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Kostant harmonic subspace of the symmetric algebra

Definition

Let $g$ be a finite-dimensional complex semisimple Lie algebra with the symmetric adjoint action of Finite semisimple Lie algebras and the symmetric adjoint action. Its Killing form $B (x, y) = tr ({ad}_{x} {ad}_{y})$ is nondegenerate and invariant by Engel, the trace criterion, and Killing nondegeneracy. Thus $x \mapsto B (x, \cdot)$ identifies $S (g)$ with polynomial functions on $g$ ; a finite basis and its Killing-dual basis give inverse linear substitutions.

For $x \in g$ define the constant-coefficient derivation $D_{x}$ of $S (g)$ by $D_{x} (y) = B (x, y)$ on generators. These derivations commute, so for $p \in S$ substitution extends multiplicatively and linearly to an operator $D_{p} = p (\partial)$ . Under the polynomial-function identification, $D_{x}$ is directional differentiation along $x$ , since it differentiates $B (y, \cdot)$ to $B (y, x)$ . The Kostant harmonic subspace is $H = {h \in S : D_{p} h = 0 for every p \in S_{+}^{g}} .$ Equivalently one may test only homogeneous positive-degree invariants, since their finite homogeneous sums exhaust that ideal. This is a graded subspace. With $I = S S_{+}^{g}$ , its degree- $d$ part is the annihilator $I_{d}^{⊥}$ for the perfect bilinear Fischer pairing $(p, q)_{B} = (D_{p} q) (0)$ on $S_{d}$ , as verified below. It is stable under the adjoint action. The direct-sum complement assertion $S_{d} = H_{d} \oplus I_{d}$ is a separate theorem; it is not inferred from complex bilinear nondegeneracy.

Facts & Assumptions

Given: The finite-dimensional semisimple Lie algebra and the displayed polynomial and differential constructions.

[F1]

The symmetric algebra, its finite-degree pieces and adjoint invariant ideal are defined in Finite semisimple Lie algebras and the symmetric adjoint action.

[F2]

The Killing form is symmetric, invariant and nondegenerate by Engel, the trace criterion, and Killing nondegeneracy.

Proof

1.1F1F2givenalgebra

The derivations $D_{x}, D_{y}$ commute because their commutator is a derivation vanishing on each linear generator. Hence the polynomial assignment $p \mapsto D_{p}$ is well-defined. Choose a basis $e_{1}, \dots, e_{n}$ and its $B$ -dual basis $f_{1}, \dots, f_{n}$ . For multi-indices $a, b$ of total degree $d$ , the product rule gives $(D_{e^{a}} f^{b}) (0) = {\begin{cases} \prod_{i} a_{i}! & a = b, \\ 0 & a \neq b . \end{cases}$ Both monomial lists are bases of $S_{d}$ , so the pairing is nondegenerate in both variables. It is symmetric: on products of linear generators its value is the sum over all bijections of the products of the corresponding $B$ -pairings, unchanged on interchanging the two products by symmetry of $B$ . This also verifies that degree zero has the usual nondegenerate scalar product.

2.1step 1.1F1givenalgebra

For homogeneous $p, q, h$ with $\deg p + \deg q = \deg h$ , commuting constant-coefficient differentiation gives $(p q, h)_{B} = (q, D_{p} h)_{B}$ . If $h \in S_{d}$ is harmonic, it therefore pairs to zero with every degree- $d$ product of a positive invariant and any polynomial, hence with $I_{d}$ . Conversely if it annihilates $I_{d}$ , fix a homogeneous invariant $p$ of degree $k > 0$ . If $k > d$ then $D_{p} h = 0$ by degree; if $k \leq d$ , the displayed identity and nondegeneracy on $S_{d - k}$ force $D_{p} h = 0$ . This proves $H_{d} = I_{d}^{⊥}$ exactly. Homogeneous decomposition also shows that testing every such $p$ is equivalent to testing every positive-degree invariant and that $H$ is graded.

3.1step 1.1F1F2givenalgebra∎

On a generator $y$ , invariance of $B$ gives $[{ad}_{z}, D_{x}] (y) = - B (x, [z, y]) = B ([z, x], y)$ . Both sides are derivations, so $[{ad}_{z}, D_{x}] = D_{[z, x]}$ on $S$ . The commutator product rule extends this to $[{ad}_{z}, D_{p}] = D_{{ad}_{z} p}$ for every polynomial $p$ . For invariant $p$ the commutator is zero; hence $D_{p} h = 0$ implies $D_{p} ({ad}_{z} h) = 0$ . This proves adjoint stability of $H$ . Constants are harmonic, $I_{0} = 0$ , and in dimension zero $S = H = C$ with $I = 0$ . No complementary-subspace conclusion has been used, and all chosen dual bases are finite.

Depends on

Used by

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Sources

Pavel Etingof, Lie Groups and Lie Algebras, §§15–17; finite-dimensional local trace proof (standard reference, not scraped)