Fix $\mathbb{C}$ as the base field, and reductive groups are assumed to be connected.

Consider the example $SO_N\subset SL_N$. $SO_N$ is its own normalizer in $SL_N$, and the rank is much smaller than the rank of $SL_N$. How can this be reflected in the Dynkin diagram of $SL_N$?

In Reductive subgroup corresponding to a subdiagram of the Dynkin diagram of a simple group one sees that Levi subgroups $L$ of a given semi-siimple group $G$ can be recovered from Dynkin subdiagrams of the diagram of $G$, and subgroups of maximal rank can be recovered similarly from extended Dynkin diagrams. As is commented below, having maximal rank means being of the same rank as $G$ is, namely containing a maximal torus of $G$.

Contrary to the discussions in loc.cit, one considers a reductive subgroup $H\subset G$, such that the normalizer $N(H,G)$ of $H$ in $G$ is of lower rank than $G$. Its connected component, denoted as $N^\circ$, is not covered in loc.cit. Say $SO_{2n}\subset SL_{2n}$, the extended Dynkin diagram of $SL_{2n}$ looks like a loop with dots, while the one for $SO_{2n}$ contains branching vertices. It is not clear that the latter is produced from the former by removing vertices.

And furthermore when one passes to a general base field, say perfect or of characteristic zero for simplicity, with separable closure $\bar{k}$, it becomes more complicated if one restricts to the notion of $k$-rank. Consider a reductive $k$-subgroup $H\subset G$ that is self-normalizing, in the sense that it equals the neutral connected component of $N(H,G)$. By comparing the $k$-ranks and $\bar{k}$-ranks of $H$ and $G$, one is led to several different cases. Does one still have arguments similar to the operations on Dynkin diagrams as is in Reductive subgroup corresponding to a subdiagram of the Dynkin diagram of a simple group ?

Sorry for any misunderstanding about the cited mathoverflow discussions above, and references on extended Dynkin diagrams are also welcome.

`$N$`

. – Jim Humphreys Oct 1 '10 at 17:00