Cartan Subalgebras and Root Space Decompositions

Suppose $x=x_s+x_n$ is an abstract Jordan decomposition. Then ${\mathrm{ad}}_x={\mathrm{ad}}_{x_s}+{\mathrm{ad}}_{x_n}$ by linearity of $\mathrm{ad}$, the two summands commute because $[{\mathrm{ad}}_{x_s},{\mathrm{ad}}_{x_n}]={\mathrm{ad}}_{[x_s,x_n]}=0$, and by [L4] the first is semisimple while the second is nilpotent. The uniqueness assertion of [L1] therefore identifies them with the additive Jordan–Chevalley parts of ${\mathrm{ad}}_x$. This proves (i).

Now let ${\mathrm{ad}}_x=S+N$ be the additive Jordan–Chevalley decomposition of the derivation $T={\mathrm{ad}}_x$; by [L1] there are polynomials $p,q\in \mathbb{C}[t]$ with $S=p(T)$ and $N=q(T)$. We show that $S$ is a derivation and that $S$ acts on the generalized eigenspace of $T$ for $\lambda$ as multiplication by $\lambda$. The generalized eigenspaces ${\mathfrak{g}}_{\lambda }=\{y:(T-\lambda {)}^{k}y=0\text{ for some }k\}$ satisfy $\mathfrak{g}={⨁}_{\lambda }{\mathfrak{g}}_{\lambda }$ and $[{\mathfrak{g}}_{\lambda },{\mathfrak{g}}_{\mu }]\subseteq {\mathfrak{g}}_{\lambda +\mu }$: the second claim follows from the binomial expansion $(T-\lambda -\mu {)}^n[y,z]={\sum }_{i+j=n}\left(\genfrac{}{}{0px}{}{n}{i}\right)[(T-\lambda {)}^{i}y,(T-\mu {)}^{j}z]$. Each ${\mathfrak{g}}_{\lambda }$ is invariant under $T$, hence under $p(T)=S$. On ${\mathfrak{g}}_{\lambda }$ the operator $T$ equals $\lambda \cdot 1$ plus a commuting nilpotent operator, so $p(T)=p(\lambda )\cdot 1+(\text{nilpotent})$ there; since $S$ is semisimple and restricts semisimply to the invariant subspace ${\mathfrak{g}}_{\lambda }$, this forces $S{\mathrm{\mid }}_{{\mathfrak{g}}_{\lambda }}=p(\lambda )\cdot 1$. On the other hand $S-\lambda \cdot 1=(T-\lambda \cdot 1)-N$ is a difference of two commuting nilpotent operators on ${\mathfrak{g}}_{\lambda }$, hence nilpotent; comparing with the scalar operator $(p(\lambda )-\lambda )\cdot 1$ gives $p(\lambda )=\lambda$. Therefore $S{\mathrm{\mid }}_{{\mathfrak{g}}_{\lambda }}=\lambda \cdot 1$, and for $y\in {\mathfrak{g}}_{\lambda }$, $z\in {\mathfrak{g}}_{\mu }$ one has $S[y,z]=(\lambda +\mu )[y,z]=[\lambda y,z]+[y,\mu z]=[Sy,z]+[y,Sz]$ because $[y,z]\in {\mathfrak{g}}_{\lambda +\mu }$. Thus $S$ is a derivation; $T$ is a derivation by the Jacobi identity, so $N=T-S$ is a derivation too.

By [L2] there are unique $y_s,y_n\in \mathfrak{g}$ with $S={\mathrm{ad}}_{y_s}$ and $N={\mathrm{ad}}_{y_n}$. Then ${\mathrm{ad}}_{y_s+y_n}=S+N={\mathrm{ad}}_x$, so $y_s+y_n=x$ by injectivity of $\mathrm{ad}$ [L3]; also ${\mathrm{ad}}_{[y_s,y_n]}=[S,N]=0$, so $[y_s,y_n]=0$ by [L3]; and ${\mathrm{ad}}_{y_s}=S$ is semisimple while ${\mathrm{ad}}_{y_n}=N$ is nilpotent. Hence $x=y_s+y_n$ is an abstract Jordan decomposition by [L4].

