Suppose we have a metric $$(g_{ij})_x=(g_{ij})_{x=0}+(g'_{ij}) x+\frac{(g''_{ij})}{2!} x^2+\frac{(g'''_{ij})}{3!}x^3 \dots $$ Now consider $$\frac{(\mathrm{det} (g_{ij})_x)}{(\mathrm{det} (g_{ij})_{x=0})}$$ What will be the coefficient of $x^3$?

I am getting the coefficient of $x^3$ to be $$\frac{1}{3!}g'''_{ij}+\frac{1}{2!}(g^{ij}g''_{ij})(g^{ij}g'_{ij})-\frac{1}{2}(g')^{ij}(g'')_{ij}+\frac{1}{6}(g^{ij}g'_{ij})^3$$ but I am being told this is incorrect. In particular, I should be getting $-\frac{1}{6}(g')^{ij}(g'')_{ij}$, instead of $-\frac{1}{2}(g')^{ij}(g'')_{ij}$. I have no idea where the $\frac{1}{3}$ is coming from.

  • $\begingroup$ What does $g’$ mean? It’s a function of more than one variable. $\endgroup$
    – Deane
    Sep 25, 2022 at 16:06
  • $\begingroup$ @Deane- $g'$ here refers to $x$ derivative. $\endgroup$ Sep 25, 2022 at 16:07

1 Answer 1


It's hard to make sense of your notation; I'm assuming by $(g_{ij})_x$ you mean $[(g\circ \gamma)(x)]_{ij}$ where $\gamma(x)$ is some one-dimensional curve? Also why are indices appearing in the argument of $\det$?

In any case the fact that $g$ is a metric is a red herring here; we can consider an arbitrary nonsingular-symmetric-matrix-valued one-dimensional function $M(x): \mathbb{R}\to \mathbb{R}^{n\times n}$. First apply the Jacobi formula, $$\frac{d}{dx} \det[M(x)] = \det[M(x)] M^{ij} M'_{ij}$$ after which calculating the third derivative is highly unpleasant (so many product rules!) but purely mechanical. You'll need to make repeated, iterated use of the matrix inverse derivative $$\frac{d}{dx} M^{ij} = -M^{ik}M'_{k\ell}M^{\ell j}.$$


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