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This is probably a dumb question; so far, I have been using $g^{ij}g_{ij} = n$ as granted, but I am probably doing something wrong here. Suppose that $g_{ij}(t)$ is a time-evolving metric. Let $h_{ij} = \partial_t g_{ij}(t)$. In order to find $\partial_t(g^{ij})$, I need to do:

\begin{alignat*}{2} &\quad \partial_t (g^{ij}g_{jk}) &&= 0 \\ \Leftrightarrow& \quad g_{jk}(\partial_tg^{ij}) + g^{ij}(\partial_t g_{jk}) &&= 0 \\ \Leftrightarrow& \quad g_{jk}(\partial_tg^{ij}) &&= -g^{ij}(\partial_t g_{jk}) = -g^{ij} h_{jk} = -h^i_k \end{alignat*} At this step, standard thing would be to multiply each side by $g^{kl}$ so that I get $\partial_t g^{il} = -h^{il}$. However, what happens if I multiply by $g^{jk}$?

\begin{alignat*}{2} & g^{jk}g_{jk}(\partial_tg^{ij}) && = -g^{jk}h^i_k \\ \Leftrightarrow \quad& n(\partial_t g^{ij}) = -h^{ij} \end{alignat*}

I am assuming the former is clearly right. But I don't see an error with multiplying both sides by $g^{jk}$ instead of $g^{kl}$. What is the error that I am making?

James C
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  • you are already summing over $j$, so you cannot just mulitply by $g^{jk}$ ($j$ is no longer a free index) – Thomas May 22 '22 at 18:41
  • @thomas So is it a general rule of thumb to avoid multiple contractions over one index? (for example, I am doing two contractions over $k$ at the same time) – James C May 22 '22 at 22:35
  • One way to check whether you are doing the contractions correctly is to write out the formulas in dimensions $n=2$ and $n=3$ without using either the Einstein summation convention or summation signs. – Deane May 23 '22 at 04:15
  • If you have an expression like $t^{ij}s_{jk}$, then the resulting expression (which would look something like $r^{i}_k$) has no longer any dependency on the index which was used in the sum. If that is not clear to you, then, as suggested by @Deane it might be a good idea to write this down explicitly in small dimensions. – Thomas May 23 '22 at 05:54
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    The above comments said it all on your mistake in the computations. But I would like to add that since $g^{ij}(t)$ are the coefficients of the inverse matrix of $g_{ij}(t)$, you can find out the result by looking at the derivative of $[M(t)]^{-1}$ where $M$ is a $C^1$ family of invertible matrices. This is something like $-M(t)^{-1} M'(t)M(t)^{-1}$ if I remember well. – Didier May 23 '22 at 10:48

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