If $x=u+v$ is any abstract Jordan decomposition, then ${\mathrm{ad}}_u,{\mathrm{ad}}_v$ is a commuting semisimple-plus-nilpotent decomposition of ${\mathrm{ad}}_x$ by step 1.1's computation, so uniqueness in [L1] gives ${\mathrm{ad}}_u=S={\mathrm{ad}}_{y_s}$ and ${\mathrm{ad}}_v=N={\mathrm{ad}}_{y_n}$; injectivity of $\mathrm{ad}$ [L3] gives $u=y_s$ and $v=y_n$. Hence the decomposition of (ii) is unique. If $\mathfrak{g}=0$ then $x=0=y_s=y_n$ and every assertion holds with the zero endomorphism, which is both semisimple and nilpotent; no nonempty choice is made anywhere in this argument, the Axiom of Choice being used only through the appeal to [L1].

The endomorphism ${\mathrm{ad}}_x$ of the finite-dimensional complex vector space $\mathfrak{g}$ has an additive Jordan–Chevalley decomposition by [L1]. Applying clause (ii) of [L2] to it produces elements $y_s,y_n\in \mathfrak{g}$ such that $x=y_s+y_n$, $[y_s,y_n]=0$, ${\mathrm{ad}}_{y_s}$ is semisimple and ${\mathrm{ad}}_{y_n}$ is nilpotent, and such that ${\mathrm{ad}}_{y_s}$, ${\mathrm{ad}}_{y_n}$ are the additive parts of ${\mathrm{ad}}_x$. By [L3] this is an abstract Jordan decomposition of $x$.

Let $x=u+v$ be any abstract Jordan decomposition. Clause (i) of [L2] identifies ${\mathrm{ad}}_u$ and ${\mathrm{ad}}_v$ with the additive Jordan–Chevalley parts of ${\mathrm{ad}}_x$, which are the parts ${\mathrm{ad}}_{y_s}$ and ${\mathrm{ad}}_{y_n}$ produced in step 1.1; uniqueness of the abstract decomposition in [L2] then gives $u=y_s$ and $v=y_n$. For $\mathfrak{g}=0$ the only element is $0$ and the decomposition $0=0+0$ satisfies the definition vacuously; the Axiom of Choice enters only through [L1] and [L2].

${\mathfrak{g}}^x$ is a Lie subalgebra: for $y,z\in {\mathfrak{g}}^x$, Jacobi and [L5] give $[x,[y,z]]=[[x,y],z]+[y,[x,z]]=0$.

${\mathfrak{g}}^x$ equals its normalizer. Let $y\in N_{\mathfrak{g}}({\mathfrak{g}}^x)$. Since $x\in {\mathfrak{g}}^x$, we have $[y,x]\in {\mathfrak{g}}^x$, so $[x,[y,x]]=0$. Decompose $y={\sum }_{\lambda }{y}_{\lambda }$ in the eigenspaces of the semisimple operator ${\mathrm{ad}}_x$ [L2]. Then $[x,y]={\sum }_{\lambda }\lambda y_{\lambda }$ and $[x,[y,x]]=-{\sum }_{\lambda }{\lambda }^2{y}_{\lambda }=0$; the summands lie in distinct eigenspaces, so ${\lambda }^2{y}_{\lambda }=0$, and since the field is $\mathbb{C}$ we get $y_{\lambda }=0$ for $\lambda \mathrm{\ne }0$. Hence $y=y_0\in {\mathfrak{g}}^x$, so $N_{\mathfrak{g}}({\mathfrak{g}}^x)={\mathfrak{g}}^x$.

