# Expanding the volume element of a Riemannian manifold as a Taylor series.

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.

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

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 $$\frac{d}{dx} M^{ij} = -M^{ik}M'_{k\ell}M^{\ell j}.$$