# Taylor expansion of $g^{-1/2}$ where $g$ is riemannian metric in normal coordinates

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.

• Could you share the link to the paper? What have you tried? Commented Mar 8, 2019 at 22:49
• 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 $1 - \frac{1}{2} \epsilon$. Commented Mar 9, 2019 at 11:36
• @Blazej yes you are right, there should be a $\frac{1}{3}$ in the first expression Commented Mar 9, 2019 at 16:51

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.