A strong relationship between $\mathrm{ad}(X)$ and $1-\mathrm{Ad}_g$ when $\mathrm{Ad}_gX=X$

Question

Let $G$ be a linear algebraic semisimple group over $\mathbb{C}$ with Lie algebra $\mathfrak{g}$, and let $(g,X)\in G\times\mathfrak{g}^{\operatorname{reg}}$ be such that $\mathrm{Ad}_gX=X$ (where $X\in\mathfrak{g}^{\operatorname{reg}}$ means that $Z_{\mathfrak{g}}(X)$ has minimal dimension).

Conjecture. We have $$\operatorname{im}(1-\mathrm{Ad}_g)\subseteq\operatorname{im}\mathrm{ad}(X), \quad\text{and}\quad\ker\mathrm{ad}(X)\subseteq\ker(1-\mathrm{Ad}_g).$$

I already know that the second identity is true. Indeed, using the accepted answer to this question, $Z_G(X)$ is abelian and hence $Z_G(X)\subseteq Z_G(g)$. Passing to Lie algebras gives that $\ker\mathrm{ad}(X)\subseteq\ker(1-\mathrm{Ad}_g)$.

I believe the first identity is also true. At least, it is true when $G=\mathrm{SL}(2,\mathbb{C})$, as I verified it by brut force. It also seems true for $G=\mathrm{SL}(n,\mathbb{C})$, as I verified the identity in many cases using a symbolic mathematical computation program. But I don't have any conceptual reason for why this happens.

I would be interested to know if the conjecture is true, or if we can determine a subclass of the linear algebraic semisimple groups for which it holds (for all pairs $(g,X)$).

Answer 1

Consider the Killing form on $\mathfrak g$. Since $Ad_g$ is orthogonal, it is easy to see that $im(1-Ad_g)^\perp=ker(1-Ad_g)$. Since $ad_g$ is anti-selfadjoint, $im(ad(x))^\perp=ker(ad(x))$. Therefore, your two statements are equivalent. Since you know the second holds, so does the first.

Answer 2

Your conjecture that ${\rm im} (1-{\rm Ad}_g)\subseteq {\rm im}( {\rm ad}\,X)$ is indeed true. Let me sketch the argument.

First we reduce to the case where the regular element $X\in \mathfrak{g}$ is nilpotent. In $\mathfrak g$, we have the Jordan-Chevalley decomposition $X=X_s+X_n$ of $X$ and the centraliser $\mathfrak{l}={\rm Lie}(L)$ of $X_s$ in $\mathfrak g$ is a Levi subalgebra of $\mathfrak g$. The centraliser $L=Z_G(X_s)$ is a Levi subgroup of $G$; in particular, it is connected and reductive. Since ${\rm ad}\,X$ acts invertibly on the eigenspaces $\mathfrak{g}_\lambda$ of ${\rm ad}\,X_s$ with $\lambda\ne 0$ and ${\rm Ad}_g-1$ preserves each $\mathfrak{g}_\lambda$ we just need to show that $({\rm Ad}_g-1)(\mathfrak{l})\subseteq [X,\mathfrak{l}]$. The subspace on the right equals $[X_n,\mathfrak{l}]$ and $X_n$ is a regular nilpotent element of $\mathfrak{l}$. Since $\mathfrak{l}=\mathfrak{z}(\mathfrak{l})\oplus [\mathfrak{l},\mathfrak{l}]$ and $L=Z(L)\cdot (L,L)$ we may assume without loss that $L=G$ and $X=e$ is regular nilpotent. The centraliser $Z_G(e)$ is generated by $Z(G)$ and the connected component $Z_G(e)^\circ$ which is unipotent. So no generality will be lost by assuming that $g$ is unipotent.

So far the argument works under very mild assumptions on the characteristic of the base field, but to make this post shorter I will now use exponentiation and hence rely on the fact that ${\rm char}(K)=0$. Since $g\in Z_G(e)$ is unipotent we can write ${\rm Ad}_g =\exp\,{\rm ad}\,n$ for some nilpotent element $n\in \mathfrak{g}_e$ where $\mathfrak{g}_e$ denotes the centraliser of $e$ in $\mathfrak g$. Moreover, ${\rm im}({\rm Ad}_g-1)={\rm im}({\rm ad}\,n)$. So we just need to show that $[n,\mathfrak{g}]\subseteq [e,\mathfrak{g}]$ for any $n\in \mathfrak{g}_e$.

As the Killing for $\kappa$ of $\mathfrak g$ is nondegenerate and $\mathfrak{g}$-invariant, it is easy to see that $\mathfrak{g}_e$ coincides with the orthogonal complement of $[e,\mathfrak{g}]$ with respect to $\kappa$. So we just need to check that $\kappa([n,\mathfrak{g}],\mathfrak{g}_e)=0$. By the $\mathfrak{g}$-invariance of $\kappa$, this is equivalent to $[n,\mathfrak{g}_e]=0$. The latter holds because for $e$ regular the centraliser $\mathfrak{g}_e$ is abelian.