ExampleConstruction: Literature-sourcedVerification: AI-adaptedPipeline-generatedverified 2026-10-15 (gpt-6-luna, gpt-6.1-sol, and delegated Codex review/repair agents; see completed run evidence)

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Two physical photon fibers without global frame

Example

Assume Countable Choice. For each nonzero $k$ , the physical photon fiber is the real plane $n^{⊥}$ (or its complexification), where $n = k / ∣ k ∣$ , and therefore has dimension two. Locally in direction space one can choose two continuous orthonormal polarization vectors. There is no continuous real orthonormal polarization frame on all of $S^{2}$ . The global subsidiary quotient $K_{0} / N$ of Lorenz auxiliary subsidiary null quotient is isometric to the transverse space $H_{T}$ of square-integrable sections from Positive transverse photon one-particle space. At a fixed direction $n$ , the auxiliary fiber $K_{0, n} = {(a, v) \in C \times C^{3} : n \cdot v = a}$ , with form $[(a, v), (b, w)]_{η} = - \overline{a} b + \overline{v} \cdot w$ , has radical $N_{n} = C (1, n)$ and quotient isometric to the complex transverse plane. Thus four auxiliary components do not represent four positive physical polarizations; the global section space is not a single fiber.

Facts & Assumptions

Given: Countable Choice; the transverse projection, the subsidiary quotient, and the no-global-frame lemma.

[A1]

Assume Countable Choice. The Axiom of Countable Choice ( ${A C}_{ω}$ )

[F1]

The range of $P_{T} (n) = I - n n^{T}$ is the rank-two plane $n^{⊥}$ . Positive transverse photon one-particle space

[F2]

The global subsidiary quotient $K_{0} / N$ is isometric to $H_{T}$ by $[ϵ] \mapsto ϵ - n ϵ^{0}$ ; these are spaces of square-integrable sections. Lorenz auxiliary subsidiary null quotient

[F3]

The transverse planes over $S^{2}$ admit no global continuous real orthonormal frame. No global continuous real polarization frame

[F4]

The cross product is orthogonal to both factors and satisfies $∥ u \times v ∥^{2} = ∥ u ∥^{2} ∥ v ∥^{2} - ⟨ u, v ⟩^{2}$ . The cross product in $R^{3}$ The cross product is bilinear, alternating, and orthogonal to both factors The squared cross-product norm is the Gram determinant of two vectors

Verification

Given: [A1] and [F1]–[F3].

1.1A1F1F4givenconstruct

Since $n$ is unit, $P_{T} (n) = I - n n^{T}$ has kernel $R n$ and range $n^{⊥}$ , of dimension $3 - 1 = 2$ , by [F1]. For any fixed unit vector $u$ not parallel to $n_{0}$ , $P_{T} (n) u$ is nonzero on a neighborhood of $n_{0}$ ; there define $e_{1} (n) = P_{T} (n) u / ∣ P_{T} (n) u ∣$ and $e_{2} (n) = n \times e_{1} (n)$ . By [F4], $e_{2}$ is orthogonal to both unit vectors and $∣ e_{2} ∣^{2} = 1 - 0 = 1$ . It is therefore a continuous unit vector in $n^{⊥}$ , orthogonal to $e_{1}$ , so the two vectors form a local frame.

2.1F1F2F3step 1.1algebra∎

For the fixed-direction fiber in the example, define $T_{n} (a, v) = v - n a$ . Since $n \cdot v = a$ and $n \cdot n = 1$ , its image lies in $n^{⊥} \subset C^{3}$ , and $T_{n} (0, z) = z$ for every $z \in n^{⊥}$ . Writing $v = T_{n} (a, v) + n a$ and $w = T_{n} (b, w) + n b$ gives $[(a, v), (b, w)]_{η} = \overline{T_{n} (a, v)} \cdot T_{n} (b, w)$ . Its radical is therefore $\ker T_{n} = C (1, n)$ : a radical vector pairs with itself to give $∥ T_{n} (a, v) ∥^{2} = 0$ , and every kernel vector pairs to zero with all others. Thus $[(a, v)] \mapsto T_{n} (a, v)$ is a positive isometry of $K_{0, n} / N_{n}$ onto the two-complex-dimensional plane of step 1.1. In contrast, [F2] identifies the global quotient with $H_{T}$ , the space of square-integrable sections of these planes. A global continuous real frame would contradict [F3], so the local construction cannot extend over the whole sphere.

Two physical photon fibers without global frame

Example

Facts & Assumptions

Verification

Depends on

Dependency tree · 2 levels

Sources