Let $\mathfrak{g}$ be a (finite dimensional) semi-simple Lie algebra over a field $k$ and let $x \in \mathfrak{g}$. By definition, we have the equivalence:
$$ \mathrm{rk}(\mathrm{ad}_x) = 0 \iff x = 0,$$
where $\mathrm{rk}(\mathrm{ad}_x)$ is the rank of $\mathrm{ad}_x$ seen as an element of $\mathrm{End}(\mathfrak{g})$. I would like to know if there is a classification of elements $x \in \mathfrak{g}$ such that $\mathrm{rk}(\mathrm{ad}_x) \leq 1$? I am primarily interested in the case where $k = \mathbb{C}$ and $\mathfrak{g}$ is simple of classical type.
Lie Algebra – Rank One Adjoint Operators on a Lie Algebra
lie-algebraslie-groupsmatricesrt.representation-theory
Best Answer
Another approach.
To show it's impossible (the rank can't be 1), it is enough to show this when the field (assumed of char 0) is algebraically closed, and in turn it's enough to show the result in case $\mathfrak{g}$ is simple. If $x$ has $\mathrm{ad}(x)$ of rank 1, $x$ has centralizer of codimension 1. It is known (see e.g. this MathSE answer) that $\mathfrak{g}$ has no subalgebra of codimension 1, unless $\mathfrak{g}$ is 3-dimensional. But for $\mathfrak{sl}_2$, the operator $\mathrm{ad}_x$ has rank 2 for every nonzero $x$ (alternatively, all 2-dimensional subalgebras have a trivial centralizer, so can't be centralizer of an element).
A similar approach can be used (with a little further work) to classify which $\mathfrak{g}$ admit an $x$ with $\mathrm{ad}(x)$ of rank 2, and classify such elements $x$.