Intuitive reason that the regular representation is a uniform function

Question

Corollary 12.14 of Digne-Michel's book Representations of finite groups of Lie type gives various decompositions of the regular representation $\operatorname{reg}_G$ in terms of the Deligne-Lusztig characters. The two I am primarily interested in are $$\operatorname{reg}_G = \frac{1}{|G^F|_p} \sum_{T \in \mathcal{T}} \epsilon_G \epsilon_T R^G_T(\operatorname{reg}_T) = \frac{1}{|G^F|_p} \sum_{\substack{T \in \mathcal{T}\\ \theta \in \operatorname{Irr}(T^F)}} \epsilon_G \epsilon_T R^G_T(\theta).$$

My question is: can someone give an intuitive reason for why these decompositions exist? I am not asking for a proof, but for a moral reason that this should be true.

For context: the Deligne-Lusztig characters form a core part of my thesis on the representation theory of finite groups of Lie type, but I do not have the space to completely develop all of the results. I have therefore chosen a few proofs which I feel to be illustrative of the techniques in the field, and for the rest I am trying to offer an intuitive explanation. If I had to guess a decomposition of the regular representation then this would certainly be it (possibly modulo the signs), but I am having difficulty coming up with a way to justify this to readers.

An example of the type of argument I am looking for: the Mackey decomposition for Deligne-Lusztig induction/restriction can be interpreted as `pushing forward' the Bruhat decomposition of a finite group of Lie type onto its representations. Indeed, you can make this argument rigorous (at least on the level of characters, I'm not focusing on the actual representations) by looking at the action of $G^F$ on a certain finite affine variety related to the double cosets, then using properties of the Lefschetz number to make combinatorial simplifications, so to me this gives a good intuition for the Mackey decomposition.

Answer 1

This is an interesting question, but the kind of geometric or structural intuition your are looking for may not exist. To put it another way, the reason behind the fact in the OP is a non-trivial combination of several other central facts, but can't conceptually be reduced to either of them.

If we want to guess that some virtual character such as $$ \frac{1}{\lvert G^F\rvert_p} \sum_{T, \theta} \epsilon_G \epsilon_T R^G_T(\theta)$$ equals the regular character $\operatorname{reg}_G$, then, as a first step, we had better make sure that they have the same degree. That this is true is non-trivial and the proof actually uses some of the same steps as the proof that the two characters are equal. More precisely, to compute the degree of the above character, one has to, as far as we know, at some point use the fact that the Steinberg character $\mathrm{St}$ is zero on all non-semisimple elements. This is based on properties of Bruhat decomposition/BN-pairs and Curtis's alternating sum formula for $\mathrm{St}$.

The above fact about the values of $\mathrm{St}$ together with the fact that values of $\sum_{\theta} R^G_T(\theta)$ are Lefschetz numbers and hence zero on non-trivial semisimple elements, gives an easy expression for the inner product of $\mathrm{St}$ and $\sum_{\theta} R^G_T(\theta)$. The alternating sum formula for $\mathrm{St}$ together with the fact that $\mathrm{St}(1)=|G^F|_p$, can then be used to show that $$\sum_{\theta} \epsilon_G \epsilon_T R^G_T(\theta)(1) = \frac{|G^F|}{\lvert G^F\rvert_p}.$$ Finally, a non-trivial result of Steinberg says that there are $|G^F|_p^2$ $F$-stable maximal tori, so $\sum_{T,\theta} \epsilon_G \epsilon_T R^G_T(\theta)$ indeed has degree $|G^F|$.

One might ask whether lifting the question to isomorphism of representations or some other categorified objects would make the fact in the OP more 'geometric'. It is not clear to me that this can be done and the fact about the values of the character $\mathrm{St}$ highlighted above is used in all of the proofs I know of the statement in the OP. Note that the proof of 12.14 in Digne--Michel uses this fact (when it refers to 9.4), so in particular I am not aware of any character-free proof (as in, does not use anything about character values).

