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](https://mathoverflow.net/q/253456/90299) 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 $(g,X)$).



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


