Alphabeta Math
LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07
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.

Finite torus fourier orthogonality and affine change

Statement

For the normalized negative-exponent Fourier transform on (Z/mZ)2, the characters are an orthonormal basis, and f=bf^(b)χb,f2=bf^(b)2,xf(x)=0    f^(0)=0. For every invertible matrix T over Z/mZ, and g(x)=f(Tx+a), g^(y)=ωyT1af^(TTy).

Facts & Assumptions

Given: the objects and hypotheses in the statement above.

[F1]

On V=(Z/mZ)2, m1, let ω=e2πi/m and χb(x)=ωb1x1+b2x2. Residue representatives do not affect these values. With inner product f,g=m2xf(x)g(x), define f^(b)=f,χb=m2xf(x)ωbx. The norm is f2=m2xf(x)2. At m=1 there is one character, the constant function one. The sign in the exponent is part of this convention. (Finite torus fourier transform).

Proof

1.1

For m>1, j=0m1ωcj is m if c=0 modulo m, and otherwise is zero because multiplication by 1ωc0 telescopes to 1ωcm=0. Applying this to each coordinate shows χb,χc is one for b=c and zero otherwise. At m=1 the one character has norm one directly.

F1
2.1

There are m2 orthonormal characters in the m2-dimensional function space, hence they form a basis: linear independence follows by taking inner products, and an independent list of that length spans by elementary elimination. Expansion in this basis gives inversion and, on taking its squared norm, Parseval. The zero coefficient is exactly the normalized sum, establishing both directions of the mean-zero criterion.

step 1.1algebra
3.1

Substitute u=Tx+a in the defining sum. The exponent becomes yT1(ua)=(TTy)u+yT1a. Bijection of this substitution preserves the sum and its normalization, giving the positive phase in the displayed formula. It covers a=0, constant and zero functions, and the singleton torus as well.

F1step 2.1algebra

Depends on

Used by

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