Manuscript XV-F@3 · XV-F-v3.tex · lemma · d4e74afa

Lemma: The derivative of XII-E's quadratic coefficient

Source manifest (Gate F0) › Verification of the reproduced displays · lines 1118–1141 · proof included or linked

XV-F@3#lem:a2, XV-F@3#eq:a2dot

With $a_{2}$ , $b_{2}$ as in (eq:a2), the directional derivative at $G$ in the direction $h$ is

a_{2} ˙^{μν} [G; h] = h^{μ}_{α} G^{α ν} + G^{μ}_{α} h^{α ν} - \frac{1}{2} (tr_{η} h) G^{μν} - \frac{1}{2} (tr_{η} G) h^{μν} + (\frac{1}{4} (tr_{η} G) (tr_{η} h) - \frac{1}{2} G_{α β} h^{α β}) η^{μν} =: d^{μν, ab, c d} G_{ab} h_{c d},

bilinear and symmetric in $(G, h)$ , with $a ˙_{2} [r; r] = 2 a_{2} [r]$ .

Proof. Differentiate each term of (eq:a2) at $s = 0$ along $G + s h$ ; $\frac{d}{d s} b_{2} [G + s h] ∣_{0} = \frac{1}{4} (tr G) (tr h) - \frac{1}{2} G_{α β} h^{α β}$ . Symmetry and the Euler identity follow from homogeneity of degree two.

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XV-F@3#eq:a2dot · line 1122
$a_{2} ˙^{μν} [G; h] = h^{μ}_{α} G^{α ν} + G^{μ}_{α} h^{α ν} - \frac{1}{2} (tr_{η} h) G^{μν} - \frac{1}{2} (tr_{η} G) h^{μν} + (\frac{1}{4} (tr_{η} G) (tr_{η} h) - \frac{1}{2} G_{α β} h^{α β}) η^{μν} =: d^{μν, ab, c d} G_{ab} h_{c d},$

Lemma: The derivative of XII-E's quadratic coefficient

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