarifying or useful for your question.

Assume $\mathbf{G}$ is a connected reductive algebraic group over an algebraic closure $\mathbb{K} = \overline{\mathbb{F}_p}$ of the finite field of prime order $p>0$. Furthermore, let us assume that $F : \mathbf{G} \to \mathbf{G}$ is a (generalised) Frobenius endomorphism of $\mathbf{G}$ admitting an $\mathbb{F}_q$-rational structure $G = \mathbf{G}^F$, which is a finite reductive group.

Now, we want to "sensibly" parameterise the irreducible characters of $G$. In general this means that the parameterisation should be given, as far as possible, in terms of data from the algebraic group $\mathbf{G}$ and should not depend on $q$. However, a priori it is not clear at all that such a classification can be achieved. It is one of the great feats of finite group representation theory that Lusztig managed to obtain such an elegant classification theory. We roughly recap this below.

Assume $\mathbf{T} \leqslant \mathbf{G}$ is an $F$-stable maximal torus of $\mathbf{G}$. To every irreducible character $\theta \in \mathrm{Irr}(T)$ ($T = \mathbf{T}^F$) Deligne and Lusztig have defined, using $\ell$-adic cohomology, a virtual character $R_{\mathbf{T}}^{\mathbf{G}}(\theta)$ of the group $G$. We call this a Deligne--Lusztig character of $G$. The following curcial fact is known about these virtual characters:

  • The character of the regular representation of $G$ is a sum of Deligne--Lusztig characters. In particular, every $\chi \in \mathrm{Irr}(G)$ occurs in some $R_{\mathbf{T}}^{\mathbf{G}}(\theta)$.

This is somewhat surprising as this is not true of every class function on $G$ (unless $\mathbf{G} = \mathrm{GL}_n(\mathbb{K})$)! With this one has a chance to at least bunch the irreducible characters together but to do this we need more information about the Deligne--Lusztig characters.

Let $\mathbf{G}^{\star}$ be a dual group of $\mathbf{G}$ and let $F^{\star} : \mathbf{G}^{\star} \to \mathbf{G}^{\star}$ be a (generalised) Frobenius endomorphism such that $G^{\star} = {\mathbf{G}^{\star}}^{F^{\star}}$ is a dual group of $G$. We will denote by $\nabla(\mathbf{G},F)$ the set of all pairs $(\mathbf{T},\theta)$ where $\mathbf{T} \leqslant \mathbf{G}$ is an $F$-stable maximal torus and $\theta \in \mathrm{Irr}(\mathbf{T}^F)$. The group $G$ acts naturally on this set by conjugation and we denote the orbits of this action by $\nabla(\mathbf{G},F)/G$. Conversely we denote by $\nabla^{\star}(\mathbf{G},F)$ the set of all pairs $(\mathbf{T}^{\star},s)$ where $\mathbf{T}^{\star} \leqslant \mathbf{G}^{\star}$ is an $F^{\star}$-stable maximal torus and $s \in T^{\star} = {\mathbf{T}^{\star}}^{F^{\star}}$ is a semisimple element. The group $G^{\star}$ also acts naturally on $\nabla^{\star}(\mathbf{G},F)$ by conjugation and we denote by $\nabla^{\star}(\mathbf{G},F)/G^{\star}$ the orbits of this action. With this notation we have the following result:

  • We have a natural bijection $\nabla(\mathbf{G},F)/G \to \nabla^{\star}(\mathbf{G},F)/G^{\star}$ which satisfies $(\mathbf{T},1_T) \mapsto (\mathbf{T}^{\star},1)$.

If $(\mathbf{T},\theta)$, $(\mathbf{T}',\theta') \in \nabla(\mathbf{G},F)$ are in the same $G$-orbit then we have $R_{\mathbf{T}}^{\mathbf{G}}(\theta) = R_{\mathbf{T}'}^{\mathbf{G}}(\theta')$. In particular, if $(\mathbf{T},\theta)$ corresponds to $(\mathbf{T}^{\star},s)$ under the above bijection then we may simply write $R_{\mathbf{T}^{\star}}^{\mathbf{G}}(s)$ for $R_{\mathbf{T}}^{\mathbf{G}}(\theta)$. With this in hand we may state one of the most important theorems concerning Deligne--Lusztig characters.

