Characterizing strong pseudoconvexity, obstructions to biholomorphisms, and Lyapunov exponents

Abstract

In this paper we consider the following question: For bounded domains with smooth boundary, can strong pseudoconvexity be characterized in terms of the intrinsic complex geometry of the domain? Our approach to answering this question is based on understanding the dynamical behavior of real geodesics in the Kobayashi metric and allows us to prove a number of results for domains with low regularity. For instance, we show that for convex domains with $C^{2,\epsilon }$ boundary strong pseudoconvexity can be characterized in terms of the behavior of the squeezing function near the boundary, the behavior of the holomorphic sectional curvature of the Bergman metric near the boundary, or any other reasonable measure of the complex geometry near the boundary. The first characterization gives a partial answer to a question of Fornæss and Wold. As an application of these characterizations, we show that a convex domain with $C^{2,\epsilon }$ boundary which is biholomorphic to a strongly pseudoconvex domain is also strongly pseudoconvex.

Appendix A. Properties of complex hyperbolic space

In this section we sketch the proof of Theorem 2.11:

Theorem A.1

If $\gamma _1,\gamma _2:{{\mathrm{{\mathbb {R}}}}}_{\ge 0} \rightarrow {{\mathrm{{\mathbb {B}}}}}_d$ are geodesic rays such that

$$\begin{aligned} \liminf _{s,t \rightarrow \infty } K_{{{\mathrm{{\mathbb {B}}}}}_d}(\gamma _1(s), \gamma _2(t)) < +\infty , \end{aligned}$$

then there exists $T \in {{\mathrm{{\mathbb {R}}}}}$ such that

$$\begin{aligned} \lim _{t \rightarrow \infty } K_{{{\mathrm{{\mathbb {B}}}}}_d}(\gamma _1(t), \gamma _2(t+T)) =0. \end{aligned}$$

Moreover, if the images of $\gamma _1$ and $\gamma _2$ are contained in the same complex geodesic then

$$\begin{aligned} \lim _{t \rightarrow \infty } \frac{1}{t} \log K_{{{\mathrm{{\mathbb {B}}}}}_d}(\gamma _1(t), \gamma _2(t+T)) = -2 \end{aligned}$$

otherwise

$$\begin{aligned} \lim _{t \rightarrow \infty } \frac{1}{t} \log K_{{{\mathrm{{\mathbb {B}}}}}_d}(\gamma _1(t), \gamma _2(t+T)) = -1. \end{aligned}$$

Proof

The first assertion is a consequence of the Kobayashi distance on ${{\mathrm{{\mathbb {B}}}}}_d$ being induced by a negatively curved Riemannian metric (it is isometric to complex hyperbolic space), see for instance [13, Proposition 4.1].

To establish the second assertion it is easiest to work with the domain

$$\begin{aligned} {{\mathrm{{\mathcal {P}}}}}_d = \left\{ (z_1, \dots , z_d) : {{\mathrm{Im}}}(z_1) > \sum _{i=2}^d \left| z_i\right| ^2\right\} \end{aligned}$$

which is biholomorphic to ${{\mathrm{{\mathbb {B}}}}}_d$.

Suppose that $\gamma _1, \gamma _2 :{{\mathrm{{\mathbb {R}}}}}_{\ge 0} \rightarrow {{\mathrm{{\mathcal {P}}}}}_d$ are geodesic rays with

$$\begin{aligned} \lim _{t \rightarrow \infty } K_{{{\mathrm{{\mathcal {P}}}}}_d}(\gamma _1(t), \gamma _2(t)) =0. \end{aligned}$$

Using the fact that the biholomorphism group ${{\mathrm{Aut}}}_0({{\mathrm{{\mathcal {P}}}}}_d)$ of ${{\mathrm{{\mathcal {P}}}}}_d$ acts transitively on the set of geodesic rays in ${{\mathrm{{\mathcal {P}}}}}_d$, we can assume that

$$\begin{aligned} \gamma _1(t) = ie^{2t}e_1. \end{aligned}$$

Then we must have

$$\begin{aligned} \gamma _2(t) = v+\left( \alpha + i(e^{2t} + \left\| v\right\| ^2)\right) e_1 \end{aligned}$$

for some $v \in {{\mathrm{Span}}}_{{{\mathrm{{\mathbb {C}}}}}}\{e_2,\dots , e_d\}$ and $\alpha \in {{\mathrm{{\mathbb {R}}}}}$. Moreover, $\gamma _1$ and $\gamma _2$ are contained in the same complex geodesic if and only if $v=0$.

The estimates on

$$\begin{aligned} \lim _{t \rightarrow \infty } \frac{1}{t} \log K_{{{\mathrm{{\mathcal {P}}}}}_d}(\gamma _1(t), \gamma _2(t)) \end{aligned}$$

will follow from the well known fact that if $V \subset {{\mathrm{{\mathbb {C}}}}}^d$ is an affine subspace which intersects ${{\mathrm{{\mathcal {P}}}}}_d$ then

$$\begin{aligned} K_{V \cap {{\mathrm{{\mathcal {P}}}}}_d}(z,w) = K_{{{\mathrm{{\mathcal {P}}}}}_d}(z,w) \end{aligned}$$

for all $z,w \in V \cap {{\mathrm{{\mathcal {P}}}}}_d$.

