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) 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.


1 Answer 1


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.

  • $\begingroup$ Thanks so much. Yes my mistake was exactly the omission of the contributions of the quadratic part. $\endgroup$ Mar 2, 2020 at 8:49

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