For an $n$-dimensional Lie algebra, is there always a matrix representation $\{M\}$ and a single vector $v$ such that $\{Mv\}$ is $n$-dimensional?

For an $$n$$-dimensional Lie algebra $$L$$, is there always a matrix representation $$\rho:L\to\mathfrak{gl}(V)$$ and a single vector $$v\in V$$ such that $$\{\rho(x)v\mid x\in L\}$$ is an $$n$$-dimensional subspace of $$V$$?

This would necessarily be a faithful representation.

I'm focusing on Lie algebras over $$\mathbb R$$, but more general answers are welcome.

This might have something to do with weights or Whitehead's lemma, but I don't know enough about representation theory to be sure.

• That only gives a faithful representation, a set of matrices $\{M\}$. It doesn't give the vector $v$. There are some faithful representations such that $\{Mv\}$ is less than $n$-dimensional for any $v$. For example, the 2D Lie algebra spanned by $\begin{bmatrix}1&0\\0&0\end{bmatrix}$ and $\begin{bmatrix}0&1\\0&0\end{bmatrix}$. – mr_e_man Sep 11 '19 at 18:12
• So that $v,\rho(x)v,\ldots ,\rho(x)^{n-1}v$ are linearly independent and $\rho(x)^n=0$ for all $x\in L$? There is a version by Block on Ado's theorem doing this, I think. – Dietrich Burde Sep 11 '19 at 18:28
• I think I should clarify the question... – mr_e_man Sep 11 '19 at 18:31
• Do you insist that the underlying module should also be finite dimensional? By Poincaré-Birkhoff-Witt $L$ embeds into the universal enveloping algebra $V=U(L)$. So with $v=1_{U(L)}$ we should be in business. If you don't accept infinite size matrices, then this will not work :-) – Jyrki Lahtonen Sep 11 '19 at 18:49
• Yes, $V$ is finite-dimensional. – mr_e_man Sep 11 '19 at 18:51

First observe that for given $$v$$, the map $$L_{\rho,v}:x\mapsto\rho(x)v$$ is linear. The goal is to show that there exists $$(\rho,v)$$ for which this map is injective. Indeed choose $$(\rho,v)$$ for which this map has kernel $$K(\rho,v)$$ of minimal dimension.
Assume by contradiction that there exists a nonzero $$x$$ in $$K(\rho,v)$$. Let $$\rho'$$ be a faithful finite-dimensional representation (as ensured by Ado's theorem). Then there exists $$v'$$ in the space of $$\rho'$$ such that $$\rho'(x)v'\neq 0$$. Hence $$(\rho\oplus\rho',v\oplus v')$$ contradicts the minimality.
The argument, suitably reformulated, shows that $$n$$ being the dimension of $$\mathfrak{g}$$, then $$\rho^{\oplus n}$$ possesses a vector with the required property for any faithful representation $$\rho$$. Also note that the proof is completely self-contained (i.e., doesn't use Ado's theorem) if we assume beforehand that $$\mathfrak{g}$$ has a faithful linear representation, which for instance is a trivial fact when it has a trivial center.
• Yes exactly I meant $\neq 0$, I indeed messed up with Lie group intuition. I fixed. – YCor Sep 12 '19 at 6:32
• Nice +1. The last phrase is a little strange: if $\mathfrak{g}$ has trivial center, then no additional argument is needed at all, since the adjoint representation is an $n$-dimensional faithful representation. – Stephen Sep 12 '19 at 12:40
• @Stephen well, this is exactly what I'm meaning (when the center is trivial, Ado's theorem is immediate). But still in this case one has to pass to a power of the adjoint representation (namely $k$ times, where $k$ is the minimal cardinal of a subset with trivial centralizer) to get a representation as desired. – YCor Sep 12 '19 at 12:45