# Slodowy slice intersecting a given orbit "minimally"?

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!

• Although it's a different question, you may be interested in @AlexanderPremet's answer to my question Commutativity and Kostant sections. Mar 18, 2022 at 19:59
• Thank you! Yeah I had been attracted to your question when I looked for an answer to this 1.5 years ago :) Mar 19, 2022 at 2:01
• 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. Mar 19, 2022 at 2:02
• 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). Mar 19, 2022 at 2:04
• 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! Mar 19, 2022 at 14:33

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 ….

• 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 ... Mar 19, 2022 at 6:09