Finite order automorphisms of semi simple lie algebras (Kac Lemma 8.1) I am currently reading Kac's book on infinite dimensional Lie algebras and have some trouble with Lemma 8.1. The setup is as follows:
Let $\mathfrak{g}$ be a finite dimensional semisimple Lie algebra over $\mathbb{C}$ and $\sigma \in \text{Aut}(\mathfrak{g})$ be a finite order automorphism i.e $\sigma^m=id_{\mathfrak{g}}$. Then $\sigma$ acts semisimply with eigenvalues the $m$-th roots of unity and hence we can put a $\mathbb{Z}/m \mathbb{Z}$   grading on $\mathfrak{g}= \bigoplus_{n \in \mathbb{Z}/m \mathbb{Z}} {\mathfrak{g}_\bar{n}} $. Letting $(-,-)$ denote the killing form Kac proves that $(\mathfrak{g}_\bar{i},\mathfrak{g}_\bar{j})=0$  if $i+j \neq 0 \mod m$ and that the killing form gives a non-degenerate pairing between $\mathfrak{g}_\bar{i}$ and $\mathfrak{g}_\bar{j}$ otherwise. I can follow the proof of this fine but I suspect it may help in proving the statement I am having trouble with. Now letting $\mathfrak{h}_\bar{0}$ be a maximal $ad$ (here this is with respect to the adjoint representation on $\mathfrak{g}$) diagonalizable subalgebra of $\mathfrak{g}_\bar{0}$ one can see that the centralizer  $C_\mathfrak{g}(\mathfrak{h}_\bar{0})= \mathfrak{h} + \Sigma_{\alpha} \mathfrak{g}_{\alpha}$ where $\mathfrak{h}$ is a cartan subalgebra of $\mathfrak{g}$ and $\mathfrak{g}_{\alpha}$ are the root spaces with respect to $\mathfrak{h}$ such that $\alpha(h_0)=0 \ \forall \ h_0 \in \mathfrak{h}_\bar{0}  $. Indeed as $\mathfrak{h}_0$ is ad diagonalizable it is contained in some Cartan subalgebra.
Kac now claims that $C_\mathfrak{g}(\mathfrak{h}_0)= \mathfrak{h} + \mathfrak{s}$ where $\mathfrak{s}$ is semisimple and $\mathfrak{g}_\bar{0} \cap \mathfrak{s} = 0$. I guess $\mathfrak{s}$ is just the semisimple Lie algebra generated by the $\mathfrak{g}_{\alpha}$ but I cannot see how the intersection with $\mathfrak{g}_\bar{0}$ is trivial. I tried showing that  $\mathfrak{g}_\bar{0}$  cannot contain a root space $\mathfrak{g}_\alpha$ and showing it would have to be contained in the cartan as a conseqeunce but even this I cannot prove as $\mathfrak{g}_\bar{0}$ is not semisimple. (Kac claims $\mathfrak{g}_\bar{0}$ need only be reductive which should follow from the aforementioned facts about the centralizer; I don't see how this works either and would be grateful if someone could explain it to me) I also tried somehow making use of the facts about the killing form mentioned earlier but this also does not really seem to help besides that fact that if $\mathfrak{g}_\bar{0}$ contains a root space it also contains the negative root space. EDIT: As kindly pointed out by Torsten Schöneberg in the comments this is not true.
Any help and/or hint as to show this intersection is trivial and how it implies $\mathfrak{g}_\bar{0}$ reductive would be greatly appreciated.
 A: It doesn't seem obvious to me. Here is one approach: by using exactly the same scheme of proof as Proposition 8.2 from Humphrey's book Introduction to Lie algebras and representation theory one can show that

*

*The restriction  of the Killing form to $\mathfrak{h}_\overline{0}$ is non-degenerate, and


*The centralizer $C$ of $\mathfrak{h}_\overline{0}$ in $\mathfrak{g}_\overline{0}$ is $\mathfrak{h}_\overline{0}$.
Moreover,


*The Lie algebra $\mathfrak{g}_\overline{0}$ is reductive since the restriction of the Killing form on $\mathfrak{g}$ to it is non-degenerate: this follows from the general fact that if a Lie algebra has a representation such that the associated trace form is non-degenerate then the Lie algebra is reductive.

