Let $\mathfrak{g}$ be a complex semisimple Lie algebra and $(e,f,h)$ a principal $\mathfrak{sl}_2$-triple (see below). Let $\mathfrak{g}_e$ be the centralizer of $e$ and let $x\in f+\mathfrak{g}_e$. Is it true that
$$\mathfrak{g}=\mathfrak{g}_e\oplus[x,\mathfrak{g}]?$$
**Remarks.**

(i) The affine space $f+\mathfrak{g}_e$ is sometimes called the Kostant section, Kostant slice, or principal Slodowy slice. We have $f+\mathfrak{g}_e\subseteq\mathfrak{g}^{reg}$.

(ii) Since $x,e\in\mathfrak{g}^{reg}$ we have $\dim\mathfrak{g}_e+\dim[x,\mathfrak{g}]=\dim\mathfrak{g}$ so it suffices to show that $\mathfrak{g}=\mathfrak{g}_e+[x,\mathfrak{g}]$.

(iii) When $\mathfrak{g}$ is of type $A_1$ or $A_2$, I verified by brute force that indeed $\mathfrak{g}=\mathfrak{g}_e\oplus[x,\mathfrak{g}]$.

**Definition.** A *principal $\mathfrak{sl}_2$-triple* in $\mathfrak{g}$ is a triple $(e,f,h)$ of elements of $\mathfrak{g}$ such that $[h,e]=2e,[h,f]=-2f,[e,f]=h$ and $e$ is *regular*, i.e. its centralizer has dimension equal to the rank of $\mathfrak{g}$.

transverse to the adjoint orbits, meaning that if $x\in f+\mathfrak{g}_e$, then $T_x\mathfrak{g}=T_x(f+\mathfrak{g}_e)\oplus T_x(G\cdot x)$ where $G$ is the adjoint group of $\mathfrak{g}$. Under the identification $T_x\mathfrak{g}=\mathfrak{g}$, we recover the identity in question. $\endgroup$ – SHP May 18 '18 at 8:21Quantization of Slodowy slices, Section 2.2. But I think it goes back to Slodowy's work. $\endgroup$ – SHP May 18 '18 at 13:51