The space of convex domains in complex Euclidean space

Abstract

In this mostly expository article, we describe some properties of the space of convex domains in complex Euclidean space (endowed with the local Hausdorff topology). In particular, we give careful proofs that the Kobayashi metric, the Bergman kernel/metric, and the Kähler–Einstein metric are all continuous on the space of convex domains. The group of affine automorphisms acts on this space and we also describe the orbit closures for some special classes of domains.

Appendix A: Computing Orbit Closures

In this appendix we sketch the proof of Theorem 7.3 and prove Theorem 7.4. We begin by making the following observation.

Proposition A.1

Suppose $\Omega \subset {{\,\mathrm{\mathbb {C}}\,}}^d$ is a bounded convex domain and $z_n \in \Omega$, $A_n \in {{\,\mathrm{Aff}\,}}({{\,\mathrm{\mathbb {C}}\,}}^d)$ are sequences where $A_n(\Omega ,z_n)$ converges to some $(D,z_\infty )$ in ${{\,\mathrm{\mathbb {X}}\,}}_{d,0}$. Then the following are equivalent:

• (1) $A_n \rightarrow \infty$ in ${{\,\mathrm{Aff}\,}}({{\,\mathrm{\mathbb {C}}\,}}^d)$ (that is, the sequence $A_n$ leaves every compact subset of ${{\,\mathrm{Aff}\,}}({{\,\mathrm{\mathbb {C}}\,}}^d)$),

• (2) $\delta _\Omega (z_n) \rightarrow 0$.

Proof

With $\Omega \in {{\,\mathrm{\mathbb {X}}\,}}_d$ fixed, for any $\epsilon > 0$ the set

$$\begin{aligned} \{ (\Omega , z) : \delta _\Omega (z) \ge \epsilon \} \end{aligned}$$

is compact in ${{\,\mathrm{\mathbb {X}}\,}}_{d,0}$. So the proposition follows immediately from the fact that the action of ${{\,\mathrm{Aff}\,}}({{\,\mathrm{\mathbb {C}}\,}}^d)$ on ${{\,\mathrm{\mathbb {X}}\,}}_{d,0}$ is continuous and proper. $\square$

1.1 A.1 Strongly Pseudoconvex and Finite Type Domains

It will be convenient to introduce the notion of line type.

Given a function $f: {{\,\mathrm{\mathbb {C}}\,}}\rightarrow {{\,\mathrm{\mathbb {R}}\,}}$ with $f(0)=0$ let $\nu (f)$ denote the order of vanishing of f at 0.

Definition A.2

Suppose that $\Omega \subset {{\,\mathrm{\mathbb {C}}\,}}^d$ is a bounded convex domain with ${{\,\mathrm{\mathcal {C}}\,}}^m$ boundary and $r : {{\,\mathrm{\mathbb {C}}\,}}^d \rightarrow {{\,\mathrm{\mathbb {R}}\,}}$ is a defining function for $\Omega$, that is r is a ${{\,\mathrm{\mathcal {C}}\,}}^m$ function, $\Omega = \{ r < 0\}$, and $\nabla r \ne 0$ in a neighborhood of $\partial \Omega$. Then the line type of$x \in \partial \Omega$ is

$$\begin{aligned} \ell (\Omega ,x):= \sup \left\{ \mu (r \circ \ell ) : \ell : {{\,\mathrm{\mathbb {C}}\,}}\rightarrow {{\,\mathrm{\mathbb {C}}\,}}^d \text { is a non-trivial affine map with } \ell (0)=x \right\} . \end{aligned}$$

Then the line type of$\Omega$ is

$$\begin{aligned} \sup _{x \in \partial \Omega } \ell (\Omega , x). \end{aligned}$$

McNeal [29] proved that if $\Omega$ is a bounded convex domain with ${{\,\mathrm{\mathcal {C}}\,}}^\infty$ boundary, then $x \in \partial \Omega$ has line type m if and only if the DAngelo type at x is also m (see also [7]).

