If $\chi\in\operatorname{Irr}(L_J)$ and $^\ast R^{L_J}_T(\chi)\neq 0$, does $(\mu,R^{L_I}_{L_J}(\chi))=0$ when $^\ast R^{L_I}_T(\mu)=0$?

Question

Suppose $G$ is a finite group with a split $BN$-pair, satisfying the commutator relations, and such that representatives in $N$ of elements of the Weyl group can be chosen in a nice way: if $W\ni w=s_{i_1}\cdots s_{i_r}$, then $N\ni\dot{w}=\dot{s}_{i_1}\cdots\dot{s}_{i_r}$. (I'm essentially assuming $G$ behaves like a finite group of Lie type $G^F$.)

Let $(W,S)$ denote the corresponding Coxeter system, and $T$ be a maximal torus. I'm assuming all characters are complex characters in what follows. Suppose for $J\subseteq S$, you take an irreducible character $\chi\in\operatorname{Irr}(L_J)$ on the standard Levi $L_J$ such that the generalized restriction/Harish-Chandra restriction $^\ast R^{L_J}_T(\chi)\neq 0$. For $J\subseteq I\subseteq S$, are the irreducible constituents of the generalized induction character $R^{L_I}_{L_J}(\chi)$ again characters $\lambda\in\operatorname{Irr}(L_I)$ such that $^\ast R^{L_I}_T(\lambda)\neq 0$?

Taking $\mu\in\operatorname{Irr}(L_I)$ such that $^\ast R^{L_I}_T(\mu)=0$, I wondering if $(\mu,R^{L_I}_{L_J}(\chi))=0$, but the various formulas I know like the Mackey Intertwining Number don't seem to lend themselves well here.

Answer 1

By that adjointness we have that $(\mu,R^{L_I}_{L_J}\chi)=({}^*R^{L_I}_{L_J}\mu,\chi)$. If this value is not zero, then it means that ${}^*R^{L_I}_{L_J}\mu$ contains a non-zero multiple of $\chi$, say it is $m\chi$. Then by the transitive property we see that ${}^*R^{L_I}_{T}\mu$ contains $m({}^*R^{L_J}_{T}\chi)\neq0$, a contradiction.