Every $z\in {\mathfrak{g}}^x$ acts by zero on ${\mathfrak{g}}^x$. Since $[x,z]=0$, the operators $A={\mathrm{ad}}_x$ and $B={\mathrm{ad}}_z$ commute [L2], and $A$ is semisimple [L2], so $\mathfrak{g}={⨁}_{\lambda }{V}_{\lambda }$ with $V_{\lambda }=\ker (A-\lambda )$ and each $V_{\lambda }$ is $B$-invariant. For $t\in \mathbb{C}$ put $M_t=A+tB={\mathrm{ad}}_{x+tz}$. For $t\mathrm{\ne }0$ an element $v={\sum }_{\lambda }{v}_{\lambda }$ with $v_{\lambda }\in V_{\lambda }$ is killed by $M_t$ exactly when $\lambda v_{\lambda }+tB{v}_{\lambda }=0$ for every $\lambda$, so $\dim \ker M_t=\dim \ker (B{\mathrm{\mid }}_{V_0})+{\sum }_{\lambda \mathrm{\ne }0}{m}_{\lambda }(t)$, where $m_{\lambda }(t)=\dim \ker (B+\frac{\lambda }{t}){\mathrm{\mid }}_{V_{\lambda }}$ is the multiplicity of the eigenvalue $-\frac{\lambda }{t}$ of the endomorphism $B{\mathrm{\mid }}_{V_{\lambda }}$ and therefore vanishes for all but finitely many $t$. Choosing $t$ outside this finite exceptional set and using [L1] at the element $x+tz$ gives $\dim \ker (B{\mathrm{\mid }}_{V_0})=\dim \ker M_t\ge m=\dim V_0$, hence $\ker (B{\mathrm{\mid }}_{V_0})=V_0$ and $B{\mathrm{\mid }}_{V_0}=0$; by definition $V_0={\mathfrak{g}}^x$.

Since $z\in {\mathfrak{g}}^x$ was arbitrary, step 1.3 shows that ${\mathrm{ad}}_z{\mathrm{\mid }}_{{\mathfrak{g}}^x}=0$ for every $z\in {\mathfrak{g}}^x$, so ${\mathfrak{g}}^x$ is abelian and in particular nilpotent; alternatively, every adjoint operator of ${\mathfrak{g}}^x$ is nilpotent on ${\mathfrak{g}}^x$ and [L4] applies. By step 1.2, ${\mathfrak{g}}^x$ equals its normalizer, so [L3] makes ${\mathfrak{g}}^x$ a Cartan subalgebra. When $\mathfrak{g}=0$ we have $x=0$, ${\mathfrak{g}}^x=0$, and the zero subalgebra is a Cartan subalgebra of the zero algebra; no nonempty choice occurs.

If $\mathfrak{g}=0$, the zero subalgebra is nilpotent and equals its own normalizer, hence is a Cartan subalgebra. Assume now $\mathfrak{g}\mathrm{\ne }0$. There is a nonzero semisimple element: if every element of $\mathfrak{g}$ had nilpotent adjoint operator, then [L2] would make $\mathfrak{g}$ nilpotent, hence solvable, so its radical would be $\mathfrak{g}\mathrm{\ne }0$, contradicting semisimplicity [L6]. Choose an element $x$ whose adjoint operator is not nilpotent and let $x=x_s+x_n$ be its decomposition from [L1]; if $x_s=0$ then ${\mathrm{ad}}_x={\mathrm{ad}}_{x_n}$ is nilpotent, contrary to the choice of $x$, so $x_s\mathrm{\ne }0$ is semisimple.

A toral subalgebra of maximal dimension exists and is maximal by inclusion. Let $\mathfrak{t}$ be any maximal toral subalgebra; it is nonzero since the zero subalgebra is contained in $\mathbb{C}{x}_s$ from step 1.1. By [L7] the subalgebra $\mathfrak{t}$ is abelian, so the operators ${\mathrm{ad}}_h$, $h\in \mathfrak{t}$, are pairwise commuting, and each is semisimple by [L7]; hence by [L4] and [L8] they are simultaneously diagonalisable and $\mathfrak{g}={⨁}_{\lambda \in {\mathfrak{t}}^{\ast }}{\mathfrak{g}}_{\lambda },{\mathfrak{g}}_{\lambda }=\{x\in \mathfrak{g}:[h,x]=\lambda (h)x\text{ for all }h\in \mathfrak{t}\}\mathrm{.}$ Put $\mathfrak{l}={\mathfrak{g}}_0=C_{\mathfrak{g}}(\mathfrak{t})$.