Proposition A.3

Suppose m is a positive integer and $\Omega \subset {{\,\mathrm{\mathbb {C}}\,}}^d$ is a bounded convex domain with line type m (in particular, $\Omega$ has ${{\,\mathrm{\mathcal {C}}\,}}^m$ boundary). If $D \in \mathrm{AL}(\Omega )$, then there exists $A \in {{\,\mathrm{Aff}\,}}({{\,\mathrm{\mathbb {C}}\,}}^d)$ such that

$$\begin{aligned} A D = \left\{ (z_1,\dots ,z_d) \in {{\,\mathrm{\mathbb {C}}\,}}^d : { \mathrm Im}(z_1) > P(z_2,\dots ,z_d) \right\} , \end{aligned}$$

where $P: {{\,\mathrm{\mathbb {C}}\,}}^{d-1} \rightarrow {{\,\mathrm{\mathbb {R}}\,}}$ is non-degenerate non-negative convex polynomial with $P(0)=0$ and $\mathrm{deg}(P) \le m$.

For a careful proof of Proposition A.3 see for instance [40, Theorem 10.1] which is based on arguments in [6, 17, 30].

Using Proposition A.3, one can deduce the following.

Proposition A.4

Suppose $\Omega \subset {{\,\mathrm{\mathbb {C}}\,}}^d$ is a bounded convex domain with strongly pseudoconvex boundary. If $D \in \mathrm{AL}(\Omega )$, then there exists $A \in {{\,\mathrm{Aff}\,}}({{\,\mathrm{\mathbb {C}}\,}}^d)$ such that

$$\begin{aligned} A D = \left\{ (z_1,\dots ,z_d) \in {{\,\mathrm{\mathbb {C}}\,}}^d : { \mathrm Im}(z_1) > \sum _{j=2}^d \left| z_j\right| ^2 \right\} . \end{aligned}$$

Proof

Notice that $\Omega$ has line type 2 so we can use Proposition A.3. Suppose that $D \in \mathrm{AL}(\Omega )$. By Proposition A.3 there exists $A_0 \in {{\,\mathrm{Aff}\,}}({{\,\mathrm{\mathbb {C}}\,}}^d)$ such that

$$\begin{aligned} A_0 D = \left\{ (z_1,\dots ,z_d) \in {{\,\mathrm{\mathbb {C}}\,}}^d : { \mathrm Im}(z_1) > P(z_2,\dots ,z_d) \right\} , \end{aligned}$$

where $P: {{\,\mathrm{\mathbb {C}}\,}}^{d-1} \rightarrow {{\,\mathrm{\mathbb {R}}\,}}$ is non-degenerate non-negative convex polynomial with $P(0)=0$ and $\mathrm{deg}(P) \le 2$.

Since P is non-negative and $P(0)=0$, we must have $\nabla P(0) = 0$. So P is a homogeneous polynomial of degree two. Since P is real valued, it must be Hermitian and since P is non-degenerate, it must be positive definite. So by changing A we may assume that $P(z) = \sum _{j=2}^d \left| z_j\right| ^2$. $\square$

1.2 A.2 Smoothly Bounded Domains

Proposition A.5

Suppose $\Omega \subset {{\,\mathrm{\mathbb {C}}\,}}^d$ is a bounded convex domain. If $\Omega \subset {{\,\mathrm{\mathbb {C}}\,}}^d$ has ${{\,\mathrm{\mathcal {C}}\,}}^1$ boundary and $D \in \mathrm{AL}(\Omega )$, then ${{\,\mathrm{Aut}\,}}(D)$ contains a one-parameter subgroup.

The following argument is essentially due to Frankel [14].

