A lower bound of the number of conjugacy classes in the automorphism group of a simple Lie algebra $\mathfrak{s}$, of finite dimension over an arbitrary field $\mathbb{F}$, can be the size of the image of the automorphism group by the trace.


I think this map $\operatorname{tr}:\operatorname{Aut}(\mathfrak{s})\longrightarrow\mathbb{F}$ must be surjective. However, I have no clue. (For splitable Lie algebras, there is an explicit way).



Conjugacy classes in the automorphism group of a simple Lie algebra
1. If $\mathfrak{s}$ is $K$-anisotropic, where $K$ is a real or $p$-adic field (this is equivalent to $\mathfrak{s}$ not containing $\mathfrak{sl}\_2$, and also to the corresponding group be compact), then the trace is bounded, hence non-surjective.
2. Suppose that $\mathrm{Aut}(\mathfrak{s})^0$ is Zariski-dense in the underlying algebraic group (this is automatic if $K$ is perfect, by Rosenlicht's theorem). In this case, by Zariski density, if $\mathrm{Aut}(\mathfrak{s})^0$ has infinitely many traces (resp. characteristic polynomials), then the same holds over the algebraic closure. So we can reduce (in this case) to the algebraically closed case.
3. In the algebraically closed case, if the number of characteristic polynomials of $\mathrm{Aut}(\mathfrak{s})^0$ is finite, it is (by connectedness) reduced to one, and hence it follows that every element in $\mathrm{Aut}(\mathfrak{s})^0$ is unipotent. In characteristic zero, this is not possible. I'm not sure about the modular case.


The same conclusion holds for the number of traces, although the characteristic zero is then used in a stronger way.


4. If $\mathfrak{s}$ is the Lie algebra of a simple algebraic group, the restrictions on the characteristic should be dropped (still needing Rosenlicht), to infer that the number of characteristic polynomials is infinite.


