Let $p$ be a point in a Riemannian manifold $M$ and $d_p$ be the distance from the point $p$. Prove that $\lim_{x\rightarrow p}\Delta d_p(x)=\infty$
I can easily prove it in $\mathbb{R}^n$. But for a general Riamannian manifold I am not sure what to do. This is what I think so far.
I know that in a coordinate system, the laplacian is written as
$$\Delta f=
\sum_{ij}g^{ij}\big(\partial_i\partial_jf
-\sum_k\Gamma_{ij}^k\partial_kf \big)$$
This can be simplified a little if we use polar coordinates, because the basic vector fields are normal. It becomes
$$\Delta f=
\sum_ig^{ii}\big(\partial_i^2f
-\sum_k\Gamma_{ii}^k\partial_kf \big)$$
So for the distance, which is $x_1$ for polar, we have $\partial_ix_1=\delta_{i1}$ and thus $\partial_i^2x_1=0$ for all $i$. Therefore we have
$$\Delta x_1=
-\sum_ig^{ii}\Gamma_{ii}^1$$
This is as far as I can go. I don't see how this would tend to infinity (or anything else)
Please let me know if I miss something. Any help or hints are welcome. Thank you
Best Answer
If you use normal coordinate, the computation along this line will be easier: here we have $d=\sqrt{x_1^2+...+x_n^2}$, $\Gamma_{ij}^k(p)=0$ so $\Gamma_{ij}^k=O(d)$; and $\frac{\partial g_{ij}}{\partial x_k}(p)=0$, so $g_{ij}=\delta_{ij}+O(d^2)$. Take inverse we get also $g^{ij}=\delta_{ij}+O(d^2)$. Now from $\Delta d=g^{ij}(d_{ij}-\Gamma_{ij}^kd_k)$, by direct calculation we note $d_k=O(1)$, so $\Gamma_{ij}^kd_k\to 0$. Also by direct computation we see $d_{ij}=O(d^{-1})$; so $g^{ij}d_{ij}=\delta_{ij}d_{ij}+O(d^2)O(d^{-1})$, with $ O(d^2)O(d^{-1})\to 0$. Thus at distance $d$, $\Delta d$ differs from the Euclidean version by $O(d)\to 0$, the result follows from the Euclidean case.