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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.

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It isn't really a question of relating Dynkin diagrams (whether over an algebraically closed or smaller field), since this subgroup of the special linear group isn't of maximal rank and since its intrinsic root system isn't so closely related to the bigger one. There may be problems such as branching rules of representations where it's useful to compare the group and subgroup here. But a vast number of smaller reductive groups can live in a given special linear group of large rank. In any case, your rank description is off a bit for even $N$. –  Jim Humphreys Oct 1 '10 at 17:00
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P.S. Perhaps you are thinking about the technique of "folding" a Dynkin diagram which plays a role in the work of Tits and others on forms of algebraic groups over various fields. (A basic article by Tits is freely available from AMS online in the old Proc. Symp. Pure Math. 9 at e-math.org.) But this seems far away from the question you ask in your next-to-last paragraph, so clarification would help. In any case, a subgroup of maximal rank has the same rank as the big group (thus shares a maximal torus). –  Jim Humphreys Oct 1 '10 at 17:26
    
@Jim:thanks a lot for the comments and the reference! I modify a little bit the statements above, and is starting reading the article by Dynkin mentioned below. –  genshin Oct 2 '10 at 14:08

1 Answer 1

Not easy!. If you want to know classification of $A_1$-subalgebras then it is the same as classification of nilpotent elements and it can be reduced to classifying all distinguished parabolics in all your Levi subalgebras.

For bigger subalgebras, you should read original Dynkin's paper that contains more or less complete answer if I understand things correctly...

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