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.

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.