8
$\begingroup$

I have a naive question on complex representations of finite groups of Lie type. Let $\bf G$ be a reductive group (say connected, with connected center, for safety) defined over a finite field $\mathbb F_q$, and let $G={\bf G}(\mathbb F_q)$. Assume first that $G$ is split over $\mathbb F_q$, so there is a maximal torus $\bf T$ in $\bf G$ defined and split over $\mathbb F_q$. Let $g \in T = {\bf T}(\mathbb F_q)$ be a regular element.

Then is it true that if $\pi$ is an irreducible complex representation of $G$, of character $\chi = tr\ \pi$, then $\chi(g) \neq 0$ implies $\pi$ is a principal series ?

By a principal series, one means of course a sub representation of $Ind_B^G \psi$, where $B$ is a Borel containing $T$ and $\psi: T \rightarrow \mathbb C^\ast$ a one-dimensional character.

A glance at the table of characters found for example in Fulton-Harris shows that this is true for $GL_2$, so my intuition tells me that it should be true, and probably easy, in general, but the argument escapes me and I haven't seen this statement in the literature (e.g. in Curtis' survey or Digne and Michel's book).

Also if the answer is yes, then can one remove the assumption that $G$ is split (that is we work with a general $\bf G$ as above, and take for $\bf T$ a quasi-split group, and ask the same question) ?

$\endgroup$
3
  • 3
    $\begingroup$ Dear Joel, It looks like you might have good answers below, but just in case you want another approach: my memory is that if $R(\theta)$ is a DL rep'n attached to some character $\theta$ of a non-split torus $T$, then while $R(\theta)$ can be hard to desribe explicitly, we can describe its tensor product with the Steinberg in a simpler way: e.g. for $GL_2$, there is a formula something like $ st \otimes R(\theta) = Ind_{T}^{GL_2} \theta$. Since passing to characters converts tensor products to just pointwise products of characters, it should now be easy to use this to compute the char. of ... $\endgroup$
    – Emerton
    Commented Jun 6, 2013 at 22:51
  • 2
    $\begingroup$ $R(\theta)$ on a regular split $g$, because Steinberg and $\Ind_T^{GL_2} \theta$ are pretty concrete things. (The basic point should be that the trace of an induction is just the number of fixed points in the corresponding permutation representation.) I forget the analogues of this formula for more general groups $G$, but my memory is that they are in the DL paper. Cheers, Matt $\endgroup$
    – Emerton
    Commented Jun 6, 2013 at 22:53
  • $\begingroup$ Thanks to all three of you, Matt, Jim and Jay. After a crash self-course on Deligne-Luztig theory (how beautiful!), I now understand your arguments, which are basically the same. $\endgroup$
    – Joël
    Commented Jun 7, 2013 at 14:37

2 Answers 2

6
$\begingroup$

So, I think the answer to your question is yes. This may not be the slickest proof but I think it works. Firstly let $\mathrm{pr}_G$ be the projection map from the space of all class functions to the subspace of uniform functions, i.e. the subspace spanned by Deligne-Lusztig virtual characters, (see Digne and Michel - 12.11 and 12.12). Now for a pair $(\mathbf{S},\theta)$ consisting of an $F$-stable maximal torus and character $\theta$ of $\mathbf{S}^F$ let $c_{\mathbf{S},\theta} \in \mathbb{C}$ be such that

$$\mathrm{pr}_G(\chi) = \sum_{}c_{\mathbf{S},\theta}R_{\mathbf{S}}^{\mathbf{G}}(\theta)$$

where this sum is taken up to conjugacy. The characteristic function of a semisimple element is a uniform class function, (see DM - Proposition 12.20), in particular we have $(\mathrm{pr}_G(\chi))(g) = \mathrm{pr}_G(\chi(g))$. Hence if $\chi(g)$ is non-zero this implies that

$$\mathrm{pr}_G(\chi)(g) = \sum_{}c_{\mathbf{S},\theta}R_{\mathbf{S}}^{\mathbf{G}}(\theta)(g) \neq 0$$

However $R_{\mathbf{S}}^{\mathbf{G}}(\theta)(g) \neq 0$ implies that the conjugacy class of $g$ intersected with $\mathbf{S}^F$ is non-empty (see Carter - Proposition 7.5.3). As your element is regular this implies that $\mathbf{S}$ is rationally conjugate to your chosen torus $\mathbf{T}$. Now $c_{\mathbf{S},\theta} \neq 0$ implies that $\chi$ occurs in $R_{\mathbf{S}}^{\mathbf{G}}(\theta)$ with non-zero multiplicity, hence in $R_{\mathbf{T}}^{\mathbf{G}}(\theta')$ with non-zero multiplicity where $(\mathbf{T},\theta')$ is any pair rationally conjugate to $(\mathbf{S},\theta)$. However this says that $\chi$ is in the principal series as $R_{\mathbf{T}}^{\mathbf{G}}(\theta')$ is simply the Harish-Chandra induction of $\theta'$.

Note that this argument does not require that $F$ is split.

$\endgroup$
0
2
$\begingroup$

[REVISION]: After looking closer at the literature related to Deligne-Lusztig's Annals of Math. 103 (1976) paper, I'm more persuaded that the answer to your question is yes. In their paper, see in particular the formula (7.6.2) describing the value of an arbitrary irreducible character at a regular semisimple element. This applies in particular to such an element lying in the split maximal torus, but is more general.

I got misled at first while trying to understand some of the earlier published examples in rank 2; there it's tricky to work out which of the characters are in the principal series. The special cases are well worth studying, but preferably translated into the Deligne-Lusztig framework (as presented in the large book by Carter and the smaller one by Digne-Michel). As I mentioned in my earlier attempt at an answer, the character table of $\mathrm{Sp}_4(\mathbb{F}_q)$ was worked out by Bhama Srinivasan in Trans. Amer. Math. Soc. 131 (1968), available online here. In her notation, the regular semisimple elements in a split maximal torus belong to the classes labelled $B_3(i,j)$. (There are a handful of minor errors in the character table, mostly related to parameters. But note that the degree of $\theta_1$ is $\frac{1}{2} q^2 (1+q^2)$. It does lie in the principal series.) The characters of finite groups of type $G_2$ were worked out in a 1974 conference paper by Chang and Ree. .

Another related result based on Deligne-Lusztig theory which I quoted earlier only shows why your question "almost" has an affirmative answer: .see Prop. 7.5.3 in Roger Carter's 1985 book Finite Groups of Lie Type. (Srinivasan's special case predates the 1976 paper but along with Green's work on finite general linear groups helped to motivate the later developments.) In any case, I hope my reference above to the D-L paper is more helpful.

$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .