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

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

The Harish-Chandra projection computes the highest-weight scalar

Statement

Let $M$ be a cyclic highest-weight module with highest vector $v$ of weight $λ$ , and let $z \in Z (U (g))$ . Then

$z v = pr (z) (λ) v,$

so the scalar by which $z$ acts on $M$ is obtained by evaluating the Harish-Chandra projection at $λ$ .

Facts & Assumptions

Given: A cyclic highest-weight module $M = U (g) v$ of highest weight $λ$ and a central element $z \in Z (U (g))$ .

Proof

technique · direct

1.1

By Central elements lie in the zero-weight subspace of $U (g)$ , the central element $z$ lies in $U (g)_{0}$ , so write $z = pr (z) + a$ using The Harish-Chandra projection, with $a \in n^{-} U (g) + U (g) n^{+}$ .

givenconstruct

2.1

Every summand of $a$ either has a factor from $n^{+}$ on the right, which kills the highest vector $v$ , or has a nontrivial factor from $n^{-}$ on the left, which lowers the weight and therefore cannot contribute to the highest-weight line. Hence $a v = 0$ .

step 1.1algebra

3.1

The element $pr (z)$ lies in $U (h)$ , so it acts on $v$ by the scalar obtained from the polynomial $pr (z)$ by evaluation at the weight $λ$ . Combining this with step 2.1 gives $z v = pr (z) (λ) v$ .

step 1.1step 2.1∎

Depends on

Used by

Central characters are dot-Weyl orbits Corollary
Harish-Chandra isomorphism for the center Theorem

Dependency tree · two levels

8 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 4 (standard reference, not scraped)