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.
Best Answer
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.