On Margulis cusps of hyperbolic $4$-manifolds

Abstract

We study the geometry of the Margulis region associated with an irrational screw translation $g$ acting on the $4$-dimensional real hyperbolic space. This is an invariant domain with the parabolic fixed point of $g$ on its boundary which plays the role of an invariant horoball for a translation in dimensions ${\le }$3. The boundary of the Margulis region is described in terms of a function $\fancyscript{B}_{\alpha }: [0,\infty ) \rightarrow {\mathbb {R}}$ which solely depends on the rotation angle $\alpha \in {\mathbb {R}}/{\mathbb {Z}}$ of $g$. We obtain an asymptotically universal upper bound for $\fancyscript{B}_{\alpha }(r)$ as $r \rightarrow \infty$ for arbitrary irrational $\alpha$, as well as lower bounds when $\alpha$ is Diophantine and the optimal bound when $\alpha$ is of bounded type. We investigate the implications of these results for the geometry of Margulis cusps of hyperbolic $4$-manifolds that correspond to irrational screw translations acting on the universal cover. Among other things, we prove bi-Lipschitz rigidity of these cusps.

Appendix: On arithmetical characterization of presence

The problem we investigate here is when a given denominator $q_n$ in the continued fraction expansion of an irrational number $\alpha$ is present in the boundary function $\fancyscript{B}_{\alpha }$ (see Sect. 3). We need only consider the case where $a_{n+1}=1$ since Corollary 3.8 guarantees that $q_n$ is present when $a_{n+1} \ge 2$. Assuming $a_{n+1}=1$, the same corollary and the definition of fair triples show that

$$\begin{aligned} q_{n} \ \text {is present} \quad \Longleftrightarrow \quad r_{n-1,n}<r_{n,n+1}. \end{aligned}$$

By the formula (13), this condition can be written as

$$\begin{aligned} q_{n} \ \text {is present} \quad \Longleftrightarrow \quad \frac{\delta _{n}-\delta _{n+1}}{\delta _{n-1}-\delta _{n}} < \frac{q_{n+1}^2-q_{n}^2}{q_{n}^2-q_{n-1}^2}. \end{aligned}$$

(24)

The right side of the inequality in (24) is easily computed:

$$\begin{aligned} \frac{q_{n+1}^2-q_{n}^2}{q_{n}^2-q_{n-1}^2} = \frac{(q_{n}+q_{n-1})^2-q_{n}^2}{q_{n}^2-q_{n-1}^2} = \frac{2q_{n} q_{n-1} + q_{n-1}^2 }{q_{n}^2-q_{n-1}^2} = \frac{2 \mu +1}{\mu ^2-1}, \end{aligned}$$

where

$$\begin{aligned} a_{n} < \mu = \frac{q_{n}}{q_{n-1}} < a_{n}+1. \end{aligned}$$

(25)

To estimate the left side of the inequality in (24), we use the inequalities

$$\begin{aligned} 0.95 x^2 \le \sin ^2 x \le x^2 \quad \text {for} \ |x| \le \frac{\pi }{12} \end{aligned}$$

which can be easily proved using calculus. Since the denominator $q_6$ is always $\ge 13$, by Lemma 2.7(ii),

$$\begin{aligned} \pi \Vert q_n \alpha \Vert < \frac{\pi }{q_{n+1}} \le \frac{\pi }{12} \quad (n \ge 5). \end{aligned}$$

It follows that

$$\begin{aligned} 3.8 \pi ^2 \Vert q_n \alpha \Vert ^2 \le \delta _n = 4 \sin ^2 (\pi \Vert q_n \alpha \Vert ) \le 4 \pi ^2 \Vert q_n \alpha \Vert ^2 \quad (n \ge 5). \end{aligned}$$

Introduce the quantity

$$\begin{aligned} \lambda = \frac{\Vert q_{n} \alpha \Vert }{\Vert q_{n+1} \alpha \Vert } \end{aligned}$$

which by Lemma 2.7(iii) satisfies

$$\begin{aligned} a_{n+2} < \lambda < a_{n+2}+1. \end{aligned}$$

(26)

Note that since $a_{n+1}=1$, we have $\Vert q_{n-1} \alpha \Vert = \Vert q_{n} \alpha \Vert + \Vert q_{n+1} \alpha \Vert$, which shows

$$\begin{aligned} \frac{\Vert q_{n-1} \alpha \Vert }{\Vert q_{n} \alpha \Vert } = 1 + \lambda ^{-1}. \end{aligned}$$

