Minimal dimension of a Lie algebra of matrices, with a restrictive property

Question

Let $\mathfrak{g}$ be a sub-Lie-algebra of $\mathfrak{gl}_n(\mathbb{C})$, the Lie algebra of complex $n\times n$ square matrices.

Let us call $(H)$ the hypothesis: for all $x, y\in\mathbb{C}^n$, whenever $x$ and $y$ are linearly independent, we have $\langle \mathfrak{g}(x)\cup \mathfrak{g}(y)\rangle=\mathbb{C}^n$. Here, $\mathfrak{g}(x)$ is the linear subspace of $\mathbb{C}^n$ that consists of all the images of $x$ by a $g\in \mathfrak{g}$ (idem for $\mathfrak{g}(y)$). I think of $(H)$ as "no hyperplane is conjugate to a pair of linearly independent vectors".

My problem is to determine the minimal dimension that $\mathfrak{g}$ must have, in order for $\mathfrak{g}$ to be able to have the property $(H)$.

It is clear that we must have $\dim(\mathfrak{g})>\frac{n-1}{2}$. Indeed, if $\dim(\mathfrak{g})\leq\frac{n-1}{2}$, we take two arbitrary linearly independant vectors $x, y$, and we have $\dim(\langle\mathfrak{g}(x)\cup \mathfrak{g}(y)\rangle)\leq n-1$.

We should be able to prove in fact that $\dim(\mathfrak{g})\geq n-1$ (I know it from an assertion of Hermann Weyl). But, in order to reach this new lower bound, we cannot anymore take two arbitrary linearly independent vectors.

I think, to reach that goal, that we must start from a vector basis that is correctly adapted to all the transformations of $\mathfrak{g}$. But I don't see for the moment how to get the point.

Could anyone give me some advice?

Answer 1

Assume, for the moment, that the representation on $\mathbb{C}^n$ is irreducible. (This is the main case.) Then $\mathfrak{g}$ must be reductive. Let $\mathfrak{b}$ be a Borel subalgebra of $\mathfrak{g}$, let $\mathfrak{b}^-$ be an opposite Borel subalgebra, let $x$ be a highest-weight vector in $\mathbb{C}^n$, and let $y$ be a lowest-weight vector. Then $\mathfrak{b}(x) = \langle x \rangle$ and $\mathfrak{b}^-(y) = \langle y \rangle$, so $$n = \dim \langle \mathfrak{g}(x), \mathfrak{g}(y) \rangle \le \dim \mathfrak{g}(x) + \dim \mathfrak{g}(y) \le 2 \bigl(1 + \dim(\mathfrak{g}/\mathfrak{b}) \bigr) \le 1 + \dim \mathfrak{g}. $$

We may now assume that the representation is reducible, so we may let $M$ be a nontrivial, proper submodule of $\mathbb{C}^n$. Clearly, we must have $\dim M = 1$. (Otherwise, we could choose $x$ and $y$ to both be in $M$.) Choose $x \in M$ and $y \notin M$. Then, since $\mathfrak{g}(x) \subseteq M$ and $\langle \mathfrak{g}(x), \mathfrak{g}(y) \rangle = \mathbb{C}^n$, the subspace $\mathfrak{g}(y)$ must project onto all of $\mathbb{C}^n/M$, so $$ \dim \mathfrak{g} \ge \dim \mathfrak{g}(y) \ge \dim (\mathbb{C}^n/M) = n-1 .$$