I'm looking for a source of properties for semisimple Lie algebra elements, specifically finite dimensional classical Lie algebras.


I start with the assumption that I have a complexified Lie algebra $\mathfrak{g}^\mathbb{C}$, with a semisimple element $\Lambda \in \mathfrak{g}^\mathbb{C}$, and that it's conjugate $\bar{\Lambda}$ (under the complex structure of $\mathfrak{g}^\mathbb{C}$) has the property that,
$$\left[ \Lambda , \bar{\Lambda}\right] = 0$$
Does this imply that $ker\> ad \> \Lambda = ker\> ad \> \bar{\Lambda}$, I know that in many cases I've worked with, it does, yet is there a reason why in general this is or isn't true?



Properties of semisimple Lie algebra elements
First the counterexample: Let $\mathfrak g=\mathfrak{sl}(3,\mathbb C)$ and
$$
\Lambda:=\left[ \begin {array}{ccc} 1&-3\,i&0\\3\,i&1&0
\\0&0&-2\end {array} \right]
$$
Then $\Lambda$ and $\overline\Lambda$ commute and are semisimple. But only $\Lambda$ commutes with
$$
\left[ \begin {array}{ccc} 0&0&i\\0&0&1
\\0&0&0\end {array} \right]
$$
Now the explanation: Write $\Lambda=\Lambda\_{re}+i\Lambda\_{im}$ the decomposition into real and imaginary part. Then $\Lambda\_{re},\Lambda\_{im}$ span a semisimple commutative subalgebra of $\mathfrak g$. Therefore they lie in a Cartan subalgebra $\mathfrak t$ of $\mathfrak g$. The eigenvalues of $\mathfrak t$ on $\mathfrak g$ are the roots. If $\alpha$ is a root then its conjugate $\overline\alpha$ is a root, as well. So starting with $\mathfrak t$ we are done if we can find $\Lambda\in\mathfrak t$ and a root $\alpha$ with $\alpha(\Lambda)=0$ and $\overline\alpha(\Lambda)\ne0$. This is possible iff $\alpha$ and $\overline\alpha$ are linearly independent iff $\overline\alpha\ne\pm\alpha$. Roots with this property are called complex. In the example above, I chose
$$
\left[ \begin {array}{ccc} x&-y&0
\\y&x&0\\0&0&-2
\,x\end {array} \right] 
$$
The eigenvalues are $x\pm iy$ and $-2x$. The roots are their differences. Then $(x+iy)-(-2x)=3x+iy$ is a complex root. So choose $x=1$ and $y=3i$.


