Are the elements of a Lie algebra separated by its finite-dimensional representations? Let $\mathbb{k}$ be some field and let $\mathfrak{g}$ be a $\mathbb{k}$-Lie algebra.

Does there exist for every nonzero element $x$ of $\mathfrak{g}$ a finite-dimensional representation of $\mathfrak{g}$ on which $x$ acts nonzero?

In other words, do the finite-dimensional representations of $\mathfrak{g}$ separate the elements of $\mathfrak{g}$?

*

*If $\mathfrak{g}$ is finite-dimensional, then this is true by Ado’s theorem.


*As pointed out in an answer to a similar question, the finite-dimensional representations of $\mathfrak{g}$ even separate the points in the universal enveloping algebra $\operatorname{U}(\mathfrak{g})$ if $\mathfrak{g}$ is finite-dimensional and $\mathbb{k}$ is of characteristic zero.
(This seems to be Theorem 2.5.7 in Dixmier’s Enveloping Algebras.)
Both of the above arguments show even stronger assertions, but also need some additional assumptions and quite a bit of work.
 A: The Wikipedia says there are no finite-dimensional representations of (non-trivial) affine Lie algebras, which implies the desired result is false:
https://en.wikipedia.org/wiki/Affine_Lie_algebra
A: There are many possible answers. Here's one Edit: Here are a few:
Consider the Lie algebra (over a fixed field of characteristic zero) with presentation
$$\mathfrak{g}=\langle x,y,z\mid [x,y]=y,\;[y,z]=z\rangle.$$
(1) It's easy to check that $f(z)=0$ for every finite-dimensional representation.
(2) However, $z\neq 0$. This is because this is by definition amalgam of two 2-dimensional Lie algebras $\langle x,y\mid [x,y]=y\rangle$ and $\langle y,z\mid [y,z]=z\rangle$ the common 1-dimensional subalgebra $Ky$, and it's known (see Encyclopedia of Math.) that subalgebras embed into their amalgam.
For (1), it's a simple consequence of the study of finite-dimensional representations of the 2-dimensional non-abelian Lie algebra $\langle x,y\mid [x,y]=y\rangle$, which we can assume to be over an algebraically closed field. Every such representation maps $y$ to a nilpotent matrix. Now consider a finite-dimensional representation of  $\mathfrak{g}$, mapping $x,y,z$ to $X,Y,Z$.. Using the first subalgebra, $Y$ is nilpotent. Also, we can make the second subalgebra act as upper triangular matrices, and $Z$ is nilpotent. So both $Y,Z$ are strictly upper triangular, and $[Y,Z]=Z$ forces $Z=0$.
Similarly we can deduce that the analogue of Higman group, the Lie algebra
$$\langle x_0,x_1,x_2,x_3\mid [x_{i-1},x_i]=x_i: i\in\mathbf{Z}/4\mathbf{Z}\rangle$$
has no non-trivial finite-dimensional representation. I guess one can elaborate using amalgams (but haven't checked details) that it's not trivial (hence infinite-dimensional).

$\DeclareMathOperator\h{\mathfrak{h}}$Here's now an example that is completely self-contained.
Consider the Lie algebra $\h$ with basis $u$, $(e_n)_{n\in\mathbf{Z}}$, law $[e_i,e_j]=(i-j)e_{i+j}$, $[u,e_i]=ie_i$, over a field $K$ of characteristic zero.
I claim that every finite-dimensional representation of $\h$ kills all $e_i$. Indeed, consider operators $U$, $E_n$ of a finite-dimensional vector space satisfying the same relations. Since $[U,E_n]=nE_n$, the $E_n$ are in distinct eigenspaces for $\mathrm{ad}(U)$, and hence the $KE_n$ generate their direct sum. Since the dimension is finite, there exists $n$ such that $E_n=0$. Then for $m\neq 2n$, $E_m=\frac{1}{2n-m}[E_n,E_{m-n}]=0$. In turn $E_{2n}=\frac{1}{2-2n}[E_1,E_{2n-1}]=0$, so $E_m=0$ for all $m\in\mathbf{Z}$.

Actually, in this example the subalgebra $\mathfrak{r}$ already has the property that every finite-dimensional representation is trivial, but using a slightly more elaborate argument, which however works in arbitrary characteristic $\neq 2$.
Let $W_n$ be the subspace generated by $\{E_k:k\ge n\}$, and $W_\infty=\bigcap_n W_n$, so $W_\infty=W_n$ for large enough $n$, say $n\ge n_0$. Then $[E_n,W_\infty]=W_{\infty}$ for all $n$.
Suppose by contradiction that $W_\infty\neq 0$. Choose $n\ge n_0$. Take a block-diagonal decomposition of $E_n$. Then the sum $M$ of characteristic subspaces for nonzero eigenvalues of $\mathrm{ad}(E_n)$ consists of those matrices in this block decomposition all of whose diagonal blocks are zero. The condition $[E_n,W_\infty]=W_\infty$ forces $W_\infty\subset M$. In particular, $E_n$ has this form. But by definition $E_n$ is block-diagonal. So $E_n=0$, and this works for all $n\ge n_0$.
So $W_{\infty}=0$, that is, $E_n=0$ for all large $n$. Similarly $E_{-n}=0$ for all large $n$. Using commutators we deduce that $E_n=\frac{1}{n+2q}[E_{n+q},E_{-q}]=0$ (choosing $q$ such that $n+2q\neq 0$ in $K$).
A: As noted by Qiaochu Yuan in the comments, every infinite-dimensional simple Lie algebra does the trick.
(Since the finite-dimensional representations of such a Lie algebra are trivial.)
We may, for example, regard $\mathfrak{sl}_n(ℂ)$ as a Lie subalgebra of $\mathfrak{sl}_{n + 1}(ℂ)$ and then form the union $\mathfrak{sl}_∞(ℂ) = ⋃_{n = 2}^∞ \mathfrak{sl}_n(ℂ)$.
Every nonzero ideal $I$ of $\mathfrak{sl}_∞(ℂ)$ must intersect some $\mathfrak{sl}_N(ℂ)$ in a nonzero way.
It then follows for every $n ≥ N$ that the intersection $I ∩ \mathfrak{sl}_n(ℂ)$ is a nonzero ideal of $\mathfrak{sl}_n(ℂ)$, from which it further follows that $I ∩ \mathfrak{sl}_n(ℂ) = \mathfrak{sl}_n(ℂ)$ because $\mathfrak{sl}_n(ℂ)$ is simple.
Thus $I = \mathfrak{sl}_∞(ℂ)$, showing that $\mathfrak{sl}_∞(ℂ)$ is simple.
