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 $G$?



Dynkin diagram of the centralizer of a semisimple element in a Levi subgroup
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](http://www.numdam.org/item/CM_1972__25_1_1_0) ([MR](http://www.ams.org/mathscinet-getitem?mr=318337)) 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](http://lanl.arxiv.org/abs/1308.4056) gives some amateurish examples.


