TITLE: Dynkin diagram of the centralizer of a semisimple element in a Levi subgroup
QUESTION [6 upvotes]: Let $G$ be a connected reductive group over an algebraically closed field and consider a semisimple element $s \in G$ and let $L$ be a Levi subgroup containing $s$. 
My question is about the two ways we can look at $C_L(s)^\circ$. On the one hand, it is the connected centralizer of $s$ in the Levi subgroup $L$, on the other hand, it is a Levi subgroup in the connected centralizer $C_G(s)^\circ$. 
The Dynkin diagram of $C_{G}(s)^\circ$ can be obtained from the extended Dynkin diagram of $G$ by removing nodes. By removing appropriate nodes from the Dynkin diagram of $C_{G}(s)^\circ$, we obtain the Dynkin diagram of the Levi subgroup $C_{L}(s)^\circ$ of $C_{G}(s)^\circ$.
On the other hand, the Dynkin diagram of $C_L(s)^\circ$ can be obtained from the extended Dynkin diagram of $L$ and the Dynkin diagram of $L$ can be obtained by removing nodes from the Dynkin diagram of $G$.
For example let $G$ have type $B_{10}$ and choose $s$ such that $C_{G}(s)^\circ$ has type $D_4 \times B_6$, then I can find a Levi subgroup of $C_{G}(s)^\circ$ of type $D_4 \times A_2 \times B_3$. This Levi is of the form $C_L(s)^\circ$ for some Levi $L$ of $G$. 
However, even though there exists a Levi subgroup of $G$ of type $A_2 \times B_7$ and one of its subgroups of maximal rank has the desired type $A_2 \times D_4 \times B_3$, the bases of the corresponding root systems seem to be completely different to me. In fact, I am not even sure if that subgroup of type $A_2 \times D_4 \times B_3$ lies in $C_G(s)$ or not.
Also, it seems to me that the other Levi subgroups of $G$ do not even have a subgroup of the desired type.
Did I do anything wrong here? If not, why is it not possible for me to obtain the same basis of the Levi of $C_{G}(s)^\circ$ when considering it as a maximal rank subgroup of a Levi subgroup of $G$?

REPLY [3 votes]: In the Bourbaki numbering ($\alpha_i = \epsilon_i - \epsilon_{i + 1}$ for $i < 10$ and $\alpha_{10} = \epsilon_{10}$), I believe that you can take $s = \alpha_1^\vee(-1)\alpha_3^\vee(-1)$ (with connected centraliser of type $D_4 \times B_6$, where the base for the $D_4$ piece is $\{\alpha_3, \alpha_2, \alpha_1, -\mu\}$ and that for the $B_6$ piece is $\{\alpha_5, \dotsc, \alpha_{10}\}$) and $L$ to be the centraliser of the image of $\alpha_5^\vee + 2\alpha_6^\vee + 3(\alpha_7^\vee + \dotsb + \alpha_{10}^\vee)$, which, as you predicted, is of type $A_2 \times B_7$.  The base for the $A_2$ piece is $\{\alpha_5, \alpha_6\}$, and that for the $B_7$ piece is $\{\alpha_1, \alpha_2, \alpha_3, \alpha_4 + \dotsb + \alpha_7, \alpha_8, \alpha_9, \alpha_{10}\}$.
I found Carter's paper Conjugacy classes in the Weyl group (MR) very useful for understanding this kind of calculation with what he calls admissible diagrams (Section 4 in his paper).  Section 4 in my paper On counting orbits in root systems gives some amateurish examples.