Context: Let $\pi: \widehat{G} \rightarrow G$ be a surjective morphism between connected reductive groups defined over $\mathbb{F}_q$ whose kernel is a central torus. Then $\pi : \widehat{G}^F \rightarrow G^F$ is surjective, where $F$ denotes the Frobenius morphisms inducing the $\mathbb{F}_q$-rational structures. We thus have an inflation map $- \circ \pi: \text{Irr}(G^F) \rightarrow \text{Irr}(\widehat{G}^F)$. We also have a dual morphism $\pi^* : G^* \rightarrow \widehat{G}^*$ which is injective under the above assumptions.

Question: Is the inflation map compatible with Lusztig series in the sense that for a semisimple $s \in (G^*)^F$ and any $\chi$ in the Lusztig series $\mathcal{E}(G^F,[s])$ associated to $s$ we have $\chi \circ \pi \in \mathcal{E}(\widehat{G}^F,[\pi^*(s)])$?

Motivation: It is easy to see that every element of $\mathcal{E}(\widehat{G}^F,[\pi^*(s)])$ is inflated from $G^F$, so if the answer to my question is positve, this would result in a bijection between $\mathcal{E}(G^F,[s])$ and $\mathcal{E}(\widehat{G}^F,[\pi^*(s)])$.

My Approach: It seems natural to me to try to show that inflation is compatible with Lusztig induction, i.e. $(R^G_T \theta) \circ \pi = R^{\widehat{G}}_{\pi^{-1}(T)} (\theta \circ \pi)$ for any rational maximal torus $T$ and any $\theta \in \text{Irr}(T^F)$. To show this one could try to compare the Deligne-Lusztig varieties $Y_U = \{g \in G \:|\: g^{-1}F(g) \in U\}$ and $Y_{\pi^{-1}(U)}$ for $U$ the unipotent radical of a Borel subgroup of $G$ containing $T$. This looks promising to me as $\pi$ induces a surjective morphism $Y_{\pi^{-1}(U)} \rightarrow Y_{U}$ whose fibres are cosets of $\ker(\pi)^F$. As I see it it would suffice to show that this morphism gives an isomorphism between the $\ker(\pi)^F$-quotient of $Y_{\pi^{-1}(U)}$ and $Y_{U}$ but I am unable to prove it.

I tried to mimic the proof of Proposition 13.20 in Representations of Finite Groups of Lie Type by Digne and Michel where the case $s = 1$ is studied for a more general $\pi$. But unfortunately and contrary to what is claimed there the proof given there does not establish an isomorphism like the above as far as I can tell. Any help is appreciated.

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    $\begingroup$ What you're trying to show is Proposition 13.22 in Digne-Michel. The statement is proved for Deligne-Lusztig induction from any Levi subgroup. In the statement they (accidentally) assume that the parabolic is F-stable but this assumption is not used in the proof. $\endgroup$ – Jay Taylor Mar 2 '15 at 11:14

Your approach is correct and is proven in the book by Digne-Michel (in fact a more general statement is proven there). Indeed, by Proposition 13.22 in Digne-Michel we know that

$$R_{T\subseteq B}^G(\theta)\circ \pi = R_{\widehat{T}\subseteq \widehat{B}}^{\widehat{G}}(\theta\circ\pi)$$

where $\widehat{T} = \pi^{-1}(T)$ and $\widehat{B} = \pi^{-1}(B)$. Now you just need to show that the conjugacy class of $\theta\circ\pi$ corresponds to the conjugacy class of $\pi^*(s)$ under duality.

By the definition of duality we have an isomorphism $\lambda : X(T) \to Y(T^*)$ between the character group of $T$ and the cocharacter group of $T^* \leqslant G^*$, which is a dual torus. Furthermore, $\lambda$ is compatible with the Frobenius endomorphism.

According to Proposition 13.7 in Digne-Michel there exists a character $\chi \in X(T)$ of $T$ such that $\theta = \chi|_{T^F}$, i.e., $\theta$ is the restriction of $\chi$. Now let $\gamma = \lambda(\chi)$ be the cocharacter corresponding to $\chi$. If we assume that $n \geqslant 1$ is large enough so that $T$ is a split torus with respect to the Frobenius endomorphism $F^n$ then we have $s = N_{F^n/F}(\gamma(\zeta))$ where $\zeta$ is a $(q^n-1)$th root of unity and $N_{F^n/F} : T^* \to T^*$ is the norm map.

Now let us assume that we have chosen an isomorphism $\widehat{\lambda} : X(\widehat{T}) \to Y(\widehat{T}^*)$ extending $\lambda$. In other words, we have $\widehat{\lambda}(\chi\circ\pi) = \pi^*\circ\lambda(\chi)$. Now, clearly $\theta\circ\pi = (\chi\circ\pi)|_{T^F}$ and $\widehat{\lambda}(\chi\circ\pi) = \pi^*\circ\lambda(\chi) = \pi^*\circ\gamma$. As $\pi^*$ clearly commutes with the norm map this gives us that $\theta\circ\pi$ corresponds to $\pi^*(s) = N_{F^n/F}(\pi^*(\gamma(\zeta)))$. Hence

$$R_{T^*}^G(s)\circ \pi = R_{\widehat{T}^*}^{\widehat{G}}(\pi^*(s))$$,

which is what you want.

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  • $\begingroup$ Thank you very much for your answer. I believe it was really the identity $(R^G_T \theta) \circ \pi = R^{\widehat{G}}_{\pi^{-1}(T)}(\theta \circ \pi) $ that I had looked for, the rest was more or less clear to me. I think it is strange that the case $\theta = 1$ in 13.20 is proved that easily on the geometric level, whereas in the general case one has to do actual calculations (even though they will simplify in my case) or is it just as I feared that the proof of 13.20 is incomplete? $\endgroup$ – Matthias Klupsch Mar 2 '15 at 12:39
  • $\begingroup$ Well, the thing that is easier in the $\theta=1$ case is that you can just look at the cohomology of the quotient variety $\mathcal{L}^{-1}(U)/T^F$. This will be the same as taking the cohomology of $\mathcal{L}^{-1}(U)$ and tensoring with the trivial character. However, for arbitrary $\theta$ one cannot simply replace the tensor product with the cohomology of some variety. This is what makes it more complicated. $\endgroup$ – Jay Taylor Mar 2 '15 at 13:30
  • $\begingroup$ Also the proof of 13.20 looks fine to me. I think the point is that they're implicitly using Proposition 10.10. $\endgroup$ – Jay Taylor Mar 2 '15 at 13:51
  • $\begingroup$ The main issue I am having with the proof of 13.20 is that the morphism from $\mathcal{L}^{-1}(U)/T^F$ to $\mathcal{L}^{-1}(U_1)/T_1^F$ (in the notation used there) is only proven to be a bijection and not an isomorphism. I my situation I can prove that I have a bijective morphism $\mathcal{L}^{-1}(\pi^{-1}(U))/\ker(\pi)^F \rightarrow \mathcal{L}^{-1}(U)$ in a way absolutely analogously to what in the proof of 13.20 is called the construction of an inverse morphism. It is not clear to me why the inverse should be a morphism of varieties in any of the two cases. $\endgroup$ – Matthias Klupsch Mar 2 '15 at 14:19
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    $\begingroup$ Sorry, just one final comment. The fact that the map is a bijective morphism of varieties is enough (you don't need it to be an isomorphism). Indeed Proposition 10.12(ii) in DM ensures that the Leftschetz traces of the quotients will coincide. Sorry this took me a while to remember this fact. $\endgroup$ – Jay Taylor Mar 3 '15 at 16:53

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