Manuscript III@2 · III.tex · lemma · 82dad3e5

Lemma: Sign and differential identities

Conventions, signs, and support › Geometry and operators · lines 411–457 · proof included or linked

III@2#lem:signs, III@2#eq:commutator, III@2#eq:sign-lock

For smooth $ξ$ and $u$ ,

D K ξ = □_{η} ξ, G^{(1)} (K ξ) = 0, \partial^{μ} G^{(1)} (h)_{μν} = 0.

Moreover,

[P, X_{ξ}] u = (□_{η} ξ^{ν}) \partial_{ν} u + 2 (\partial^{μ} ξ^{ν}) \partial_{μ} \partial_{ν} u,

and therefore

L_{K ξ} = - [P, X_{ξ}] .

Proof. In $D K ξ$ , the derivatives of $\partial_{μ} ξ^{μ}$ cancel, leaving $□_{η} ξ$ . Insert this and

(r K ξ)_{μν} = \partial_{μ} ξ_{ν} + \partial_{ν} ξ_{μ} - η_{μν} \partial_{α} ξ^{α}

into (eq:Glin) to obtain $G^{(1)} K ξ = 0$ .

Taking the divergence of (eq:Glin) gives

- \frac{1}{2} □_{η} (D h)_{ν} + \frac{1}{2} □_{η} (D h)_{ν} + \frac{1}{2} \partial_{ν} \partial^{α} (D h)_{α} - \frac{1}{2} \partial_{ν} \partial^{α} (D h)_{α} = 0.

Apply the product rule twice to $□_{η} (ξ^{ν} \partial_{ν} u)$ . The constant mass term commutes with $X_{ξ}$ , proving (eq:commutator). Substituting $K ξ$ in (eq:Lh) gives its negative.

Displayed formulas in this passage

III@2#eq:commutator · line 420
$[P, X_{ξ}] u = (□_{η} ξ^{ν}) \partial_{ν} u + 2 (\partial^{μ} ξ^{ν}) \partial_{μ} \partial_{ν} u,$
III@2#eq:sign-lock · line 427
$L_{K ξ} = - [P, X_{ξ}] .$
unlabeled · line 414
$D K ξ = □_{η} ξ, G^{(1)} (K ξ) = 0, \partial^{μ} G^{(1)} (h)_{μν} = 0.$
unlabeled · line 437
$(r K ξ)_{μν} = \partial_{μ} ξ_{ν} + \partial_{ν} ξ_{μ} - η_{μν} \partial_{α} ξ^{α}$
unlabeled · line 445
$- \frac{1}{2} □_{η} (D h)_{ν} + \frac{1}{2} □_{η} (D h)_{ν} + \frac{1}{2} \partial_{ν} \partial^{α} (D h)_{α} - \frac{1}{2} \partial_{ν} \partial^{α} (D h)_{α} = 0.$

Lemma: Sign and differential identities

Displayed formulas in this passage

Nearby in this section