Finite semisimple Cartan, root and string structure

The center of $\mathfrak{g}$ is an abelian ideal, so is zero. Thus $\mathrm{ad}:\mathfrak{g}\to \mathrm{End}(\mathfrak{g})$ is injective. Let $\mathcal{D}$ be its finite-dimensional derivation algebra; $[D,{\mathrm{ad}}_x]={\mathrm{ad}}_{Dx}$ makes $\mathrm{ad}\mathfrak{g}$ an ideal in it. The invariant trace form $(D,T)\mapsto {\mathrm{tr}}_{\mathfrak{g}}(DT)$ restricts to the nondegenerate Killing form there. Consequently $\mathcal{D}=\mathrm{ad}\mathfrak{g}\oplus J$ as vector spaces, where $J$ is the perpendicular complement. Trace cyclicity and the ideal property make $J$ an ideal as well. Two complementary ideals commute, since their bracket lies in their intersection. Hence $D\in J$ satisfies ${\mathrm{ad}}_{Dx}=[D,{\mathrm{ad}}_x]=0$ for every $x$, whence $D=0$. Every derivation is therefore inner.

Choose a toral subalgebra $\mathfrak{h}$ of maximal dimension; the zero subalgebra starts the finite dimension search. Commuting diagonalizable operators admit simultaneous eigenspaces: successively decompose by a finite basis of operators, each preserving the earlier eigenspaces. Write $\mathfrak{g}={⨁}_{\lambda }{\mathfrak{g}}_{\lambda }$ for these joint weights, with ${\mathfrak{g}}_0=C_{\mathfrak{g}}(\mathfrak{h})$. Invariance gives $(\lambda (h)+\mu (h))B(x,y)=0$ for $x\in {\mathfrak{g}}_{\lambda },y\in {\mathfrak{g}}_{\mu }$. Thus distinct opposite-weight pairs are the only possible nonzero pairings, and F1 makes $B{\mathrm{\mid }}_{{\mathfrak{g}}_0}$ nondegenerate.

For the generalized eigenspaces of ${\mathrm{ad}}_x$, the derivation identity gives $({\mathrm{ad}}_x-\lambda -\mu {)}^N[y,z]={\sum }_{j=0}^N\left(\genfrac{}{}{0px}{}{N}{j}\right)[({\mathrm{ad}}_x-\lambda {)}^{j}y,({\mathrm{ad}}_x-\mu {)}^{N-j}z]\mathrm{.}$ If $y,z$ have generalized weights $\lambda ,\mu$, a sufficiently large $N$ makes this zero. Thus their bracket has generalized weight $\lambda +\mu$. The operator $S$ acting by $\lambda$ on each such space is therefore a derivation. By step 1.1 it is ${\mathrm{ad}}_{x_s}$, and $x_n=x-x_s$ has nilpotent adjoint action and commutes with $x_s$. If $y$ commutes with $x$, its adjoint operator preserves the generalized eigenspaces and hence commutes with $S$; center-freeness gives $[y,x_s]=[y,x_n]=0$. This constructs internal semisimple and nilpotent parts and their centralizer property.

For $x\in {\mathfrak{g}}_0$, step 2.1 puts $x_s,x_n$ in ${\mathfrak{g}}_0$. The commuting diagonalizable operators of $\mathfrak{h}$ and $x_s$ are simultaneously diagonalizable, so maximality forces $x_s\in \mathfrak{h}$. Therefore ${\mathrm{ad}}_x{\mathrm{\mid }}_{{\mathfrak{g}}_0}={\mathrm{ad}}_{x_n}{\mathrm{\mid }}_{{\mathfrak{g}}_0}$ is nilpotent. F1 makes the adjoint image of ${\mathfrak{g}}_0$ solvable, and its central kernel is abelian; the derived series then shows ${\mathfrak{g}}_0$ itself solvable. F2 triangularizes its representation on all of $\mathfrak{g}$. The commutators are strictly upper triangular, so $B([{\mathfrak{g}}_0,{\mathfrak{g}}_0],{\mathfrak{g}}_0)=0$. Nondegeneracy from step 1.2 gives $[{\mathfrak{g}}_0,{\mathfrak{g}}_0]=0$. Now ${\mathrm{ad}}_{x_n}$ commutes with every ${\mathrm{ad}}_y$, $y\in {\mathfrak{g}}_0$; their product is nilpotent since one factor is nilpotent. Hence $B(x_n,y)=0$ for all such $y$, and nondegeneracy forces $x_n=0$. Every $x\in {\mathfrak{g}}_0$ is thus in $\mathfrak{h}$, proving self-centralization. This applies to any maximal toral subalgebra.

Call the nonzero weights $\mathrm{\Phi }$. Step 1.2 and F1 show $-\alpha \in \mathrm{\Phi }$ and give perfect opposite-root pairings, together with nondegeneracy on $\mathfrak{h}$. Jacobi gives $[{\mathfrak{g}}_{\alpha },{\mathfrak{g}}_{\beta }]\subseteq {\mathfrak{g}}_{\alpha +\beta }$, interpreting absent weights as zero. For $e\in {\mathfrak{g}}_{\alpha },f\in {\mathfrak{g}}_{-\alpha }$, invariance gives $[e,f]=B(e,f)H_{\alpha }$, since pairing with $h\in \mathfrak{h}$ gives $B(e,[f,h])=\alpha (h)B(e,f)$. Choose $e,f$ with $B(e,f)\mathrm{\ne }0$. If $\alpha (H_{\alpha })=0$, the nonzero element $z=[e,f]$ commutes with both $e$ and $f$. Its adjoint action is diagonalizable because $z\in \mathfrak{h}$. Each eigenvalue space of ${\mathrm{ad}}_z$ is preserved by ${\mathrm{ad}}_e,{\mathrm{ad}}_f$, and the trace of their commutator on that space is both zero and its dimension times the eigenvalue of ${\mathrm{ad}}_z$. Thus every eigenvalue is zero, contradicting center-freeness and $z\mathrm{\ne }0$. Hence $c_{\alpha }\mathrm{\ne }0$. Rescaling $f$ so $B(e,f)=2\mathrm{/}{c}_{\alpha }$ gives exactly the stated rank-one relations.

