# Every Borel contains a Cartan, and conjugacy theorems: A simple proof?

## Conjugacy of Borel subalgebras $$\newcommand{\ad}{\mathrm{ad}\,}$$

Let $$\mathfrak{g}$$ be a semisimple Lie algebra over $$\mathbb{C}$$. A Borel subalgebra $$\mathfrak{b} \subseteq \mathfrak{g}$$ is a maximal solvable subalgebra. It is a classic theorem that all Borel subalgebras of semisimple $$\mathfrak{g}$$ are conjugate under

• the action of $$\langle \exp(\ad x): x \in \mathfrak{g} \text{ ad-nilpotent}\rangle$$, or
• the adjoint action of an algebraic group $$G$$ with Lie algebra $$\mathfrak{g}$$,

the first sense implying the second. The proof I know of (e.g. in Humphrey's Introduction to Lie Algebras and Representation Theory) are very involved, involving a complicated induction on the dimension of intersection of two Borel subalgebras, as well as on the dimension of the ambient Lie algebra $$\mathfrak{g}$$.

## Conjugacy of Cartan subalgebras

In contrast, there is a short, geometric proof that all Cartan subalgebras of a semisimple Lie algebra $$\mathfrak{g}$$ are conjugate. One uses the following lemma from algebraic geometry:

Lemma: If $$f: \mathbb{A}^m \to \mathbb{A}^n$$ has $$df_x$$ surjective for some $$x \in \mathbb{A}^m$$, then the image of $$f$$ contains a dense open subset of $$\mathbb{A}^n$$.

Now given a root space decomposition $$\mathfrak{g} = \mathfrak{h} \oplus \bigoplus_{\alpha \in \Phi} \mathfrak{g}_\alpha$$ with root vectors $$x_\alpha \in \mathfrak{g}_\alpha$$, define $$f: \mathfrak{g} \to \mathfrak{g}$$ by $$f(h + \sum_{\alpha \in \Phi} t_\alpha x_\alpha) = \left(\prod_{\alpha \in \Phi} \exp (t_\alpha \ad x_\alpha)\right) h,$$ where $$h \in \mathfrak{h}$$ and the product is taken in some fixed order. The differential at any regular element of $$\mathfrak{h}$$ is nonzero, which shows that the conjugates of $$\mathfrak{h}$$ under the exponentials of nilpotent elements of $$\mathfrak{g}$$ contains a dense open subset of $$\mathfrak{g}$$, and thus intersects the conjugates of any other Cartan subalgebra $$\mathfrak{h'}$$. Further, as regular semisimple elements are dense in $$\mathfrak{g}$$, every Cartan contains a regular semisimple element and thus is equal to the centralizer of a regular semisimple element.

## Hope for a better life

Further, one can show that if a Borel subalgebra $$\mathfrak{b}$$ contains the Cartan subalgebra $$\mathfrak{h} \subseteq \mathfrak{g}$$, then $$\mathfrak{b}$$ must be one of the "standard" Borel subalgebras associated to $$\mathfrak{h}$$, formed by taking the sum of $$\mathfrak{h}$$ with all positive roots (after fixing such a choice of positive roots). The problem of showing that all Borel subalgebras are conjugate thus reduces to showing that every Borel subalgebra contains a Cartan. This leads to the question:

Is there a geometric argument for that every Borel subalgebra of $$\mathfrak{g}$$ contains a Cartan subalgebra of $$\mathfrak{g}$$, akin to the geometric argument for Cartan subalgebras above?

It would be helpful even to have a geometric argument showing that every Borel subalgebra of $$\mathfrak{g}$$ contains a regular semisimple element.