LemmaStatement: AI-generatedProof: AI-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.

A scalar normalization survives a two-point functional

Statement

Let $A$ be a complex vector space with a complex-bilinear multiplication $A \times A \to A$ , written by juxtaposition; no associativity or multiplicative identity is required. Let $ℓ : A \to C$ and $τ_{t} : A \to A$ be complex-linear maps, where $t$ is an arbitrary fixed label. For a real number $γ > 0$ , let $\sqrt{γ}$ denote its unique nonnegative real square root. Given $F_{1}, F_{2} \in A$ , define $B_{j} = - (\sqrt{γ})^{- 1} F_{j}$ for $j = 1, 2$ . Then

$ℓ (τ_{t} (B_{1}) B_{2}) = γ^{- 1} ℓ (τ_{t} (F_{1}) F_{2}) .$

In particular, for $F \in A$ and $B = - F / \sqrt{γ}$ , one has $ℓ (τ_{t} (B) B) = γ^{- 1} ℓ (τ_{t} (F) F)$ . These identities require no limiting operation or convention for the time coordinate.

Facts & Assumptions

Given: The complex vector space and its bilinear product, the two complex-linear maps, the positive real number $γ$ , and the elements $F_{1}, F_{2}$ specified in the Statement.

[F1]

Scalar multiplication in a complex vector space is closed and satisfies $a (b u) = (a b) u$ . Vector space over a field

[F2]

A complex-linear map preserves scalar multiplication and finite sums. Linear map between vector spaces over the same field

[F3]

A complex-bilinear product is complex-linear in each argument separately. Balanced maps from a right module and a left module, and bilinear maps over a commutative ring

[F4]

In a Dedekind-complete ordered field, every $x \geq 0$ has a unique $r \geq 0$ with $r^{2} = x$ . Nonnegative real square roots exist uniquely

[F5]

The real numbers in their Cauchy-sequence realization form a complete ordered field: every nonempty bounded-above subset has a least upper bound. Thus they satisfy the Dedekind-complete ordered-field hypotheses of [F4]. The Cauchy-sequence reals have the least-upper-bound property Dedekind-complete ordered real field

[F6]

The complex numbers are the field $C = R [x] / (x^{2} + 1)$ , with the constant-class embedding $ι : R \to C$ . This embedding is injective and preserves addition, multiplication and the identity. Real scalars in complex scalar actions are identified with their images under $ι$ . The complex numbers as ℝ[x]/(x²+1), with the real embedding and imaginary unit i 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

Goal: Prove the two-component identity and its equal-component specialization.

1.1givenF1F4F5F6algebra

By [F5], the real field satisfies the hypotheses of [F4], so $r = \sqrt{γ}$ exists and satisfies $r \geq 0$ and $r^{2} = γ$ . Since $γ > 0$ , $r = 0$ would imply $γ = 0$ , so $r > 0$ . In the real field $a = - r^{- 1}$ is therefore defined and $a^{2} = (r^{2})^{- 1} = γ^{- 1}$ . By [F6], $ι (r) ι (r^{- 1}) = ι (1) = 1$ and $ι (r) \neq 0$ , so the real inverse maps to the complex inverse. Preservation of addition and multiplication also gives $ι (a)^{2} = ι (γ^{- 1})$ . With the stated identification this is $a^{2} = γ^{- 1}$ in $C$ , and $B_{j} = a F_{j}$ belongs to $A$ by [F1].

2.1F1F2F3step 1.1algebra

By [F2], $τ_{t} (B_{1}) = a τ_{t} (F_{1})$ . Both this element and $B_{2} = a F_{2}$ lie in $A$ . Applying [F3] separately in the two product arguments and composing the scalar actions by [F1] gives $τ_{t} (B_{1}) B_{2} = a^{2} τ_{t} (F_{1}) F_{2} = γ^{- 1} τ_{t} (F_{1}) F_{2}$ , an equality in $A$ .

3.1F2step 2.1algebra∎

Apply $ℓ$ to the equality in step 2.1 and use [F2] to obtain $ℓ (τ_{t} (B_{1}) B_{2}) = γ^{- 1} ℓ (τ_{t} (F_{1}) F_{2})$ . Taking $F_{1} = F_{2} = F$ gives the stated diagonal identity. The argument is algebraic for each fixed label $t$ , so it involves neither limits nor any time-coordinate convention.

Depends on

Used by · 1

Isotropic thermal bath time correlator

Dependency tree · 2 levels

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