First suppose that $v=0$. Then

$$\begin{aligned} \lim _{t \rightarrow \infty } \frac{1}{t} \log K_{{{\mathrm{{\mathcal {P}}}}}_d}(\gamma _1(t), \gamma _2(t)) = \lim _{t \rightarrow \infty } \frac{1}{t} \log K_{{{\mathrm{{\mathcal {H}}}}}}(ie^{2t}, \alpha +ie^{2t}) \end{aligned}$$

where ${{\mathrm{{\mathcal {H}}}}}= \{ z \in {{\mathrm{{\mathbb {C}}}}}: {{\mathrm{Im}}}(z) > 0\}$. Then

$$\begin{aligned} K_{{{\mathrm{{\mathcal {H}}}}}}(ie^{2t}, \alpha +ie^{2t}) = \frac{1}{2} {{\mathrm{arcosh}}}\left( 1 + \frac{\alpha ^2}{2e^{4t}} \right) \end{aligned}$$

and using the fact that ${{\mathrm{arcosh}}}(x) = \log (x +\sqrt{x^2-1})$ we then have

$$\begin{aligned} K_{{{\mathrm{{\mathcal {H}}}}}}(ie^{2t}, \alpha +ie^{2t}) = \frac{1}{2} \log \left( 1 + \frac{\alpha ^2}{2e^{4t}} + \frac{\left| \alpha \right| }{\sqrt{2}e^{2t}}\right) = \frac{\left| \alpha \right| }{2\sqrt{2}}e^{-2t} + \mathrm{O} \left( e^{-4t} \right) . \end{aligned}$$

So

$$\begin{aligned} \lim _{t \rightarrow \infty } \frac{1}{t} \log K_{{{\mathrm{{\mathcal {P}}}}}_d}(\gamma _1(t), \gamma _2(t)) = -2. \end{aligned}$$

Next suppose that $v \ne 0$. Then let ${\overline{\gamma }}_2(t) = v + i(e^{2t}+\left\| v\right\| ^2)e_1$. Since

$$\begin{aligned} \lim _{t \rightarrow \infty } \frac{1}{t} \log K_{{{\mathrm{{\mathcal {P}}}}}_d}(\gamma _2(t), {\overline{\gamma }}_2(t)) = -2 \end{aligned}$$

it is enough to show that

$$\begin{aligned} \lim _{t \rightarrow \infty } \frac{1}{t} \log K_{{{\mathrm{{\mathcal {P}}}}}_d}(\gamma _1(t), {\overline{\gamma }}_2(t)) = -1. \end{aligned}$$

Next for t sufficiently large let

$$\begin{aligned} s_t = t +\frac{1}{2}\log \left( 1 -\frac{\left\| v\right\| ^2}{e^{2t}} \right) . \end{aligned}$$

Then

$$\begin{aligned} K_{{{\mathrm{{\mathcal {P}}}}}_d}({\overline{\gamma }}_2(t), {\overline{\gamma }}_2(s_t)) = \frac{1}{2}\left| \log \left( 1 -\frac{\left\| v\right\| ^2}{e^{2t}} \right) \right| = \frac{\left\| v\right\| ^2}{2}e^{-2t} + \mathrm{O}\left( e^{-4t} \right) \end{aligned}$$

so it is enough to show that

$$\begin{aligned} \lim _{t \rightarrow \infty } \frac{1}{t} \log K_{{{\mathrm{{\mathcal {P}}}}}_d}(\gamma _1(t), {\overline{\gamma }}_2(s_t)) = -1. \end{aligned}$$

Now since ${\overline{\gamma }}_2(s_t) = v + ie^{2t}e_1$ and

$$\begin{aligned} {{\mathrm{{\mathcal {P}}}}}_d \cap \left( ie^{2t} + {{\mathrm{{\mathbb {C}}}}}\cdot v\right) = \left\{ ie^{2t} + z\frac{v}{\left\| v\right\| } : z \in {{\mathrm{{\mathbb {C}}}}}, \left| z\right| \le e^t \right\} . \end{aligned}$$

we have

$$\begin{aligned}&\lim _{t \rightarrow \infty } \frac{1}{t} \log K_{{{\mathrm{{\mathcal {P}}}}}_d}(\gamma _1(t), {\overline{\gamma }}_2(s_t)) = \lim _{t \rightarrow \infty } \frac{1}{t} \log K_{e^t {{\mathrm{{\mathbb {D}}}}}}(0, \left\| v\right\| ) \\&\quad = \lim _{t \rightarrow \infty } \frac{1}{t} \log K_{{{\mathrm{{\mathbb {D}}}}}}(0, e^{-t} \left\| v\right\| ) = -1 \end{aligned}$$

where in the last equality we used the fact that $K_{{{\mathrm{{\mathbb {D}}}}}}(0,z) = \left| z\right| + { \mathrm O}\left( \left| z\right| ^2\right)$ for z close to 0. $\square$