Proof

We will show that there exists $z_0 \in D$ and a non-zero vector $v \in {{\,\mathrm{\mathbb {C}}\,}}^d$ such that $z_0 + {{\,\mathrm{\mathbb {R}}\,}}\cdot v \subset D$. Then, since D is open and convex,

$$\begin{aligned} z+{{\,\mathrm{\mathbb {R}}\,}}\cdot v \subset D \end{aligned}$$

for every $z \in D$. So ${{\,\mathrm{Aut}\,}}(D)$ contains the one-parameter group

$$\begin{aligned} u_t(z) = z+tv. \end{aligned}$$

By definition there exists a sequence $A_n \in {{\,\mathrm{Aff}\,}}({{\,\mathrm{\mathbb {C}}\,}}^d)$ such that $A_n \rightarrow \infty$ in ${{\,\mathrm{Aff}\,}}({{\,\mathrm{\mathbb {C}}\,}}^d)$ and $A_n \Omega$ converges to D in ${{\,\mathrm{\mathbb {X}}\,}}_d$. Fix some $z_0 \in D$. By passing to tail of $(A_n)_{n \in {{\,\mathrm{\mathbb {N}}\,}}}$ we can assume that $z_0 \in A_n \Omega$ for every n. Then define $z_n : = A_n^{-1}(z_0) \in \Omega$. Notice that Proposition A.1 implies that $\lim _{n \rightarrow \infty } \delta _\Omega (z_n) = 0$.

Next pick $\xi _n \in \partial \Omega$ such that $\left\| \xi _n-z_n\right\| = \delta _\Omega (z_n)$. Notice that the real affine line

$$\begin{aligned} \xi _n + {{\,\mathrm{\mathbb {R}}\,}}\cdot i(\xi _n-z_n) \end{aligned}$$

is tangent to $\partial \Omega$ at $\xi _n$. Then, since $\Omega _n$ has ${{\,\mathrm{\mathcal {C}}\,}}^1$ boundary, there exists $r_n \rightarrow \infty$ such that

$$\begin{aligned} \{ z_n + it(\xi _n-z_n) : t \in (-r_n, r_n) \} \subset \Omega . \end{aligned}$$

Let

$$\begin{aligned} v_n : = \frac{i}{\left\| A_n(\xi _n)-z_0\right\| } (A_n(\xi _n)-z_0). \end{aligned}$$

By passing to a subsequence we can assume that $\lim _{n \rightarrow \infty } v_n = v$. We claim that $z_0 + {{\,\mathrm{\mathbb {R}}\,}}\cdot v \subset D$. Since $A_n$ is a complex affine transformation

$$\begin{aligned} A_n\Big (z_n + it(\xi _n-z_n) \Big )&= A_n(z_n) + itA_n(\xi _n) - it A_n(z_n) = z_0 + it (A_n(\xi _n)-z_0) \\&= z_0 + t \left\| A_n(\xi _n)-z_0\right\| v_n. \end{aligned}$$

Since $A_n\xi _n \in \partial A_n \Omega$ we have

$$\begin{aligned} 0< \delta _D(z_0) = \lim _{n \rightarrow \infty } \delta _{A_n\Omega }(z_0) \le \liminf _{n \rightarrow \infty } \left\| A_n(\xi _n)-z_0\right\| . \end{aligned}$$

So

$$\begin{aligned} \epsilon :=\inf _{n \ge 1}\left\| A_n(\xi _n)-z_0\right\| \end{aligned}$$

is positive and

$$\begin{aligned} \{ z_0 + tv_n : t \in (-r_n\epsilon , r_n\epsilon ) \} \subset A_n\Omega . \end{aligned}$$

Thus $z_0 + {{\,\mathrm{\mathbb {R}}\,}}\cdot v \subset D$ which completes the proof by the remarks above. $\square$

1.3 A.3 Proof of Theorem 7.4

A result of Frankel will allow us to reduce to lower dimensional cases.

