In Highlights 12, we used some of the equivalent conditions for a connected algebraic group $ G$ over a field $ k=\overline{k}$ to have semisimple rank 1 in the study of reductive groups (these are the groups whose unipotent radical $ R(G)_u$ is trivial).

Precisely, we showed that such a $ G$ must have a semisimple commutator $ [G,G]$ subgroup whose dimension is three, and that we can write

$ G = Z(G)^\circ\cdot [G,G]$

where the $ -\cdot-$ denotes that this is an almost direct product: in other words, the multiplication map $ Z(G)^\circ\times[G,G]\to G$ is surjective with finite kernel.

Let $ T\subseteq G$ be a maximal torus. We will show in this post that $ C_G(T) = T$ for a connected reductive group $ G$ of semisimple rank 1.
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Lately I've been reading a bit of Robert Steinberg's book, "Endomorphisms of Algebraic Groups", so I've decided to explain what I am reading for this series on linear algebraic groups, which will take us away from Lie algebras for a bit of time.

Let $ G$ be a linear algebraic group over an algebraically closed field $ k$. In this case we can choose a Borel subgroup $ B$ and a maximal torus $ T\subseteq B\subseteq G$; in fact, all the maximal tori reside in Borel subgroups because tori are solvable. What about if we are given an automorphism $ \sigma:G\to G$? This prompts a fairly natural question:

Is it possible to choose $ B$ and $ T$ such that they are stable under $ \sigma$?

If so, this would be very convenient in working with automorphisms, and pretty useful for working with $ \theta$-groups for instance (more on this later!). It turns out that if $ \sigma$ can be realised as conjugation by a semisimple element after an embedding of $ G\hookrightarrow G'$, then this is possible. This was proven by Robert Steinberg and also by David J. Winter.
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