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Cauchy kernel through bidegree three

Example

Let Ω≤3:=∑d=03Ωd,d be the part of the Cauchy kernel in bidegrees (d,d) with 0≤d≤3. Its complete–monomial and Schur expansions are computed below. Pairing the x-factor with s21(x) gives s21(y).

Facts & Assumptions

Given: The degreewise stable ring, the finite complete and monomial conventions, their stable bases, the Cauchy expansions, the Hall form, and Jacobi–Trudi.

[F1]

In the bidegree completion, Ω(x,y)=∑λhλ(x)mλ(y)=∑λsλ(x)sλ(y); both sums are taken by diagonal bidegree (Power-sum, complete, and Schur expansions of the Cauchy kernel).

[F2]

Each Λd is the inverse limit of the rank-N homogeneous symmetric-polynomial parts, and multiplication is induced by rankwise polynomial multiplication (The stable graded ring of symmetric functions).

[F3]

In rank N, hk is the sum of all monomials of total degree k, with h0=1 (Power sums pk and complete homogeneous symmetric polynomials hk).

[F4]

In rank N, mλ is the sum of the distinct monomials whose exponent tuples are permutations of the padded tuple λ (Monomial symmetric polynomials indexed by partitions).

[F5]

In rank N, the mλ indexed by partitions of length at most N form a Z-basis of the symmetric polynomials (Monomial symmetric polynomials form an R-basis of the symmetric-polynomial ring).

[F6]

The stable orbit sums mλ, for λ⊢d, form a Z-basis of Λd and project to the finite orbit sums whenever N≥d (The monomial symmetric functions form the integral stable basis).

[F7]

The finite hr specialize compatibly to stable elements, and hλ=∏ihλi denotes their stable product (Elementary and complete families freely generate the stable ring).

[F8]

The Hall form is graded and satisfies ⟨hλ,mμ⟩H=δλμ (The Hall inner product on symmetric functions).

[F9]

For r≥ℓ(λ), sλ=det⁡(hλi−i+j)1≤i,j≤r, with h0=1, negative subscripts zero, and the empty determinant equal to 1 (Jacobi–Trudi and dual Jacobi–Trudi identities).

[F10]

The stable Schur functions form an orthonormal basis for the Hall form: ⟨sλ,sμ⟩H=δλμ (Schur functions form an orthonormal integral basis).

[F11]

The completed tensor product has diagonal bidegree kernel Ω=∏i,j(1−xiyj)−1 (Bidegree completion of two symmetric-function rings).

Verification

technique · direct
1.1F2F5F6

At rank 3, the projection Λd→A3d is an isomorphism for each 0≤d≤3: for d>0, every partition of d has length at most d≤3, and [F5] and [F6] identify the stable and finite monomial bases; for d=0, both components are Z with basis 1. Thus finite rank-3 coefficient calculations determine the stable coefficients in all degrees used here.

2.1F3F4F7step 1.1algebra

At rank 3, grouping monomials by distinct exponent orbits and counting ordered products gives the complete-function identities below; m2m1=m3+m21 and m11m1=m21+3m111, while types (3),(2,1),(1,1,1) occur in h13 with multiplicities 1,3,3!=6.

h1=m1,h2=m2+m11,h11=h12=m2+2m11.

h3=m3+m21+m111,h21=h2h1=m3+2m21+3m111,h111=h13=m3+3m21+6m111.

m2m1=m3+m21,m11m1=m21+3m111.

3.1F9step 2.1algebra

Jacobi–Trudi evaluates the Schur functions through degree three as shown; substituting step 2.1 gives the monomial expressions, including s111=h13−2h2h1+h3.

s1=h1,s2=h2,s11=h12−h2,s3=h3,s21=h2h1−h3,s111=h13−2h2h1+h3.

s1=m1,s2=m2+m11,s11=m11,s3=m3+m21+m111,s21=m21+2m111,s111=m111.

s111=(m3+3m21+6m111)−2(m3+2m21+3m111)+(m3+m21+m111)=m111.

4.1F1step 1.1step 2.1step 3.1

The partitions of degrees 0,1,2,3 are respectively {∅}, {(1)}, {(2),(1,1)}, and {(3),(2,1),(1,1,1)}, so [F1] gives these complete–monomial and Schur components of Ω≤3 in bidegrees (d,d).

Ω0,0=1⊗1,Ω1,1=h1(x)⊗m1(y),Ω2,2=h2(x)⊗m2(y)+h11(x)⊗m11(y).

Ω3,3=h3(x)⊗m3(y)+h21(x)⊗m21(y)+h111(x)⊗m111(y).

Ω0,0=1⊗1,Ω1,1=s1(x)⊗s1(y),Ω2,2=s2(x)⊗s2(y)+s11(x)⊗s11(y).

Ω3,3=s3(x)⊗s3(y)+s21(x)⊗s21(y)+s111(x)⊗s111(y).

4.2F8step 2.1step 3.1algebra

Since s21=h21−h3, duality in [F8] gives ⟨s21,h3⟩H=⟨h21−h3,m3+m21+m111⟩H=1−1=0.

4.3F8step 2.1step 3.1algebra

Similarly, ⟨s21,h21⟩H=⟨h21−h3,m3+2m21+3m111⟩H=2−1=1.

4.4F8step 2.1step 3.1algebra

Also, ⟨s21,h111⟩H=⟨h21−h3,m3+3m21+6m111⟩H=3−1=2.

5.1F1F8F10step 3.1step 4.1step 4.2step 4.3step 4.4

Gradedness removes degrees below three, so contracting the first factor in the complete–monomial expansion gives m21(y)+2m111(y)=s21(y); contraction of the Schur expansion gives the same result by [F10].

6.1F1F3F4F5F6F11step 2.1algebra∎

Degree 0 gives 1⊗1 (and [F11] specializes to 1 when either alphabet is zero); in degree 1, h1=m1=s1 has coefficient one. Degree 3 is the retained upper endpoint and rank 3 is its threshold rank; repeated parts in h11 and h111 have the coefficients from step 2.1, while each mλ lists distinct monomials. All counts are finite, so no choice is used, and no equivalence is asserted.

Depends on

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