Let $(M,g)$ be a riemannian manifold.
In normal coordinates for any $q\in M$, there is a Taylor expansion of $g_q$ given by $$(g_q)_{ij}(x)=\delta_{ij}+\frac{1}{3}R_{kijl}(q)x^k x^l+O(|x|^3)$$ Now here is my question:
How does one derive from the above expression, that $$\left(\sqrt{g_q}^{-1}\right)^{ij}(x) =\delta^{ij}-\frac{1}{6}R_{kijl}(q)x^k x^l+O(|x|^3)$$ I know how one derives the Taylor expansion of $g$, using a geodesic variation.
Here $\sqrt{g_q}$ denotes the positive square root of $g$ (as a matrix).
I am kinda clueless here, any help would be very much appreciated!
I read this in a paper, but there is no further explanation.

  • $\begingroup$ Could you share the link to the paper? What have you tried? $\endgroup$ Commented Mar 8, 2019 at 22:49
  • $\begingroup$ I think the coefficient is wrong. Inverse of a matrix $1+ \epsilon$ is (up to terms of order $\epsilon^2$) equal to $1- \epsilon$ for a small matrix $\epsilon$. Square root of $1-\epsilon$ is clearly $1 - \frac{1}{2} \epsilon$. $\endgroup$
    – Blazej
    Commented Mar 9, 2019 at 11:36
  • $\begingroup$ @Blazej yes you are right, there should be a $\frac{1}{3}$ in the first expression $\endgroup$ Commented Mar 9, 2019 at 16:51

1 Answer 1


Let us start with another description of the setting. We have $0\in U\subset\mathbb{R}^n$ an open set and a map $g:U\to\mathrm{Sym}^2\left(\mathbb{R}^n\right),$ where $\mathrm{Sym}^2\left(\mathbb{R}^n\right)$ denotes the space of symmetric matrices. The Taylor expansion you quote can also be written in terms of matrix-valued maps as $$g(x)=I+\frac{1}{3}R_{kl}(0)x^kx^l+O\left(|x|^3\right),$$ where $R_{kl}(0)$ is the matrix whose $ij$-entry is $R_{kijl}(0)$. By Taylor's theorem, this is equivalent to saying that the map $g$ is twice differentiable at $0$ and satisfies $$g(0)=I,\quad \frac{\partial g}{\partial x^i}=0,\quad\frac{\partial^2g}{\partial x^k\partial x^l}=\frac{2}{3}R_{kl}(0).$$ Finally, the map $\sqrt{\;\;}:\mathrm{Sym}^2_+\left(\mathbb{R}^n\right)\to\mathrm{Sym}^2_+\left(\mathbb{R}^n\right)$ is differentiable, and so is the inverse map for matrices. This means that the map $\sqrt{g}^{-1}$ is also differentiable, and you can compute its derivatives (that is, its Taylor expansion) using the chain rule.


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