In the Wikipedia article on Ricci curvature there is a formula, the third of the paragraph "Direct geometric meaning", that reads: $$ d\mu_g = \Big[ 1 - \frac{1}{6}R_{jk}x^jx^k+ O(|x|^3) \Big] d\mu_{{\rm Euclidean}}\,. $$

The article says that this is computed from: $$ g_{ij} = \delta_{ij} - \frac{1}{3}R_{ikj\ell}x^kx^\ell + O(|x|^3)\,, $$

which is easily found (e.g. in John Lee, Riemannian Manifolds). But how does one go from the latter to the former? And, more important, is there a text in which this is done, possibly with some context?



2 Answers 2


Ok, I have found a trick! For a positive definite matrix $M$ close enough to the identity (as in our case), we have: $$ \log(\det M) = \mbox{Tr}(\log M)\;. $$

In my case: $$ \sqrt{g} = \sqrt{\det(g_{ij})}=\exp\bigg(\dfrac{1}{2}\mbox{Tr}(\log(g_{ij}))\bigg) \approx 1 +\dfrac{1}{2}\mbox{Tr}\bigg(-\dfrac{1}{3} R_{ikjl}\,x^kx^l\bigg)\;. $$

All the terms missing would be $O(x^3)$ or higher, so we just get: $$ 1 -\dfrac{1}{6} \sum_i R_{ikil}\,x^kx^l=1 -\dfrac{1}{6} R_{kl}\,x^kx^l\;. $$


My guess is that since the covariant volume $dV=\sqrt{-g} d^n x$, where $g=|g_{ij}|$ is the determinant of the metric, is conserved we have that: $$\sqrt{-g} dV_g=dV_{Euclidian}$$ since the Euclidean metric is $g_{ij}=\delta_{ij}$ and its determinant is one.

Using the expansion $$g_{ij} = \delta_{ij} - \frac{1}{3}R_{ikj\ell}x^kx^\ell + O(|x|^3)\,,$$ to compute $\sqrt{-g}$ and rearranging should give the desired result.


  • $\begingroup$ Yes, it's a determinant. But a very complicated one. That's why I would like to see the calculation done somewhere, or at least some tricks to simplify the math... $\endgroup$
    – geodude
    Mar 13, 2016 at 18:15

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