Suppose I have a simple, simply connected (linear) algebraic group $\mathcal{G}$ over an algebraically-closed field $k$, which could have any characteristic. In fact, to keep things simple, let's imagine $\mathcal{G}$ is simply-laced. Let $\mathcal{B}$ be a choice of Borel, and $\mathcal{T}$ be a maximal torus for $\mathcal{B}$ (and hence for $\mathcal{G}$). Our $\mathcal{B}$ gives a choice of simple positive roots. I'll write $\mathbb{G}\_M$ for the multiplicative group of $k$.


Now, suppose we look at the Dynkin diagram for $\mathcal{G}$, and imagine removing a single node from the diagram; for simplicity let's imagine this doesn't disconnect the diagram. Then I believe we should be able to find a subgroup $\mathbb{G}\_M\times\mathcal{G}'$ in $\mathcal{G}$ and a Borel subgroup $\mathcal{B}'$ in $\mathcal{G}'$, such that the inclusion $\mathbb{G}\_M\times\mathcal{B}'\hookrightarrow \mathcal{G}$ factors through $\mathcal{B}$; such that the Dynkin diagram for $\mathcal{G}'$ is just the Dynkin diagram for $\mathcal{G}$ with our chosen node removed; and indeed such that when we use the inclusion $\mathcal{G'}\hookrightarrow\mathcal{G}$ to map the simple roots of $\mathcal{G}'$ determined by $\mathcal{B'}$ into roots of $\mathcal{G}$, and hence map the Dynkin diagram of $\mathcal{G}'$ into the Dynkin diagram of $\mathcal{G}$, we get the obvious inclusion of 'the diagram with the node removed' into 'the original diagram'.


Moreover, if instead I did disconnect the diagram, I should get a similar picture, with a subgroup $\mathbb{G}\_M\times\mathcal{G}\_1\times\dots\times\mathcal{G}\_k$ where the groups $\mathcal{G}\_k$ have Dynkin diagrams corresponding to the components of the picture I get by removing the node from the original Dynkin diagram.


My question is: a) am I right that this kind of operation is all legal and above board, b) are there any caveats one should be aware of, and c) what are good references to appeal to to make this kind of thing rigorous.



Chopping up Dynkin diagrams
Brian's comment does what you want, and describes the *almost* direct product caveat.


A standard and excellent reference for all things of this nature is Demazure's "Sous-groupes Paraboliques des groupes réductifs," Exposè XXVI of SGA3. I believe that Borel and Tits earlier work "Groupes Réductifs" has similar results covered, in the case of a base field instead of base scheme. Maybe this would be better for your question.


There's no need for the assumptions simple, simply-connected, simply-laced, nor the assumption that you're working over an algebraically closed field. The general setting is a reductive group $G$ over a base scheme $S$.


For a (connected) reductive group $G$ over an algebraically closed field, once you fix $T \subset B \subset G$ as you wish, there's a bijection between subsets of the vertex-set of the Dynkin diagram and parabolic subgroups $P$ of $G$ containing $B$ (so-called standard parabolics). Each parabolic subgroup has a standard Levi decomposition $P = LU$, where $U$ is the unipotent radical of $P$, and $L$ is a reductive group containing $T$.


When $G$ is a simply-connected group, the group $L$ has simply-connected semisimple derived subgroup $L'$. The Dynkin diagram of $L'$ is the subdiagram induced by subset of the vertex-set of the Dynkin diagram of $G$. 


Getting the isogeny class right -- e.g. identifying a precise surjective map from $G\_m^r \times L' \rightarrow L$ with finite kernel -- is kind of a pain, in my experience. Typically, I just "know the answer" for classical groups (e.g., Jim's example, in $SL\_3$, one finds a Levi $L \cong GL\_2$) and for certain exceptional groups. Occasionally, I have to do the tedious work of determining the root datum (in the sense of Borel) of the Levi to figure it out (or more commonly looking through the literature to find someone else who's done it). Oh how I wish that this were implemented in SAGE. 


