1
$\begingroup$

$\DeclareMathOperator\div{div}$Let $(X,d)$ be a metric space and let $\gamma$ be a geodesic in $X$. Roughly speaking, the divergence of $\gamma$ at a point $x\in \gamma$ is a function $\div:\mathbb{R}_+\to \mathbb{R}_+$, where $\div(r)$ is defined as the infimal length of a path avoiding a ball of radius $r$ around $x$.

The following is maybe well-known to experts in the area. Consider a right-angled Artin group $G$ to simplify things, although what I'm going to say can probably be asked for all co-compactly cubulated groups. Behrstock and Charney proved in here that if a geodesic crosses an infinite number of strongly separated hyperplanes, it has super-linear divergence at any point $x$. However, the super-linear function heavily depends on the geodesic and on $x$, more precisely, it depends on the distance between successive hyperplanes around $x$. For some applications, it can be interesting to have a fixed super-linear function.

There is no hope in general to get a fixed super-linear function. For instance, consider the group $G=\mathbb{Z}*\mathbb{Z}^2$, where $\mathbb{Z}$ is generated by $a$, and $\mathbb{Z}^2$ by $b$ and $c$. Then, take $\gamma$ of the form $ab^2ab^4ab^8....$. One can pick a family of strongly separated hyperplanes, choosing one hyperplane in each flat crossed by $\gamma$. But one can actually take those hyperplanes anywhere in each of the flats. If the hyperplanes are right in the middle of each crossed flat, there is no way to guaranty a uniform super-linear divergence. On the other hand, in this example, at "good" points, i.e. those at the jonctions of two flats, one has a fixed exponential divergence. This holds more generally at transition points in relatively hyperbolic groups.

So my question is : can a notion of "good" points can be defined (for examples, points within a uniformly bounded distance of both a hyperplane on the left and a hyperplane on the right, among the family of strongly separated hyperplanes) ? And if so, is the super-linear divergence uniform at these points ? It seems to me that Behrstock and Charney's argument do not apply directly to prove such a statement.

Improve this question
$\endgroup$
4
  • $\begingroup$ What is "fixed exponential divergence" or "infimal length"? $\endgroup$
    – markvs
    Commented Dec 8, 2020 at 0:19
  • $\begingroup$ @MarkSapir Well, about infimal length first : I didn't write a precise definition of divergence on purpose, because there are several (non-equivalent ones) in litterature. Usually, it is defined as the minimum length of a path from $\gamma(-r)$ to $\gamma(r)$ avoiding the ball around $x$ (assuming here that $\gamma(0)=x$). Anyway, whatever precise definition you take, you always a super-linear divergence for a geodesic that crosses an infinite number of strongly separated hyperplanes. $\endgroup$
    – M. Dus
    Commented Jul 6, 2021 at 15:16
  • $\begingroup$ @MarkSapir About the fixed exponential divergence, I mean an exponential divergence not depending on the geodesic. To be more specific, I would hope to find an exponential divergence depending only on the distance from x to hyperplanes one the left and on the right, among the family of strongly separated hyperplanes. $\endgroup$
    – M. Dus
    Commented Jul 6, 2021 at 15:18
  • $\begingroup$ I would suggest you rewrite the question making it more readable. $\endgroup$
    – markvs
    Commented Jul 6, 2021 at 16:24

0

Reset to default

You must log in to answer this question.