For $x\in \mathfrak{l}$ the Jordan parts lie in $\mathfrak{l}$: by [L1] and [L9] there is a polynomial $p$ with ${\mathrm{ad}}_{x_s}=p({\mathrm{ad}}_x)$, and $[{\mathrm{ad}}_x,{\mathrm{ad}}_h]={\mathrm{ad}}_{[x,h]}=0$ for $h\in \mathfrak{t}$ [L6, L8], so $[{\mathrm{ad}}_{x_s},{\mathrm{ad}}_h]=0$ and therefore ${\mathrm{ad}}_{[x_s,h]}=0$, that is, $[x_s,h]\in Z(\mathfrak{g})=0$; thus $x_s\in \mathfrak{l}$, and $x_n=x-x_s\in \mathfrak{l}$ as well. Moreover $\mathfrak{t}+\mathbb{C}{x}_s$ is toral: it is a subalgebra because $[x_s,\mathfrak{t}]=0$, it is abelian, and each of its elements has semisimple adjoint operator by [L4] since ${\mathrm{ad}}_{x_s}$ and the commuting operators ${\mathrm{ad}}_h$ are semisimple. By the maximality of $\mathfrak{t}$ we get $x_s\in \mathfrak{t}$.

Invariance of $B$ gives $(\lambda (h)+\mu (h))B(y,z)=B([h,y],z)+B(y,[h,z])=0$ for $y\in {\mathfrak{g}}_{\lambda }$, $z\in {\mathfrak{g}}_{\mu }$ and $h\in \mathfrak{t}$; if $\lambda +\mu \mathrm{\ne }0$ some $h$ has $(\lambda +\mu )(h)\mathrm{\ne }0$, so $B({\mathfrak{g}}_{\lambda },{\mathfrak{g}}_{\mu })=0$.

For $x\in \mathfrak{l}$ we have ${\mathrm{ad}}_{\mathfrak{l}}(x)={\mathrm{ad}}_{\mathfrak{l}}(x_n)$, because $x_s\in \mathfrak{t}$ centralises $\mathfrak{l}$; by [L1] the operator ${\mathrm{ad}}_{x_n}$ is nilpotent, so every adjoint operator of $\mathfrak{l}$ is nilpotent on $\mathfrak{l}$ and [L2] makes $\mathfrak{l}$ nilpotent.

The algebra $\mathfrak{l}$ is abelian. It is nilpotent by step 4.1, hence solvable, so [L3] supplies a common eigenvector in the nonzero module $\mathfrak{g}$. Its line is invariant. Apply [L3] to the quotient by that line and then to successive nonzero quotients; the dimension drops at each step, and lifting the resulting invariant flag gives a basis of $\mathfrak{g}$ in which all ${\mathrm{ad}}_x$, $x\in \mathfrak{l}$, are upper triangular. For $x\in [\mathfrak{l},\mathfrak{l}]$ the operator ${\mathrm{ad}}_x$ is a sum of commutators of upper triangular operators, hence strictly upper triangular, hence nilpotent; consequently $B(x,y)=\mathrm{tr}({\mathrm{ad}}_x{\mathrm{ad}}_y)=0$ for every $y\in \mathfrak{l}$, since a strictly upper triangular operator times an upper triangular operator stays strictly upper triangular.

The restriction $B{\mathrm{\mid }}_{\mathfrak{l}}$ is nondegenerate: if $x\in \mathfrak{l}$ satisfies $B(x,\mathfrak{l})=0$, then for every nonzero weight $\lambda$ we have $B(x,{\mathfrak{g}}_{\lambda })=0$ by step 3.2, and $B(x,\mathfrak{l})=0$ by hypothesis, so $B(x,\mathfrak{g})=0$ and [L5] gives $x=0$. Applying this to step 5.1 yields $[\mathfrak{l},\mathfrak{l}]=0$.

