Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-6.1-sol)
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.

Orthogonality of the characters x↦e2πikx/N on Z/NZ

Statement

Let N≥1 and let k,ℓ∈Z. Then

∑x=0N−1e2πi(k−ℓ)x/N={N,k≡ℓ(modN),0,k≢ℓ(modN).

At N=1 the congruence holds for all k,ℓ and the sum is 1. The sum is the finite sum of the complex family x↦e2πi(k−ℓ)x/N over the von Neumann natural N={0,…,N−1} (A finite sum in a commutative monoid indexed by an arbitrary finite set), and N on the right is the natural number N read in C as the additive multiple N⋅1C, that is, the value at N of the canonical embedding N→C (The canonical natural ι(n)=n⋅1F of a field, In a field, the additive multiple n⋅1F is the canonical natural ι(n): the additive power of the group-power definition and the canonical natural are the same function, both being the unique one given by the recursion ι(0)=0F, ι(σ(n))=ι(n)+1F).

Facts & Assumptions

Given: A natural number N≥1, integers k,ℓ, the complex number ω:=e2πi(k−ℓ)/N, the partial sums Sn:=∑x<nωx for n∈N, and S:=SN.

[F1]

k≡ℓ(modN) means N∣(k−ℓ), that is, k−ℓ=Nm for some integer m; the relation is an equivalence relation and is defined for every integer modulus, including N≥1 (Congruence modulo an integer: a≡b(modn) when n∣(a−b), including the moduli 0 and 1, Congruence modulo every integer is an equivalence relation on Z).

[F2]

For complex z,w: exp⁡z=1 exactly when z∈2πiZ, and exp⁡z=exp⁡w exactly when z−w∈2πiZ (ker⁡(exp⁡)=2πiZ, and exp⁡z=exp⁡w exactly when z−w∈2πiZ, The complex exponential by its power series).

[L1]

exp⁡(z+w)=exp⁡z exp⁡w for all complex z,w (exp⁡(z+w)=exp⁡z exp⁡w, and the complex exponential extends the real exponential).

[L2]

A finite sum in a commutative monoid is computed from any enumeration of its finite index set and does not depend on it; it is unchanged by reindexing along a bijection, additive over disjoint splittings, and subject to the finite Fubini rule (A finite sum in a commutative monoid indexed by an arbitrary finite set, Finite commutative-monoid sums are invariant under bijective reindexing, split over disjoint unions, and satisfy the finite Fubini rule).

[L3]

Powers in C: ω0=1 and ωn+1=ωnω for every n∈N (Integer powers in the complex field), and ωm+n=ωmωn for all naturals m,n (Laws of integer exponents, claim 1). Induction is available (The principle of mathematical induction).

[L4]

Additive natural powers: in the additive group of C, the element N⋅1C defined by 0⋅1C=0 and σ(n)⋅1C=n⋅1C+1C equals the image of the natural number N under the canonical embedding N→C (Powers gn: natural exponents in a monoid and integer exponents in a group, with g0=e read additively, In a field, the additive multiple n⋅1F is the canonical natural ι(n): the additive power of the group-power definition and the canonical natural are the same function, both being the unique one given by the recursion ι(0)=0F, ι(σ(n))=ι(n)+1F).

[L5]

Field laws of C: multiplication is associative and commutative and distributes over addition, every nonzero element has an inverse, and 1≠0 (C=R[x]/(x2+1) is a field, every element is uniquely a+bi, and every nonzero element has inverse (a−bi)/(a2+b2)).

Proof

technique · cases
1.1F2L1L2L3given

The summands are the powers of ω: e2πi(k−ℓ)x/N=ωx for every x∈N. Indeed, for x=0 both sides are 1 by [F2] (as 0∈2πiZ) and [L3]; and if the identity holds at x, then the addition law [L1] and the power recursion [L3] give e2πi(k−ℓ)(x+1)/N=e2πi(k−ℓ)x/Ne2πi(k−ℓ)/N=ωxω=ωx+1, so induction [L3] proves it for every x. Consequently S=∑x<Nωx by [L2], since the finite sum depends only on the listed values.

1.2L2L3L5given

The geometric identity: (1−ω)Sn=1−ωn for every n∈N. For n=0 the sum S0 is empty, hence 0 by [L2], and 1−ω0=0 by [L3]. If the identity holds at n, then Sn+1=Sn+ωn by the recursion clause of [L2], so distributivity [L5] gives (1−ω)Sn+1=(1−ω)Sn+(1−ω)ωn=(1−ωn)+(ωn−ωnω)=1−ωn+1, using the hypothesis, distributivity, associativity and the power recursion [L3]. Induction [L3] gives the identity at every n.

2.1F1F2step 1.1

The two alternatives for ω: ω=1 exactly when k≡ℓ(modN), and ωN=1 always. For the first, [F2] gives ω=1  ⟺  2πi(k−ℓ)/N∈2πiZ  ⟺  (k−ℓ)/N∈Z  ⟺  N∣(k−ℓ)  ⟺  k≡ℓ(modN) by [F1]; for the second, e2πi(k−ℓ)=1 by [F2] because 2πi(k−ℓ)∈2πiZ, and step 1.1 identifies e2πi(k−ℓ) with ωN.

2.2assume-case congruentF1F2step 1.1L2L4given

Case k≡ℓ(modN): then k−ℓ=Nm for some integer m by [F1], so every exponent 2πi(k−ℓ)x/N=2πimx lies in 2πiZ and every summand ωx equals 1 by step 1.1 and [F2]. The sum therefore consists of N copies of 1C, and the recursion clause of [L2] computes it as the additive natural power N⋅1C of [L4]. In particular at N=1, where every pair k,ℓ is congruent, the sum is the single term 1, the image of the natural number 1 under the canonical embedding.

3.1assume-case noncongruentstep 1.2step 2.1L5

Case k≢ℓ(modN): then ω≠1 and ωN=1 by step 2.1, so the geometric identity of step 1.2 at n=N gives (1−ω)S=1−ωN=0. Since 1−ω≠0, the field laws [L5] give S=(1−ω)−1⋅0=0.

4.1casesF1step 2.2step 3.1∎

The alternatives of [F1] are exhaustive and mutually exclusive, so steps 2.2 and 3.1 cover every pair (k,ℓ): the sum equals the natural number N read in C in the congruent case and 0 otherwise, which is the stated formula.

Remarks

  • The case ωm=1 is exactly the case split used by Taylor. In Taylor's proof of Proposition 11.2 the sum Sm of the powers of ω satisfies Sm=ωmSm, so it vanishes whenever ωm≠1; the coincident case ωm=1 is separated first. Here the split is made on k≡ℓ(modN) and the noncoincident case is settled by the geometric identity, which is the same computation in explicit finite-sum form.

  • No dependence on the representation-theoretic orthogonality. The published orthogonality lemma for finite abelian groups and the real-variable factorisation lemma for bn−an are stated outside the complex-sum setting used here, so this lemma proves the complex geometric identity directly from the recursion instead of importing them. They are independent cross-checks, not prerequisites.

Depends on

Used by

Dependency tree · two levels

64 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