  • Assume $(\mathbf{T}^{\star},s)$ and $({\mathbf{T}^{\star}}',s')$ are not in the same $G^{\star}$-orbit then $R_{\mathbf{T}^{\star}}^{\mathbf{G}}(s)$ and $R_{{\mathbf{T}^{\star}}'}^{\mathbf{G}}(s')$ have no irreducible constituent in common.

Now, for any semisimple element $s \in G^{\star}$ we denote by $\mathcal{E}(G,s)$ the set of all irreducible characters $\chi \in \mathrm{Irr}(G)$ such that $\chi$ is a constituent of $R_{\mathbf{T}^{\star}}^{\mathbf{G}}(s)$ for some $F^{\star}$-stable maximal torus $\mathbf{T}^{\star}$ containing $s$. The set $\mathcal{E}(G,s)$ is called a Lusztig series of $G$ and by the above we have a disjoint union

$$\mathrm{Irr}(G) = \bigsqcup_{(s)} \mathcal{E}(G,s)$$

where the union runs over all $G^{\star}$-conjugacy classes of semisimple elements of $G^{\star}$.

Let us now assume that the centre $Z(\mathbf{G})$ of $\mathbf{G}$ is connected (similar statements hold when $Z(\mathbf{G})$ is not connected but this is more complicated to state). One of the most amazing parts of Lusztig's classification result is that, under this assumption, there exists a bijection

$$\mathcal{E}(G,s) \to \mathcal{E}(C_{G^{\star}}(s),1)$$

where $C_{G^{\star}}(s)$ is the centraliser of $s$ in $G^{\star}$. Note that, as $Z(\mathbf{G})$ is connected we have $C_{\mathbf{G}^{\star}}(s)$ is a connected reductive algebraic group. As $F^{\star}(s) = s$ we have $F^{\star}$ induces a (generalised) Frobenius endomorphism of $C_{\mathbf{G}^{\star}}(s)$ and so $C_{G^{\star}}(s) = C_{\mathbf{G}^{\star}}(s)^{F^{\star}}$ is a finite reductive group.

The crucial part of Lusztig's classification result is given by the following result:

  • Assume $\mathbf{H}$ is a connected reductive algebraic group and $F : \mathbf{H} \to \mathbf{H}$ is a (generalised) Frobenius endomorphism of $\mathbf{H}$. Let $\mathbf{T}_0 \leqslant\mathbf{B}_0\leqslant \mathbf{H}$ be an $F$-stable maximal torus and Borel subgroup of $\mathbf{H}$. This data determines a Coxeter system $(\mathbf{W},\mathbb{S})$ where $\mathbf{W} = N_{\mathbf{H}}(\mathbf{T}_0)/\mathbf{T}_0$ and $\mathbb{S}$ is a set of Coxeter generators determined by $\mathbf{B}_0$. By our choices $F$ induces an automorphism $\gamma : \mathbf{W} \to \mathbf{W}$ which stabilises $\mathbb{S}$ (i.e. it is an automorphism of the Coxeter system $(\mathbf{W},\mathbb{S})$). Lusztig has then shown that there is a bijection $$\mathcal{E}(H,1) \to X(\mathbf{W},\gamma)$$ where $X(\mathbf{W},\gamma)$ is a set whose definition depends only on $\mathbf{W}$ and $\gamma$.

This is somehow the truly amazing thing. That these irreducible characters $\mathcal{E}(H,1)$ can be parameterised "independently of $q$". This gives us the classification that we desire. In fact, it is really only the classification of the semisimple conjugacy classes of $G^{\star}$ that really depends on $q$. Note that this mimics the classification of the conjugacy classes of $G$. In general a conjugacy class of $G$ is a product $(s)\mathcal{O}$ where $(s)$ is a semisimple conjugacy class of $G$ and $\mathcal{O}$ is a unipotent conjugacy class of $C_G^{\circ}(s) = C_{\mathbf{G}}^{\circ}(s)^F$. The unipotent conjugacy conjugacy classes are parameterised "independently of $q$" and depend only upon the action of $F$ on the root system of $C_{\mathbf{G}}(s)$.

