# Highlights in Linear Algebraic Groups 10: G/B is Projective

In Highlights 9 of this series, we showed that for an algebraic group $G$ and a closed subgroup $H\subseteq G$, we can always choose a representation $G\to\rm{GL}(V)$ with a line $L\subseteq V$ whose stabiliser is $H$. In turn, this allows us to identify the quotient $G/H$ with the orbit of the class $[L]$ in the projective space $\mathbf{P}(V)$, which satisfies the universal property for quotients, thereby giving us a sensible variety structure on $G/H$.

In this post, we specialise to the case of a Borel subgroup $B\leq G$; that is $B$ is maximal amongst the connected solvable groups. Such a subgroup is necessarily closed!

The fact that will allow us to study Borel subgroups is the fixed point theorem: a connected solvable group that acts on a nonempty complete variety has a fixed point. By choosing a representation $G\to \rm{GL}(V)$ with a line $L\subseteq V$ whose stabiliser is $B$, we get identify $G/B$ with a quasiprojective variety. However, in this case $G/B$ is actually projective. Here is a short sketch:
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# Highlights in Linear Algebraic Groups 8: Borel Subgroups I

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.
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