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 $\mathfrak{g}$.

  • $\begingroup$ Something is wrong here: $e\in[x,\mathfrak g]\cap\mathfrak g_e$ for e. g. $x=f$ $\endgroup$ – მამუკა ჯიბლაძე May 17 '18 at 11:35
  • $\begingroup$ @მამუკაჯიბლაძე No: $e\notin [f,\mathfrak{g}]$ [$:=\mathrm{Im}(ad(f))$] in $\mathfrak{sl}_2$. $\endgroup$ – YCor May 17 '18 at 11:37
  • $\begingroup$ @YCor Oh yes sorry $\endgroup$ – მამუკა ჯიბლაძე May 17 '18 at 11:38
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    $\begingroup$ This is one of the main reasons why Slodowy slices are studied. They are 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 8:21
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    $\begingroup$ @მამუკაჯიბლაძე It is stated, for example, in Gan--Ginzburg Quantization of Slodowy slices, Section 2.2. But I think it goes back to Slodowy's work. $\endgroup$ – SHP May 18 '18 at 13:51

This is indeed the case. To see this, filter $\mathfrak{g}$ by $Ker(ad_e^i)$. Let $Gr(\mathfrak{g})$ be the associated graded. We have a symbol map $sym : \mathfrak{g} \to Gr(\mathfrak{g})$ given by sending an element to the "leading term" w.r.t. this filtration. It would suffice to show that it is generated by the symbols of elements of $\mathfrak{g}_e + [x,\mathfrak{g}]$, by a standard descent argument (look at the least element not in your subspace in terms of this filtration) But, after passage to the associated graded, $ad_x$ acts on symbols like $ad_f$ modulo previous graded pieces. More precisely, we have $sym(ad_x(u)) = sym(ad_f(u))$ for every $u$. This shows that the image of the symbol don't change if you replace $x$ by $f$, so we can reduce to this case. But for $x = f$ the statement is a standard result on representations of $\mathfrak{sl}_2$ with no trivial components (which follows from the triple being principal in this case).

Edit: As mensioned in the comment below, the condition on non-triviality of the representation is irrelevant here. It is just true for every $\mathfrak{sl}_2$ representation.

  • $\begingroup$ Should not $\mathfrak g_e + [x,\mathfrak g]=\mathfrak g$ be true for any nilpotent $e$ whatsoever? I mean, $\mathfrak g=\mathfrak g_e\oplus[f,\mathfrak g]$ seems to be true even if there are trivial components. $\endgroup$ – მამუკა ჯიბლაძე May 17 '18 at 15:39

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