Theorem A.6

(Frankel [15, Theorem 9.3]) Suppose $\Omega \in {{\,\mathrm{\mathbb {X}}\,}}_d$ and V is a complex affine k-plane intersecting $\Omega$. Let $D = \Omega \cap V$ and suppose there exists affine maps $A_n \in {{\,\mathrm{Aff}\,}}(V)$ such that $A_n(D)$ converges to a ${{\,\mathrm{\mathbb {C}}\,}}$-properly convex domain $D_\infty$ in V in the local Hausdorff topology. Then there exist affine maps $B_n \in {{\,\mathrm{Aff}\,}}({{\,\mathrm{\mathbb {C}}\,}}^d)$ such that $B_n(\Omega )$ converges to $\Omega _\infty$ in ${{\,\mathrm{\mathbb {X}}\,}}_d$ with

$$\begin{aligned} \Omega _\infty \cap V = D_\infty . \end{aligned}$$

We will also use the following observation of Fu and Straube.

Lemma A.7

(Fu-Straube [16, Theorem 1.1]) Suppose $\Omega \in {{\,\mathrm{\mathbb {X}}\,}}_d$. If there exists a non-constant holomorphic map ${{\,\mathrm{\mathbb {D}}\,}}\rightarrow \partial \Omega$, then there exists a non-constant affine map ${{\,\mathrm{\mathbb {D}}\,}}\rightarrow \partial \Omega$.

Lemma A.8

If $\Omega \subset {{\,\mathrm{\mathbb {C}}\,}}$ is convex and $\Omega \ne {{\,\mathrm{\mathbb {C}}\,}}$, then ${{\,\mathrm{\mathcal {H}}\,}}\subset \mathrm{AL}(\Omega )$.

Proof

Since the boundary is differentiable almost everywhere, by applying an initial affine transformation we can assume that $0 \in \partial \Omega$ is a ${{\,\mathrm{\mathcal {C}}\,}}^1$ point of $\partial \Omega$ and the real axis is tangent to $\Omega$ at 0. Then let $A_n \in {{\,\mathrm{Aff}\,}}({{\,\mathrm{\mathbb {C}}\,}})$ be the affine map $z \rightarrow nz$. Then $A_n \Omega =n\Omega$ converges to ${{\,\mathrm{\mathcal {H}}\,}}$ in ${{\,\mathrm{\mathbb {X}}\,}}_1$. $\square$

Proposition A.9

Suppose $\Omega \in {{\,\mathrm{\mathbb {X}}\,}}_d$ and there exists a non-constant holomorphic map $\varphi : {{\,\mathrm{\mathbb {D}}\,}}\rightarrow \partial \Omega$, then there exists a domain $D \in \mathrm{AL}(\Omega )$ such that

$$\begin{aligned} D \cap \{ (z_1,z_2,0,\dots ,0) : z_1,z_2 \in {{\,\mathrm{\mathbb {C}}\,}}\} = {{\,\mathrm{\mathcal {H}}\,}}\times {{\,\mathrm{\mathcal {H}}\,}}\times \{(0,\dots ,0)\}. \end{aligned}$$

The following argument comes from [5, Sect. 5].

Proof

By Lemma A.7 there exists a non-constant affine map ${{\,\mathrm{\mathbb {D}}\,}}\rightarrow \partial \Omega$. Then we can find a complex affine 2-plane V intersecting $\Omega$ such that there exist a non-constant affine map ${{\,\mathrm{\mathbb {D}}\,}}\rightarrow \partial (\Omega \cap V)$. Then by Theorem A.6 we can assume that $d=2$.

By applying an initial affine transformation to $\Omega$, we can assume that

• (1) $\Omega \subset \{ (z_1,z_2) \in {{\,\mathrm{\mathbb {C}}\,}}^2 : { \mathrm Im}(z_1) > 0\}$,

• (2) $\{0\} \times {{\,\mathrm{\mathbb {D}}\,}}\subset \partial \Omega$, and

• (3) $(i,0) \in \Omega$.