Every element of $\mathfrak{l}$ is semisimple: for $x\in \mathfrak{l}$ we have $x_n\in \mathfrak{l}$ by step 3.1, and $[x_n,y]=[x,y]-[x_s,y]=0$ for every $y\in \mathfrak{l}$ because $x\in \mathfrak{l}$ and $x_s\in \mathfrak{t}$; hence ${\mathrm{ad}}_{x_n}$ commutes with ${\mathrm{ad}}_y$ and the product ${\mathrm{ad}}_{x_n}{\mathrm{ad}}_y$ is nilpotent, so $B(x_n,y)=0$ for all $y\in \mathfrak{l}$. Nondegeneracy from step 6.1 forces $x_n=0$. Thus $\mathfrak{l}$ is abelian and consists of semisimple elements, i.e. $\mathfrak{l}$ is toral; since $\mathfrak{t}\subseteq \mathfrak{l}$, maximality gives $\mathfrak{l}=\mathfrak{t}$.

Finally $N_{\mathfrak{g}}(\mathfrak{t})=\mathfrak{t}$: if $x\in N_{\mathfrak{g}}(\mathfrak{t})$ and $x={\sum }_{\lambda }{x}_{\lambda }$ is its decomposition from step 2.1, then for every $h\in \mathfrak{t}$ we have $[h,x]={\sum }_{\lambda }\lambda (h)x_{\lambda }\in \mathfrak{t}={\mathfrak{g}}_0\mathrm{.}$ Uniqueness of the direct weight-space decomposition forces every nonzero-weight component $\lambda (h)x_{\lambda }$ of this sum to vanish. For each $\lambda \mathrm{\ne }0$, choose $h\in \mathfrak{t}$ with $\lambda (h)\mathrm{\ne }0$; then $x_{\lambda }=0$. Therefore $x\in {\mathfrak{g}}_0=\mathfrak{l}=\mathfrak{t}$. Since $\mathfrak{t}$ is abelian and hence nilpotent and equals its normalizer, [L7] makes it a Cartan subalgebra. The Axiom of Choice was inherited through [L1] and [L9].

Let $H\in \mathfrak{h}$ and $N=\dim \mathfrak{g}$. By [L3] applied to ${\mathrm{ad}}_H$, the spaces $V_{\lambda ,H}=\ker ({\mathrm{ad}}_H-\lambda {)}^N$ are ${\mathrm{ad}}_H$-invariant and $\mathfrak{g}={⨁}_{\lambda }{V}_{\lambda ,H}$. Moreover $\mathfrak{h}\subseteq {\mathfrak{g}}_0$: for $H^{\mathrm{\prime }}\in \mathfrak{h}$, [L1] gives $({\mathrm{ad}}_H{)}^n{H}^{\mathrm{\prime }}=0$ for some $n$, so $H^{\mathrm{\prime }}\in {\mathfrak{g}}_0$ because the condition defining ${\mathfrak{g}}_0$ is vacuous at $\alpha =0$. This is (iii).

We claim each $V_{\lambda ,H}$ is stable under ${\mathrm{ad}}_Y$ for every $Y\in \mathfrak{h}$. By [L1] the operator ${\mathrm{ad}}_H$ is nilpotent on $\mathfrak{h}$, so there is $m\ge 0$ with $({\mathrm{ad}}_H{)}^{m}Y=0$; put $N=\dim \mathfrak{g}$, so that $({\mathrm{ad}}_H-\lambda {)}^{N}X=0$ for $X\in V_{\lambda ,H}$ by the definition of $V_{\lambda ,H}$ and [L3]. The operator identity $({\mathrm{ad}}_H-\lambda {)}^n[Y,X]={\sum }_{k=0}^n\left(\genfrac{}{}{0px}{}{n}{k}\right)[({\mathrm{ad}}_H{)}^{n-k}Y,({\mathrm{ad}}_H-\lambda {)}^{k}X]$ holds for every $n\ge 0$ by induction on $n$, because ${\mathrm{ad}}_H$ is a derivation: the case $n=0$ is trivial and the induction step applies ${\mathrm{ad}}_H-\lambda$ to both sides and uses Pascal's rule. Taking $n=m+N$, every summand vanishes, since either $n-k\ge m$, so $({\mathrm{ad}}_H{)}^{n-k}Y=0$, or $k\ge N$, so $({\mathrm{ad}}_H-\lambda {)}^{k}X=0$. Hence $({\mathrm{ad}}_H-\lambda {)}^{m+N}[Y,X]=0$ and $[Y,X]\in V_{\lambda ,H}$. As $Y\in \mathfrak{h}$ was arbitrary, every $V_{\lambda ,H}$ is stable under ${\mathrm{ad}}_{\mathfrak{h}}$.

