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.
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,
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 $C=\mathfrak{h}_{\overline{0}}$.