Let $X$ be a CAT(0) space, $x \in X$ and $\xi$ be a point in the boundary at infinity of $X$. Denote by $\rho(t)$ the geodesic ray (parameterized by arc length) starting from $x$ and asymptotic to $\xi$. Let $y$ be a point close to $x$ and let $\overline{xy}$ denote the geodesic segment connecting $x$ and $y$. I would like to estimate the angle (at $x$) between $\overline{xy}$ and $\rho$, in terms of the Busemann function $$b_{\xi}(y) = \lim_{t \to \infty} d(y, \rho(t)) -t. $$ Precisely, I would like to say that if $b_{\xi}(y)>0$, then the angle is obtuse.

Can this be true? I’m interested in the case where $X$ is a Euclidean building, but I was hoping it is a known fact (either true or false) about CAT(0) spaces.

  • $\begingroup$ If "close" means that x and y lie in a common (closed) chamber then you can take an apartment that contains that chamber and contains xi in its boundary. Then you can use that what you want is a "known fact" for Euclidean spaces (specifically this apartment). $\endgroup$ Jul 2, 2021 at 10:47
  • $\begingroup$ In view of HJRW’s answer, is the statement true if I change obtuse with not acute (so I allow $\pi/2$)? $\endgroup$
    – AMath91
    Jul 2, 2021 at 12:03
  • $\begingroup$ I wanted to use an argument like this, but I’m worried that this apartment will not contain the whole ray $\rho$. Assuming $x$ and $y$ are in the a common chamber, can I always find an apartment containing $y$, $x$ and $\rho$? $\endgroup$
    – AMath91
    Jul 2, 2021 at 12:08
  • $\begingroup$ yes given a chamber and a point at infinity you can find an apartment that contains the apartment and the point at infinity. This apartment being convex and containing $x$ as well as the endpoint of $\rho$, it will contain all of $\rho$. $\endgroup$ Jul 2, 2021 at 14:11
  • 1
    $\begingroup$ Perhaps it's really time you asked a new question. How would $x$ and $y$ be contained in a common chamber in a non-discrete building? I think the statement that is still true (and implies the above in the discrete case) is that for a point $z \in X$ and a point $\xi \in \partial X$ there is an apartment containing both. (You recover the above by taking $z$ to be an interior point of the chamber.) I guess, what you want is an apartment that contains a germ from $x$ (toward y) as well as $\xi$ in the boundary and that could well be true, but I don't have a reference ready for that. $\endgroup$ Jul 2, 2021 at 14:51

2 Answers 2


The answer is "no" even in the hyperbolic plane $\mathbb{H}^2$.

Consider the horocycle about $\xi$ through $x$: this is the set of points such that $b_{\xi}(z)=0$. Let $\gamma$ be the geodesic through $x$ tangent to the horocycle, and take $y$ to be any point on $\gamma$ (except $x$ itself). Then the angle between $\overline{xy}\subset\gamma$ and $\rho$ is $\pi/2$, but $b_{\xi}(y)>0$ since $y$ is outside the horodisc enclosed by the horocycle.

However, similar geometric arguments show that a slightly weaker statement is true in $\mathbb{H}^2$, which may be covered by your assumption that $y$ is 'close enough' to $x$, as follows. For any $\theta<\pi/2$ there is an $\epsilon>0$ such that, if $d(x,y)<\epsilon$ and the angle between $\rho$ and $\overline{xy}=\theta$, then $b_{\xi}(y)<0$.

  • $\begingroup$ So, finally, is the claim in OP correct for $\mathbb{H}^2$ or not? $\endgroup$
    – markvs
    Jul 2, 2021 at 10:15
  • $\begingroup$ @MarkSapir No: the level set of the Busemann function is a horocycle, the set perpendicular to the ray is geodesic line. The point y can lie between the two. $\endgroup$ Jul 2, 2021 at 10:57
  • $\begingroup$ @StefanWitzel: So the second paragraph of this answer is false? $\endgroup$
    – markvs
    Jul 2, 2021 at 11:02
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    $\begingroup$ @MarkSapir: What Henry is saying is that in H^2 if y is close to x relative to the value of b(y) (so not uniformly) then the angle is obtuse (while the implicit quantification in the OP suggests "close" to be independent of b(y)). In any case this is due to a lower curvature bound of H^2. I think, this possibility can be ruined by gluing rescaled hyperbolic planes along the ray, making the non-uniform bound go to 0. I can extend on that later if nobody does it before me. $\endgroup$ Jul 2, 2021 at 11:21
  • $\begingroup$ Thanks for your answer. What if I weaken the statement by only requiring the angle not to be acute, so allowing $\pi/2$? $\endgroup$
    – AMath91
    Jul 2, 2021 at 12:04

This is an elaboration on HJRWs answer. The OP would want that every point satisfies at least one of $\beta_\xi(y) \le 0$ or $\angle_x(\xi,y) \ge \pi/2$ (I'll call these the desired inequalities). So counterexamples are points $y$ with $\beta_\xi(y) > 0$ but $\angle_x(\xi,y) < \pi/2$. Consider the picture of $\mathbb{H}^2$.

enter image description here

The locus of points $y$ such that $\angle_x(\xi,y) = \pi/2$ is the geodesic line drawn in red so the ones with non-acute angle are the ones below (or on) it. The locus of points with $\beta_\xi(y) = 0$ is the blue horocycle, so the ones with $\beta_\xi(y) \ge 0$ are the ones outside of it. Visibly there is plenty of room between for counterexamples. The picture also shows that this is not about strictness of the inequalities. The example point $y$ shown violates both desired inequalities strictly.

Note also that either one of the desired inequalities can be violated arbitrarily badly by taking a point near $\xi$ outside the blue circle respectively near the right end just above the red line.

Now HJRW points out that, in the specific example of $\mathbb{H}^2$ if one considers a small ball $B_r(x)$ around $x$ then it is not possible to violate the desired inequalities arbitrarily badly: the boundary of $B_r(x)$ will meet the blue and the red line at a finite angle.

However, we can scale the metric on $\mathbb{H}^2$ changing its curvature. In doing so, the picture above is unaffected and angles are unaffected. So picking a point that violates the angle-inequality as badly as desired, one can scale the metric to make the point as close to $x$ as one wants.

To turn this into a single counterexample, take $H_r$ to be the hyperbolic plane with metric scaled by $r \in \mathbb{R}_{>0}$ and glue $(H_r)_{r >0}$ along the geodesic ray $\rho$ (in the respective $H_r$). The result is CAT(0) and it contains arbitrarily closely to $x$ a point $y$ with $\beta_\xi(y) > 0$ but $\angle_x(\xi,y)$ as close to $0$ as desired (look inside the appropriate $H_r$).


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