Thus, for $n \ge 5$,

$$\begin{aligned} \frac{\delta _{n}-\delta _{n+1}}{\delta _{n-1}-\delta _{n}}&< \frac{4 \Vert q_{n} \alpha \Vert ^2 - 3.8 \Vert q_{n+1} \alpha \Vert ^2}{3.8 \Vert q_{n-1} \alpha \Vert ^2 - 4 \Vert q_{n} \alpha \Vert ^2} \\&= \frac{1-0.95 \lambda ^{-2}}{0.95(1+\lambda ^{-1})^2-1} = \frac{\lambda ^2-0.95}{-0.05 \lambda ^2+1.9 \lambda + 0.95}. \end{aligned}$$

and

$$\begin{aligned} \frac{\delta _{n}-\delta _{n+1}}{\delta _{n-1}-\delta _{n}}&> \frac{3.8 \Vert q_{n} \alpha \Vert ^2 - 4 \Vert q_{n+1} \alpha \Vert ^2}{4 \Vert q_{n-1} \alpha \Vert ^2 - 3.8 \Vert q_{n} \alpha \Vert ^2} \\&= \frac{0.95 - \lambda ^{-2}}{(1+\lambda ^{-1})^2-0.95} = \frac{0.95 \lambda ^2-1}{0.05 \lambda ^2+2 \lambda + 1}. \end{aligned}$$

Introducing the rational functions

$$\begin{aligned} X(t)&=\frac{2t +1}{t^2-1} \\ Y(t)&=\frac{t^2-0.95}{-0.05 t^2+1.9 t + 0.95} \\ Z(t)&=\frac{0.95 t^2-1}{0.05 t^2+2 t + 1}, \end{aligned}$$

the condition (24) and the above estimates can be summarized as

$$\begin{aligned} q_{n} \ \text {is present} \quad \Longleftrightarrow \quad \frac{\delta _{n}-\delta _{n+1}}{\delta _{n-1}-\delta _{n}} < X(\mu ) \end{aligned}$$

(27)

and

$$\begin{aligned} Z(\lambda ) < \frac{\delta _{n}-\delta _{n+1}}{\delta _{n-1}-\delta _{n}} < Y(\lambda ) \quad (n \ge 5), \end{aligned}$$

(28)

where $\mu , \lambda$ satisfy (25) and (26).

The following can be deduced from (27) and (28) (compare the graphs of $X,Y,Z$ in Fig. 5).

• If $a_{n} \ge 3$ and $a_{n+2} \ge 3$, then $\mu , \lambda >3$ and

  $$\begin{aligned} \frac{\delta _{n}-\delta _{n+1}}{\delta _{n-1}-\delta _{n}} > Z(\lambda )>Z(3)>X(3)>X(\mu ). \end{aligned}$$

  so $q_{n}$ is absent.

• If $a_{n}=2$ and $a_{n+2} \ge 5$, then $2<\mu <3, \lambda >5$ and

  $$\begin{aligned} \frac{\delta _{n}-\delta _{n+1}}{\delta _{n-1}-\delta _{n}} > Z(\lambda )>Z(5)>X(2)>X(\mu ), \end{aligned}$$

  so $q_{n}$ is absent.

• If $a_{n} \ge 5$ and $a_{n+2}= 2$, then $\mu >5, 2<\lambda <3$ and

  $$\begin{aligned} \frac{\delta _{n}-\delta _{n+1}}{\delta _{n-1}-\delta _{n}} > Z(\lambda )>Z(2)>X(5)>X(\mu ), \end{aligned}$$

  so $q_{n}$ is absent.

• If $a_{n}=1$ and $a_{n+2} \le 2$, then $1<\mu <2, 1<\lambda <3$ and

  $$\begin{aligned} \frac{\delta _{n}-\delta _{n+1}}{\delta _{n-1}-\delta _{n}} < Y(\lambda )<Y(3)<X(2)<X(\mu ), \end{aligned}$$

  so $q_{n}$ is present.

• Finally, if $a_{n}=2$ and $a_{n+2}=1$, then $2<\mu <3, 1<\lambda <2$ and

  $$\begin{aligned} \frac{\delta _{n}-\delta _{n+1}}{\delta _{n-1}-\delta _{n}} < Y(\lambda )<Y(2)<X(3)<X(\mu ), \end{aligned}$$

  so $q_{n}$ is present.

Fig. 5

Graphs of the rational functions $X, Y$, and $Z$. Note that $Y$ has a singularity at $t \approx 38.5$ (not shown here) but that does not interfere with our estimates on the interval $[1,3]$

These findings are summarized in Fig. 6. In all other cases, the presence or absence of $q_n$ also depends on other partial quotients such as $a_{n-1}, a_{n+3}$, etc.

Fig. 6

The locus of presence (blue) and absence (red) of $q_n$ in the $(a_{n},a_{n+2})$-plane when $a_{n+1}=1$. Here we assume $n \ge 5$. The white cells can go either blue or red depending on other partial quotients. (Color figure online)