ExampleConstruction: Literature-sourcedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-13

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.

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First weight layers of the basic affine sl2 character

Example

For the basic level-one module $L (Λ_{0})$ of untwisted affine ${s l}_{2}$ , with $q = e^{- δ}$ and $z = e^{- α}$ , $e^{- Λ_{0}} ch L (Λ_{0}) = 1 + (z^{- 1} + 1 + z) q + (z^{- 1} + 2 + z) q^{2} + O (q^{3}) .$ The notation records relative loop degree; a change in the complementary value of $Λ_{0}$ cancels under normalization. The displayed coefficients are Laurent polynomials, and the remainder has degree at least three in the formal completion.

Facts & Assumptions

Given: The normalized loop realization with $h = h_{1}$ , $h_{0} = c - h$ , $α (h) = 2$ and $B (h, h) = 2$ . Fix the basic dominant weight by its simple-coroot labels $Λ_{0} (h_{0}) = 1$ and $Λ_{0} (h) = 0$ , with an arbitrary complementary Cartan value.

[F1]

Weyl Kac character formula gives the normalized numerator divided by the positive-root product.

[F2]

Kac moody integral and dominant integral weights permits the supplied labels $1, 0$ and an arbitrary complementary value. Affine central coroot from the transpose null ray identifies the central coroot here as $c = h_{0} + h_{1}$ ; hence the supplied labels give $Λ_{0} (c) = 1$ and level one.

[F3]

Roots of an untwisted affine Lie algebra gives real roots $\pm α + n δ$ of multiplicity one and imaginary roots $n δ$ of multiplicity one here.

[F4]

Affine Weyl group is a coroot lattice semidirect product gives the unique forms $t_{m h}$ and $t_{m h} s$ and displays the full translation formula in its Statement.

Verification

1.1

Put $λ = Λ_{0} + ρ$ . Then $λ (h) = 1$ and $λ (c) = 3$ , by F2 and $ρ (h_{0}) = ρ (h) = 1$ in F1. F4's translation formula, with $ν (h) = α$ and $B (m h, m h) / 2 = m^{2}$ , gives $t_{m h} λ - λ = 3 m α - (3 m^{2} + m) δ, t_{m h} s λ - λ = (3 m - 1) α - (3 m^{2} - m) δ .$ Translations have sign plus, because $t_{h} = s_{0} s$ and powers have even sign; the second family has sign minus. Thus the normalized alternant in F1 is $U = \sum_{m \in Z} (z^{- 3 m} q^{3 m^{2} + m} - z^{1 - 3 m} q^{3 m^{2} - m})$ . Only $m = 0$ and the degree-two terms from $m = - 1$ in the first family and $m = 1$ in the second contribute below degree three. Indeed both quadratic expressions are at least four for the other nonzero choices. Hence $U = (1 - z) + (z^{3} - z^{- 2}) q^{2} + O (q^{3})$ .

F1F2F4algebra

1.2

By F3 the product is $P = (1 - z) \prod_{n \geq 1} (1 - q^{n}) (1 - z q^{n}) (1 - z^{- 1} q^{n})$ . Put $S = 1 + z + z^{- 1}$ . The $n = 1$ triple is $1 - S q + S q^{2} - q^{3}$ and the $n = 2$ triple is $1 - S q^{2} + O (q^{4})$ ; all later triples begin at degree three or more. Thus $P = (1 - z) (1 - S q + O (q^{3}))$ , with zero coefficient at degree two inside the parentheses.

F3algebra

2.1

Polynomial division gives $(z^{3} - z^{- 2}) / (1 - z) = - (z^{- 2} + z^{- 1} + 1 + z + z^{2}) = - T$ . By step 1.1, $U / (1 - z) = 1 - T q^{2} + O (q^{3})$ . By step 1.2, the inverse of $P / (1 - z)$ through degree two is $1 + S q + S^{2} q^{2} + O (q^{3})$ . Consequently F1 gives $U / P = 1 + S q + (S^{2} - T) q^{2} + O (q^{3})$ . Direct multiplication gives $S^{2} = z^{2} + z^{- 2} + 3 + 2 z + 2 z^{- 1}$ and $S^{2} - T = 2 + z + z^{- 1}$ , proving the statement. Cancellation of $1 - z = 1 - e^{- α}$ is valid in the downward completion by its geometric inverse; division never lowers $q$ degree. At each fixed degree the numerator has finitely many terms by the quadratic bounds and the denominator has finitely many relevant positive-degree factors. This also justifies every displayed truncation without an analytic identity or AC.

F1step 1.1step 1.2algebra∎

Depends on

Used by

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Dependency tree · two levels

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Sources

Kleshchev, Sections 6.4 and 11.2 (standard reference, not scraped)
Perrin, Chapter 12 (standard reference, not scraped)