Alphabeta Math
Pipeline-generated
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.

✓ 6 results · all verified · 5 also independently AI-judged
Every result on this page is machine-checked by a proof checker and read in full and owner-audited; the judge is an additional, independent cross-model AI review of the proofs. The 1 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.

Weyl Character and Multiplicity Formulas — Examples

1 · Prerequisites

2 · Summary

These leaves check the formulas on the smallest systems. The A2 example expands both sides of the denominator identity over the eight subsets of the positive roots and matches the six Weyl translates term by term. The sl2 example runs the telescoping quotient (e(m+1)ω−e−(m+1)ω)/(eω−e−ω) and the dimension specialization, including the boundary case m=0.

The sl3 adjoint module is used twice: Kostant's formula computes the zero weight with P(θ)=2, and Freudenthal's recursion recovers the same multiplicity from the six extremal weights, with the indeterminate case 0=0 occurring only at the top weight. The dimension formula for a fundamental sl3 module checks the normalisation against the three-dimensional defining representation, and the counterexample shows that dropping the μ↦μ+ρ shift in Kostant's formula returns the wrong multiplicity already for L(2ω) in sl2.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passjudge pass (gpt-6.1-sol)Open item page →

The A2 Weyl denominator expansion

Example

Assume the Axiom of Choice (The Axiom of Choice). Take g=sl3 with simple roots α1,α2 realized as in Root systems of the classical complex Lie algebras, positive roots Φ+={α1,α2,α1+α2}, Weyl vector ρ=α1+α2 and Weyl group W=S3={1,s1,s2,s1s2,s2s1,w0} with lengths 0,1,1,2,2,3. Expanding both sides of the denominator identity (The Weyl denominator identity) gives eρ(1−e−α1)(1−e−α2)(1−e−α1−α2)=eρ−es1ρ−es2ρ+es1s2ρ+es2s1ρ−ew0ρ=A(ρ), a finite identity between polynomials. The left side expands over the eight subsets S⊆Φ+ with signs (−1)∣S∣eρ−∑α∈Sα, and the unit terms from S={α1,α2} and S={α1+α2} cancel; the remaining monomials are eρ−eα2−eα1+e−α1+e−α2−e−ρ, which agrees term by term with the six Weyl translates ρ,α2,α1,−α1,−α2,−ρ of the right side.

Facts & Assumptions

Given: The Axiom of Choice, the realization of sl3 with diagonal Cartan subalgebra and roots ±α1,±α2,±(α1+α2), the positive system Φ+={α1,α2,α1+α2}, the Weyl vector ρ, the Weyl group W=S3 with its elements and lengths, and the alternant A(ρ).

[A1]

The Axiom of Choice is assumed; it enters through the root-system and denominator suppliers below (The Axiom of Choice).

[F1]

In this realization the positive roots are α1,α2 and α1+α2, the Weyl group acts by the simple reflections s1,s2 with s1α1=−α1, s1α2=α1+α2, s2α2=−α2, s2α1=α1+α2, so s1ρ=α2, s2ρ=α1, s1s2ρ=−α1, s2s1ρ=−α2 and w0ρ=−ρ, while ρ=12∑α∈Φ+α=α1+α2 (Root systems of the classical complex Lie algebras, Classical complex matrix Lie algebras, The Weyl vector rho for a chosen positive system, The root set is a reduced crystallographic root system).

[F2]

The Weyl group of A2 is S3={1,s1,s2,s1s2,s2s1,w0} with lengths 0,1,1,2,2,3, length being the number of inversions and the least number of simple reflections in an expression (Weyl length equals inversion number).

[F3]

The denominator identity states A(ρ)=eρ∏α∈Φ+(1−e−α), and A(ν)=∑w∈W(−1)ℓ(w)ewν (The Weyl denominator identity, The Weyl alternation operator).

Verification

1.1F1F2F3A1

The right side of the identity is the alternant A(ρ)=∑w∈W(−1)ℓ(w)ewρ=eρ−es1ρ−es2ρ+es1s2ρ+es2s1ρ−ew0ρ by [F3] and the length table [F2], and substituting the translates computed in [F1] gives A(ρ)=eρ−eα2−eα1+e−α1+e−α2−e−ρ.

1.2F1algebra

The left side eρ(1−e−α1)(1−e−α2)(1−e−α1−α2) expands over the eight subsets S of Φ+ as ∑S(−1)∣S∣eρ−∑α∈Sα; the exponents are ρ for S=∅, α2 for S={α1}, α1 for S={α2}, 0 for S={α1+α2} and for S={α1,α2}, −α1 for S={α1,α1+α2}, −α2 for S={α2,α1+α2} and −ρ for S=Φ+, with the signs +,−,−,−,+,+,+,− in this order.

