Let $(M,g)$ be an $n$-dimensional Riemannian manifold. Let $p\in M$, and let $\{x^i\}\_{i=1}^n$ be normal coordinates centered around $p$. 


Using Jacobi field, one can show that the metric $g$ has the following Taylor expansion 
\begin{align}
g\_{ij}(x)&=\delta\_{ij}-\frac{1}{3}R\_{ipqj}(p)x^px^q-\frac{1}{6}\nabla\_rR\_{ipqj}(p)x^px^qx^r \\
&\qquad+\left(-\frac{1}{20}\nabla\_r\nabla\_rR\_{ipqj}(p)+\frac{2}{45}g^{kl}R\_{ipqk}R\_{jrsl}(p)x^px^qx^rx^s\right)+O(|x|^5)
\end{align}
where $\nabla$ is the Levi-Civita connection of $(M,g)$, $R\_{ijkl}$ are the (components of the) Riemann curvature tensor, while $x$ is a point near $p$ with coordinates $x^i$, and $|x|:=d(x,p)$, the radial distance from $p$. 


Using this, together with the Jacobi formula for derivative of determinant function, one should be able to obtain the Taylor expansion of $\det(g\_{ij})$. It is claimed (e.g. in [Hamilton's Ricci flow page 59](https://bookstore.ams.org/gsm-77/)) that 
\begin{align}
\det(g\_{ij})(x)&=1-\frac{1}{3}R\_{ij}(p)x^ix^j-\frac{1}{6}\nabla\_kR\_{ij}(p)x^ix^jx^k \\
&\quad-\left(\frac{1}{20}\nabla\_l\nabla\_kR\_{ij}(p)+\frac{1}{90}g^{pq}g^{rs}R\_{pijr}R\_{qkls}(p)-\frac{1}{18}R\_{ij}R\_{kl}(p)\right)x^ix^jx^kx^l \\
&\quad+O(|x|^5)
\end{align}
where $R\_{ij}$ are the (components of the) Ricci curvature tensor. 


My question is that



> 
> How do we obtain the term 
>  \begin{align}
> \frac{1}{90}g^{pq}g^{rs}R\_{pijr}R\_{qkls}(p)-\frac{1}{18}R\_{ij}R\_{kl}(p)
> \end{align}
> 
> 
> 


I believe it should come from the term $\displaystyle\frac{2}{45}g^{kl}R\_{ipqk}R\_{jrsl}(p)$ in the expansion of $g\_{ij}$. By using Jacobi's formula and evaluating at $p$ (since $p$ is the point where $x=0$, many terms will vanish), one should see that the coefficient of $x^ix^jx^kx^l$ is just 
\begin{align}
\frac{1}{4!}g^{ab}\partial\_i\partial\_j\partial\_k\partial\_lg\_{ab}(p)
\end{align}
where $\partial\_i=\frac{\partial}{\partial x^i}$. In other word, I expect that we only need to take the trace (and take care of the counting of possibly repeated terms). But then tracing $$\displaystyle\frac{2}{45}g^{kl}R\_{ipqk}R\_{jrsl}(p)$$ doesn't seem to yield the desired result. In particular, I am not quite sure how $R\_{ij}R\_{kl}$ pops out. 


Any comment, hint or answer are greatly appreciated. 



Taylor expansion of determinant of Riemannian metric in normal coordinates up to higher order
After dropping the first order terms using the normal coordinate condition, 


$$\partial^4\_{ijkl} \det(g) = \partial^3\_{ijk} (g^{-1} \partial\_l g) = g^{-1} \partial^4\_{ijkl} g + ( \partial^2\_{ij} g^{-1} \partial^2\_{kl} g + \partial^2\_{ik} g^{-1} \partial^2\_{jl} + \partial^2\_{jk} g^{-1} \partial^2\_{il} g)$$


Note that


$$ \partial^2\_{ij} g^{-1} = -g^{-1} (\partial^2\_{ij} g) g^{-1} $$


after dropping the first order terms. So you should get contributions both from the quartic part of the expansion of $g$ and the quadratic part of the expansion of $g$, the latter of which you don't seem to have accounted for. 