In summary, I think the result in the OP can be made plausible by combining the observation in LSpice's answer with a summary of why the two characters have the same degree (e.g., as given above), but this doesn't reduce the result to any straightforward conceptual or geometric principle. Instead, like many non-trivial proofs, it is a combination of several main ingredients. In this case the main ingredients are the existence and properties of the Steinberg character (the alternating sum formula, character values), Bruhat decomposition, the character formula for Deligne--Lusztig characters, etc.

Answer 2

I think the two equalities are the same, since $\operatorname{reg}_T$ equals $\sum_\theta \theta$; so I'll focus on the equality $$\operatorname{reg}_G = \frac{1}{\lvert G^F\rvert_p} \sum_{T, \theta} \epsilon_G \epsilon_T R^G_T(\theta).$$ Let's call the virtual character on the right $\rho_G$.

I don't know if this counts as intuition or only as a crude proof sketch, but, at least for me, the fundamental fact in harmonic analysis is that the integral of a non-trivial character of a group is $0$. From this point of view, when I evaluate $\rho_G$ at an element $g = s u$, I get (up to constants) $$ \sum_{T, \theta} \epsilon_G\epsilon_T\sum_{g s g^{-1} \in T} \theta(g s g^{-1})Q_{g^{-1}T g}^{G_s}(u) = \sum_{g \in G} \sum_{T \subseteq g G_s g^{-1}} \epsilon_G\epsilon_T Q_{g^{-1}T g}^{G_s}(u)\sum_\theta \theta(g s g^{-1}) = 0 = \operatorname{reg}_G(s u) $$ unless $s = 1$. That, of course, doesn't tell you what happens on the unipotent set—and, even if you knew that you also had vanishing on (the complement of $1$ in) the unipotent set, it wouldn't tell you which multiple of the regular representation you've recovered—but at least this makes it plausible that some equality like the one you propose holds.

(If that's too much computation, maybe "Fubini + Abelian harmonic analysis" is a pithier way of saying it.)

Answer 3

I think the argument of this formula in Digne--Michel's book can be viewed as a process unpacking properties (of duality functors and Deligne--Lusztig inductions) around the probably more fundamental formula $$(*)\quad \mathrm{St}_G=\sum_{P_I} (-1)^{|I|}\mathrm{Ind}_{P_I^F}^{G^F}1,$$ where $\mathrm{St}_G$ is the Steinberg character. (Note that, Digne--Michel's book uses this formula to define the Steinberg character, however, this is not how initially it was defined; more details on the history can be found in this survery of Humphreys.) Then an important middle step from $(*)$ to the regular char formula is to realise that (see 12.13, and 12.8, in Digne--Michel) $$\sum_{P_I} (-1)^{|I|}\mathrm{Ind}_{P_I^F}^{G^F}1=\frac{1}{|W|}\sum_T\epsilon_T\epsilon_GR_T^G(1).$$ In the below, instead of the original formula, I will focus on $(*)$ which admits a very nice geometric explanation:

Every finite group of Lie type is equipped with a geometric/combinatorial construction called spherical building, which admits a simplicial complex structure, hence a chain complex $$C_n\rightarrow C_{n-1}\rightarrow...\rightarrow C_0;$$ for example, in the case of type $\mathsf{A}$, the highest simplexes correspond to the complete flags and the lower simplexes correspond to partial flags. The original finite group of Lie type then acts on the cycles/boundaries/homologies etc. Now the Hopf trace formula tells you that one can use the alternating sum of the characters of the group actions on $C_r$ (which turns out to be the parabolic induction from the trivial character) to obtain the alternating sum of the characters on the homology groups.

On the other hand, the building is (topologically) a bouquet of equidimensional spheres; in particular, its integral homology is zero except the $0$-th-degree and the highest degree, and the highest degree part turns out to be $\mathrm{St}_G$ by the Hopf formula. This gives a geometric explanation of the formula $(*)$!

To track the details of this story, you may search the names Curtis and Solomon--Tits. A very nice book containing all these stuffs is Ronan's Lectures on buildings

Remark. Another fantastic aspect of the above picture is it suggests that the sign character of the Weyl group should be viewed as an analogue of the Steinberg character. Indeed, there is a subcomplex of the building called a Coxeter complex, and doing the above process for this subcomplex gives a similar formula for the sign character.