Manuscript Em01-G@5 · Em01-G-v5.tex · lemma · 3fbfe951

Lemma: Projected supported coefficients

Supported algebraic coefficients · lines 725–761 · proof included or linked

Em01-G@5#lem:projection, Em01-G@5#eq:projection0, Em01-G@5#eq:projection1

One has

I \sum (e^{I})^{μν} ℓ_{μ} c_{0, 2}^{I, J} I \sum (e^{I})^{μν} ℓ_{μ} c_{ζ, 2}^{I, J} = - \frac{m f ( b )}{8} P_{J} (n) ℓ^{ν}, = 0, ∣ ζ ∣ = 1.

Proof. Contract with an arbitrary $t_{ν}$ . Completeness replaces the first input by $H = (ℓ_{♭} \otimes t + t \otimes ℓ_{♭}) /2$ . Equation (eq:gauge-slot) kills both source terms in (eq:c02). Direct substitution in the density array gives

ℓ_{α} ℓ_{μ} D_{dens}^{α μ} (e_{J}, H) = \frac{1}{2} (ℓ \cdot t) P_{J} .

This proves (eq:projection0).

For the mixed coefficient, set $w = 1$ in (eq:top). The total source momentum is $K = - h$ , and $T_{H} (K) = 0$ . As $z \to 0$ ,

T_{e_{J}} (z) = O (∣ z ∣^{2}), C_{e_{J}} (z) = O (∣ z ∣), T_{H} (K - z) = O (∣ z ∣) .

The mixed divergence identity makes the contraction with $K$ of the source $O (∣ z ∣^{2})$ . Its linear term is zero, proving (eq:projection1).

Displayed formulas in this passage

Em01-G@5#eq:projection0 · line 728
$I \sum (e^{I})^{μν} ℓ_{μ} c_{0, 2}^{I, J} I \sum (e^{I})^{μν} ℓ_{μ} c_{ζ, 2}^{I, J} = - \frac{m f ( b )}{8} P_{J} (n) ℓ^{ν}, = 0, ∣ ζ ∣ = 1.$
Em01-G@5#eq:projection1 · line 728
$I \sum (e^{I})^{μν} ℓ_{μ} c_{0, 2}^{I, J} I \sum (e^{I})^{μν} ℓ_{μ} c_{ζ, 2}^{I, J} = - \frac{m f ( b )}{8} P_{J} (n) ℓ^{ν}, = 0, ∣ ζ ∣ = 1.$
unlabeled · line 744
$ℓ_{α} ℓ_{μ} D_{dens}^{α μ} (e_{J}, H) = \frac{1}{2} (ℓ \cdot t) P_{J} .$
unlabeled · line 754
$T_{e_{J}} (z) = O (∣ z ∣^{2}), C_{e_{J}} (z) = O (∣ z ∣), T_{H} (K - z) = O (∣ z ∣) .$

Lemma: Projected supported coefficients

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