Fix a basis $H_1,\dots ,H_r$ of $\mathfrak{h}$; iterating step 1.2 over the pairwise compatible decompositions $\mathfrak{g}={⨁}_{\lambda }{V}_{\lambda ,H_j}$ refines the direct sum decomposition to $\mathfrak{g}={⨁}_{({\lambda }_1,\dots ,{\lambda }_r)}(V_{{\lambda }_1,H_1}\cap \cdots \cap V_{{\lambda }_r,H_r})$, and each summand is stable under ${\mathrm{ad}}_H$ for every $H\in \mathfrak{h}$. For a tuple $({\lambda }_1,\dots ,{\lambda }_r)$ with nonzero summand $W$ let $\alpha \in {\mathfrak{h}}^{\ast }$ be the linear functional with $\alpha (H_j)={\lambda }_j$. By [L2], the solvable algebra $\mathfrak{h}$ acts triangularly on $W$ in a suitable basis $v_1,\dots ,v_s$; the diagonal entries of such a triangular form are eigenvalues of ${\mathrm{ad}}_{H_j}$ on $W$ for each $j$, and since $W\subseteq V_{{\lambda }_j,H_j}$ the only eigenvalue of ${\mathrm{ad}}_{H_j}$ there is ${\lambda }_j$, so every diagonal entry equals $\alpha$. Hence each $v_i$ satisfies $({\mathrm{ad}}_H-\alpha (H){)}^i{v}_i=0$ for all $H\in \mathfrak{h}$, and therefore $W\subseteq {\mathfrak{g}}_{\alpha }$. In particular only finitely many ${\mathfrak{g}}_{\alpha }$ are nonzero.

Since each ${\mathfrak{g}}_{\alpha }$ is a linear subspace by definition and stable under every ${\mathrm{ad}}_H$ by step 1.2, and since an element of ${\mathfrak{g}}_{\alpha }$ satisfies the generalized eigenvalue condition for each $H_j$ with value $\alpha (H_j)$, we have ${\mathfrak{g}}_{\alpha }\subseteq V_{\alpha (H_1),H_1}\cap \cdots \cap V_{\alpha (H_r),H_r}$; combined with step 2.1 and the injectivity of the map $({\lambda }_1,\dots ,{\lambda }_r)\mapsto {\sum }_j{\lambda }_j{e}_j$ on the dual basis, this gives ${\mathfrak{g}}_{\alpha }=V_{\alpha (H_1),H_1}\cap \cdots \cap V_{\alpha (H_r),H_r}$ and the direct sum decomposition (ii), with only finitely many nonzero terms. This proves (i) and (ii).

For $X\in {\mathfrak{g}}_{\alpha }$, $Y\in {\mathfrak{g}}_{\beta }$ and $H\in \mathfrak{h}$, the binomial expansion gives $({\mathrm{ad}}_H-(\alpha +\beta )(H){)}^n[X,Y]={\sum }_{k=0}^n\left(\genfrac{}{}{0px}{}{n}{k}\right)[({\mathrm{ad}}_H-\alpha (H){)}^{k}X,({\mathrm{ad}}_H-\beta (H){)}^{n-k}Y]$; choosing $n\ge 2P$ where $P$ bounds the two vanishing exponents for $X$ and $Y$, so that for every $k$ either $k\ge P$ or $n-k\ge P$, every summand is zero, hence $[X,Y]\in {\mathfrak{g}}_{\alpha +\beta }$, which is (iv). If $\mathfrak{g}=0$ all spaces are zero and every assertion is vacuous.