For every n, let $z_n = (i/n,0) \in \Omega$. Then pick

$$\begin{aligned} \xi _n \in (\{i/n\} \times {{\,\mathrm{\mathbb {C}}\,}}) \cap \partial \Omega \end{aligned}$$

such that

$$\begin{aligned} \left\| \xi _n-z_n\right\| = \inf \left\{ \left\| \xi -z_n\right\| : \xi \in (\{i/n\} \times {{\,\mathrm{\mathbb {C}}\,}}) \cap \partial \Omega \right\} . \end{aligned}$$

Since $\overline{\Omega }$ contains no complex affine line, we must have

$$\begin{aligned} \limsup _{n \rightarrow \infty } \left\| \xi _n-z_n\right\| < +\infty . \end{aligned}$$

Suppose $\xi _n = ( i/n, a_n)$. By passing to a subsequence we can suppose that $a_n \rightarrow a$. Then

$$\begin{aligned} \lim _{n \rightarrow \infty } \left\| \xi _n-z_n\right\| =\lim _{n \rightarrow \infty } \left| a_n\right| = \left| a\right| \end{aligned}$$

and $(0,a) \in \partial \Omega$. Since $\{0\} \times {{\,\mathrm{\mathbb {D}}\,}}\subset \partial \Omega$ and $\Omega$ is convex, we also have $\left| a\right| \ge 1$.

Then consider the matrix

$$\begin{aligned} A_n = \begin{pmatrix} n &{} \quad 0 \\ 0 &{} \quad a_n^{-1} \end{pmatrix}. \end{aligned}$$

Let $T \in {{\,\mathrm{Aff}\,}}({{\,\mathrm{\mathbb {C}}\,}}^2)$ be the affine map

$$\begin{aligned} T(z_1,z_2) = (i(z_1-i), z_2). \end{aligned}$$

By construction, $TA_n(\Omega , z_n) \in {{\,\mathrm{\mathbb {K}}\,}}_2$ where ${{\,\mathrm{\mathbb {K}}\,}}_2 \subset {{\,\mathrm{\mathbb {X}}\,}}_2$ is the subset from Proposition 4.4. So by passing to a subsequence we can assume that $A_n\Omega$ converges to some $\Omega _1$ in ${{\,\mathrm{\mathbb {X}}\,}}_2$.

Let $C_2 \subset {{\,\mathrm{\mathbb {C}}\,}}$ be the open convex set such that

$$\begin{aligned} \{0\} \times \overline{C_2} = (\{0\} \times {{\,\mathrm{\mathbb {C}}\,}}) \cap \partial \Omega . \end{aligned}$$

Then define $D_2 = a^{-1} \cdot C_2$.

Claim$\{0\} \times D_2 \subset \partial \Omega _1$and$\Omega _1 \subset {{\,\mathrm{\mathcal {H}}\,}}\times D_2$. $\square$

Proof of Claim

If $(x,y) \in \Omega _1$, then there exists $(x_n, y_n) \in \Omega$ such that $A_n(x_n,y_n) \rightarrow (x,y)$. Thus $nx_n \rightarrow x$ and $y_n/a_n \rightarrow y$. So $x_n \rightarrow 0$ and $y_n \rightarrow ay$. Thus $y \in a^{-1} \cdot C_2$. So $\Omega _1 \subset {{\,\mathrm{\mathbb {C}}\,}}\times D_2$. Since $\Omega \subset {{\,\mathrm{\mathcal {H}}\,}}\times {{\,\mathrm{\mathbb {C}}\,}}$ we also have $\Omega _1 \subset {{\,\mathrm{\mathcal {H}}\,}}\times {{\,\mathrm{\mathbb {C}}\,}}$. So

$$\begin{aligned} \Omega _1 \subset ({{\,\mathrm{\mathbb {C}}\,}}\times D_2) \cap ({{\,\mathrm{\mathcal {H}}\,}}\times {{\,\mathrm{\mathbb {C}}\,}}) = {{\,\mathrm{\mathcal {H}}\,}}\times D_2. \end{aligned}$$

