LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-05

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The leading PBW symbol of a central element is invariant

Statement

If $z \in Z (U (g))$ has leading PBW symbol $σ_{m} (z) \in {gr}_{m} U (g) ≅ S^{m} (g)$ , then $σ_{m} (z)$ is invariant under the adjoint action of $g$ .

Facts & Assumptions

Given: A central element $z \in Z (U (g))$ of PBW degree $m$ and its leading symbol $σ_{m} (z) \in S^{m} (g)$ .

Proof

technique · direct

1.1

Centrality means $[x, z] = 0$ for every $x \in g$ . For $x \in F_{1} U (g)$ and $z \in F_{m} U (g)$ , the commutator estimate in The associated graded algebra of the PBW filtration is commutative gives $[x, z] \in F_{m} U (g)$ . Thus the adjoint action of $x$ induces an operator on ${gr}_{m} U (g)$ .

given

2.1

Under the PBW identification with $S (g)$ , that induced operator is the derivation extending $x \cdot y = [x, y]$ on $y \in g$ . It therefore sends $σ_{m} (z)$ to the degree- $m$ symbol of $[x, z]$ . The latter commutator is zero by step 1.1, so $x \cdot σ_{m} (z) = 0$ for every $x \in g$ .

step 1.1algebra

3.1

Therefore the leading PBW symbol of $z$ lies in $S (g)^{g}$ .

step 2.1∎

Depends on

Used by

Harish-Chandra isomorphism for the center Theorem

Dependency tree · two levels

7 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Pavel Etingof, Representations of Lie Groups (standard reference, not scraped)
Lin Chen, Geometric Representation Theory I, Lecture 5 (standard reference, not scraped)