Borel subgroups are an important type of subgroup that will allow us to gain insight into the mysterious structure of algebraic groups. We shall look at the definition and some basic examples in this post. As usual, algebraic group means some linear algebraic group defined over an algebraically closed field $ k$.

A Borel subgroup $ B\subseteq G$ of an algebraic group $ G$ is a maximal connected solvable subgroup amongst the solvable subgroups of $ G$. Notice the absense of the adjective “closed” here: although $ G$ may contain solvable groups that are not closed, one that is maximal amongst the connected solvable ones must be closed. Indeed, if $ B$ is a subgroup maximal amongst the connected solvable subgroups of $ G$ then its closure $ \overline{B}$ is also connected and solvable.

Before we go any further, it’s helpful to have an example. Since all linear algebraic groups are closed subgroups of some $ \mathrm{GL}_n(k)$, let’s do $ \mathrm{GL}_n(k)$. We claim that a Borel subgroup of $ \mathrm{GL}_n(k)$ is $ B = \mathrm{T}_n(k)$, the closed subgroup of upper triangular matrices. Let’s now sketch the proof that $ B$ is actually a Borel.

It’s Solvable!

Solvability is easily but perhaps somewhat tediously proved. It’s not too hard to do for $ \mathrm{GL}_2$ though. If

$ \begin{pmatrix} a_1 & a_2 \\ 0 & a_3\end{pmatrix},\begin{pmatrix} b_1 & b_2 \\ 0 & b_3\end{pmatrix}$

are two elements of $ \mathrm{GL}_2$ then their commutator is

$ \begin{pmatrix} 1 & \tfrac{1}{a_3b_3}(-a_3b_2 – a_2b_1 + a_1b_2 + a_2b_3) \\ 0 & 1\end{pmatrix}$

Then $ [B,B]\cong\mathbb{G}_a$, which is solvable, so $ B$ is solvable. With generators, the general case $ \mathrm{GL}_n$ is handled in a similar manner. Thus $ B$ is solvable. We juts have to show that it is connected, and maximal amongst the connected solvable groups. Let’s do connected next:

It’s Connected!

A typical method to show that an algebraic group is connected is to show that it is generated as an abstract group by connected subgroups. One connected subgroup of $ B$ is the subgroup of diagonal matrices. This is just $ \mathbb{G}_m\times\mathbb{G}_m$ so it’s connected. Then we already remarked that the commutator subgroup is just $ \mathbb{G}_a$ so it’s also connected, and evidently these are enough to generate $ B$.

It’s Maximal!

So, the final step is to show that $ B$ is actually maximal amongst the connected solvable groups. This follows from the Lie-Kolchin theorem, which states that if $ H$ is any connected solvable group of $ \mathrm{GL}(V)$ for a vector space $ V$ then there is a common eigenvector for $ H$. In particular, $ H$ is isomorphic to a subgroup of upper triangular matrices.

Let $ C$ be a connected maximal solvable subgroup $ C\subseteq G$ such that $ B\subseteq C$. Because $ C$ is connected and solvable, there is some $ g\in G$ such that $ gCg^{-1} \subseteq B = \mathrm{T}_n$. But then $ B\subseteq C$ so therefore

$ gBg^{-1}\subseteq B.$

This implies that $ g\in B$ so $ gCg^{-1} = C\subseteq B$ so $ B = C$. In case this part seems a litte terse: it becomes clear once the matrices are written down, that if $ gBg^{-1}\subseteq B$ then $ g\in B$. Actually, $ gBg^{-1} = B$ (it can’t be a proper subgroup for dimension reasons) implies $ g\in B$ for any Borel: in other words, $ B$ is its own normaliser. But these and other theoretical facts we shall encounter in future posts, when we start to explore the structure of reductive groups more closely.

1 Comment

  1. Hello,

    Thanks for the post. I’m trying to generalize the connected affirmation to GLn. The idea must be the same, but I cannot prove that in this case the commutator subgroup is also connected. I hope you can give a hand.

    Thank you.

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