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

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

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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

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This theorem from SGA 3 is an efficient reference for Lemma 2, though the background and setting are more than needed to describe the structure theory over an algebraically closed field. Concretely, the desired subgroup is generated by a maximal torus together with pairs of root subgroups (or the rank 1 semisimple groups they generate). The standard commutation relations show that no further roots appear if the given set of roots is closed and symmetric. Here as in SGA 3, the Lie algebra reflects accurately what is going on since the subgroup is of maximal rank. –  Jim Humphreys Mar 20 '10 at 14:29
I will look further at literature later this weekend, but meanwhile I think the essential point coming from the 1949 Borel-de Siebenthal paper (as generalized to the algebraic group situation in any characteristic) is that the connected reductive subgroups of maximal rank are generated by a maximal torus along with root subgroups belonging to a root system found via the extended Dynkin diagram. Sets of simple roots only generate the root systems in Levi subgroups; the others are usually called "pseudo-Levi", like the copy of $A_2$ inside $G_2$. –  Jim Humphreys Mar 19 '10 at 21:02