5
$\begingroup$

For an open complete Riemannian manifold $M$ with non-negative sectional curvature, the Busemann function defined below is a convex exhaustion function (by Cheeger-Gromoll's proof of soul theorem)

The Buesemann function: $$b(x)=\sup_{\gamma} b_{\gamma}(x)$$ where the $sup$ is taken over all rays from a given point and $b_{\gamma}$ is the Busemann function associated with ray $\gamma$:

$$b_{\gamma}(x)=\lim_{t\to \infty}(t-d(x, \gamma(t))$$

A function $f:M\to \mathbb [a,\infty)$ is called an exhaustion function on $M$ if its sublevel set $\Omega_c:=f^{-1}((-\infty, c])$ is compact for all $c$ and $M=\cup_c \Omega_c$.

I am wondering that when we assume Ricci curvature is non-negative, is there any example where the Busemann function $b$ is not an exhaustion? (In this case the Busemann function is subharmonic.)

$\endgroup$
8
  • $\begingroup$ What is "exaustion"? $\endgroup$ Oct 5, 2011 at 18:56
  • $\begingroup$ sorry, should be exhaustion. $\endgroup$
    – user16750
    Oct 5, 2011 at 19:01
  • $\begingroup$ But could you define what an "exhaustion function" is? $\endgroup$
    – Deane Yang
    Oct 5, 2011 at 19:19
  • 1
    $\begingroup$ Ok, what is "exhaustion"? If this means a function whose sublevels are compact (the only definition I could find), then your first claim is false. For example, consider a Euclidean space. $\endgroup$ Oct 5, 2011 at 19:19
  • $\begingroup$ Sorry for any misleading due to my typo and not describe the question clearly. The Buesemann function is $b(x)=\sup b_{\gamma}(x)$ where the sup is taken over all rays and $b_{\gamma}$ is the usual busemann function. $\endgroup$
    – user16750
    Oct 5, 2011 at 19:22

2 Answers 2

4
$\begingroup$

This is an interesting question. I don't have an answer but I want to clarify what is being asked as there seems to be some confusion on the issue.

Given a point $p$ and a ray $\gamma$ starting at $p$ define its Busemann function $b_\gamma(x)$ by the formula $b_\gamma(x)=\lim_{t\to\infty}(t-d(x,\gamma(t)))$. This function is known to be convex if sectional curvature is nonnegative (it's an easy consequence of Toponogov comparison). Then one can define $b=\sup_\gamma b_\gamma$ where the supremum is taken over all rays starting at $p$. If $sec\ge 0$ then the sublevel sets of $b$ are convex. If a sublevel set of $b$ were noncompact it would contain a ray starting at $p$ which contradicts the definition of $b$.

This argument doesn't work for $Ric\ge 0$ because in this case the Busemann function $b$ is only known to be subharmonic rather than convex.

So the question is whether it's still true that sublevel sets of $b$ are compact for manifolds of nonnegative Ricci curvature.

Sorry for posting this as an answer but it was too long for the comment field.

$\endgroup$
4
  • $\begingroup$ Thanks. Do you have any idea what will the minimum set of $b$ look like? $\endgroup$
    – user16750
    Oct 5, 2011 at 22:26
  • $\begingroup$ Thanks! Isn't the minimum set of $b$ always just $p$ itself? $\endgroup$
    – Deane Yang
    Oct 5, 2011 at 22:47
  • 2
    $\begingroup$ @Deane, no, for example, $M=S^1\times \mathbb R$, then the minimum set is $S^1$. $\endgroup$
    – user16750
    Oct 5, 2011 at 23:17
  • $\begingroup$ So the first thing I would try is to see if the Omori-Yau maximum principle could be used here. Have you (unknown, not Vitali) tried this? $\endgroup$
    – Deane Yang
    Oct 6, 2011 at 1:45
4
$\begingroup$

Sorry it's not an answer. I just came across a similar question recently and found a reference.

If the manifold has Ricci nonnegative and Euclidean volume growth, then the Busemann function is an exhaustion function. This is a result due to Zhongmin Shen.

See P400 Lemma 3.4 in Shen, Zhongmin, Complete manifolds with nonnegative Ricci curvature and large volume growth. Invent. Math. 125 (1996), no. 3, 393–404.

But I did not check the details of this result.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.