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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.

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Consider the (identity component of the) centralizer of the kernel of the simple root corresponding to that node (viewed as a character of the torus), and its derived group and maximal central torus. That does the job, except for the caveat that the direct product feels too strong (just "almost direct product"). Indeed, only in the adjoint case do the simple roots give a basis of the character group of the chosen maximal torus. –  BCnrd May 20 '10 at 15:24
    
To amplify BCnrd's comment, it's helpful to start with the smallest example SL`$_3(k)$` and look at a Levi subgroup of type GL`$_2(k)$` which has only an almost direct product decomposition of the type you want. (Textbooks called Linear Algebraic Groups may be helpful sources for structure of parabolic subgroups relative to Dynkin diagrams.) –  Jim Humphreys May 20 '10 at 15:35
    
I'm a little confused by this 'almost direct product' comment. If I'm working with $SL_3$, can't I just take diag($t^{-2}$, $tM$) where $M$ is in $SL_2$ and $t$ is in $\mathbb{G}_M$? This subgroup seems to do everything I asked for, and it really is $\mathbb{G}_M\times SL_2$. Or am I missing something? Sorry to be thick! –  blt May 20 '10 at 18:45
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The difference is that the Levi subgroup referred to is determined by G,B,T, and the choice of a subset of the Dynkin diagram. The Levi subgroup is often an "almost direct product". The subgroup you refer to is contained in the Levi subgroup $GL_2$, but it's not "natural", and there are many possible choices. I think one can always embed an honest direct product -- like you want -- into the natural almost direct product. It's an exercise fiddling with isogenies and algebraic tori. –  Marty May 20 '10 at 19:58
    
Ah! I see. Thanks :) –  blt May 20 '10 at 20:53

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up vote 4 down vote accepted

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 r\'eductifs," Expos\'e XXVI of SGA3 (How does one do French accents here properly?). I believe that Borel and Tits earlier work "Groupes R\´eductifs" 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.

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You'll find at tea.mathoverflow.net/discussion/221/… a discussion about a couple of ways to get diacritics. –  Mariano Suárez-Alvarez May 20 '10 at 17:40

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