To analyse the structure of a group G

you will need the radical and a torus T.

The group of Weyl may also may also suit

to prevent the scattering of many a root.

Functors are nice including the one of Lie

Parabolics bring in the ge-o-metry!

The theory of weights may seem oh so eerie

Until you start representation theory!

To analyse the structure of a group G

you will need the radical and a torus T.

The group of Weyl may also may also suit

to prevent the scattering of many a root.

Functors are nice including the one of Lie

Parabolics bring in the ge-o-metry!

The theory of weights may seem oh so eerie

Until you start representation theory!

The structure of reductive and semisimple groups over an algebraically closed field will be pinnacle of this post series. After we have finished with this, this series will end and we will start to learn about algebraic groups from the perspective of group schemes, and we shall use some of the results we have seen so far by using that we really have just been studying the $ \overline{k}$-points of group schemes (classical algebraic geometry).

The topic for today is the radical and unipotent radical, that will allow us to define the concept of semisimple and reductive group. We will then use the roots, which are certain characters of a maximal torus. These will give us a root system, so we will take a break to study these, and classification of root systems will enable us to classify algebraic groups.

### The Unipotent Radical

Let $ G$ be an connected algebraic group over a field $ k = \overline{k}$. We start with the following observation that is not difficult to prove: if $ A$ and $ B$ are two normal solvable subgroups of $ G$ then $ AB$ is a normal solvable subgroup of $ G$, evidently containing $ A$ and $ B$. Thus, there is a unique maximal normal solvable subgroup $ N\subseteq G$. Moreover, since $ N$ is maximal, it is closed. Of course, $ N$ might not necessarily be connected.

The radical of $ G$, denoted by $ R(G)$, is the connected component $ N^\circ$. The radical is the largest connected normal solvable subgroup: indeed, it is certainly normal and solvable, and any connected normal solvable subgroup is contained in $ N$.

The *unipotent radical* of $ G$ is the subgroup $ R(G)_u$ of unipotent elements of $ R(G)$. For a nontrivial group $ G$, if $ R(G)$ is trivial then we call $ G$ semisimple. If $ R(G)_u$ is trivial then we call $ G$ reductive. So semisimple groups are a specific kind of reductive group. That we want $ G$ to be nontrivial in the definition is just a convention that really doesn't make much difference.

### Radical Facts

In the previous post Highlights 11, we looked at several equivalent conditions for a group $ G$ to have semisimple rank 1. We would like to see what these conditions reveal if we assume also that $ G$ is reductive. What we will do is essentially the content of Corollary 25.3 in Humphrey's book, although the proof there is a little terse so I shall try to be more self-contained and leisurely. In order to do this, we need more information in the radical $ R(G)$.

**Theorem**. If $ G$ is connected reductive then $ R(G) = Z(G)^\circ$ where $ Z(G)$ is the center of $ G$. Moreover, $ R(G)$ is a torus.

*Proof.*The unipotent radical $ R(G)$ is connected and solvable, and hence $ G$ can be embedded in $ \mathrm{GL}_n(k)$ for some $ n$ such that $ R(G)$ is a subgroup of the upper triangular matrices. Since $ G$ is reductive, $ R(G)_u$ is trivial so the exact sequence $ 1\to U_n\to T_n\to D_n\to 1$ restricted to $ R(G)$ shows that $ R(G)$ must be a torus.

Now, since $ R(G)$ is a torus, $ N_G(R(G))^\circ = C_G(R(G))^\circ$, and these are both $ G$, so $ C_G(R(G))^\circ = G$, so $ C_G(R(G)) = G$, and thus $ R(G)\subseteq Z(G)$ and since $ R(G)$ is connected, $ R(G)\subseteq Z(G)^\circ$. Conversely, the connected component $ Z(G)^\circ$, being central, is certainly normal solvable, and connected, and so $ Z(G)^\circ\subseteq R(G)$. QED

**Theorem**. If $ G$ is connected reductive then $ R(G)\cap [G,G]$ is finite where $ [G,G]$ is the commutator subgroup of $ G$.