2.1F3step 1.1step 1.2algebra∎

The two unit contributions in step 1.2, namely −e0 from S={α1+α2} and +e0 from S={α1,α2}, cancel, so the left side equals eρ−eα2−eα1+e−α1+e−α2−e−ρ, the same six monomials with the same signs as the right side computed in step 1.1; hence both sides of the denominator identity agree term by term in Z[P].

CounterexampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-6.1-sol)Open item page →

Omitting the rho shift breaks Kostant's formula

Statement refuted

False claim. In Kostant's weight multiplicity formula the second shift can be omitted, that is, the multiplicity of μ as a weight of L(λ) equals m^λ(μ):=∑w∈W(−1)ℓ(w)P(w(λ+ρ)−μ) as well.

Facts & Assumptions

Given: The Axiom of Choice, g=sl2 with positive root α, fundamental weight ω=α/2, Weyl group W={1,s}, Weyl vector ρ=ω, the weight λ=2ω, the weight μ=0, and the Kostant partition function P.

[A1]

The Axiom of Choice is assumed; it enters through the Kostant formula used in the counterexample (The Axiom of Choice).

[F1]

For sl2 the positive root satisfies α=2ω and sω=−ω, so ρ=α/2=ω and λ+ρ=3ω (Finite Weyl root system, lattice and chamber conventions, Fundamental weights, The Weyl vector rho for a chosen positive system).

[F2]

L(2ω) is the irreducible three-dimensional sl2-module whose weights are 2ω,0,−2ω, each with multiplicity one (Finite-dimensional representations of sl_2, Integral, dominant, and strictly dominant weights).

[F3]

P(β) counts the families (nα) with nαα=β, so P(α)=P(2ω)=1, while P(−4ω)=P(3ω)=0 because −4ω and 3ω=32α are not nonnegative integral multiples of the simple root α; the multiplicity formula of the Statement is mλ(μ)=∑w∈W(−1)ℓ(w)P(w(λ+ρ)−(μ+ρ)) (The Kostant partition function, Kostant's weight multiplicity formula).

Counterexample

technique · direct
1.1F1F2F3A1

The correct formula of [F3] gives m2ω(0)=P(3ω−ω)−P(−3ω−ω)=P(2ω)−P(−4ω)=1−0=1, in agreement with the weight string of [F2].

1.2F1F3

The modified expression omitting the shift μ↦μ+ρ gives m^2ω(0)=P(3ω−0)−P(−3ω−0)=P(3ω)−P(−3ω)=0−0=0, because 3ω=32α and −3ω are not nonnegative integral multiples of α.

2.1step 1.1step 1.2∎

Steps 1.1 and 1.2 exhibit a weight, namely μ=0 in L(2ω), whose true multiplicity is 1 while the modified formula returns 0; hence the modified formula is false, and the shift μ↦μ+ρ inside the partition argument is not a convention that can be dropped.

ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passjudge pass (gpt-6.1-sol)Open item page →

Kostant multiplicity in the sl3 adjoint module

Example

Assume the Axiom of Choice (The Axiom of Choice). Take g=sl3 with Φ+={α1,α2,α1+α2} and let θ=α1+α2 be the highest root, so that the adjoint module is L(θ) (The adjoint highest weight is the highest root, Highest-weight classification). Compute the multiplicity of μ=0 from Kostant's formula (Kostant's weight multiplicity formula): with ρ=θ, the terms are P(w(2θ)−θ) for w∈W=S3, and w(2θ)−θ∉Q+ for every w≠1, so only w=1 contributes and mθ(0)=P(θ)=2, the two partitions being θ itself and α1+α2; this matches the adjoint weights ±α1 (multiplicity one each), ±α2 (multiplicity one each), ±θ (multiplicity one each) and the zero weight of multiplicity two, for dim⁡L(θ)=8.

Facts & Assumptions

Given: The Axiom of Choice, the realization of sl3 with diagonal Cartan subalgebra and roots ±α1,±α2,±θ, where θ=α1+α2, the Weyl group W=S3 with simple reflections s1,s2, the Weyl vector ρ, and the adjoint module L(θ).

[A1]

The Axiom of Choice is assumed; it enters through the Kostant formula and the highest-weight suppliers below (The Axiom of Choice).

