According to this question , I want to calculate $DE(g_{ij})\widetilde g_{ij}$. If I treat $g^{ij}$ as irrelevant variable ,then I can get the same result as picture below. But I doubt that $g^{ij}$ has some connection with $g_{ij}$, so in the calculation should consider $g^{ij}$.

Consider the linearization of $\frac{\partial}{\partial x^i}\{g^{kl}\frac{\partial}{\partial x^j}g_{kl}\}$ , let $$ H=\frac{\partial}{\partial x^i}\{g^{kl}\frac{\partial}{\partial x^j}g_{kl}\} $$ Then $$ DH(g_{ij})\widetilde g_{ij}=\frac{d}{ds}|_{s=0}H(g_{ij}+s\widetilde g_{ij}) \\ =\frac{d}{ds}|_{s=0}\{\frac{\partial}{\partial x^i} ((g^{kl}+\frac{1}{s}\widetilde g^{kl})\frac{\partial}{\partial x^j}(g_{kl}+s\widetilde g_{kl}))\} \\ =\frac{d}{ds}|_{s=0}(\frac{1}{s}\widetilde g^{kl})\frac{\partial}{\partial x^i} \frac{\partial}{\partial x^j}(g_{kl}) + ... $$ In above calculation ,I am not sure it should be $\frac{1}{s}\widetilde g^{kl}$ or $s\widetilde g^{kl}$ in the term of $\widetilde g^{kl}$.But obviously, there are 2 order term different to the results in picture below.

enter image description here

  • $\begingroup$ Note that $(g + s\tilde{g})^{-1} \not = g^{-1} + s^{\pm 1} \tilde{g}^{-1}$ for both $\pm$. Therefore, your calculation is wrong. $\endgroup$ – mcd Oct 27 '16 at 21:56
  • $\begingroup$ @mcd I know there are some wrong , but it is important there will has 2 order term . $\endgroup$ – lanse7pty Oct 29 '16 at 2:28

You are right that one has to consider the derivatives of $g^{kl}$. The text only computes the second order terms. So if you have a derivative falling on some $g^{kl}$ term then the order is 1, because the Christoffel-symbols do not involve derivatives of $g^{kl}$.

By Jacobi's formula (cf. https://en.wikipedia.org/wiki/Jacobi%27s_formula ) we have that $\mathrm{tr} (A(x)^{-1} \partial_x A(x)) = \partial_x \log \det A(x)$. This implies that $H(g) = \partial_{ij} \log \det g$. Thus, $$\partial_s H(g + s\tilde{g})|_{s=0} = \partial_{ij} \mathrm{tr}(g^{-1}\tilde{g}).$$

You can also calculate $\partial_s (g + s\tilde{g})^{-1}|_{s=0} = -g^{-1}\tilde{g}g^{-1}$ and use this to calculate the terms that are not a trace.

  • $\begingroup$ Thanks ,I have edit my question , there are 2 order term different to the result of picture in my calculation. Could you give a detail process of derivative falling on $g^{kl}$ term ? $\endgroup$ – lanse7pty Oct 27 '16 at 2:49

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