Let us now consider the case where $\mathbf{U}_n$ is the group of unitriangular matrices in $\mathrm{GL}_n(\mathbb{K})$ and $F : \mathbf{U}_n \to \mathbf{U}_n$ is the Frobenius endomorphism $F(x_{ij}) = (x_{ij}^q)$ so that $\mathbf{U}_n^F = \mathrm{U}_n(q)$. One may now take the strategy for finite reductive groups and apply it to our situation here. In particular we would like to do the following:

  • Break $\mathrm{Irr}(\mathrm{U}_n(q))$ into series $\mathcal{F}_i$ such that the irreducible characters in each $\mathcal{F}_i$ can be parameterised "independently of $q$".

The theory of supercharacters gives a way to determine such series $\mathcal{F}_i$ (see "Supercharacters and Superclasses for Algebra Groups" by Diaconis and Isaacs). In particular, these are given by the transitive closure of the condition that two irreducible characters occur as constituents of a supercharacter. These series have a nice combinatorial description but in contrast to the case of finite reductive groups once cannot parameterise $\mathcal{F}_i$ independently of $q$. In this sense the problem is wild. See this paper by Marberg for a recap on the supercharacter theory of $\mathrm{U}_n(q)$


Another approach to the character theory of $\mathrm{U}_n(q)$ is given by the theory of character sheaves on $\mathbf{U}$. This theory (conjectured by Lusztig) was developed by Boyarchenko and Drinfeld. In Theorem 1.13 of


Boyarchenko gives a classification of the irreducible characters of $\mathrm{U}_n(q)$ in terms of minimal idempotents of the $\mathbf{U}$-equivariant bounded derived category of $\overline{\mathbb{Q}_{\ell}}$-constructible sheaves on $\mathbf{U}$. This may also give a way to see why the classification of such characters is wild by showing that the classification of such minimal idempotents is also a wild problem.

One also has a way to break the irreducible characters up into series based on the theory of $L$-packets. This approach is similar to the Lusztig series encountered in the case of finite reductive groups.

  • $\begingroup$ While Lusztig's Jordan decomp of reps of reductive groups over finite fields is important, and the independence of the unipotent reps of $q$ is remarkable, it is not the lack of this which makes a classification problem wild in the technical sense of the original question. E.g., $\mathrm{GL}_2(\mathbb{Z}/p^r)$ has a Jordan decomp of reps (due to Hill) and the 'nilpotent' reps depend on $q$ and even on $r$, but classifying the reps of this group (over any finite local PIR) is a tame problem. $\endgroup$ Feb 3, 2014 at 8:37
  • $\begingroup$ By the way, doesn't the existence of a good character sheaf theory for $\mathbf{U}(\mathbb{F}_q)$ mean that the irreps are parametrised by objects related to the infinite algebraic group $\mathbf{U}$ in a way which is similar to the case of reductive groups? $\endgroup$ Feb 3, 2014 at 8:49
  • $\begingroup$ This is true. I somehow didn't make it clear how starkly different the situations are. The point I wanted to make was that the wildness is somehow contained in parameterising the families $\mathcal{F}_i$. If you could parameterise these families depending on $q$ then this would probably turn out to be a tame problem, as in your example. I suppose the point to make is that you have no hope of parameterising the families $\mathcal{F}_i$ even though the families themselves are a nice way to bunch up the characters. $\endgroup$
    – Jay Taylor
    Feb 4, 2014 at 6:17
  • $\begingroup$ Well, things are slightly different between the unipotent and reductive cases. In the case of $U_n(q)$, according to Boyarchenko, the characteristic functions of character sheaves are (up to multiplication by a power of $q$) irreducible characters. This is not true in the reductive case. In the unipotent case one only has a labelling by objects in the bounded derived category of constructible sheaves. It is a miracle that in the reductive case you can label the characters (and character sheaves) by objects living in the algebraic group. This seems not (or not known) to be the case in $U_n(q)$. $\endgroup$
    – Jay Taylor
    Feb 4, 2014 at 6:26

I am surprised that no one has answered so far this interesting question. I am not an expert at all, and don't know the answer, but I'll give my thoughts here. Perhaps me saying something stupid will encourage an expert to weigh in, if only to contradict me.

