# Relating the components of the Riemann curvature tensor to the second partials of the components of the metric

I am currently reading through a proof of Proposition 6 in

Chernoff's theorem and discrete time approximations of Brownian motion on manifolds OG Smolyanov, H Weizsäcker, O Wittich - Potential Analysis

which is relating the geodesic distance in a Riemannian manifold $$L$$ to the geodesic distance in a Riemannian manifold $$M$$ with $$L$$ embedded in $$M$$ via $$\phi:L\rightarrow M$$.

Let $$\xi$$ denote Riemannian normal coordinates in $$L$$ and $$\eta$$ denote Riemannian normal coordinates in $$M$$. Throughout, all indexing variables using latin characters (e.g. $$a,b,u,v$$) will range from $$1$$ to $$\dim(L)$$ and all greek characters (e.g. $$\alpha,\beta, \rho,\mu$$) range from $$1$$ to $$\dim(M)$$.

In one of the final lines of the proof, we are trying to show that $$\left(\frac{\partial^2g_{ab}^L}{\partial\xi^u\partial\xi^v}-\frac{\partial^2g_{\alpha\beta}^M}{\partial\xi^\rho\partial\xi^\mu}\frac{\partial\phi^\rho}{\partial\xi^u} \frac{\partial\phi^\mu}{\partial\xi^v} \frac{\partial\phi^\alpha}{\partial\xi^a} \frac{\partial\phi^\beta}{\partial\xi^b}\right)(0)\xi^a\xi^b \xi^u\xi^v=0$$.

It is stated that by relating the partial derivatives of the metric tensor to curvature using the Taylor Expansion: $$g_{ab}(\xi)=\delta_{ab}+\frac{1}{3} R_{auvb}(0)\xi^u\xi^v +O(|\xi|^3)$$

we obtain $$\left(\frac{\partial^2g_{ab}^L}{\partial\xi^u\partial\xi^v}-\frac{\partial^2g_{\alpha\beta}^M}{\partial\xi^\rho\partial\xi^\mu}\frac{\partial\phi^\rho}{\partial\xi^u} \frac{\partial\phi^\mu}{\partial\xi^v} \frac{\partial\phi^\alpha}{\partial\xi^a} \frac{\partial\phi^\beta}{\partial\xi^b}\right)(0)\xi^a\xi^b \xi^u\xi^v=2\left(R_{auvb}-R_{\alpha\rho\mu\beta}\frac{\partial\phi^\rho}{\partial\xi^u} \frac{\partial\phi^\mu}{\partial\xi^v} \frac{\partial\phi^\alpha}{\partial\xi^a} \frac{\partial\phi^\beta}{\partial\xi^b}\right)(0)\xi^a\xi^b \xi^u\xi^v.$$

I see that $$2R_{auvb}$$ is the second order term in the Taylor expansion of $$g^L_{ab}$$ and thus should be the same as $$\frac{\partial^2g_{ab}^L}{\partial\xi^u\partial\xi^v}(0)$$.

However, this seems strange to me, given the classical formula for the components of the curvature tensor (where Christoffel symbols vanish): $$R_{auvb}=\frac{1}{2}\left(\frac{\partial^2g_{ab}}{\partial\xi^u\partial\xi^v}+ \frac{\partial^2g_{uv}}{\partial\xi^a\partial\xi^b}- \frac{\partial^2g_{av}}{\partial\xi^b\partial\xi^u}- \frac{\partial^2g_{ub}}{\partial\xi^a\partial\xi^v}\right)$$

This should suggest that the final three terms cancel each other out, but I can't see a reason why that would occur.

• In normal coordinates, more holds than just the Christoffel symbols vanishing at a point. In particular, if $\xi = \xi^a\partial_a$, then, since the integral curves of $\xi$ are geodesics, $\nabla_\xi\xi = 0$ on a neighborhood of $L$. Therefore, $\nabla_{\partial_u}(\nabla_\xi\xi = 0$. The missing equations might follow from this. Note also that you don't need the last three terms of the last display to vanish. You only need them, contracted with 4 copies of the vector $\xi$ to vanish. – Deane Yang Feb 19 at 19:08
• In terms of the "final three terms cancelling each other out": almost certainly this is a manifestation of the first Bianchi identity. – Willie Wong Feb 19 at 19:10