I am trying to find all irreducible representations of $G = \operatorname{GL}_3(\mathbb{F}_q)$. I know that the order of $G$ is $(q^3-1)(q^3 - q)(q^3 - q^2)$ and the number of conjugacy classes is $q(q-1)(q+1)$.

My teacher suggested the following:

By Bushnell and Henniart's "The local Langlands conjecture for GL(2)", chapter 2, we know all irreducible representations of $\operatorname{GL}_2(\mathbb{F}_q)$.

Now look at $f: \operatorname{GL}_2(\mathbb{F}_q) \times \mathbb{F}_q^* \to B \subset \operatorname{GL}_3(\mathbb{F}_q)$ $$ (A, k) \mapsto \begin{pmatrix} A & 0 \\ 0 & k \\ \end{pmatrix} $$ Now for irreducible representations $\varphi$ of $\operatorname{GL}_2(\mathbb{F}_q)$ and characters $\chi$ of $\mathbb{F}_q^*$ look at $\operatorname{Ind}_P^G(\varphi \bigotimes \chi)$ where $P \subset G$ are all matrices of the form $$\begin{pmatrix} * & * & * \\ * & * & * \\ 0 & 0 & * \\ \end{pmatrix} $$ where we used the injection $B \to P$ and let $\varphi \bigotimes \chi$ work trivial on the upperright part.

Unfortunately, I do not know how to proceed from here. Does anybody know of a good source that would explain this approach (I recall my teacher using the words Levi-subgroups) or a better approach to constructing the irreducible representations of $G$?

  • $\begingroup$ Have a look here. $\endgroup$ – Dietrich Burde Jun 24 '15 at 18:26
  • $\begingroup$ That is indeed a good source, although quite different from my (suggested) approach. I will keep it as a plan B! $\endgroup$ – Krijn Jun 24 '15 at 19:48

This is the approach known as parabolic induction. It's the standard way to approach the problem of classifying the representations of (the rational points of) algebraic groups. I suspect that you aren't familiar with the theory of algebraic groups, so let me explicitly talk about general linear groups over finite fields. To simplify notation, let me fix a $q=p^n$ and write $G_n=GL_n(\mathbb{F}_q)$.

Inside $G_n$, there are natural occurances of subgroups $G_{n_1}\times\cdots\times G_{n_r}$, where $\sum_{i=1}^r n_i=n$, where the groups are the block-diagonal embeddings (and their $G_n$-conjugates, but we can just forget about this and assume that everything is in the standard block-diagonal form). These are Levi subgroups. In general, the idea of a Levi subgroup of a reductive group (something which behaves reasonably similarly to $GL_n$) is that it's a subgroup with a smaller "semisimple rank", which is somehow a measure of the expected complexity of the representation theory. So hopefully you classify the irreducible representations of these Levi subgroups -- which all external tensor products of irreducible representations of the individual $G_{n_i}$ factors -- and then try to build representations of $G_n$ from these.

There's a natural way of doing this. To a Levi subgroup $M=G_{n_1}\times\cdots\times G_{n_r}$, there is a larger parabolic subgroup $P$ which contains $M$ as a quotient in a natural way -- $P$ has a unipotent radical $N$, which is a certain normal subgroup, and $P=MN$. Then one can view an irreducible representation $\sigma$ of $M$ as an irreducible representation of $P$ by composing with the projection $P\rightarrow N$. The reason for doing this is that $P$ is now a large enough subgroup of $G_n$ that one can hope to describe the irreducible subquotients of the induced representation $Ind_P^G\ \sigma$. Explicitly in your case, the parabolic subgroup $P$ associated to $M=G_{n_1}\times\cdots\times G_{n_r}$ is the group of upper-triangular matrices obtained by "filling in" the $0$ entries above the block-diagonal of $M$. In principle, once you know the representations of $M$, you should be able to use character theory to describe the irreducible subquotients of this induced representation.

You might then hope that, having done this, you've obtained all representations of $G_n$, meaning that (after enough work) the classification boils down to understanding the representations of $\mathbb{F}_q^\times$. Unfortunately, this isn't the case. There are cuspidal representations, which are by definition those representations which aren't obtained by the process I just described. In fact, it's really the classification of the cuspidal representations which is the difficult part of the classification.

I think that Dietrich's link should be a good reference for this, although I can't say that I've ever read it. Basically any approach to the construction of the representations of $G_n$ will proceed by parabolic induction (apart from case-specific methods relying on isomorphisms, e.g. writing $GL_2(\mathbf{F}_2)\simeq S_6$ and then classifying representations in this way). You'd be best off first trying to understand the classification of representations of $GL_2$, which will be simpler as there is a single Levi subgroup (up to conjugacy) which is abelian.

  • $\begingroup$ +1; You are most likely aware: for the complex characters of general linear groups, you can construct them directly using symmetric functions (see the reference in my answer). $\endgroup$ – Stephen Jun 25 '15 at 18:05
  • $\begingroup$ Using chapter 2 of Bushnell and Henniart, I feel like I understand how they classify the representations of $GL_2$. However, if I try this approach to $GL_3$ I cannot even find the non-cuspidal representations (principal series?), because I would like to use Mackey's restriction-induction formula on $Res(Ind(\psi \bigotimes \chi )$, but Bruhat's decomposition is way more complex in $GL_3$. Can this be done easier? $\endgroup$ – Krijn Jun 29 '15 at 10:20
  • $\begingroup$ I think that a similar approach is the way to go here; it doesn't really become that much worse. Note that there are two kinds of proper Levi subgroup in $GL_3$: the torus $GL_1\times GL_1\times GL_1$, and $GL_2\times GL_1$. The Weyl group of $GL_3$ is naturally the symmetric group $S_3$ (generated by the three natural transpositions when one thinks of $S_3$ as a permutation group inside of $GL_3$), and the Bruhat decomposition is over this. It might look like this will become unpleasant as you have to consider six cases rather than two, but it's not too bad really... (continued) $\endgroup$ – PL. Jun 29 '15 at 15:52
  • $\begingroup$ ... you can try conjugating Levi subgroups by elements of this Weyl group by hand, and you'll see that things remain fairly simple. The upshot of this is that you can perform the same kind of restriction-induction analysis and make the following conclusions. Let $M$ be a Levi in $GL_3$ and let $\zeta$ be a cuspidal rep of $M$. Then: 1) if $M=GL_2\times GL_1$, then $Ind \zeta$ is irreducible; 2) if $M$ is the torus, then $Ind \zeta_1\otimes\zeta_2\otimes\zeta_3$ is irreducible, unless there are $w,w'\in W$ such that $\zeta_2=\zeta_1^w$, $\zeta_3=\zeta_1^{w'}$... (continued) $\endgroup$ – PL. Jun 29 '15 at 15:57
  • $\begingroup$ ... and 3) two cuspidal data $(M_1,\zeta_1)$ and $(M_2,\zeta_2)$ are isomorphic if and only if $M_1$ is conjugate to $M_2$ and the tensor factors of $\zeta_1$ and $\zeta_2$ are isomorphic, up to conjugacy within the Weyl group. $\endgroup$ – PL. Jun 29 '15 at 15:58

If you are willing to settle for the calculation of the characters of the irreducible complex representations of $\mathrm{GL}_n(F_q)$, then this is written down in section 6. of Chapter IV of Macdonald's book Symmetric functions and Hall polynomials, in an exposition parallel to (but more complicated than) the calculation of the characters of the symmetric groups.


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