I do not see how Kac's argument proves that $\mathfrak{g}_\overline{0}$ is reductive, but the above argument is quite direct so I won't worry about it too much.
For your convenience, I will reproduce the proof of 1. and 2. below, with some minor differences to what Humphreys writes. First let's see how 1. and 2. imply Kac's remaining assertion.
By 2. $C_\mathfrak{g}(\mathfrak{h}_{\overline{0}}) \cap \mathfrak{g}_{\overline{0}}=\mathfrak{h}_{\overline{0}}$. Now let $\mathfrak{s}$ be the subalgebra of $\mathfrak{g}$ generated by the set of $\mathfrak{g}_\alpha$'s with $\alpha(\mathfrak{h}_{\overline{0}})=0$. By the preceding, the intersection of $\mathfrak{s}$ with $\mathfrak{g}_{\overline{0}}$ is contained in $\mathfrak{h}_{\overline{0}}$. On the other hand, we have $(x,y)=0$ for all $x \in \mathfrak{s}$ and $y \in \mathfrak{h}_\overline{0}$: this follows from the fact that $\mathfrak{h}$ orthogonal to all $\mathfrak{g}_\alpha$'s and
$$([x_1,x_2],y)=(x_1,\alpha(y) x_2) \quad \hbox{for $x_1 \in \mathfrak{g}_\alpha$ and $x_2 \in \mathfrak{g}_{-\alpha}$.}$$ Now use 1. to conclude that $\mathfrak{s} \cap \mathfrak{h}_\overline{0}=0$ and hence $\mathfrak{s} \cap \mathfrak{g}_\overline{0}=0$.
Proof of 1. and 2.:
Let $C$ be the centralizer of $\mathfrak{h}_{\overline{0}}$ in $\mathfrak{g}_{\overline{0}}$. Since the restriction of the Killing form to $\mathfrak{g}_{\overline{0}}$ is non-degenerate, the same holds for its restriction to $C$ (by considering the eigen-space decomposition for $\mathfrak{g}_{\overline{0}}$ with respect to $\mathrm{ad}(\mathfrak{h}_{\overline{0}})$).
We next prove that in fact $C=\mathfrak{h}_{\overline{0}}$. Given $x \in \mathfrak{g}$ we will write $x_s$ and $x_n$ for its semi-simple and nilpotent parts, which are the unique elements of $\mathfrak{g}$ such that $\mathrm{ad}(x)=\mathrm{ad}(x_s)+\mathrm{ad}(x_n)$ is the Jordan decomposition of $\mathrm{ad}(x)$. We note that if $x \in C$, then $x_s$ and $x_n$ both belong to $C$ as well (since the semi-simple and nilpotent components of a linear transformation $T$ are polynomials in $T$). By maximality of $\mathfrak{h}_{\overline{0}}$ each semi-simple element of $C$ belongs to $\mathfrak{h}_{\overline{0}}$. Therefore it suffices to show that each nilpotent $x \in C$ belongs to $\mathfrak{h}_{\overline{0}}$.
To do this, we first observe that for all $x \in C$, the adjoint representation $\mathrm{ad}_C(x)$ of $x$ on $C$ is a nilpotent operator: writing $x=x_s+x_n$ we have $x_s \in \mathfrak{h}_{\overline{0}}$ by the above, and therefore
$$\mathrm{ad}_C(x)=\mathrm{ad}_C(x_s)+\mathrm{ad}_C(x_n)=\mathrm{ad}_C(x_n)$$ is nilpotent. It follows that $C$ is a nilpotent Lie algebra.
We next observe that the restriction of the Killing form $(\cdot,\cdot)$ to $\mathfrak{h}_{\overline{0}}$ is non-degenerate: if $h \in \mathfrak{h}_{\overline{0}}$ and $(h,\mathfrak{h}_{\overline{0}})=0$ then $(h,x)=0$ for all semi-simple $x \in C$ by the above, and on the other hand $(h,x)=0$ for all nilpotent $x \in C$ since $\mathrm{ad}(h) \mathrm{ad}(x)$ is nilpotent, hence of trace $0$. Thus $(h,C)=0$ and it follows that $h=0$, proving our claim.
Now we show that $C$ is abelian: if not, then since $\mathrm{ad}_C(x)$ is nilpotent for all $x \in C$ we have $Z(C) \cap [C,C] \neq 0$. Suppose that $0 \neq z \in Z(C) \cap [C,C]$ and let $z_n$ be the nilpotent part of $z$. We have $z_n \in Z(C)$ and hence $(z_n,x)=0$ for all $x \in C$, whence $z_n=0$. Thus $z$ is semi-simple and hence $z \in \mathfrak{h}_{\overline{0}}$. But now for $h \in \mathfrak{h}_{\overline{0}}$ we have $(h,z)=0$ since $z \in [C,C]$, contradicting $(\cdot,\cdot)$ non-degenerate on $\mathfrak{h}_{\overline{0}}$. We have proven that $C$ is abelian.
Finally, given $x \in C$ nilpotent, we have $(x,y)=0$ for all $y \in C$ since $C$ is abelian, which implies $x=0$ and hence $C=\mathfrak{h}_{\overline{0}}$.