The implication "maximal toral $\Rightarrow$ Cartan" is [L1]. For the converse, let $\mathfrak{h}$ be a Cartan subalgebra; by [L3] it is nilpotent and $N_{\mathfrak{g}}(\mathfrak{h})=\mathfrak{h}$. Let $\mathfrak{g}={⨁}_{\alpha }{\mathfrak{g}}_{\alpha }$ be its generalized weight decomposition from [L2], so that ${\mathfrak{g}}_0=\{x\in \mathfrak{g}:({\mathrm{ad}}_H{)}^{k}x=0\text{ for all }H\in \mathfrak{h}\text{ and large }k\}$.

We have ${\mathfrak{g}}_0=\mathfrak{h}$: [L2] gives $\mathfrak{h}\subseteq {\mathfrak{g}}_0$; conversely every $X\in N_{\mathfrak{g}}(\mathfrak{h})$ lies in ${\mathfrak{g}}_0$, because $({\mathrm{ad}}_H{)}^{k}X=({\mathrm{ad}}_H{)}^{k-1}[H,X]$ with $[H,X]\in \mathfrak{h}$ and ${\mathrm{ad}}_H$ nilpotent on $\mathfrak{h}$ by [L5]; so $N_{\mathfrak{g}}(\mathfrak{h})\subseteq {\mathfrak{g}}_0$. If ${\mathfrak{g}}_0\mathrm{\ne }\mathfrak{h}$, then ${\mathfrak{g}}_0\mathrm{/}\mathfrak{h}$ is a nonzero module for the solvable algebra $\mathfrak{h}$, so by [L5] there is $X\mathrm{\notin }\mathfrak{h}$ whose image in ${\mathfrak{g}}_0\mathrm{/}\mathfrak{h}$ is a common eigenvector for all ${\mathrm{ad}}_H$; the diagonal functional has value $0$ on every $H$, because ${\mathrm{ad}}_H$ is nilpotent on ${\mathfrak{g}}_0$, so $[H,X]\in \mathfrak{h}$ for all $H$ and $X\in N_{\mathfrak{g}}(\mathfrak{h})\setminus \mathfrak{h}$, contradicting $N_{\mathfrak{g}}(\mathfrak{h})=\mathfrak{h}$. Hence ${\mathfrak{g}}_0=\mathfrak{h}$.

For $x\in \mathfrak{h}$ we have $x_s,x_n\in \mathfrak{h}$ and ${\mathrm{ad}}_{x_s}$ acts on each ${\mathfrak{g}}_{\alpha }$ as the scalar $\alpha (x)$: by [L4] write ${\mathrm{ad}}_{x_s}=p({\mathrm{ad}}_x)$; on ${\mathfrak{g}}_{\alpha }$ the operator ${\mathrm{ad}}_x$ is $\alpha (x)$ plus a commuting nilpotent operator, so $p({\mathrm{ad}}_x)=p(\alpha (x))$ plus a commuting nilpotent operator, while ${\mathrm{ad}}_{x_s}$ restricts semisimply to the invariant subspace ${\mathfrak{g}}_{\alpha }$; hence ${\mathrm{ad}}_{x_s}{\mathrm{\mid }}_{{\mathfrak{g}}_{\alpha }}=p(\alpha (x))\cdot 1$, and since $x_s=x-x_n$ with ${\mathrm{ad}}_{x_n}$ nilpotent on ${\mathfrak{g}}_{\alpha }$, comparison of scalar parts gives $p(\alpha (x))=\alpha (x)$. In particular ${\mathrm{ad}}_{x_s}$ commutes with every ${\mathrm{ad}}_H$, $H\in \mathfrak{h}$, so by [L7] $[x_s,H]=0$ and, $\mathfrak{g}$ being centerless [L6], $x_s\in C_{\mathfrak{g}}(\mathfrak{h})\subseteq N_{\mathfrak{g}}(\mathfrak{h})=\mathfrak{h}$; then $x_n=x-x_s\in \mathfrak{h}$ as well.