Since $a_n^{-1}\cdot C_2 \times \{ 0 \} \subset A_n \partial \Omega$ and $a_n^{-1} \rightarrow a^{-1}$, the definition of the local Hausdorff topology implies that $\{0\} \times D_2 \subset \overline{\Omega }_1$. Since $\Omega _1 \subset {{\,\mathrm{\mathcal {H}}\,}}\times D_2$, we must have $\{0\} \times D_2 \subset \partial \Omega _1$. $\square$

Let $C_1 \subset {{\,\mathrm{\mathbb {C}}\,}}$ be the open convex set such that

$$\begin{aligned} C_1 \times \{0\} = ({{\,\mathrm{\mathbb {C}}\,}}\times \{ 0\}) \cap \Omega . \end{aligned}$$

Next define $D_1 = \cup _{n =1}^\infty n C_1$. Then $D_1$ is a non-empty convex open cone since $0 \in \partial C_1$.

Claim:$D_1 \times D_2 \subset \Omega _1$.

Proof of Claim

By construction

$$\begin{aligned} nC_1 \times \{0\} \subset A_n \Omega \end{aligned}$$

so, by the definition of the local Hausdorff topology, $D_1 \times \{0\} \subset \overline{\Omega }_1$. Now suppose that $(x,y) \in D_1 \times D_2$. Since $D_1$ is a cone, $(nx,0) \in \overline{\Omega }_1$ for all n. Further, the previous claim implies that $(0,y) \in \overline{\Omega }_1$. Thus by convexity

$$\begin{aligned} (x,y) = \lim _{n \rightarrow \infty } \frac{1}{n}(nx,0) + \frac{n-1}{n}(0,y) \in \overline{\Omega }_1. \end{aligned}$$

Thus $D_1 \times D_2 \subset \overline{\Omega }_1$. Since $\Omega _1$ has complex dimension 2, $D_1 \times D_2 \subset \Omega _1$. $\square$

Next consider the matrices

$$\begin{aligned} B_n = \begin{pmatrix} n &{} \quad 0 \\ 0 &{} \quad 1 \end{pmatrix}. \end{aligned}$$

Then since $D_1$ and ${{\,\mathrm{\mathcal {H}}\,}}$ are cones we have

$$\begin{aligned} D_1 \times D_2 \subset B_n\Omega _1 \subset {{\,\mathrm{\mathcal {H}}\,}}\times D_2. \end{aligned}$$

So by passing to a subsequence we can assume that $B_n\Omega _1$ converges to some $\Omega _2$ in ${{\,\mathrm{\mathbb {X}}\,}}_2$.

Claim:$\Omega _2 = D_1 \times D_2$.

Proof of Claim

Notice that $D_1 \times D_2 \subset \Omega _2$ since $D_1 \times D_2 \subset B_n\Omega _1$ for any n.

For every $z \in D_2$ let $S_{z} \subset {{\,\mathrm{\mathbb {C}}\,}}$ be the convex open set such that

$$\begin{aligned} S_{z} \times \{z\}= ({{\,\mathrm{\mathbb {C}}\,}}\times \{z\} ) \cap \Omega _1. \end{aligned}$$

Then define ${{\,\mathrm{\mathcal {C}}\,}}_z = \bigcup _{n \in {{\,\mathrm{\mathbb {N}}\,}}} n \cdot S_{z}$. Then ${{\,\mathrm{\mathcal {C}}\,}}_z$ is a convex open cone since $0 \in \overline{S_z}$. Further

$$\begin{aligned} {{\,\mathrm{\mathcal {C}}\,}}_z \times \{z\}= ({{\,\mathrm{\mathbb {C}}\,}}\times \{z\} ) \cap \Omega _2. \end{aligned}$$

Since $D_1 \times D_2 \subset \Omega _2$ we see that $\overline{D}_1 \subset \overline{{{\,\mathrm{\mathcal {C}}\,}}}_z$. Suppose, for a contradiction, that $\overline{D}_1 \ne \overline{{{\,\mathrm{\mathcal {C}}\,}}}_z$ for some $z \in D_2$. Then there exists some $w \in {{\,\mathrm{\mathcal {C}}\,}}_z \setminus D_1$. Then $(tw,z) \in \Omega _2$ for all $t > 0$. Then by convexity

$$\begin{aligned} (w,0) = \lim _{n \rightarrow \infty } \frac{1}{n}(nw, z) + \frac{n-1}{n}(0,0) \in \overline{\Omega }_2. \end{aligned}$$