[F1]

In this realization the positive roots are α1,α2,θ and θ=α1+α2 is the highest root; the simple reflections act by s1θ=α2, s2θ=α1, s1α1=−α1, s2α2=−α2, and ρ=α1+α2=θ (Root systems of the classical complex Lie algebras, Classical complex matrix Lie algebras, Height and highest root, The Weyl vector rho for a chosen positive system).

[F2]

W=S3={1,s1,s2,s1s2,s2s1,w0} with lengths 0,1,1,2,2,3, and the adjoint module is the irreducible module L(θ) of highest weight θ; its weights are the roots together with 0, with the root spaces gα one-dimensional and h of dimension 2 (The adjoint highest weight is the highest root, Finite semisimple Cartan, root and string structure, Weyl length equals inversion number).

[F3]

P(θ)=2, because a family (nα1,nα2,nθ) with nα1α1+nα2α2+nθθ=θ is either (0,0,1) or (1,1,0), while P(ν)=0 for ν∉Q+ (The Kostant partition function).

[F4]

Kostant's formula reads mλ(μ)=∑w∈W(−1)ℓ(w)P(w(λ+ρ)−(μ+ρ)), and θ is dominant integral (Kostant's weight multiplicity formula, Integral, dominant, and strictly dominant weights).

Verification

1.1F1F2F3F4A1

With λ=θ, ρ=θ and μ=0 the arguments of [F4] are w(2θ)−θ; for w=1 this is θ with P(θ)=2 by [F3], while for w=s1 it is 2α2−θ=α2−α1, for w=s2 it is α1−α2, for w=s1s2 it is −2α1−θ, for w=s2s1 it is −2α2−θ and for w=w0 it is −3θ, none of which lies in Q+, so those partitions vanish by [F3].

1.2F3

The two partitions counted by P(θ)=2 are the family with nθ=1 and the family with nα1=nα2=1, both summing to θ.

2.1F2step 1.1step 1.2∎

Steps 1.1 and 1.2 give mθ(0)=P(θ)=2; the adjoint weights are the six roots, each with multiplicity one by [F2], and the zero weight whose multiplicity is the dimension 2 of h, so the weighted count is 6⋅1+2=8=dim⁡L(θ), matching the computed zero multiplicity.

ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passOpen item page →

Freudenthal recursion for the sl3 adjoint zero weight

Example

Assume the Axiom of Choice (The Axiom of Choice). Keep the setting of Kostant multiplicity in the sl3 adjoint module: g=sl3, λ=θ=α1+α2=ρ, μ=0. Then (λ+ρ,λ+ρ)−(μ+ρ,μ+ρ)=(2θ,2θ)−(θ,θ)=3(θ,θ), and in the right side of Freudenthal's weight multiplicity recursion only j=1 contributes, since 2α,3α,… are not weights of the adjoint module L(θ); hence the recursion reads 3(θ,θ)mλ(0)=2∑α∈Φ+(α,α)mλ(α). In the realization α1=e1−e2, α2=e2−e3, θ=e1−e3 of Root systems of the classical complex Lie algebras the three positive roots have the same length, and mλ(α1)=mλ(α2)=mλ(θ)=1, so 3(θ,θ)mλ(0)=2⋅3(θ,θ) and mλ(0)=2, recovering Kostant multiplicity in the sl3 adjoint module. The extremal weights ±α1,±α2,±θ=Wλ are the Weyl orbit of the top weight and each has multiplicity one (Extremal Weyl-orbit weights); among them only the top weight θ=λ has the vanishing recursion coefficient of the indeterminate case 0=0 in Freudenthal recursion terminates from the highest weight, whose base value mλ(λ)=1 is stated there.

Facts & Assumptions

Given: The Axiom of Choice, the realization of sl3 with positive roots α1,α2,θ=α1+α2 of equal length, the Weyl vector ρ=θ, the top weight λ=θ, the weight μ=0, the adjoint module L(θ), and the multiplicities mθ(ν).

[A1]

The Axiom of Choice is assumed; it enters through the Freudenthal recursion and the highest-weight suppliers below (The Axiom of Choice).

[F1]

In the realization α1=e1−e2, α2=e2−e3 the positive roots are α1,α2,θ with (α1,α1)=(α2,α2)=(θ,θ), and ρ=α1+α2=θ (Root systems of the classical complex Lie algebras, Classical complex matrix Lie algebras, The Weyl vector rho for a chosen positive system).

