Why is this paragraph so short?

Question

$G$ is a connected, reductive linear algebraic group.The reference is Springer, Linear Algebraic Groups. I am having trouble making sense out of anything in this paragraph.

Proposition 7.31(ii) just says that $(G,G) \cap C$ is finite. From there, he deduces that $(G,G)$ is semisimple of rank one. Why should that imply that $(G,G)$ have rank one (i.e. any nontrivial torus of $(G,G)$ has rank one), or that $(G,G)$ is semisimple?

7.2.3(ii) says that if $H$ is a semisimple group of rank one, with maximal torus $S$, there is a character $\alpha$ of $S$ such that $\mathfrak h = \mathfrak s \oplus \mathfrak g_{\alpha} \oplus \mathfrak g_{-\alpha}$, where $\mathfrak g_{\alpha} = \{ X \in \mathfrak h : \textrm{Ad }x(X) = \alpha(x)X, \textrm{ for all } x \in S \}$.

The letter $P$ denotes the set of nonzero characters of $G$ which show up as weights (with respect to the rational representation $\textrm{Ad }: T \rightarrow \textrm{GL}(\mathfrak g)$. If $\beta \in P$, $G_{\beta} = Z_G((\textrm{Ker } \beta)^0)$ can be shown to be connected, and 7.13(ii) says that $G$ is solvable if and only if all the $G_{\beta} : \beta \in P$ are solvable.

I don't get why $P$ has only two elements. I know a similar result is true for $(G,G)$ (once we show it is semisimple of rank one) and for $G/C$. I don't see why anything transfers over to $G$ though.

Answer 1

Here is why $(G,G)$ is semisimple of rank one.

Lemma: Let $G$ be a connected algebraic group, let $N$ be a zero-dimensional closed, solvable, normal subgroup of $G$ such that $G/N$ is semisimple. Then $G$ is semisimple. If $N$ consists of semisimple elements (e.g. in characteristic zero, zero dimensional algebraic group has this property), then $G$ and $G/N$ have the same rank.

A connected group is semisimple if and only if it has no nontrival closed, connected, normal solvable subgroups. If $G$ were not semisimple, it would have such a group $K$. The image of $K$ under the canonical homomorphism $G \rightarrow G/N$, which is $(N.K)/N$, is closed, connected, normal in $G/N$, and solvable. This forces $N.K = N$ or $N.K = G$. In the first case, we have $K \subseteq N$, whence $K = \{1_G\}$.

In the second case, the fact that $N,K$ are both normal gives us that $N.K.$ is the image under a morphism of algebraic groups with domain $N \times K$. So $\textrm{Dim } G \leq \textrm{Dim } N + \textrm{Dim } K = 0 + \textrm{Dim } K$, whence $G = K$.

Finally if $N$ consists of semisimple elements, it is contained in a maximal torus $T$. Hence $T/N$ is a maximal torus of $G/N$. But $T$ and $T/N$ have the same dimension. $\blacksquare$

Now when $G$ is connected, reductive, and of semisimple rank one, the commutator subgroup of the quotient $G/R(G)$ is equal to itself; otherwise the semisimple group $G/R(G)$ would have a proper closed, connected, normal, solvable subgroup (if $H$ is any group, then $H/(H,H)$ is abelian, so $H$ is solvable if and only if $(H,H)$ is). Also, the surjective map $G \rightarrow G/R(G)$ maps the commutator subgroup onto the commutator subgroup.

It follows that the restriction of $G \rightarrow G/R(G)$ to $(G,G)$ is still surjective. The kernel is $R(G) \cap (G,G)$, which is finite, and abelian. Also, $R(G)$ is a torus, so this kernel consists of semisimple elements. So we can apply the lemma.

Answer 2

Okay I think I have all of it now. Since $G/C$ is semisimple of rank one, it has dimension $3$. So $C$ has codimension $3$ in $G$. Since $C$ is a torus, it is contained in a maximal torus, hence contained in all maximal tori. So we have inclusions $$C \subseteq T \subseteq Z_G(T) \subseteq G$$ the first of which must be proper: if $C$ were a maximal torus, there would be only finitely many Borel subgroups of $G$, because there are only finitely many Borel subgroups which contain a given maximal torus. There cannot be finitely many Borel subgroups, because their union is $G$, with $G$ irreducible and not solvable.

The torus $C$ lying in the center of $G$, we conclude that the Weyl groups $W(G,T), W(G/C,T/C)$ are isomorphic. The latter has order $2$. Let $\alpha$ be a root of $T$ in $G$. Then $-\alpha$ is also a root of $T$ in $G$, and $\alpha, -\alpha$ are the only nonzero weights. By a dimension argument can conclude that the direct sum $$\mathfrak t \oplus \mathfrak g_{\alpha} \oplus \mathfrak g_{-\alpha}$$ is all of $G$, that $T = Z_G(T)$ with $\textrm{Dim } T = \textrm{Dim } C + 1$, and that the spaces $\mathfrak g_{\alpha}$ and $\mathfrak g_{-\alpha}$ are one dimensional. This is because $T$ has codimension at most $2$ in $G$, hence $\mathfrak t$ has codimension at most $2$ in $\mathfrak g$.