Let $\mathfrak{g}$ be a complex semisimple Lie algebra. Is it true that for any $X\in\mathfrak{g}$, there exists an $\mathfrak{sl}_2$-triple $(e,h,f)$ in $\mathfrak{g}$ such that

  1. We have $X\in e+Z_{\mathfrak{g}}(f)$, and
  2. $\dim Z_{\mathfrak{g}}(X)=\dim Z_{\mathfrak{g}}(e)$?

($Z_{\mathfrak{g}}(-)$ always the centralizer in $\mathfrak{g}$. One can deduce from the above conditions that the conjugacy orbit of $X$ must then meet the slice transversally, like in classical Kostant section situation.)

When $X$ is regular this is the well-known result about Kostant section. When $\mathfrak{g}=\mathfrak{gl}_n$ this is also true and can be deduced from the rational form of $X$. It might be that this question can be answered by a simple reference, but thanks a lot in advance in all cases!

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    $\begingroup$ Although it's a different question, you may be interested in @AlexanderPremet's answer to my question Commutativity and Kostant sections. $\endgroup$
    – LSpice
    Mar 18, 2022 at 19:59
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    $\begingroup$ Thank you! Yeah I had been attracted to your question when I looked for an answer to this 1.5 years ago :) $\endgroup$ Mar 19, 2022 at 2:01
  • $\begingroup$ Btw I think you will be amused by the following proof: choose $X=X_s+X_n$ in a split $G$ over $\mathbb{Q}_p$ with $X_s$ split. Consider the Shalika germ expansion (do we need $p\gg\operatorname{rank}(G)$ to have germs for arbitrary $X$?) for $X$. We claim that for any nilpotent $e$ with non-zero germ $\Gamma_e(X)$ we have the asserted property. $\endgroup$ Mar 19, 2022 at 2:02
  • $\begingroup$ I then turn the above into AG following inspiration from $\S5$ of Langlands-Shelstad '87 (... transfer factor) which is based on Langlands '83 (Orbital Integrals on Forms of SL(3), I). $\endgroup$ Mar 19, 2022 at 2:04
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    $\begingroup$ Explicit asymptotic expansions in $p$-adic harmonic analysis II gives fairly explicit conditions for existence of germs, but not much weaker than you'd expect from DeBacker+Kim–Murnaghan-type results. I don't think I follow the second step of your proof, but I like the first step! $\endgroup$
    – LSpice
    Mar 19, 2022 at 14:33

1 Answer 1


Woooo, I think I finally find a proof!$ \newcommand{\Lie}{\operatorname{Lie}} \newcommand{\Lg}{\mathfrak{g}} \newcommand{\Ll}{\mathfrak{l}} \newcommand{\Ad}{\operatorname{Ad}}$

Let $G=\operatorname{Aut}(\Lg)^\circ$ be a corresponding complex Lie group. Write $X=X_s+X_n$ the Jordan decomposition into semisimple and nilpotent part. The centralizer $L:=Z_G(X_s)$ is a Levi subgroup. Denote by $\Ll=\Lie L$ and $Z(\Ll)=\Lie Z(L)$ the center of $\Ll$. Write $\mathcal{O}:=Z(\Ll)\cdot\Ad(L)X_n$, a locally closed subvariety in $\Ll$. Note that $\mathcal{O}\cong Z(\Ll)\times\Ad(L)X_n$ has an obvious map to $Z(\Ll)$.

Pick a parabolic subgroup $P$ containing $L$ as its Levi subgroup. Let $U$ be the unipotent radical of $P$ so that $P=LU$. There exists a nilpotent $G$-orbit in $\Lg$ that intersects $X_n+\Lie U$ at a dense open subset of $X_n+\Lie U$. Let $e$ be any element in the intersection. We have to prove the two asserted property in the question, i.e.

  1. $\Ad(G)X$ meets the Slodowy slice $e+Z_{\Lg}(f)$ and
  2. $\dim Z_G(X)=\dim Z_G(e)$.

We have $Z_G(X)=Z_L(X_n)$ and also $\dim Z_G(e)=\dim Z_L(X_n)$ [LS79, Theorem 1.3]. This proves (2).

Consider the “generalized Grothendieck-Springer alternation” given by $$ \tilde{\mathfrak{g}}_X:=\{(g,\gamma)\in (P\backslash G)\times\mathfrak{g}\;|\;\Ad(g)\gamma\in\Lie P\text{ is such that its image under}\Lie P\twoheadrightarrow\Lie L\text{ lies in }\mathcal{O}\}. $$ It is the usual Grothendieck–Springer alternation if $X$ is regular semisimple, hence the name. There is a natural smooth map $\mu:\tilde{\mathfrak{g}}_X\rightarrow Z(\Ll)$ via $\mathcal{O}\twoheadrightarrow Z(\Ll)$ and another natural $G$-equivariant map $\pi:\tilde{\Lg}_X\rightarrow\Lg$ sending $(g,\gamma)\mapsto\gamma$. For $\zeta\in Z(\Ll)$ that are $(G/L)$-regular (this means $Z_G(\zeta)=L$), we have $\pi(\mu^{-1}(\zeta))=\Ad(G)(\zeta+X_n)$. For the central fiber we have instead $\pi(\mu^{-1}(0))\supset\Ad(G)e$.

Since $e+Z_{\Lg}(f)$ meets $\Ad(G)e$ transversally, there exists a neighborhood (analytically or algebraically, whichever) $U$ of $e$ in $\Lg$ such that for any orbit $\Ad(G)Y$ we have $\Ad(G)Y\cap U\not=\emptyset\implies\Ad(G)Y\cap (e+Z_{\Lg}(f))\ne\emptyset$. The preimage $\pi^{-1}(U)$ intersects $\mu^{-1}(0)$. Thanks to smoothness of $\mu$, there exists a neighborhood $U_{Z(\Ll)}$ of $0\in Z(\Ll)$ such that $\pi^{-1}(U)$ intersects $\mu^{-1}(\zeta)$ for any $\zeta\in U_{Z(\Ll)}$. For $\zeta\in U_{Z(\Ll)}\cap Z(\Ll)^{\text{$(G/L)-reg$}}$ this means $U$ intersects $\Ad(G)(\zeta+X_n)$, and thus the asserted property (1) is true for ($X$ replaced by) $\zeta+X_n$.

Since nilpotent orbits are stable under scalar scaling, that (1) is true for $\zeta+X_n$ implies that it is true for $c(\zeta+X_n)$ for all $c\in\mathbb{C}^{\times}$ and $\zeta\in U_{Z(\Ll)}\cap Z(\Ll)^{\text{$(G/L)-reg$}}$, and thus also for $c\zeta+X_n$ since $c(\zeta+X_n)$ and $c\zeta+X_n$ are also conjugate. Since such $c\zeta$ covers all of $Z(\Ll)^{\text{$(G/L)-reg$}}$, the assertion for $X=X_s+X_n$ is proved.

[LS79] Lusztig, G.; Spaltenstein, N., Induced unipotent classes, J. Lond. Math. Soc., II. Ser. 19, 41-52 (1979). ZBL0407.20035.

Sadly, I can't recall why I needed this cute property ….

  • $\begingroup$ p.s. I now remember my motivation: I wanted to study the affine Springer fiber over $X$, and wanted a definition of a "regular locus" (and have it non-empty!) like in the situation when $X$ is regular semisimple. As I commented under the question, this answer will lead to a definition at least when $X_s$ is split. Analogous construction might or might not work in the non-split case ... $\endgroup$ Mar 19, 2022 at 6:09

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