So first, one thing I am pretty sure of is that when one consider the problem of classification of complex representations of $U_n(\mathbb F_q)$, we consider it for $n$ fixed, but $q$ variable among the prime powers. In other words, you consider $U_n$ as a group scheme over spec $\mathbb Z$, and you consider the complex representations of its group of points over all finite field. This is how, at least, the problem of classifying representation of G(\mathbb F_q) for $G$ a reductive (as opposed as unipotent) is presented and studied.

No in what sense is the problem wild for G=U_n but not wild for G=GL_n ? I don't know. At first, I expected the results to be of a completely different nature in the two cases. But, after doing some bibliographical research, I am not so sure, and I am wondering if what people mean is not just that the results for U_n are just more complicated to prove, rather than more complicated to state.

For $GL_2(\mathbb F_q)$, for instance, it has long been known that the degree of irreducible complex representations are $1, q, q-1, q+1$. The number of representations of each of those degrees is $q-1, q-1, 1/2q(q-1), 1/2(q-1)(q-2)$ respectively. Note that all those functions of $q$ are polynomials, which at first may be surprising. All this can be found in many textbooks, such as Fulton-Harris. We have similar results for all reductive group G, although they are more complicated to prove (one needs for instance the Deligne-Luztig theory of generalized induction, which uses étale cohomology). But this is now well-known. A general reference is Carter's book "representation of finite groups of Lie type", actually, of "reductive Lie type".

What is the situation for $U_n$? Well, it turns out that conjecturally (and superficially at least) it doesn't look so different. I have found the following survey reference: Le and Magaard, Representations of Unitriangular Groups, in Buildings, Finite Geometries, and groups, Spirnger 2012. First about the degrees of irreducible complex representations of $U_n(\mathbb F_q)$ are exactly the $q^i$ for $i=0,1,\dots,m$, with $m = \lfloor n/2 \rfloor^2$. This was proved in 1995 by Issacs and Huppert. Note that a prori it is only obvious that the degrees are power of $p$, not of $q$ (where $p$ is the prime of which $q$ is a power), and it was not at all obvious that the power of $q$ appearing were bounded independently on $q$. But the result is here and is similar to the case of reductive group. What is striking however, is the recent date (1995) where he was proved, and just for the simplest class of unipotent group. Apparently, the similar question for general unipotent group is still open. Let us now consider the number of representations of each degree $q^i$. It seems that it is conjectured, but not known yet, that this number is a polynomial $P_{i,n}(q)$ with rational coefficients. So here again, the situation seems similar to the case of reductive groups, but much harder to prove.

So to restate my point (which is a guess and may be false), it seems that what makes the theory of representation of $U_n(\mathbb F_q)$ wild is not the nature of the results, but the difficulty of their proofs. In this sense, it is different from other classification problems, as this one : the classification of semi-simple group over $\mathbb C$ of a given dimension is tame, because there are only finitely many such groups up to isomorphism, but the same question is wild for unipotnet group, because there are continuous family of such groups (except in low dimension). I was expecting something of this nature for representation of $U_n$ vs $Gl_n$, but apparently not.

  • $\begingroup$ I think the difference between the rep. theories of $U_n$ and $\mathrm{GL}_n$ over finite fields is that the irreps. of $U_n(\mathbb{F}_q)$ are inherently more difficult to parametrise (as per Geoff Robinson's comment above). Just knowing the dimensions and multiplicities of the irreps. does not imply that we have a reasonable classification or parametrisation. $\endgroup$ Feb 1, 2014 at 15:13

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