The following lemma is from Fisher, Marsden, and Moncrief's paper: the structure of the space of solutions of Einstein's equations:1

1.1. Lemma.

If Ein( $\left.{ }^{(4)} g\right)=0$, and ${ }^{(4)} h$ is any symmetric two tensor, then $$ \delta\left[\operatorname{DEin}\left({ }^{(4)} g\right) \cdot ^{4} h\right]=0 $$ where $\delta=\delta_{(4) g}$ is the divergence with respect to ${ }^{(4)} g$.

They gave the following proof:

The contracted Bianchi identities assert that $\delta \operatorname{Ein}\left({ }^{(4)} g\right)=0$. Differentiation gives the identity $$ **\left[\text { D } \delta\left({ }^{(4)} g\right) \cdot ^{4}h\right] \cdot \operatorname{Ein}\left({ }^{(4)} g\right)+\delta\left[D \operatorname{Ein}\left({ }^{(4)} g\right) \cdot^{4} h\right]=0** $$ where $\delta\left({ }^{(4)} g\right)=\delta_{(4)}$ indicates the functional dependence of $\delta$ on ${ }^{(4)} g$, and $\left[\mathrm{D} \delta\left({ }^{(4)} g\right) \cdot ^{4} \mathrm{h}\right]$ is the linearized divergence operator acting on Ein $\left({ }^{(4)} g\right)$. The lemma follows since $\operatorname{Ein}\left({ }^{(4)} g\right)=0$.

Sadly, I cannot see how they found the identity in the second sentence of the proof. Could you give me some hints/suggestions on how to derive the identity in the second sentence of Lemma 1.1?

Any help/suggestions would be highly appreciated. Thanks so much.

1Fischer, Arthur E.; Marsden, Jerrold E.; Moncrief, Vincent, The structure of the space of solutions of Einstein’s equations. I: One Killing field, Ann. Inst. Henri Poincaré, Nouv. Sér., Sect. A 33, 147-194 (1980); eudml. ZBL0454.53044, MR605194.

  • $\begingroup$ You should include the statements and formulas in your question. $\endgroup$
    – Deane Yang
    Aug 4, 2022 at 4:31
  • $\begingroup$ @DeaneYang I'm sorry for the confusion. Please note that by formula, I mean the derivative of the Einstein tensor in the h direction. $\endgroup$
    – Boka Peer
    Aug 4, 2022 at 5:15
  • $\begingroup$ I don’t see any formulas in your question at all. $\endgroup$
    – Deane Yang
    Aug 4, 2022 at 5:34
  • 1
    $\begingroup$ Perhaps images are blocked; that's why it's better to include them as MathJax (our equivalent of LaTeX). $\endgroup$
    – Glorfindel
    Aug 4, 2022 at 5:53
  • 3
    $\begingroup$ I suggest writing everything using indices. $\endgroup$
    – Deane Yang
    Aug 6, 2022 at 5:11

1 Answer 1


I presume the formula you are asking about is the long one highlighed by $*$ $*$ in your question, while the standard "contracted Bianchi identity" $\delta \operatorname{Ein}\left({ }^{(4)} g\right)=0$ poses no mystery to you. The Longer equation is simply obtained by using the Leibniz rule while applying the functional derivative $\mathrm{D}$ that last identity. I would consider that a pretty clear explanation, but possibly that is not the point you are confused about.

Perhaps you are wondering why the Leinbniz rule applies to $\delta$, which is an operator. A quick way to see why is just to write the corresponding expression in coordinates. There is no need to be very explicit, the following schematic form suffices: $$ \delta \mathrm{Ein} = \partial \cdot \mathrm{Ein} + \Gamma \cdot \mathrm{Ein} . $$ That is, the covariant divergence operator $\delta(-)$ consists of a coordinate derivative part $\partial\cdot(-)$ and of the part containing multiplication by Christoffel symbols $\Gamma\cdot(-)$. Unsurprisingly, the coordinate derivative part has no dependence on the metric, hence formally $\mathrm{D}\partial = 0$. So $\mathrm{D}\delta$ comes entirely from the Christoffel part, which quite obviously obeys the Leibniz rule.

If that's still not where your confusion lies, please clarify your question further.

  • $\begingroup$ Thanks so much. Yes, you answered my question. $\endgroup$
    – Boka Peer
    Aug 6, 2022 at 7:18

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