Manuscript Oai069-A@1 · Oai069-A.tex · theorem · a4913aab

Theorem: Two-probe homogeneous stress coverage

Phase 3: an independent free-scalar consistency audit › Two probes with complete coverage inside that class · lines 1477–1514 · proof included or linked

Oai069-A@1#thm:stresscoverage

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For (eq:stressclass),

p_{1} (T^{a, b}) = \frac{a}{2}, p_{2} (T^{a, b}) = - a - 6 b .

The two probes determine $a, b$ uniquely:

a = 2 p_{1}, b = - \frac{p _{2} + 2 p _{1}}{6} .

Therefore agreement of the two probes implies equality of every member of the conserved class $W_{4}$ .

The normalized traceless choice is uniquely

a = 1, b = - \frac{1}{6}, L = - \frac{1}{2} Q .

Proof. On the probe neighborhood, $u = x^{0}$ , so

K_{0} (u) = 1, T^{can}_{00} (u) = 1/2, C_{00} (u) = 0, C^{μ}_{μ} (u) = - 6.

The unit-integral smear gives the displayed values. The coefficient matrix has nonzero determinant $- 3$ . The inverse formulas prove coverage.

The first probe fixes $a = 1$ under the declared normalization $p_{1} = 1/2$ . Zero trace then fixes $b = - 1/6$ . Within the homogeneous scalar dimension-two Wick class, $L = λ Q$ , and $□ L = 2 λ K_{0} = - K_{0}$ gives $λ = - 1/2$ .

Displayed formulas in this passage

unlabeled · line 1480
$p_{1} (T^{a, b}) = \frac{a}{2}, p_{2} (T^{a, b}) = - a - 6 b .$
unlabeled · line 1485
$a = 2 p_{1}, b = - \frac{p _{2} + 2 p _{1}}{6} .$
unlabeled · line 1493
$a = 1, b = - \frac{1}{6}, L = - \frac{1}{2} Q .$
unlabeled · line 1500
$K_{0} (u) = 1, T^{can}_{00} (u) = 1/2, C_{00} (u) = 0, C^{μ}_{μ} (u) = - 6.$

Theorem: Two-probe homogeneous stress coverage

Displayed formulas in this passage

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