Why is only assuming $\mathfrak g$ is nilpotent as a Lie algebra not sufficient for Engel's theorem?

Question

On the wikipedia page of Engel's theorem it says that "if we merely have a Lie algebra of matrices which is nilpotent as a Lie algebra, then this conclusion does not follow", meaning that it might not be possible to choose a basis in which all elements are upper triangular. Can someone give me an example and explain why is this true for the adjoint representation? Is it because its kernel is commutative?

Answer 1

Let our ground field be $\mathbb C$ (but notice nothing depends on that).

Example 1: Let $\mathfrak g_1 = \{\pmatrix{x&0\\0&x}: x \in \mathbb C \}$. You will agree that this one-dimensional Lie algebra is abelian, i.e. (more than) nilpotent, but that these matrices cannot all be brought in strictly upper triangular form.

Example 2: Let $\mathfrak g_2$ be the two-dimensional abelian (i.e. $[e_1,e_2]=0$) Lie algebra with basis $e_1, e_2$. Define a representation $\rho : \mathfrak g_2 \rightarrow \mathfrak{gl}_2(\mathbb C)$ via $\rho(e_1) := \pmatrix{1&0\\0&1}, \rho(e_2) := \pmatrix{0&0\\0&0}$. Check that the kernel pf $\rho$ is commutative, but its image is isomorphic to our first example $\mathfrak g_1$, in particular still cannot be brought to strictly upper triangular form, in spite of $\mathfrak g_2$ being nilpotent.

Exercise 1: Come up with more examples. What about a nilpotent (e.g. abelian) Lie algebra, of dimension $6$, made up of $27\times 27$-matrices some of which are diagonal, some of which have imaginary eigenvalues, some of which are upper and some lower triangular?

Exercise 2: What are the adjoint representations of $\mathfrak g_1, \mathfrak g_2$? What does the adjoint representation take into consideration which the identical representation of $\mathfrak g_1$ and the representation $\rho$ of $\mathfrak g_2$ do not care about?