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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 ${\gamma _ + } = \gamma \left| {_{[0, + \infty )}} \right.$, ${\gamma _ - } = \gamma \left| {_{[ - \infty ,0)}} \right.$, 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, ${\sigma _ + }$ and ${\sigma _ - }$ forms a line. 2, The Busemann function(about ${\gamma _ + }$ and ${\gamma _ - }$ respectively) ${b_ + }$, ${b_ - }$ has the property ${b_ + } + {b_ - } = 0$.

For a Riemannian manifold with $Ric \ge - \left( {n - 1} \right)$ and contains a line,can we get 1 and 2? Maybe the simplest example is the hyperbolic space but I don't know how to compute.

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    $\begingroup$ This post needs editing $\endgroup$
    – David White
    Commented Aug 23, 2013 at 12:29

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Yes, the hyperbolic plane is a counter example. If you take a geodesic asymptotic to points $p, q$ at infinity, then the Busemann functions $b_p, b_q$ diverge to infinity at any ideal point $s$ different from $p, q$. Therefore, you cannot have $b_p$ equal the negative of $b_q$.

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  • $\begingroup$ :You choose s on the ideal boundary?(For the upper half-plane model it's the x-axis). But I think the ideal boundary does not belong to the manifold. $\endgroup$
    – wang mu
    Commented Aug 23, 2013 at 13:05
  • $\begingroup$ Yes, $s$ is an ideal point, i.e. is on the ideal boundary, hence, it does not belong to the manifold. To evaluate Busemann function at such a point, draw any geodesic ray $r(t)$ asymptotic to $s$ and consider the limit of your function as $t$ goes to infinity. The model you are using is irrelevant for this discussion. $\endgroup$
    – Misha
    Commented Aug 23, 2013 at 13:08
  • $\begingroup$ I don't understand what you meant. I ask whether ${b_ + }\left( x \right) + {b_ - }\left( x \right) = 0$ for $x \in M$. Why do you discuss points not belong to M? $\endgroup$
    – wang mu
    Commented Aug 24, 2013 at 5:24
  • $\begingroup$ @wang mu: One way to show that two functions are different is to verify that they have different asymptotic behavior along certain divergent sequences. For instance, if you have a function $f(t)$ which diverges to infinity as $t$ goes to infinity, then you know that $f(t)$ is not identically zero, right? Now, think, about your situation, which function should we use to apply this principle? $\endgroup$
    – Misha
    Commented Aug 24, 2013 at 9:29

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