# Relationship of variation of the metric and its inverse

I'm reading Tong's notes on GR http://www.damtp.cam.ac.uk/user/tong/gr.html and i cannot understrand how he derived the equation that relates the varation of the metric with its inverse in page 141 under equation 4.2.

A different approach could be that $$g_{\mu\nu}g^{\mu\nu}=d \to \delta g_{\mu\nu}g^{\mu\nu} = -\delta g^{\mu\nu}g_{\mu\nu}$$ where $$d$$ is the spacetime dimension, but i cannot understand his calculation.

• Possible duplicates: physics.stackexchange.com/q/295005/2451 , physics.stackexchange.com/q/703909/2451 and links therein. Mar 26, 2023 at 7:38
• These answers did not helped. Mar 26, 2023 at 7:43
• Hint: since $g_{\rho\mu}g^{\mu\nu}=\delta_\rho^\nu$ is constant, $(\delta g_{\rho\mu})g^{\mu\nu}+g_{\rho\mu}\delta g^{\mu\nu}=0$. Multiply by $g^{\rho\sigma}$, relabel viz. $\mu\leftrightarrow\sigma$, then use the fact metric tensors are symmetric.
– J.G.
Mar 26, 2023 at 9:16

Your approach won't work. How are you to isolate the $$\delta g^{\mu \nu}$$? You only have one scalar equation!
In the notes, we have schematically $$AB = D$$ where $$D$$ is a constant matrix and $$B$$ is the inverse to $$A$$. Now take the variation $$(\delta A) B + A (\delta B) = 0$$ (if you prefer think of this as equations for each component). Now we can isolate $$\delta B$$ by multiplying by the inverse of $$A$$ (which is $$B$$), so $$B (\delta A) B + \delta B = 0$$.
• That $D$ is the identity matrix is important here, since your last step requires $BA=D$.
• @J.G. That $D$ is the identity matrix is not particularly important - its constancy is however. Mar 26, 2023 at 10:06
• @VladimirA No your approach won't work, as I clearly mention. $\delta g^{\mu \nu}$ is a tensor with many components. The point is to solve for $\delta g^{\mu\nu}$, not have a scalar equation relating $\delta g_{\mu \nu}$ with $\delta g^{\mu\nu}$. Mar 26, 2023 at 10:06