Manuscript XX-E@1 · XX-E.tex · theorem · cf7a8f40

Theorem: Cutoff Ward breaking

The evaluated classical Ward identity and cutoff breaking › An exact evaluated compact-source-cutoff residual · lines 639–683 · proof included or linked

XX-E@1#thm:cutoff, XX-E@1#eq:cutoffbreaking

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The complete classical breaking of this compact-source comparison is

V_{ξ} O_{f, R} = s^{2} \int f (ξ^{a} - R_{R} (K ξ)^{a}) \partial_{a} ℓ z d t d^{3} x (mod s^{3}) .

It vanishes when $χ_{R} = 1$ on the spatial support of $ξ$ . It is not a universal identity for a fixed finite cutoff and the full compact gauge group.

Proof. The difference from the native functional is

O_{f, R} - O_{f} = s^{2} \int f (R - R_{R}) z \cdot \nabla ℓ z .

Its variation is

s^{2} \int f [(R - R_{R}) K ξ \cdot \nabla ℓ z + (R - R_{R}) z \cdot \nabla ℓ (K ξ)] .

The second term vanishes, and the first is (eq:cutoffbreaking) by Lemma lem:rodgauge.

If the cutoff is one on the spatial support of $ξ$ , it is one on the source $C K ξ$ , giving equality of the rod maps.

For a failure witness at fixed finite cutoff, choose a nonzero purely spatial $ξ$ supported outside $supp χ_{R}$ . Then $R_{R} K ξ = 0$ . Choose a compact $h_{00}$ so that $ξ^{a} \partial_{a} ℓ z$ is nonzero on an open set, and choose $f$ to detect it.

Displayed formulas in this passage

XX-E@1#eq:cutoffbreaking · line 643
$V_{ξ} O_{f, R} = s^{2} \int f (ξ^{a} - R_{R} (K ξ)^{a}) \partial_{a} ℓ z d t d^{3} x (mod s^{3}) .$
unlabeled · line 659
$O_{f, R} - O_{f} = s^{2} \int f (R - R_{R}) z \cdot \nabla ℓ z .$
unlabeled · line 664
$s^{2} \int f [(R - R_{R}) K ξ \cdot \nabla ℓ z + (R - R_{R}) z \cdot \nabla ℓ (K ξ)] .$

Theorem: Cutoff Ward breaking

Displayed formulas in this passage

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