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It's well known that if M is a Riemannian manifold with $Ric \ge 0$ and contains a line $\gamma $.

Set $${\gamma _ + } = \gamma | {_{[0, + \infty )}} ,{\gamma _ - } = \gamma | {_{[ - \infty ,0)}} $$,

then for any point $x \in M$, we can construct a ray ${\sigma _ + }$(${\sigma _ - }$ ) as the limit of geodesic $x{\gamma _ + }(t)$($x{\gamma _ - }(t)$).

We can proof that:
1,The Busemann function(about ${\gamma _ + }$ and ${\gamma _ - }$ respectively) ${b_ + }$, ${b_ - }$ has the property ${b_ + } + {b_ - } = 0$. This is from the maximal Principle.

2, ${\sigma _ + }$ and ${\sigma _ - }$ forms a line. This is from 1.

For Riemannian manifolds with $Ric \ge - \left( {n - 1} \right)$ containing a line. 1 and 2 may not hold. Hyperbolic space is a counter example.

Since the hyperbolic space has only one end. What I want to ask is : For non-compact Riemannian manifolds with $Ric \ge - \left( {n - 1} \right)$ with two ends and both of the ends have infinite volume. Surely we can construct a line, then 1 and 2 holds?

(Since professor Anton Petrunin gave a counter example for the case without the assumption "non-compact" and "infinite volume". I add them now.)

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    $\begingroup$ No, take double punctured torus (complete metric with constant curvature and finite volume). $\endgroup$ Mar 24, 2014 at 18:00
  • $\begingroup$ @AntonPetrunin: What if I add conditions that M is non-compact and both of the ends have infinite volume. $\endgroup$
    – wang mu
    Mar 25, 2014 at 6:37
  • $\begingroup$ There are also infinite-volume hyperbolic metrics on the 2-punctured torus, namely the completions of metrics with geodesic boundary. $\endgroup$
    – ThiKu
    Mar 25, 2014 at 7:17

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