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

Proposition: Free-scalar improvement and currents

Phase 3: an independent free-scalar consistency audit › Field input and its limited role · lines 1332–1376 · proof included or linked

Oai069-A@1#prop:improvement

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Distributionally on the supplied Wick domain,

□ Q = 2 K_{0}, (T^{can})^{μ}_{μ} = - K_{0}, \partial^{μ} C_{μν} = 0.

Hence

L = - \frac{1}{2} Q, t_{μν} = T^{can}_{μν} - \frac{1}{6} C_{μν}

satisfy

(T^{can})^{μ}_{μ} = □ L, \partial^{μ} t_{μν} = 0, t^{μ}_{μ} = 0.

For a conformal Killing field $X$ , the current $X^{ν} t_{μν}$ is conserved.

Proof. The Wick derivative rule and $□ ϕ = 0$ give $□ Q = 2 K_{0}$ . Contraction of $T^{can}$ in four dimensions gives $- K_{0}$ . Commuting distributional derivatives gives $\partial^{μ} C_{μν} = 0$ , and

C^{μ}_{μ} = - 3 □ Q = - 6 K_{0} .

Canonical conservation follows from the free equation.

For

\partial_{μ} X_{ν} + \partial_{ν} X_{μ} = 2 σ_{X} η_{μν}, σ_{X} = \frac{1}{4} \partial_{ρ} X^{ρ},

symmetry gives

\partial^{μ} (X^{ν} t_{μν}) = X^{ν} \partial^{μ} t_{μν} + σ_{X} t^{μ}_{μ} = 0.

Displayed formulas in this passage

unlabeled · line 1335
$□ Q = 2 K_{0}, (T^{can})^{μ}_{μ} = - K_{0}, \partial^{μ} C_{μν} = 0.$
unlabeled · line 1341
$L = - \frac{1}{2} Q, t_{μν} = T^{can}_{μν} - \frac{1}{6} C_{μν}$
unlabeled · line 1346
$(T^{can})^{μ}_{μ} = □ L, \partial^{μ} t_{μν} = 0, t^{μ}_{μ} = 0.$
unlabeled · line 1360
$C^{μ}_{μ} = - 3 □ Q = - 6 K_{0} .$
unlabeled · line 1366
$\partial_{μ} X_{ν} + \partial_{ν} X_{μ} = 2 σ_{X} η_{μν}, σ_{X} = \frac{1}{4} \partial_{ρ} X^{ρ},$
unlabeled · line 1372
$\partial^{μ} (X^{ν} t_{μν}) = X^{ν} \partial^{μ} t_{μν} + σ_{X} t^{μ}_{μ} = 0.$

Proposition: Free-scalar improvement and currents

Displayed formulas in this passage

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