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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.

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  • $\begingroup$ 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 $\begin{bmatrix}0&1\\0&0\end{bmatrix}$. $\endgroup$ – mr_e_man Sep 11 '19 at 18:12
  • $\begingroup$ 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. $\endgroup$ – Dietrich Burde Sep 11 '19 at 18:28
  • $\begingroup$ I think I should clarify the question... $\endgroup$ – mr_e_man Sep 11 '19 at 18:31
  • $\begingroup$ 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 :-) $\endgroup$ – Jyrki Lahtonen Sep 11 '19 at 18:49
  • $\begingroup$ Yes, $V$ is finite-dimensional. $\endgroup$ – mr_e_man Sep 11 '19 at 18:51
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Yes, as corollary of Ado's theorem.

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.

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  • $\begingroup$ Yes exactly I meant $\neq 0$, I indeed messed up with Lie group intuition. I fixed. $\endgroup$ – YCor Sep 12 '19 at 6:32
  • $\begingroup$ 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. $\endgroup$ – Stephen Sep 12 '19 at 12:40
  • $\begingroup$ @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. $\endgroup$ – YCor Sep 12 '19 at 12:45

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