Kostant harmonics give an invariant polynomial complement

Fix a Cartan, positive roots and normalized simple triples $e_i,f_i,h_i$ from F2. Write $a_{ij}={\alpha }_j(h_i)$; these are real integers, and the ${\alpha }_i$ and $h_i$ are bases. Construct an auxiliary Lie algebra $C$ as follows, without any presentation theorem for $\mathfrak{g}$. Take the vector space on all finite formally bracketed words in the finite symbols ${\mathbf{e}}_i,{\mathbf{f}}_i,{\mathbf{h}}_i$, give it the bilinear grafting bracket, and quotient by the ideal generated by antisymmetry and Jacobi. Evaluation of words gives the free universal Lie property directly. Quotient further by $[{\mathbf{h}}_i,{\mathbf{h}}_j]=0$, $[{\mathbf{h}}_i,{\mathbf{e}}_j]=a_{ij}{\mathbf{e}}_j$, $[{\mathbf{h}}_i,{\mathbf{f}}_j]=-a_{ij}{\mathbf{f}}_j$, $[{\mathbf{e}}_i,{\mathbf{f}}_j]={\delta }_{ij}{\mathbf{h}}_i$. All these relations hold for the chosen triples: for $i\mathrm{\ne }j$, ${\alpha }_i-{\alpha }_j$ is not a root because its simple coefficients have opposite signs. Hence evaluation gives a surjection $\pi :C\to \mathfrak{g}$ by simple generation. The span $C^0$ of the ${\mathbf{h}}_i$ injects under $\pi$, because their images are a basis of $\mathfrak{h}$.

Let $C^{+},C^{-}$ be the subalgebras generated by the ${\mathbf{e}}_i,{\mathbf{f}}_i$. Jacobi expresses every bracket word in one sign as a sum of words $[{\mathbf{e}}_i,u]$, or respectively $[{\mathbf{f}}_i,u]$, with $u$ shorter: repeatedly apply $[[a,b],c]=[a,[b,c]]-[b,[a,c]]$ to shorten the left entry. The same identity and the relations in step 1.1 prove inductively that $[{\mathbf{f}}_i,C^{+}]\subseteq C^0+C^{+}$, with the bracket in $C^{+}$ for positive word length greater than one; the length-one case lies in $C^0$. Indeed $[{\mathbf{f}}_i,[{\mathbf{e}}_j,u]]=-{\delta }_{ij}[{\mathbf{h}}_i,u]+[{\mathbf{e}}_j,[{\mathbf{f}}_i,u]]$, and the induction handles the last bracket. Reversing signs proves $[{\mathbf{e}}_i,C^{-}]\subseteq C^0+C^{-}$. Bracketing with ${\mathbf{h}}_i$ preserves each half by the product rule. Thus $C^{-}+C^0+C^{+}$ is stable under every generator, contains them, and by the same bracket-word reduction contains all of $C$.

Give the generators degrees $-{\alpha }_i,{\alpha }_i,0$ in the free abelian group on the simple roots. Every defining relation is homogeneous, so the quotient has a direct grading. Step 2.1 shows all nonzero degrees have either entirely nonnegative or entirely nonpositive simple coordinates, and the degree-zero space is exactly $C^0$. Induction with Jacobi gives $[{\mathbf{h}}_i,u]=\gamma (h_i)u$ in degree $\gamma$. Distinct degrees have distinct joint weights because the simple roots form a basis of ${\mathfrak{h}}^{\ast }$. Every ideal $J\subseteq C$ is graded: for an element with finite homogeneous support, choose a linear combination of the ${\mathbf{h}}_i$ separating those finitely many weights, and use its finite interpolation polynomials to project that element onto each component within $J$. A separating combination exists because finitely many nonzero linear polynomials cannot vanish on every point: restrict them to a polynomial curve in a finite basis and avoid finitely many scalar roots.

The sum $K_{\ast }$ of all ideals $J$ satisfying $J\cap C^0=0$ is an ideal with the same property: by step 3.1 each such ideal has zero degree-zero component, as does every finite sum of its elements. This is a sum over a subset of the power set of $C$, so defines a set and uses no selection. The kernel $K$ of $\pi$ meets $C^0$ trivially, so $K\subseteq K_{\ast }$. Every nonzero ideal of $\mathfrak{g}$ meets $\mathfrak{h}$ nontrivially: project a nonzero element onto its finitely many Cartan/root weight components by the same interpolation. A nonzero Cartan component already suffices; a root component spans its one-dimensional root space and brackets with the opposite root into a nonzero coroot by F2. Now $\pi (K_{\ast })$ is an ideal disjoint from $\mathfrak{h}$. To check disjointness, if $\pi (z)=\pi (h)$ with $z\in K_{\ast }$ and $h\in C^0$, then $z-h\in K\subseteq K_{\ast }$, hence $h\in K_{\ast }\cap C^0=0$. Thus $\pi (K_{\ast })=0$ and $K_{\ast }=K$.

