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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?

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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 $y=3i$.

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  • $\begingroup$ If I asked for the elements to be outside of the cartan subalgebra, namely $\Lambda \not\in \mathfrak{t}^\mathbb{C}$ nor the real and imaginary parts in the real Cartan subalgebra, would this change the general picture at all? $\endgroup$ – Jack Moon Aug 14 '16 at 1:58

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