So $w \in \overline{D}_1$. So we have a contradiction. Thus ${{\,\mathrm{\mathcal {C}}\,}}_z = D_1$ for all $z \in D_2$ and hence $\Omega _2 = D_1 \times D_2$. $\square$

Next Lemma A.8 implies that ${{\,\mathrm{\mathcal {H}}\,}}\in \mathrm{AL}(D_j)$ for $j=1,2$. So ${{\,\mathrm{\mathcal {H}}\,}}\times {{\,\mathrm{\mathcal {H}}\,}}\in { \mathrm AL}(\Omega _2)$. Then, since

$$\begin{aligned} { \mathrm AL}(\Omega _2) \subset { \mathrm AL}(\Omega _1)\subset { \mathrm AL}(\Omega ), \end{aligned}$$

we see that ${{\,\mathrm{\mathcal {H}}\,}}\times {{\,\mathrm{\mathcal {H}}\,}}\in { \mathrm AL}(\Omega )$. $\square$

Proposition A.10

If $\Omega \subset {{\,\mathrm{\mathbb {C}}\,}}^d$ is a bounded convex domain, $\partial \Omega$ is ${{\,\mathrm{\mathcal {C}}\,}}^\infty$, and there exists a point of infinite type in $\partial \Omega$, then there exists a domain $D \in \mathrm{AL}(\Omega )$ such that

$$\begin{aligned} D \cap \{ (z_1,z_2,0,\dots ,0) : z_1,z_2 \in {{\,\mathrm{\mathbb {C}}\,}}\} = {{\,\mathrm{\mathcal {H}}\,}}\times {{\,\mathrm{\mathcal {H}}\,}}\times \{(0,\dots ,0)\}. \end{aligned}$$

The following argument comes from [40, Sect. 6].

Proof

Using Theorem A.6, it is enough to consider the case when $d = 2$ and show that

$$\begin{aligned} {{\,\mathrm{\mathcal {H}}\,}}\times {{\,\mathrm{\mathcal {H}}\,}}\in \mathrm{AL}(\Omega ). \end{aligned}$$

We first show that there exists a domain

$$\begin{aligned} \Omega _1 \in \overline{{{\,\mathrm{Aff}\,}}({{\,\mathrm{\mathbb {C}}\,}}^d)\cdot \Omega }^{{{\,\mathrm{\mathbb {X}}\,}}_d} \end{aligned}$$

with a non-constant holomorphic map ${{\,\mathrm{\mathbb {D}}\,}}\rightarrow \partial \Omega _1$.

Let

$$\begin{aligned} B : = \{ x+iy : \left| x\right| \le 1, \left| y\right| \le 1\} \subset {{\,\mathrm{\mathbb {C}}\,}}. \end{aligned}$$

By applying an initial affine transformation to $\Omega$, we can assume that $0 \in \partial \Omega$ is a point of infinite type, ${{\,\mathrm{\mathbb {R}}\,}}\times {{\,\mathrm{\mathbb {C}}\,}}$ is tangent to $\Omega$ at 0, and there exists a function $f: [-1,1] \times B \rightarrow [0,1]$ such that $f(0)=0$ and

$$\begin{aligned} \Omega \cap (B \times B) = \left\{ (x+iy, z) : y > f(x,z) \right\} . \end{aligned}$$

