# Understanding the proof of lemma 1.1 from Fisher, Marsden, and Moncrief's paper

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.

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

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.