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Positive-energy tube analyticity and discrete-symmetry proof interfaces — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

None yet.

5 · Examples, experiments and counterexamples

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passverified 2026-10-22 (gpt-6.1-sol)Open item page →

One positive frequency decays in the future imaginary direction

Example

Let $p \in {\overline{V}}_{+}$ be nonzero, let $H = C^{2}$ with standard basis $e_{0}, e_{1}$ , and set $Ω = e_{0}$ , $E_{p} (S) = diag (1_{S} (0), 1_{S} (p))$ , and $U_{p} (a) = diag (1, e^{i p \cdot a})$ . Take $A = B (C^{2})$ . These data give a positive-energy joint spectral input with a translation-invariant cyclic vacuum: the zero mode fixes $Ω$ , while $e_{1}$ carries one nonzero positive frequency. The continuation is $U_{p} (a + i b) = diag (1, e^{i p \cdot a - p \cdot b})$ . For every $b \in V_{+}$ , the $e_{1}$ component decays by $e^{- p \cdot b} < 1$ ; the vacuum component remains $1$ , so the full operator norm is exactly $1$ . Reversing the Fourier sign gives $U_{p}^{-} (a + i b) = diag (1, e^{- i p \cdot a + p \cdot b})$ , which grows on $e_{1}$ in the future imaginary direction and decays there in the past direction. This finite-spectrum example illustrates the Fourier-sign convention; it is not a claim about a physical field or a general Wightman theorem.

Example references

Positive-energy joint spectral input

Positive spectrum gives bounded future-tube continuation

Fewster and Rejzner, §§5.1–5.2

Mund, §1.3 and §2.1

Facts & Assumptions

Given: ${A C}_{ω}$ , nonzero $p \in {\overline{V}}_{+}$ , and the positive-energy joint-spectrum conventions of Positive-energy joint spectral input.

[F1]

Positive-energy joint spectral input: the Minkowski pairing is $p \cdot a = p^{0} a^{0} - p \cdot a$ , the future cone is ${\overline{V}}_{+} = {p^{0} \geq ∣ p ∣}$ , bounded Borel functions act by the PVM integral, and the input includes a translation-invariant cyclic unit vector.

[F2]

Positive spectrum gives bounded future-tube continuation: for $z = a + i b$ , $b \in V_{+}$ , the continuation is $U (z) = Φ_{E} (1_{{\overline{V}}_{+}} e^{i p \cdot z})$ with norm at most one.

[A1]

$H = C^{2}$ has inner product $(x, y) = x_{0} \overline{y_{0}} + x_{1} \overline{y_{1}}$ and standard basis $e_{0}, e_{1}$ ; $Ω = e_{0}$ ; $E_{p} (S) = diag (1_{S} (0), 1_{S} (p))$ for Borel $S$ ; $U_{p} (a) = diag (1, e^{i p \cdot a})$ ; and $A = B (C^{2})$ .

Verification

1.1A1F1F2algebra

For Borel sets $S, T$ , $E_{p} (S)$ is diagonal with entries in ${0, 1}$ , $E_{p} (R^{4}) = I$ , and $E_{p} (S \cap T) = E_{p} (S) E_{p} (T)$ . For a disjoint sequence, each of the distinct points $0, p$ belongs to at most one set, so the coordinatewise sum of the projections is $E_{p} (⋃_{n} S_{n})$ and is finite on every vector. Thus $E_{p}$ is a strongly countably additive PVM supported on ${0, p} \subseteq {\overline{V}}_{+}$ . The diagonal operators $U_{p} (a)$ form a unitary representation because $∣ e^{i p \cdot a} ∣ = 1$ and $U_{p} (a + a^{'}) = U_{p} (a) U_{p} (a^{'})$ ; continuity is ordinary continuity of the exponential. Put $P_{0} : = E_{p} ({0}) = diag (1, 0)$ and $P_{p} : = E_{p} ({p}) = diag (0, 1)$ . Since $p \neq 0$ , these are the two spectral projections and $E_{p}$ vanishes off their supports. Therefore the bounded Borel multiplier $q \mapsto e^{i q \cdot a}$ has integral $Φ_{E_{p}} (q \mapsto e^{i q \cdot a}) = e^{i 0 \cdot a} P_{0} + e^{i p \cdot a} P_{p} = diag (1, e^{i p \cdot a}) = U_{p} (a),$ by the finite-support PVM integral formula. Also $U_{p} (a) Ω = Ω$ . For every $(x, y) \in C^{2}$ , the matrix $(\begin{array}{cc} x & 0 \\ y & 0 \end{array})$ sends $Ω$ to $(x, y)$ , so $Ω$ is cyclic for $A$ .

2.1A1F1F2step 1.1algebra

The cone-restricted multiplier has values $1$ at $0$ and $e^{i p \cdot (a + i b)} = e^{i p \cdot a} e^{- p \cdot b}$ at $p$ ; hence its PVM integral is $diag (1, e^{i p \cdot a - p \cdot b})$ . For $b \in V_{+}$ , $p \cdot b \geq p^{0} (b^{0} - ∣ b ∣) > 0$ , since nonzero $p \in {\overline{V}}_{+}$ has $p^{0} > 0$ . Therefore the $e_{1}$ eigenvalue has modulus $e^{- p \cdot b} < 1$ , while the $e_{0}$ eigenvalue is $1$ . The operator norm is $1$ , in agreement with [F2].

3.1A1F1algebra∎

On the $e_{1}$ component the reversed-sign continuation is $e^{- i p \cdot (a + i b)} = e^{- i p \cdot a + p \cdot b}$ , of modulus $e^{p \cdot b} > 1$ for $b \in V_{+}$ . For $b \in - V_{+}$ , $p \cdot b < 0$ , so its modulus is less than $1$ . The $e_{0}$ vacuum component stays $1$ in both continuations. Thus the nonzero frequency's damping direction reverses with the Fourier sign.