Manuscript Em01-A@1 · Em01-A.tex · lemma · 880b67af

Lemma: Operator identities

A common classical setting › Bundles, pairing, and operators · lines 325–358 · proof included or linked

Em01-A@1#lem:identities

The operators above satisfy

D K ξ = □ ξ, L K ξ = 0, \partial^{μ} (L h)_{μν} = 0.

Moreover,

S_{dD} = {h : □ h = 0, D h = 0} .

Proof. Expanding $D K ξ$ , the two gradient-divergence terms cancel, leaving $□ ξ$ . Substitution into (eq:einstein-operator), together with commutation of the constant-coefficient operators, gives $L K ξ = 0$ .

For any one-form $d$ ,

\partial^{μ} (T K d)_{μν} = □ d_{ν} .

Taking the divergence of (eq:einstein-operator) therefore gives

2 \partial^{μ} (L h)_{μν} = - □ (D h)_{ν} + □ (D h)_{ν} = 0.

If $D h = 0$ , equation (eq:einstein-operator) reduces to $2 L h = - □ T h$ . Since $T$ is invertible and commutes with $□$ , the last assertion follows.

Displayed formulas in this passage

unlabeled · line 328
$D K ξ = □ ξ, L K ξ = 0, \partial^{μ} (L h)_{μν} = 0.$
unlabeled · line 334
$S_{dD} = {h : □ h = 0, D h = 0} .$
unlabeled · line 346
$\partial^{μ} (T K d)_{μν} = □ d_{ν} .$
unlabeled · line 351
$2 \partial^{μ} (L h)_{μν} = - □ (D h)_{ν} + □ (D h)_{ν} = 0.$

Lemma: Operator identities

Displayed formulas in this passage

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