I am reading J-P. Serre's book on complex semisimple Lie algebras. Going back and forth through different parts, I am now trying to understand the proof of the existence of Cartan subalgebras (of finite dimensional Lie algebras over the complex numbers).
I am struggling with an identity which (as written in the book) I'm supposed to be able to prove by induction (this is one small part of the proof).
Given a Lie algebra $\mathfrak{g}$, for $x \in \mathfrak{g}$, we define $\mathfrak{g}_x^0$ to be the nilspace of $ad(x)$ (i.e. the set of all $y \in \mathfrak{g}$ such that $ad(x)^p(y) = 0$ for a large enough integer $p$).
We claim that $\mathfrak{g}_x^0$ is a subalgebra of $\mathfrak{g}$. In order to prove this, we wish to show that, if $y,z \in \mathfrak{g}_x^0$, then $[y,z] \in \mathfrak{g}_x^0$.
In Serre, it's said that we may prove the following by induction (and then the above claim will follow):
$$ad(x)^n[y,z] = \sum_{p=0}^n {n\choose p}[ad(x)^py,ad(x)^{n-p}z].$$
I have basically no idea how to do this. The based case is easy, and the case for $n=1$ also follows easily from the Jacobi identity. I fail at doing anything more than that.
Assuming an inductive hypothesis that the statement is true for $n=k$ and trying to prove it for the case of $n=k+1$ has also been broadly fruitless, aside from going around in circles. For example:
$$ad(x)^{k+1}[y,z] = \sum_{p=0}^{k+1} {k+1\choose p}[ad(x)^py,ad(x)^{k+1-p}z].$$
$$ \iff [x,(ad(x)^{k}[y,z])] = \sum_{p=0}^{k+1} {k+1\choose p}[ad(x)^py,ad(x)^{k+1-p}z].$$
$$ \iff [x,\sum_{p=0}^k {k\choose p}[ad(x)^py,ad(x)^{k-p}z]] = \sum_{p=0}^{k+1} {k+1\choose p}[ad(x)^py,ad(x)^{k+1-p}z].$$
$$ \iff \sum_{p=0}^k {k\choose p}[x,[ad(x)^py,ad(x)^{k-p}z]] = \sum_{p=0}^{k+1} {k+1\choose p}[ad(x)^py,ad(x)^{k+1-p}z].$$
and here I've wrestled with it but can't make it go. I don't know if this is the right approach. My only real idea here so far is that I should cleverly use the Jacobi identity, somehow, and I can't think of anything better to do than that.
Any help or advice/hints/approaches are appreciated.
Using Jacobi property of Lie bracket: $$[x,[ad(x)^py,ad(x)^{k-p}z]] = -[ad(x)^py,[ad(x)^{k-p}z,x]]-[ad(x)^{k-p}z,[x,ad(x)^py]]$$ Then using the antisymmetry of the Lie bracket: $$=-[ad(x)^py,-[x,ad(x)^{k-p}z]]-(-[[x,ad(x)^py],ad(x)^{k-p}z])$$ And the linearity of Lie bracket: $$=[ad(x)^py,[x,ad(x)^{k-p}z]]+[[x,ad(x)^py],ad(x)^{k-p}z]$$ $$=[ad(x)^py,ad(x)^{(k+1)-p}z]+[ad(x)^{p+1}y,ad(x)^{k-p}z]$$ Putting this back into your summation term, then you use the binomial coefficient identity: $$\binom{k}{p}+\binom{k}{p-1}=\binom{k+1}{p}$$ And you are done!