Reductive subgroup corresponding to a subdiagram of the Dynkin diagram of a simple group - MathOverflow

Reductive subgroup corresponding to a subdiagram of the Dynkin diagram of a simple group

2010-03-19T15:52:20Z

I am looking for a reference for the following well-known fact: For any subdiagram $\Delta_0$ of of the Dynkin diagram $\Delta=D(G)$ of an adjoint simple group $G$ over an algebraically closed field $k$, there exists a reductive subgroup of maximal rank $G_0\subset G$ with Dynkin diagram $\Delta_0$.

To be more precise, I am looking for a reference for a proof of the following well-known lemma:

Lemma 1. Let $G$ be an adjoint, connected, simple algebraic group with Dynkin diagram $\Delta=D(G)$ over an algebraically closed field $k$ of any characteristic. Let $\Delta_0$ be a subdiagram of $\Delta$ (that is, a subset $\Pi_0$ of the set $\Pi$ of vertices of $\Delta$, together with all the edges of $\Delta$ connecting pairs of vertices of $\Pi_0$). Then there exists a connected reductive $k$-subgroup of maximal rank $G_0$ of $G$ such that the corresponding adjoint semisimple group $G_0^{ad}$ has Dynkin diagram $\Delta_0$.

I know a simple proof of Lemma 1, but I would prefer to give a reference rather than a proof.

The proof goes as follows. Let $T$ be a maximal torus of $G$, and let $R=R(G,T)$ be the root system, then our $\Pi$ is a basis of $R$. Let $S$ be the subgroup of $T$ orthogonal to $\Pi_0$, then it is a subtorus of $T$ (because $G$ is adjoint). Set $G_0=C_G(S)$, the centralizer of $S$ in $G$. Then $G_0$ is a connected reductive subgroup of $G$. It is easy to see that (the adjoint group of) $G_0$ has Dynkin diagram $\Delta_0$.

Note that Lemma 1 is a special case of the following Lemma 2, for which I would also be happy to have a reference.

Lemma 2. Let $G$ be an adjoint, connected, simple algebraic group over an algebraically closed field $k$ of any characteristic. Let $T$ be a maximal torus of $G$, and let $R=R(G,T)$ be the root system. Let $R_0$ be a closed symmetric subset of $R$. Then there exists a connected reductive $k$-subgroup of maximal rank $G_0$ of $G$ with root system $R_0$.

I will be grateful to any references, comments, etc. (also to a proof of Lemma 2).

Mikhail Borovoi

Answer by Jim Humphreys for Reductive subgroup corresponding to a subdiagram of the Dynkin diagram of a simple group

2010-03-19T16:48:48Z

Here is a short answer, which can be filled in further. Whether your group is of adjoint type or not probably makes little difference. Work going back to a fundamental paper of Borel and deSiebenthal leads to explicit information about the subgroups of maximal rank in a connected reductive group (eventually over any algebraically closed field, though at first just in characteristic 0).

For your Lemma 1, you just need a Levi subgroup of a parabolic subgroup, as described in any of the standard texts on linear algebraic groups (Borel, Springer, or my book). For Lemma 2, the foundations were laid over fairly general fields by Borel and Tits in their 1965 paper Groupes reductifs in the IHES Publ. Math. (I'll have to dig out more explicit references.) Anyway, an overview with references about subgroups of maximal rank is given in section 2.1 of my 1995 AMS book, Conjugacy Classes in Semisimple Algebraic Groups (algebraically closed case).

Concerning terminology, you are really working with subsets of the root system in each case (with the Dynkin diagram just a way of encoding data about the chosen simple roots). Bourbaki's Chapter VI has a lot of root system discussion related to Borel-Tits, for example involving symmetric subsets.

Answer by Baptiste Calmès for Reductive subgroup corresponding to a subdiagram of the Dynkin diagram of a simple group

2010-03-20T09:57:36Z

Dear Mikhail,

If I understand correctly, your Lemma 2 is implied by SGA 3 Exposé 22, Théorème 5.4.7.

Everything is on a general base S (that you may take as your algebraically closed field). The kind of subgroup you want is called "de type (R)" (see Définition 5.2.1) and a subset of R that corresponds to such a group is also called "de type (R)". Now the theorem above exactly says that when a subset of R is closed, it is "de type (R)" which exactly means that there is a corresponding connected subgroup of G. By the way, Théorème 5.4.7 does not assume the subset to be symmetric, and you get things like Borel subgroups if you take only "half" of the roots. In the symmetric case, the group is reductive by Proposition 5.10.1.

Hope this helps.

Baptiste Calmès