Pardon if this is well known. Suppose I have a (say complex) connected reductive group $G^{\vee}$ with the $\tilde{\Delta}=\Delta\cup\{\alpha_0\}$ being the simple roots plus the negative highest root of $G^{\vee}$. Any proper subset of $\tilde{\Delta}$ defines a connected reductive subgroup $H^{\vee}\subset G^{\vee}$. Let $H$ and $G$ be the corresponding dual groups. For any special nilpotent orbit $u_H$ in $H$ we can define the Lusztig-Spaltenstein induction $u$ as a nilpotent orbit in $G$.

(that is, $(u_H,triv)$ corresponds to some representation of $W_H$ by Springer correspondence, we do the $j$-induction on this to get a representation of $W$, which correspond to $(u,triv)$ for some $u$)

Question: Is it true that the codimension of $u$ in the nilpotent cone of $G$ is the same is the codimension of $u_H$ in the nilpotent cone of $H$?

Thanks!

Edited: Just to say what I understand / guess so far: For any $u_H$, say the Springer fiber $\mathcal{B}_{e_H}$ at $u_H$ has dimension $e_H$. Then the corresponding Weyl group representation appears as the component of $H^{2e_H}(\mathcal{B}_{e_H})$ on which $A_{e_H}$ acts trivially. This component receives a surjection from $H^{2e_H}(H/B_H)$, which is a quotient of $\text{Sym}^{2e_H}(std_H)$, where $std_H$ is the standard representation of $W_H$. The Lusztig-Spaltenstein $j$-induction then, I think (I cannot find a reference for why this works for $H^{2e_H}(H/B_H)$ instead of $\text{Sym}^{2e_H}(std_H)$), gives a $W$-representation in $H^{2e_H}(G/B)$, which then corresponds to some orbit in $G$ whose Springer fiber has dimension $e_H$.