Understanding the set of regular elements in a Lie algebra I was reading the book "Lie groups: Beyond an Introduction" by A. Knapp, when I came across the concept of regular elements. They are defined in the following way (in the book).
Let $\sigma: \mathfrak{g} \rightarrow \mathfrak{gl} \left( V \right)$ be a finite dimensional representation. For $X \in \mathfrak{g}$, we consider $\mathfrak{h}_X = \text{span} \left\lbrace X \right\rbrace$. Then, we define
$$V_{0, X} := \left\lbrace v \in V | \sigma \left( H \right)^n v = 0 \text{ for all } H \in \mathfrak{h}_X \text{ and some } n \in \mathbb{N} \right\rbrace,$$
which is the generalized weight space for generalized eigenvalue $0$ corresponding to the commutative subalgebra $\mathfrak{h}_X$. Then, we fix $l_{\mathfrak{g}} = \min \left\lbrace \dim V_{0, X} | X \in \mathfrak{g} \right\rbrace$, and define
$$R_{\mathfrak{g}} \left( V \right) := \left\lbrace X \in \mathfrak{g} | \dim V_{0, X} = l_{\mathfrak{g}} \left( V \right) \right\rbrace.$$
For the adjoint representation $ad: \mathfrak{g} \rightarrow \mathfrak{gl} \left( \mathfrak{g} \right)$, we similarly define $\mathfrak{g}_{0, X}$, $l_{\mathfrak{g}} \left( \mathfrak{g} \right)$, and $R_{\mathfrak{g}} \left( \mathfrak{g} \right)$, and call the elements of $R_\mathfrak{g} \left( \mathfrak{g} \right)$, regular elements.
I wish to understand what these elements will be. Later in the text, the author uses a result that $R_{\mathfrak{g}} \left( \mathfrak{g} \right)$ is dense. I was thinking that perhaps, $R_{\mathfrak{g}} \left( \mathfrak{g} \right)$ is the complement of a union of finitely many hyperplanes, which makes it dense. However, I have no idea how to prove this. Any suggestions for this will be appreciated!
Also, if there is a way to get a more general result, say for $R_{\mathfrak{g}} \left( V \right)$, I would also be happy to know it.
 A: "I wish to understand what these elements will be."
One way to understand regular elements is an example. As always, the first interesting example is the Lie algebra ${\mathfrak sl}_2(k)$.
Note that for an algebraically closed field $k$, and a Lie algebra $L$ over $k$, the space
$L_0 (h)$ is a Cartan subalgebra for every $h\in L^{reg}$.
Let $L={\mathfrak sl}_2(k)$, ${\rm char} (k)\neq 2$ and
$$
x=\begin{pmatrix} a & b \\ c & -a \end{pmatrix}
$$
be an element in $L$. Then the characteristic polynomial of $x$ is given by
$$
P_x(t)=\det 
\begin{pmatrix}
t-2a & 0 & 2b \\
0 & t+2a & -2c \\
c & -b & t 
\end{pmatrix}
= t^3+4t \det (x).
$$
Hence $a_1(x)=4 \det (x)$ for all $x\in L$, and $a_0=0$. Thus ${\rm rank}(L)=1$ and $x$ is regular if and only if
$\det (x)\neq 0$.
Even better, because of ${\rm tr} (x)=0$ we have that $x\in L$ is regular if and only if $x$ is not nilpotent. So we have
$$
\mathfrak{sl}_2(k)^{reg}=\mathfrak{sl}_2(k)\setminus \mathcal{N},
$$
where $\mathcal{N}$ denotes the cone of nilpotent matrices in $\mathfrak{sl}_2(k)$.