[F2]

The adjoint module is L(θ) and its weights are 0 with multiplicity 2, and ±α1,±α2,±θ each with multiplicity 1 (The adjoint highest weight is the highest root, Kostant multiplicity in the sl3 adjoint module).

[F3]

Freudenthal's recursion reads ((λ+ρ,λ+ρ)−(μ+ρ,μ+ρ))mλ(μ)=2∑α∈Φ+∑j≥1(μ+jα,α)mλ(μ+jα) (Freudenthal's weight multiplicity recursion).

[F4]

The extremal weights Wλ each have multiplicity one, and the recursion has the base value mλ(λ)=1 with the indeterminate case 0=0 occurring, among actual weights of L(λ), only at μ=λ (Extremal Weyl-orbit weights, Freudenthal recursion terminates from the highest weight).

Verification

1.1F2F3A1

With λ=θ=ρ and μ=0 the recursion coefficient of [F3] is (2θ,2θ)−(θ,θ)=3(θ,θ), and the weights 0+jα=jα of L(θ) are nonzero only for j=1 by the weight list [F2], so each inner sum of 2∑α∈Φ+∑j≥1(αj,α)m(jα) reduces to its j=1 term (α,α)m(α).

1.2F1F2

By [F1] all three positive roots have the same squared length, and by [F2] m(α1)=m(α2)=m(θ)=1, so the right side of the recursion is 2((α1,α1)+(α2,α2)+(θ,θ))=6(θ,θ).

2.1F2step 1.1step 1.2

Combining steps 1.1 and 1.2, the recursion reads 3(θ,θ)mθ(0)=6(θ,θ), and since (θ,θ)≠0 this gives mθ(0)=2, recovering the value computed by Kostant's formula in [F2].

3.1F4step 1.1∎

The extremal weights are the six Weyl translates of λ=θ, each of multiplicity one by [F4]; the recursion coefficient 4(θ,θ)−(μ+ρ,μ+ρ) vanishes, among actual weights of L(θ), only at μ=λ by [F4], so among the extremal weights only the top weight is the indeterminate case of the recursion, whose value mθ(θ)=1 is the stated base case; this is consistent with the recursion fixing every other multiplicity from that base.

ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passjudge pass (gpt-6.1-sol)Open item page →

Weyl character and dimension formulas for sl2

Example

Assume the Axiom of Choice (The Axiom of Choice). Take g=sl2 with positive root α, W={1,s}, ρ=α/2 and λ=mω, ω=α/2, m≥0 dominant integral. Then A(ν)=eν−e−ν for every ν, and the Weyl character formula (The Weyl character formula) gives ch⁡L(mω)=e(m+1)ω−e−(m+1)ωeω−e−ω=emω+e(m−2)ω+⋯+e−mω, a sum of m+1 terms; the Weyl dimension formula (The Weyl dimension formula) gives dim⁡L(mω)=((m+1)ω,α)/((ω,α))=m+1; the boundary case m=0 gives the one-term character ch⁡L(0)=1 and dim⁡L(0)=1.

Facts & Assumptions

Given: The Axiom of Choice, g=sl2 with its positive root α, fundamental weight ω=α/2, Weyl group W={1,s}, Weyl vector ρ=ω, and the dominant integral weights mω with m≥0.

[A1]

The Axiom of Choice is assumed; it enters through the character and dimension formulas below (The Axiom of Choice).

[F1]

α=2ω, W={1,s}, sω=−ω, ρ=ω and λ+ρ=(m+1)ω; the length of s is 1 (The Weyl vector rho for a chosen positive system, Integral, dominant, and strictly dominant weights).

[F2]

A(ν)=eν−e−ν for every ν∈h∗, and the Weyl character formula and Weyl dimension formula read ch⁡L(λ)=A(λ+ρ)A(ρ)−1 and dim⁡L(λ)=∏α∈Φ+(λ+ρ,α)/(ρ,α) (The Weyl alternation operator, The Weyl character formula, The Weyl dimension formula).

[F3]

In the completed ring R the elements e±ω are invertible with eωe−ω=e0, and eω−e−ω=eω(1−e−α) is invertible with inverse e−ω(1−e−α)−1, because 1−e−α is invertible (The completed formal character ring, Geometric series are invertible in the completed character ring, The Weyl denominator identity).

[F4]

For m≥0 the module L(mω) is the finite-dimensional simple module of highest weight mω (Highest-weight classification).