The conjugate-linear assignment ${\mathbf{e}}_i\mapsto -{\mathbf{f}}_i$, ${\mathbf{f}}_i\mapsto -{\mathbf{e}}_i$, ${\mathbf{h}}_i\mapsto -{\mathbf{h}}_i$ preserves the relations of step 1.1 because all $a_{ij}$ are real. For example $[-{\mathbf{f}}_i,-{\mathbf{e}}_j]=-{\delta }_{ij}{\mathbf{h}}_i$, and $[-{\mathbf{h}}_i,-{\mathbf{f}}_j]=-a_{ij}{\mathbf{f}}_j$. It therefore defines an involutive conjugate-linear Lie automorphism of $C$. It preserves $C^0$ and permutes the ideals disjoint from it, so preserves $K_{\ast }$ and descends by step 4.1 to an involution $\sigma$ of $\mathfrak{g}$. It sends $e_i$ to $-f_i$, $f_i$ to $-e_i$, and each real coroot to its negative. Trace in any finite basis shows $B(\sigma x,\sigma y)=\overline{B(x,y)}$: conjugation by a conjugate-linear invertible map conjugates the entries and trace of the represented complex-linear operator. Hence $⟨x,y⟩=-B(x,\sigma y)$ is a Hermitian form, linear in its first variable.

For $h\in {\mathfrak{h}}_{\mathbb{R}}$ and $x\in {\mathfrak{g}}_{\alpha }$, applying $\sigma$ to $[h,x]=\alpha (h)x$ gives $[h,\sigma x]=-\alpha (h)\sigma x$, since $\alpha (h)$ is real. Thus $\sigma {\mathfrak{g}}_{\alpha }={\mathfrak{g}}_{-\alpha }$. F2's bilinear orthogonality makes the Cartan and individual root spaces pairwise orthogonal for $⟨,⟩$. The Cartan restriction is positive definite because it is the complex Hermitian extension of the positive form $B{\mathrm{\mid }}_{{\mathfrak{h}}_{\mathbb{R}}}$. Also $⟨e_i,e_i⟩=B(e_i,f_i)=2\mathrm{/}({\alpha }_i,{\alpha }_i)>0$. Invariance and $\sigma f_i=-e_i$ give $⟨[e_i,x],y⟩=⟨x,[f_i,y]⟩\mathrm{.}$ Every positive nonsimple root space is obtained by raising a smaller positive root along a simple string, as F2 proves. If $0\mathrm{\ne }x$ is in that smaller space and raising is nonzero, F3 gives $[f_i,[e_i,x]]=cx$ with real $c>0$. Therefore $⟨[e_i,x],[e_i,x]⟩=c⟨x,x⟩$. Induction on root height proves positivity on every positive root space. Moreover $⟨\sigma x,\sigma x⟩=⟨x,x⟩$, by symmetry of $B$ and reality of a Hermitian diagonal value, so negative root spaces are positive too. Orthogonality now proves the form positive definite on all of $\mathfrak{g}$.

Extend $\sigma$ to a conjugate-linear graded algebra involution of $S$. It preserves $R$: for $p\in R$, ${\mathrm{ad}}_z(\sigma p)=\sigma ({\mathrm{ad}}_{\sigma z}p)=0$. Hence it preserves $I$ degreewise. On $S_d$ define $⟨p,q{⟩}_d=(-1{)}^d(p,\sigma q{)}_B\mathrm{.}$ On products of $d$ linear generators this equals the sum over bijections of products of the Hermitian pairings in step 6.1, by F1's explicit Fischer formula. A finite orthonormal basis of $\mathfrak{g}$ exists by successive Gram–Schmidt subtraction and division by positive real square roots. Its degree-$d$ monomials are orthogonal for this pairing, with strictly positive squared norms ${\prod }_j{a}_j!$. Thus $⟨,{⟩}_d$ is positive definite. Since $\sigma I_d=I_d$, its orthogonal complement of $I_d$ is precisely the bilinear annihilator $I_d^{\perp }={\mathcal{H}}_d$ from F1.

In a finite-dimensional positive Hermitian space, choose a finite orthonormal basis $u_1,\dots ,u_k$ of a subspace. The decomposition $v={\sum }_j⟨v,u_j⟩u_j+(v-{\sum }_j⟨v,u_j⟩u_j)$ splits it from its perpendicular complement, and positivity makes their intersection zero. Apply this to $I_d$ in step 7.1. It gives exactly $S_d=I_d\oplus {\mathcal{H}}_d$. Taking finite homogeneous sums gives the graded direct sum in the Statement. Both summands are $\mathfrak{g}$-stable by F1, so the unique projection onto either summand commutes with every adjoint operator; this proves equivariance without asserting that the positive form itself is invariant under complex $\mathfrak{g}$. In degree zero $I_0=0$ and ${\mathcal{H}}_0=\mathbb{C}$; for $\mathfrak{g}=0$ the same is the whole algebra. All actual basis and spectral choices were finite, and the free-word set and sums of ideals used explicit set constructions; no AC occurs.