Apply F2 to the adjoint representation of each root triple. Every root value $\beta (h_{\alpha })$ is an integer, and ${\mathrm{ad}}_{e_{\alpha }},{\mathrm{ad}}_{f_{\alpha }}$ are nilpotent. The subspace $\mathbb{C}{H}_{\alpha }\oplus {⨁}_{k\in \mathbb{Z}\setminus \{0\},k\alpha \in \mathrm{\Phi }}{\mathfrak{g}}_{k\alpha }$ is stable under the triple: brackets shift $k$ by one, and at opposite weights step 4.1 puts the bracket in $\mathbb{C}{H}_{\alpha }$. Its $h_{\alpha }$-weights are even and its zero-weight space has dimension one. Every simple summand supplied by F2 therefore has even highest weight and contributes a zero-weight line. There is exactly one summand, and it contains the three-dimensional adjoint triple, which is a copy of $V_2$ by its relations. The entire space is that $V_2$. Thus $\dim {\mathfrak{g}}_{\alpha }=1$ and no integer multiple $k\alpha$ with $\mathrm{\mid }k\mathrm{\mid }>1$ is a root. If $\beta =c\alpha$ is any proportional root, then $h_{\beta }=h_{\alpha }\mathrm{/}c$, so $2c$ and $2\mathrm{/}c$ are integers. Their product is four, hence $c$ is one of $\pm 1,\pm 2,\pm \frac{1}{2}$. Excluding integer doubles for either root leaves only $\pm 1$.

For nonproportional $\alpha ,\beta$, the sum of the spaces ${\mathfrak{g}}_{\beta +k\alpha }$ is stable under the root triple and has $h_{\alpha }$-weights $\beta (h_{\alpha })+2k$, each of dimension one. By step 5.1 these weights are integers of one parity. In F2's decomposition, any two simple modules of the same parity have a common weight (zero for even parity and one for odd parity), which would give multiplicity at least two. Hence there is exactly one simple summand. Its weights form a consecutive step-two string symmetric about zero. Therefore the roots in this string form a consecutive interval of $k$, reflection $\beta \mapsto \beta -\beta (h_{\alpha })\alpha$ reverses the interval, and all adjacent raising/lowering brackets and their positive compositions are exactly those of F2. For proportional roots the reflection interchanges $\alpha ,-\alpha$ by step 5.1.

The roots span ${\mathfrak{h}}^{\ast }$: an element of $\mathfrak{h}$ annihilated by all roots commutes with every weight space, so is central and zero. Nondegeneracy of $B{\mathrm{\mid }}_{\mathfrak{h}}$ implies the $H_{\alpha }$, and hence the $h_{\alpha }$, span $\mathfrak{h}$ over $\mathbb{C}$. On their real span ${\mathfrak{h}}_{\mathbb{R}}$ all roots take real values, by the integral pairings of step 5.1. If $u$ belongs to both this span and its multiple by $i$, every root value on $u$ is both real and purely imaginary, hence zero; therefore $u=0$. This proves $\mathfrak{h}={\mathfrak{h}}_{\mathbb{R}}\oplus i{\mathfrak{h}}_{\mathbb{R}}$ over $\mathbb{R}$. For $u\in {\mathfrak{h}}_{\mathbb{R}}$ the weight decomposition gives $B(u,u)={\sum }_{\beta \in \mathrm{\Phi }}\beta (u{)}^2$, strictly positive for $u\mathrm{\ne }0$, and polarization makes $B$ real there. Its real dual form identifies each $\alpha$ with $H_{\alpha }$ (the real solution of its defining linear equations), so $c_{\alpha }=(\alpha ,\alpha )>0$. Thus $h_{\alpha }$ is the Euclidean coroot. Spanning, finiteness, reducedness, integrality and reflection stability proved above verify every axiom in F4.

Apply F4 to choose a simple basis ${\alpha }_i$ and its coroot basis $h_i$. Choose normalized simple root vectors $e_i,f_i$. If a positive root $\beta$ is not simple, its expansion $\beta =\sum n_i{\alpha }_i$ and positive norm imply $(\beta ,{\alpha }_i)>0$ for some $i$ with $n_i>0$. It is not proportional to that simple root. In its string, the positive $h_i$-weight $\beta (h_i)>0$ cannot be the lowest weight, so $\beta -{\alpha }_i$ is a root. It remains positive: at least one other simple coefficient is positive and every root's coefficients have one sign. By step 6.1 the bracket $[e_i,{\mathfrak{g}}_{\beta -{\alpha }_i}]$ is the nonzero one-dimensional space ${\mathfrak{g}}_{\beta }$. Induction on positive-root height generates all positive spaces; the negative argument with the $f_i$ generates all negative spaces. The brackets $[e_i,f_i]=h_i$ span $\mathfrak{h}$. Thus the simple triples generate $\mathfrak{g}$. If $\mathfrak{g}=0$, take $\mathfrak{h}=0$, $\mathrm{\Phi }=\mathrm{\varnothing }$ and all assertions have their empty meanings. All vector-space choices and inductions above are finite, so AC is not used.