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

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?

• <X> is for the linear subspace spanned by X – Julien Bernard Mar 30 '15 at 12:27
• Can you give the link of "the assertion of Hermann Weyl" ? – Dietrich Burde Mar 30 '15 at 12:37
• The reference is Mathematische Analyse des Raumproblems, p101.: "Eine nicht-verschwindende Matrix H, deren Spur erschwindet, würde nur dann in g nicht vorzukommen brauchen, wenn g einparametrig wäre; das ist aber für n = 3 offenbar durch die Voraussetzung A ausgeschlossen (sie erfordert, daß die Gruppe mindestens n − 1 Parameter besitzt)." Of course, you have to read above to understand that "hypothesis A" means, here, exactly the same as hypothesis $H$ in my question. – Julien Bernard Mar 30 '15 at 12:51
• The reference to the page 55 concerns only the definition of what is "hypothesis A". I know the entire Weyl's text, and I can say that he did not prove in it what I've asked in my question. For the general demonstration of the theorem which is the subject of his book, he only need to know that $dim(G)>1$, and not that $dim(G)\geq n-1$. That is why he only gives the result without any proof. – Julien Bernard Mar 30 '15 at 13:20

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 $$\dim \mathfrak{g} \ge \dim \mathfrak{g}(y) \ge \dim (\mathbb{C}^n/M) = n-1 .$$

• The proof in the first paragraph shows that $n-1$ is not sharp unless $n$ is very small. This is because $\dim \mathfrak{g} = \dim (\mathfrak{b} \cap \mathfrak{b}^-) + 2 \dim (\mathfrak{g}/\mathfrak{b})$ is larger than $1 + 2 \dim (\mathfrak{g}/\mathfrak{b})$ unless $\mathfrak{g} = \mathfrak{sl}_2(\mathbb{C})$ (so $\dim \mathfrak{g} = 3$). – Dave Witte Morris Mar 30 '15 at 18:10