Verification

1.1F1F2F3A1

By [F2] the character is ch⁡L(mω)=A((m+1)ω)A(ω)−1=(e(m+1)ω−e−(m+1)ω)(eω−e−ω)−1, the quotient being the formal product with the inverse of [F3].

2.1F3step 1.1algebra

Multiplying the displayed quotient by eω−e−ω telescopes: (eω−e−ω)(emω+e(m−2)ω+⋯+e−mω)=e(m+1)ω−e−(m+1)ω, so the finite sum emω+e(m−2)ω+⋯+e−mω of m+1 terms is the product of the numerator with the inverse of eω−e−ω and hence equals ch⁡L(mω) by step 1.1.

3.1F1F2F4step 1.1∎

By [F2] the dimension formula gives dim⁡L(mω)=((m+1)ω,α)/((ω,α))=m+1, since λ+ρ=(m+1)ω by [F1] and the positive system consists of the single root α; specializing to m=0 gives A(ω)A(ω)−1=e0=1 for the character and (m+1)=1 for the dimension.

ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passjudge pass (gpt-6.1-sol)Open item page →

The Weyl dimension formula for a fundamental sl3 module

Example

Assume the Axiom of Choice (The Axiom of Choice). Take g=sl3 with simple roots α1,α2, positive roots Φ+={α1,α2,α1+α2}, fundamental weight ω1 dual to α1∨ and Weyl vector ρ=α1+α2. The Weyl dimension formula (The Weyl dimension formula) gives dim⁡L(ω1)=∏α∈Φ+(ω1+ρ,α)(ρ,α)=⟨ω1+ρ,α1∨⟩⟨ρ,α1∨⟩⋅⟨ω1+ρ,α2∨⟩⟨ρ,α2∨⟩⋅⟨ω1+ρ,θ∨⟩⟨ρ,θ∨⟩=21⋅11⋅32=3, matching the three-dimensional defining representation of sl3.

Facts & Assumptions

Given: The Axiom of Choice, the realization of sl3 with simple roots α1,α2, positive roots Φ+={α1,α2,θ} with θ=α1+α2, the fundamental weight ω1, the Weyl vector ρ and the coroots α1∨,α2∨,θ∨.

[A1]

The Axiom of Choice is assumed; it enters through the dimension formula and the highest-weight suppliers below (The Axiom of Choice).

[F1]

In this realization the positive roots are α1,α2 and θ=α1+α2, whose coroots in the simply-laced system add: θ∨=α1∨+α2∨ (Root systems of the classical complex Lie algebras, Classical complex matrix Lie algebras, The root set is a reduced crystallographic root system).

[F2]

The fundamental weight ω1 is dual to α1∨: ⟨ω1,α1∨⟩=1 and ⟨ω1,α2∨⟩=0, and the Weyl vector satisfies ⟨ρ,αi∨⟩=1 and ρ=α1+α2; moreover ⟨ρ,θ∨⟩=2 (Fundamental weights, Integral, dominant, and strictly dominant weights, The root set is a reduced crystallographic root system).

[F3]

The Weyl dimension formula states dim⁡L(λ)=∏α∈Φ+(λ+ρ,α)/(ρ,α)=∏α∈Φ+⟨λ+ρ,α∨⟩/⟨ρ,α∨⟩ (The Weyl dimension formula).

[F4]

The defining representation C3 of sl3 has weights ε1,ε2,ε3, with ε1 the highest weight ω1; it is irreducible, because for a nonzero v=∑iciei with cj≠0 the matrix units Eij with i≠j and the diagonal elements of sl3 produce all three basis vectors e1,e2,e3, so the submodule generated by v is everything; hence L(ω1)≅C3 has dimension 3 (Classical complex matrix Lie algebras, Highest-weight classification).

Verification

1.1F1F2A1

By [F2] the three coroot pairings of the numerator are ⟨ω1+ρ,α1∨⟩=1+1=2, ⟨ω1+ρ,α2∨⟩=0+1=1 and, using θ∨=α1∨+α2∨ from [F1], ⟨ω1+ρ,θ∨⟩=2+1=3, while the denominators are ⟨ρ,α1∨⟩=⟨ρ,α2∨⟩=1 and ⟨ρ,θ∨⟩=2.

2.1F3step 1.1algebra

Substituting these six values into the product of [F3] gives dim⁡L(ω1)=(2/1)(1/1)(3/2)=3.

3.1F4step 2.1∎

The defining module C3 is irreducible of highest weight ω1 by [F4], so L(ω1) has dimension 3, in agreement with the product computed in step 2.1; this checks the normalisation of the product.

Sources