Hessian of the distance function--comparison with the space form with constant sectional curvature 0

Question

Let $M$ be an $n$-dimensional complete Riemannian manifold and $r$ is the distance function to a fixed point.

The Hessian comparison theorem says that if the sectional curvature of $M$ is bounded (precisely $k\le \operatorname{sec}\le K$), then the Hessian of $r$ is bounded by the Hessians of the distance function for the space form with constant sectional curvature $k$ and $K$ (precisely $\operatorname{Hess}_Kr\le\operatorname{Hess}r\le \operatorname{Hess}_kr$).

My question: What is the difference between the Hessian of $r$ and the Hessian of the distance function for the space form with constant sectional curvature 0? It's better to have a Taylor expansion.

Following the notation in Petersen's Riemannian geometry, the Hessian of the distance function for the space form with constant sectional curvature $k$ is $\operatorname{Hess}_kr=\frac{\operatorname{sn}_k'(r)}{\operatorname{sn}_k(r)}g_r$ where $g_r=g-(\operatorname{d}r)^2$ and $\frac{\operatorname{sn}_k'(r)}{\operatorname{sn}_k(r)}=\frac{1}{r}-\frac{k}{3}r+O(r^2)$. Note that $\operatorname{Hess}_0r=\frac{1}{r}g_r$. Based on the Hessian comparison theorem, I guess that $\operatorname{Hess}r-\operatorname{Hess}_0r=-\frac{\operatorname{sec}}{3}rg_r+O(r^2)$.

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I originally guessed that the difference $\operatorname{Hess}r-\operatorname{Hess}_0r$ is a multiplication of the Ricci tensor, big thanks to MySheperd who excluded this possibility and suggested that the difference is $O(r)$ in the answer below. Then I started to ask the above question.

Thank you!

Answer 1

$dr$ is defined regardless of the metric. If $g$ and $g_{0}$ are two metrics, and $\nabla^{g}$ and $\nabla^{g_{0}}$ are their corresponding Levi-Civita connections, then the connection difference yields a tensorial operation $D:T^*M\rightarrow T^*{}M\otimes T^{*}M$ such that $\mathrm{Hess}\,r-\mathrm{Hess}_{0}\,r=\nabla^{g}dr-\nabla^{g_{0}}dr=D(dr)$. If one now assumes that $g_{0}$ is locally-flat, i.e., that its curvature tensor has 0-constant sectional cruvature, then one can pick parallel coordinates for this metric which makes $(\nabla^{g_{0}}dr)_{ij}=\partial_{i}\partial_{j}r$. Therefore, in these coordiantes $D(dr)_{ij}=-\Gamma_{ij}^{k}\partial_{k}r$, where $\Gamma_{ij}^{k}$ are the chirstoffel symbols of the metric $g$. This includes only first derivatives of the metric, hence in general it won't be e.g. a multipication of the Ricci-Tensor.

In fact, if one picks $r$ to be a distance function, and picks polar coordinates with respect to this function, then an example of a locally-flat metric is $g_{0}=dr\otimes dr+\sum_{i=1}^{d-1}dx_{i}\otimes dx_{i}$. These polar coordinates are then parallel coordinates for $g_{0}$, and $D(dr)_{ij}=-\delta_{k}^{r}\Gamma_{ij}^{k}$. Since in polar coordinates $\Gamma_{ij}^{k}=o(r)$, this would imply that the difference vanishes as r goes to zero. This example also shows that the expression would depend on the choice of the particular locally-flat metric $g_{0}$.

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Edit: Regarding @Borromean follow up question, the difference does not have to be $o(r)$ either. Let us take a unit disc $\mathcal{D}\subset \mathbb{R^{2}}$ and pick for it the usual polar coordinates $(r,\theta)$. Write two different metrics: $g=dr\otimes dr+r^{2}d\theta\otimes d\theta$ and $g_{0}=d\theta\otimes d\theta+\theta^{2}dr\otimes dr$. Note how both of these metrics are flat: they are simply eucliden metrics written in polar coordinates, with the roles of $r$ and $\theta$ exchanged. As such, both have constant sectional curvature 0. Note how $r$ is a distance function in one metric but not in the other.

However, $\mathrm{Hess}\,r=\nabla^{g} dr=-\Gamma_{\theta\theta}^{r}d\theta\,\otimes d\theta=-\frac{1}{r} d\theta\,\otimes d\theta$ while $\mathrm{Hess}_{0}r=\nabla^{g_{0}}dr=-\frac{1}{\theta}(dr\otimes d\theta+d\theta\otimes dr)$. So in this case, the difference will explode as $r\rightarrow 0$. Moreover, since the curvature of both metrics vanishes, the coefficents in the expansion of this difference in either $r$ or $\theta$ will include no expressions involving the curvature, as all curvature ingredients vanish identically.