Posted by Jason Polak on 15. April 2013 · 2 comments · Categories: algebraic-geometry, group-theory · Tags: ,

highlights11In order to understand the structure of reductive groups, we will first look at some “base cases” of groups that are quite small. These are the groups of so-called semisimple rank 1, which by definition are the algebraic groups $ G$ such that $ G/R(G)$ has rank 1, where $ R(G)$ is the connected component of the unique largest normal solvable subgroup. In this post, we shall see a detailed proof of a theorem that gives several different characterisations of these groups.

As usual, we consider algebraic groups over an algebraically closed field $ k$. The proof we follow will be Theorem 25.3 in Humphrey’s book “Linear Algebraic Groups”. The reason I will go through it here is because in the book, I found the proof a bit terse and in a few points the proof relies on exercises, so it should be instructive to write down a more self-contained proof in my own words.

The Radical

Radicals show up all over the place in algebra (see the humourous title of Mary Gray’s book), and it’s no different in the theory of algebraic groups. So, let $ G$ be an algebraic group. It contains a maximal solvable normal subgroup $ N$ and we define $ R(G) = N^\circ$, the connected component of $ N$. The closure of $ N$ also is solvable and normal so $ N$ is closed. Of course $ N$ is not necessarily connected.

From the radical comes two of the most important definitions: semisimple and reductive. We call $ G$ semisimple if $ R(G)$ is trivial, and $ G$ reductive if $ R_u(G)$ is trivial where $ R_u(G)$ is the unipotent subgroup of $ R(G)$. Our aim is to understand semisimple and reductive groups.

Let us give one characterisation of this radical $ R(G)$:

Theorem. $ R(G)$ is the identity component of the intersection $ \cap B = \cap_B B$ of all the Borel subgroups of $ G$.
Proof. Since all the Borel subgroups are conjugate, $ \cap B$ is normal, it is not hard to see that $ (\cap B)^\circ$ is also normal hence a connected normal solvable subgroup of $ G$. Thus by definition , $ \cap B\subseteq R(G)$.

On the other hand, $ R(G)$ is a connected solvable normal subgroup and hence contained in every Borel subgroup so $ R(G)\subseteq \cap B$. Thus $ R(G)\subseteq (\cap B)^\circ$.

Semisimple Rank 1

We say that a connected algebraic group $ G$ is of semisimple rank 1 if $ G/R(G)$ has rank 1. (The rank of $ G/R(G)$ is denoted $ \mathrm{rank}_{ss}(G)$ and is called the semisimple rank of $ G$.) Recall that $ \mathcal{B}$ is the set of all Borel subgroups of $ G$ and it can be identified as a set with the (closed points) of the projective variety $ G/B$. We write $ \mathcal{B}^T$ for the fixed points under the action of $ T$.

Theorem. Let $ T\subseteq G$ be a maximal torus and $ G$ a connected algebraic group with Borel subgroup $ B$. Let $ W = W(G,T) = N_G(T)/C_G(T)$ be the Weyl group of $ T$. Then the following are equivalent:

  1. $ \mathrm{rank}_{ss}(G) = 1$
  2. $ |W| = 2$
  3. $ |\mathcal{B}^T| = 2$
  4. $ \mathrm{dim} G/B = 1$
  5. $ G/B\cong\mathbf{P}^1$
  6. There is an epimorphism $ \varphi:G\to\mathrm{PGL}_2(k)$ with $ (\ker\varphi)^\circ = R(G)$.
Proof. (1)$ \Rightarrow$(2)As we have remarked, every Borel subgroup contains the radical $ R(G)$ , so via the quotient map $ G\to G/R(G)$ there is a bijective correspondence between the Borel subgroups containing the torus $ T$ and the Borel subgroups of $ G/R(G)$ containing the image of $ T$ which is $ T’= T/(T\cap R(G)$. Thus, if we let $ W = W(G,T)$ and $ W’ = W(G/R(G),T’)$, we have that $ |W| = |W’|$ (recall, the order of the Weyl groups is exactly the number of Borels containing the respective torus).

