Laplacian of a distance function on a Riemann manifold

915 Views Asked by At

For some reasons I need to show the following fact.

Let $(M, g)$ be a Riemannian manifold. Let $U \subset M$ be an open set and $r: M \to \mathbb{R}$ a smooth distance function. Let us assume also that $Ric \ge (n - 1) k$.
Then $$ \Big(\frac{\Delta r}{n-1}\Big)^2 + k \ge 0.$$

I can't see how to prove it.

EDIT: I am studying the proof of the Lemma34 (Ricci Comparison Result) at page 268 of the book "Riemannian Geometry" (second edition) by Petersen. I can't understand why the third inequality in the proof does hold. If I can prove what I wrote above, then everything would be clear.

1

There are 1 best solutions below

3
On BEST ANSWER

(1) I do not understand the argument in the book. But $F:=\frac{\Delta r}{n-1} -\lambda_k$ so that $$ F_r \leq -F (\lambda_k + \frac{\Delta r}{n-1}) $$ from the second inquality in the book.

If $F\leq 0$ on $[0,\epsilon)$, we are done. If $F>0$, then we have $$ F_r/F \leq - (\lambda_k + \frac{\Delta r}{n-1}) $$ so that by integrating on $[\delta,\epsilon)$, we have $$ F(\epsilon ) \leq A, \ A:=F(\delta)e^{\int_\delta^\epsilon - (\lambda_k + \frac{\Delta r}{n-1}) } $$

Note that $\lim_\delta A=0$ Hence we complete the proof.

(2) Show that $\lim\ A=0$ : Consider a Jacobi field $ J_i(r):=(d\exp_p)_{re_n} re_i$ where $\{e_i\}$ is orthonormal. Then $$ \lim_{r\rightarrow 0} J_i'(r)= e_i $$

Hence if $h_{ij}$ is second fundamental form, $$ h_{ij}=\frac{1}{2} \partial_r g_{ij} =(J_i',J_j),\ 1\leq i,\ j \leq n-1$$

$$ h_{ij} - \frac{g_{ij}}{r} = (J_i' -(d\exp_p)_{re_n} e_i ,J_j) =(J_i' -(d\exp_p)_{re_n} e_i, (d\exp_p)_{re_n} e_j )r = O(r) $$

That is $$ H:=h_{ij}g^{ij},\ H=\frac{n-1}{r} + O(r) $$

So $ F= \frac{ H}{n-1} -\lambda_k = O(r)$.