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Picture below is from the "Three manifolds with positive Ricci flow". I try to calculate the red line.

First, in normal coordinate, I have $$ E(g_{ij})=-2R_{ij} = \frac{\partial}{\partial x^i}(g^{kl}\frac{\partial}{\partial x^j}g_{kl}) - \frac{\partial}{\partial x^k}[ g^{kl}( \frac{\partial}{\partial x^i}g_{jl} + \frac{\partial}{\partial x^j}g_{il} - \frac{\partial}{\partial x^l}g_{ij} ) ] +(\text{lower order term}) $$ now $$ DE(g_{ij})\tilde g_{ij} = \frac{d}{ds}\Big|_{s=0} E(g_{ij}+s\tilde g_{ij}) $$ I don't know what is the inverse of $g_{ij}+s\tilde g_{ij}$. For example, to calculate $$ \frac{d}{ds}\Big|_{s=0} \frac{\partial}{\partial x^i}[(g_{kl}+s\tilde g_{kl})^{-1}\frac{\partial}{\partial x^j}(g_{kl}+s\tilde g_{kl})] $$ how should I get the elements of $(g_{kl}+s\tilde g_{kl})^{-1}$ ? But seemingly, since it is normal coordinate, $\frac{\partial}{\partial x^j}g_{kl}=0$, so there must be $$ \frac{\partial^2}{\partial x^i \partial x^j}g_{kl} $$ therefore, $(g_{kl}+s\tilde g_{kl})^{-1}$ become $g^{kl}$. But why $$ DE(g_{jk})\tilde g_{jk}=-2\tilde R_{jk} ~~~? $$ After all, the lower order terms are omitted. How to see that it is equal to $-2\tilde R_{jk}$ ? Seemingly, it is a directly calculation!

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C.F.G
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Enhao Lan
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    $g$ and $\bar{g}$ are square matrices. Do you aware of inverse of sum of matrices? – C.F.G Sep 13 '21 at 17:45
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    To calculate $$\left.\frac{d}{ds}\right|0(g+s\tilde{g})^{-1},$$ first note that, since $(I+sA)(I-sA) = I - s^2A^2$, it follows that $(I+sA)^{-1} = I - sA + s^2(\cdots)$. Since $g$ is positive definite symmetric, there exists a positive definite symmetric matrix $M$ such that $M^2=g$. Therefore,$$(g+s\tilde{g})^{-1}=(M(I+sM^{-1}\tilde{g}M^{-1})M)^{-1} =M^{-1}(I+sM^{-1}\tilde{g}M^{-1})^{-1}M^{-1}=M^{-1}(I-sM^{-1}\tilde{g}M^{-1})M^{-1}+s^2(\cdots)=g^{-1}-sg^{-1}\tilde{g}g^{-1} + s^2(\cdots)$$ Therefore, $$\left.\frac{d}{ds}\right|{0}(g+s\tilde{g})^{-1} = -g^{-1}\tilde{g}g^{-1}.$$ – Deane Sep 13 '21 at 19:01

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