Since $ T’$ is a maximal torus of $ G/R(G)$, by the definition of semisimple rank 1, the rank of $ T’$ is one, so $ T’\cong \mathbb{G}_m$, and distinct elements of $ W’$ act as distinct automorphisms on $ T’$. But $ \mathrm{Aut}(\mathbb{G}_m) \cong \mathbb{Z}/2$. Thus $ |W’| = |W| \leq 2$. Since $ G$ is not solvable, $ \mathrm{dim}(G/B) \geq 1$ so $ |W| \geq 2$ (this is a standard fact that is proved in the previous section).

(2)$ \Rightarrow$(3). This is something we have already mentioned: the order of the Weyl group is the number of Borels that contain $ T$, as is also not hard to show.

(3)$ \Rightarrow$(4). Now we assume $ |\mathcal{B}^T| = 2$ and we want to prove $ \mathrm{dim}(G/B) = 1$. Since $ T$ is contained in two distinct Borel subgroups, $ B\not= G$, so $ B$ is a proper closed subgroup and hence $ \mathrm{dim}(G/B) \geq 1$. If $ \mathrm{dim}(G/B)\geq 2$ then $ |\mathcal{B}^T| = 3$ (this again follows from a geometric argument via fixed points!).

(4)$ \Rightarrow$(5). Construct a representation of $ G$ into some $ \mathrm{GL}(V)$ so that $ B$ is the stabiliser of some $ v\in V$. Given a cocharacter $ \lambda:\mathbb{G}_m\to T$, we get an action on $ \mathbf{P}(V)$ by the multiplicative group $ \mathbb{G}_m$; this extends to a morphism $ \mathbf{P}^1\to \mathbf{P}(V)$; if we consider the orbit map of the vector $ v$ we get a nonconstant morphism $ \mathbf{P}^1\to G/B$. Since $ \mathbf{P}^1$ is complete, its image is a closed irreducible subset of $ G/B$ of dimension $ 1$, which then must be $ G/B$ because $ G/B$ is also irreducible and of dimension 1 (note that the map $ \mathbf{P}^1\to G/B$ may not be an isomorphism) but since it is a dominant map from $ \mathbf{P}^1$ to a smooth variety of dimension $ 1$, the two must be isomorphic, as they have isomorphic function fields.

(5)$ \Rightarrow$(6) Now we assume that $ G/B\cong\mathbf{P}^1$ and we want to show that there is a surjective morphism $ \varphi: G\to \mathrm{PGL}_2(k)$ with $ \ker\varphi = R(G)$. Since $ G$ acts on the variety $ G/B$, it acts by automorphisms on $ \mathbf{P}^1$ whose automorphis group is $ \mathrm{PGL}_2(k)$. The kernel of this morphism are elements $ x\in G$ that stabilise every Borel, or in other words, are contained in every Borel, and we have already shown in the section on radicals that the collection of such elements is nothing else than $ R(G)$.

However, we still have to show that the morphism $ \varphi:G\to\mathrm{PGL}_2(k)$ is actually an epimorphism. Since $ R(G)\subseteq B$, we see that $ G/R(G)$ is not solvable. We claim that any such group must have dimension at least three. By writing $ B = T\cdot U$ where $ T$ is a torus $ T\subseteq B$ and $ U\subseteq B$ is the unipotent subgroup, we see that if $ \mathrm{dim}(B) = 1$ then $ B = T$ or $ B = U$. If this is the case then $ B = G$, so in our case $ \mathrm{dim}(B) \geq 2$.

Thus we see that if $ G$ is not solvable, it must have dimension at least 3 since any Borel subgroup in it must have dimension at least 2. Thus the dimension of $ G/R(G)$ is at least 3, so the morphism $ G\to\mathrm{PGL}_2(k)$ must therefore be surjective.

(6)$ \Rightarrow$(1). This follows because the group $ \mathrm{PGL}_2(k)$ has rank $ 1$.

2 Comments

  1. Nice post! Just a small comment: The fifth of the 6 equivalent statements ( (5) G/B \cong P^{1} ) is missing.

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