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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.

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