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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.

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  • $\begingroup$ 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. $\endgroup$ – Deane Yang Feb 19 at 19:08
  • $\begingroup$ In terms of the "final three terms cancelling each other out": almost certainly this is a manifestation of the first Bianchi identity. $\endgroup$ – Willie Wong Feb 19 at 19:10

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