*Proof*. First, embed $ G$ into some $ \mathrm{GL}(V)$. We have just proved in the previous theorem that $ R(G) = Z(G)^\circ$ is a torus if $ G$ is reductive. In particular, $ R(G)$ is diagonalisable so we can decompose $ V$ into a direct sum of subspaces $ V = \oplus_\alpha V_\alpha$ where the $ \alpha: \mathbb{G}_m\to R(G)$ are the weights (characters with nontrivial eigenspace) and $ V_\alpha$ is the eigenspace of $ \alpha$.

Consider an element $ h$ in the centraliser of $ Z(G)^\circ = R(G)$ in $ \mathrm{GL}(V)$. It commutes with every element $ g\in Z(G)^\circ$, so a one-line calculation shows that if $ v\in V_\alpha$ then $ hv\in V_\alpha$. Thus the centraliser of $ R(G)$ in $ \mathrm{GL}(V)$ consists of block diagonal matrices and clearly includes $ G$. The commutator subgroup then must lie in the block diagonal matrices $ \prod_\alpha \mathrm{SL}(V_\alpha)$.

Now, images elements of $ R(G)$ in $ \mathrm{GL}(V)$ that are also in the image of $ [G,G]$ consist of block diagonal scalar matrices with determinant one: indeed, suppose $ x\in R(G) = Z(G)^\circ$. Then in each block corresponding to $ V_\alpha$, the matrix of $ x$ restricted to $ V_\alpha$ commutes with every given matrix of a linear operator on $ V_\alpha$ so $ x$ must be scalar and of determinant one; since there are only finitely many $ n$-th roots of unity for each $ n$, we see that there are only finitely many elements in $ Z(G)^\circ$ and in $ [G,G]$. QED

### Reductive Groups of SS Rank 1

In this section, we fix a connected reductive group $ G$ of semisimple rank 1 (i.e. the rank of G/R(G) is 1). Given our previous theorem about connected groups of semisimple rank 1, what can we conclude about $ G$? Recall that we say that $ G$ is an almost direct product of closed subgroups $ H$ and $ K$ if the multiplication map $ H\times K\to G$ is surjective with finite kernel (sometimes called an *isogeny* in this context).

**Theorem.**Let $ G$ be a connected reductive group of semisimple rank 1. Then $ [G,G]$ is semisimple of dimension three. Moreover, there is an almost direct product decomposition $ G = [G,G]\cdot Z(G)^\circ$.

Now we consider the multiplication map $ Z(G)^\circ\times [G,G]\to G$. Note first that if $ (x,y)\in Z(G)^\circ\times [G,G]$ then $ x\in [G,G]\cap Z(G)^\circ$, and since we have proven that this intersection is finite, the map $ Z(G)^\circ\times [G,G]\to G$ indeed has finite kernel. Thus we just need to show that it is surjective.

Now we use the handy fact that $ \mathrm{PGL}_2(k)$ is perfect and so the commutator subgroup $ [G,G]$ is mapped onto $ \mathrm{PGL}_2(k)$ via the morphism $ \varphi$. Since the kernel of the restriction $ [G,G]\to \mathrm{PGL}_2(k)$ is finite, we get that $ \mathrm{dim}([G,G]) = 3$. Moreover, its radical is contained in the radical of $ G$ so its radical is finite and connected, hence trivial. Thus $ [G,G]$ is semisimple of dimension three.

Now, the dimension of the image of $ Z(G)^\circ\times [G,G]$ inside $ G$ is $ \mathrm{dim}(Z(G)^\circ) + \mathrm{dim}([G,G])$, whereas the dimension of $ G$ is the same, using the previous paragraph. Since $ Z(G)^\circ$ and $ [G,G]$ are closed connected subgroups of $ G$ and $ Z(G)^\circ$ normalises $ [G,G]$, we get that the image of $ Z(G)^\circ\times [G,G]$ is a closed connected subgroup of $ G$ which has the same dimension of $ G$, so it must be all of $ G$. QED

## 2 Comments

Thanks for writing this up, this is a godsend as I'm struggling with the occasional terseness of Humphreys' book, especially in the proof of Corollary 25.3.

You are very welcome. I'm glad you find it useful.