Manuscript XVI-D@1 · XVI-D.tex · proposition · b73c6f4a

Proposition: Exact spatial traceless response contraction

Explicit witness response and traceless channel › Spatial traceless contraction · lines 891–932 · proof included or linked

XVI-D@1#lem:contraction, XVI-D@1#eq:contraction

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The response satisfies

h_{ij} V_{λ}^{ij} = \frac{2}{3} λ^{2} F_{λ}, h_{ij} U_{D}^{ij} = - \frac{2}{3} λ^{2} F_{λ} .

The correction $λ d_{1}$ contributes zero to this channel, including after trace reversal.

These are response contractions, not completed insertion values.

Proof. The Euclidean identities are

h_{ij} δ^{ij} = 0, h_{ij} n^{i} n^{j} = \frac{2}{3} .

Thus (eq:q) and (eq:d1) give

h_{ij} S_{λ}^{ij} = \frac{2}{3} λ^{2} σ e^{2 i λ t} .

Integrating proves the first identity.

On raised spatial components,

(P V_{λ})^{ij} = V_{λ}^{ij} + \frac{1}{2} δ^{ij} tr_{η} V_{λ} .

The added term is annihilated by $h$ . Apply (eq:identify) to obtain the second identity. Any correction entering the trace also multiplies $δ^{ij}$ and is annihilated by the same contraction.

Displayed formulas in this passage

XVI-D@1#eq:contraction · line 894
$h_{ij} V_{λ}^{ij} = \frac{2}{3} λ^{2} F_{λ}, h_{ij} U_{D}^{ij} = - \frac{2}{3} λ^{2} F_{λ} .$
unlabeled · line 910
$h_{ij} δ^{ij} = 0, h_{ij} n^{i} n^{j} = \frac{2}{3} .$
unlabeled · line 916
$h_{ij} S_{λ}^{ij} = \frac{2}{3} λ^{2} σ e^{2 i λ t} .$
unlabeled · line 923
$(P V_{λ})^{ij} = V_{λ}^{ij} + \frac{1}{2} δ^{ij} tr_{η} V_{λ} .$

Proposition: Exact spatial traceless response contraction

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