I am current reading the paper: Classification of Good Gradings of Simple Lie Algebras, here's a link!
Lemma 1.1: Let $e \in \mathfrak{g}_2$, $e \neq 0$.
Then there exists $h \in \mathfrak{g}_0$ and $f \in g_{-1}$ such that $\{e, h, f\}$ form an $sl_2$-triple,
i.e., $[h, e] = 2e, [e, f] = h, [h, f] =-2f.$
Theorem 1.1: Let $\mathfrak{g}=\bigoplus_{j\in\mathbb{Z}} \mathfrak{g}_j$ be a good $\mathbb{Z}$-grading and $e \in \mathfrak{g}_2$ a good element.
Let $H \in \mathfrak{g}$ be the element defining the $\mathbb{Z}$-grading (i.e., $\mathfrak{g}_j = \{a \in \mathfrak{g} \ | \ [H, a] = ja\}$), and let $\mathfrak{s}= \{e, h, f\}$ be an $sl_2$-triple given by above Lemma.
Then $z := H - h$ lies in the center of $\mathfrak{g}^\mathfrak{s}$.
I want to ask the following
Why such $H$ exists? My idea is left multiplication by $j$ is a derivation, then $[H,a]=ja$ since any derivation is inner, but how to show $H$ is independent of $j$?
In the proof of the theorem on P.3, why $\mathfrak{g}^s$ is reductive implies $[H, \mathfrak{g}^s] = 0$? Is it because of $\mathfrak{g}^\mathfrak{s}\subseteq\mathfrak{g}_0$?
Thank you for helping!