Timeline for Angles and Busemann function in CAT(0)
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 2, 2021 at 14:42 | vote | accept | AMath91 | ||
| Jul 2, 2021 at 14:42 | vote | accept | AMath91 | ||
| Jul 2, 2021 at 14:42 | |||||
| Jul 2, 2021 at 14:19 | comment | added | HJRW | @AMath91: I suggest you ask that as a separate question, then. The mechanism of MO doesn't work that well when you ask two questions in one. | |
| Jul 2, 2021 at 14:15 | comment | added | AMath91 | Well, you answered for CAT(0) but I also remarked that I would need this result for building. | |
| Jul 2, 2021 at 14:14 | comment | added | HJRW | @AMath91: (I assume that comment was meant for me...) I have no idea about buildings! This is the answer to the question you asked, though. | |
| Jul 2, 2021 at 14:07 | comment | added | AMath91 | @MarkSapir: Ok, you convinced me. But what about for Euclidean building, which is the context I’m actually interested in? | |
| Jul 2, 2021 at 13:49 | comment | added | HJRW | @AMath91: Even then it's not true: there is $y$ such that $\overline{xy}$ makes an acute angle with $\rho$, but $b_\xi(y)>0$. To see this, just move $y$ in the above example slightly towards the horocycle. | |
| Jul 2, 2021 at 13:46 | comment | added | HJRW | @MarkSapir: As the first sentence makes clear, want the OP wants is not correct. The second paragraph is pointing out that something weaker, with the quantifiers in another order, is true. | |
| Jul 2, 2021 at 12:04 | comment | added | AMath91 | Thanks for your answer. What if I weaken the statement by only requiring the angle not to be acute, so allowing $\pi/2$? | |
| Jul 2, 2021 at 11:21 | comment | added | Stefan Witzel | @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. | |
| Jul 2, 2021 at 11:02 | comment | added | markvs | @StefanWitzel: So the second paragraph of this answer is false? | |
| Jul 2, 2021 at 10:57 | comment | added | Stefan Witzel | @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. | |
| Jul 2, 2021 at 10:15 | comment | added | markvs | So, finally, is the claim in OP correct for $\mathbb{H}^2$ or not? | |
| Jul 2, 2021 at 7:53 | history | answered | HJRW | CC BY-SA 4.0 |