Sl2 kostant harmonic decomposition

Matrix multiplication gives $[h,e]=2e$, $[h,f]=-2f$, and $[e,f]=h$. The matrices of these adjoint maps on $(h,e,f)$ give $B(h,h)=8$, $B(e,f)=B(f,e)=4$, and all other basis pairings zero. For example ${\mathrm{ad}}_h$ has diagonal $(0,2,-2)$; ${\mathrm{ad}}_e{\mathrm{ad}}_f$ maps $h\mapsto 2h$, $e\mapsto 2e$, $f\mapsto 0$, so its trace is four. The squares of ${\mathrm{ad}}_e,{\mathrm{ad}}_f$ and mixed products with ${\mathrm{ad}}_h$ have zero trace. Thus for $X=ah+be+cf$, $B(X,X)=8(a^2+bc)$. Put $x=a$, $y=(b+c)\mathrm{/}2$, $z=(b-c)\mathrm{/}(2i)$; then $a^2+bc=x^2+y^2+z^2=q$. These are orthonormal coordinates for $B\mathrm{/}8$, and $B$ is nondegenerate by this explicit diagonal matrix.

The $h$-derivation on $\mathbb{C}[a,b,c]$ is $2b{\mathrm{\partial }}_b-2c{\mathrm{\partial }}_c$. Its kernel consists of sums of monomials with equal $b,c$ exponents, hence polynomials $F(a,u)$ with $u=bc$. The $e$-derivation is $c{\mathrm{\partial }}_a-2a{\mathrm{\partial }}_b$, which on such a polynomial is $c(F_a-2a{F}_u)$. Since the polynomial ring is a domain, its vanishing is equivalent to $F_a-2a{F}_u=0$. In the invertible polynomial change $v=a^2+u$, write $G(a,v)=F(a,v-a^2)$. The equation is ${\mathrm{\partial }}_{a}G=0$, so characteristic zero gives $G=P(v)$. Hence every invariant is a polynomial in $q$. Conversely the displayed $h,e$ derivations kill $q$, and the $f$-derivation $-b{\mathrm{\partial }}_a+2a{\mathrm{\partial }}_c$ also kills $q$. Therefore the invariant ring is exactly $\mathbb{C}[q]$.

In the orthonormal coordinates for $B\mathrm{/}8$, replacement of a linear coordinate by its corresponding directional derivative sends $q$ to $\mathrm{\Delta }$ and $q^k$ to ${\mathrm{\Delta }}^k$. Replacing $B\mathrm{/}8$ by $B$ scales the identification on each degree by a nonzero scalar, so the kernels do not change. Positive-degree elements of $\mathbb{C}[q]$ are finite linear combinations of $q^k$ with $k\ge 1$. Thus their common differential kernel is exactly $\ker \mathrm{\Delta }$: necessity uses $q$, and sufficiency uses every power of $\mathrm{\Delta }$.

For a homogeneous harmonic polynomial $u$ of degree $m$, the product rule and the monomial Euler identity $x{\mathrm{\partial }}_{x}u+y{\mathrm{\partial }}_{y}u+z{\mathrm{\partial }}_{z}u=mu$ give $\mathrm{\Delta }(q^{k}u)=2k(2m+2k+1)q^{k-1}u$ for $k\ge 1$. Indeed ${\mathrm{\partial }}_j{q}^k=2k{x}_j{q}^{k-1}$ and $\mathrm{\Delta }{q}^k=2k(2k+1)q^{k-1}$ in three variables; the cross term contributes $4km{q}^{k-1}u$ and $q^k\mathrm{\Delta }u=0$. Every displayed coefficient is nonzero when $k\ge 1,m\ge 0$. For $k=0$ the Laplacian vanishes by harmonicity.

Write $S_n$ for homogeneous polynomials of degree $n$, and $H_n=\ker (\mathrm{\Delta }\mathrm{\mid }{S}_n)$. We prove $S_n={⨁}_{k=0}^{\lfloor n\mathrm{/}2\rfloor }{q}^k{H}_{n-2k}$ by induction on $n$. For $n=0,1$, differentiation twice gives zero, so $S_n=H_n$. For $n\ge 2$, the induction decomposition of $S_{n-2}$ shows that the map $\mathrm{\Delta }:q{S}_{n-2}\to S_{n-2}$ is an isomorphism: on each summand $q^{k+1}{H}_m$ it is multiplication by the nonzero scalar in 4.1 onto $q^k{H}_m$, and multiplication by the nonzero polynomial $q$ is injective. For any $p\in S_n$, there is therefore a unique $v\in q{S}_{n-2}$ with $\mathrm{\Delta }v=\mathrm{\Delta }p$; then $p-v\in H_n$. The same injectivity shows $H_n\cap q{S}_{n-2}=0$. This proves the inductive direct decomposition.

Summing the finite decompositions of 5.1 over degrees proves that multiplication $\mathbb{C}[q]\otimes \ker \mathrm{\Delta }\to S$ is surjective and injective: each tensor has finite support in the powers of $q$, and the degreewise decomposition makes all its harmonic coefficients unique. Counting the monomials $x^a{y}^b{z}^c$ with $a+b+c=n$ gives $\dim S_n=(n+1)(n+2)\mathrm{/}2$. For $n\ge 2$, 5.1 gives $\dim H_n=\dim S_n-\dim S_{n-2}=2n+1$; for $n=0,1$ the dimensions are directly $1,3$. Constants and all linear polynomials are harmonic, zero is allowed throughout, and the nonzero coefficients in 4.1 cover every higher step. The inverse decomposition is uniquely specified by finite calculations in each degree; no AC, general Chevalley restriction or general Kostant theorem is used.