Since ${{\,\mathrm{\mathbb {R}}\,}}\times {{\,\mathrm{\mathbb {C}}\,}}$ is tangent to $\Omega$ at 0, we see that $ie_1$ is the inward pointing normal vector of $\partial \Omega$ at 0. Then, since $\partial \Omega$ is ${{\,\mathrm{\mathcal {C}}\,}}^2$ smooth, there exists $r> 0$ such that

$$\begin{aligned} rie_1 + (r{{\,\mathrm{\mathbb {D}}\,}}) \cdot e_1 \subset \Omega . \end{aligned}$$

By scaling $\Omega$ we can assume that

$$\begin{aligned} ie_1 + {{\,\mathrm{\mathbb {D}}\,}}\cdot e_1 \subset \Omega . \end{aligned}$$

(12)

Since $\Omega$ is convex and ${{\,\mathrm{\mathbb {R}}\,}}\times {{\,\mathrm{\mathbb {C}}\,}}$ is tangent to $\Omega$ at 0 we have

$$\begin{aligned} ({{\,\mathrm{\mathbb {R}}\,}}\times {{\,\mathrm{\mathbb {C}}\,}}) \cap \Omega =\emptyset . \end{aligned}$$

(13)

Finally since $0 \in \partial \Omega$ is a point of infinite type, for every $m \in {{\,\mathrm{\mathbb {N}}\,}}$

$$\begin{aligned} \lim _{z \rightarrow 0} \frac{\left| f(0,z)\right| }{\left| z\right| ^m}=0. \end{aligned}$$

Then we can pick $w_m \in B \setminus \{0\}$ and $\epsilon _m \rightarrow 0$ such that $\left| f(0,w_m)\right| = \epsilon _m \left| w_m\right| ^m$ and

$$\begin{aligned} \left| f(0,z)\right| \le \epsilon _m \left| z\right| ^m \text { for all } \left| z\right| \le \left| w_m\right| . \end{aligned}$$

If $\epsilon _m =0$ for some m, then $f(0,z) = 0$ for $\left| z\right| \le \left| w_m\right|$ and so $\partial \Omega$ contains the disk

$$\begin{aligned} \{ (0,z) : \left| z\right| \le \left| w_m\right| \}. \end{aligned}$$

Then

$$\begin{aligned} z \in {{\,\mathrm{\mathbb {D}}\,}}\rightarrow \left( 0, \frac{1}{w_m} z\right) \in \partial \Omega \end{aligned}$$

is a non-constant holomorphic map and we can simply define $\Omega _1 := \Omega$.

It remains to consider the case when $\epsilon _m > 0$ for all m. Let $T \in {{\,\mathrm{Aff}\,}}({{\,\mathrm{\mathbb {C}}\,}}^2)$ be the affine map

$$\begin{aligned} T(z_1,z_2) = (i(z_1-i), z_2). \end{aligned}$$

Then consider the affine maps $A_m \in {{\,\mathrm{Aff}\,}}({{\,\mathrm{\mathbb {C}}\,}}^2)$ given by

$$\begin{aligned} A_m(z_1, z_2) =\left( \frac{1}{f(0,w_m)} z_1, \frac{1}{w_m}z_2 \right) . \end{aligned}$$

We claim that $TA_m\Omega \in {{\,\mathrm{\mathbb {K}}\,}}_2$ for every m, that is

• (1) $ie_1+{{\,\mathrm{\mathbb {D}}\,}}\cdot e_1 \subset A_m\Omega$ and $({{\,\mathrm{\mathbb {C}}\,}}\cdot e_2)\cap A_m\Omega = \emptyset$,

• (2) $ie_1+{{\,\mathrm{\mathbb {D}}\,}}\cdot e_2 \subset A_m\Omega$ and $(i,1) \in \partial A_m\Omega$.

First, by Eq. (12)

$$\begin{aligned} ie_1+{{\,\mathrm{\mathbb {D}}\,}}\cdot e_1 = A_m(f(0,w_m) i e_1+f(0,w_m){{\,\mathrm{\mathbb {D}}\,}}\cdot e_1)\subset A_m(ie_1+{{\,\mathrm{\mathbb {D}}\,}}\cdot e_1) \subset A_m \Omega \end{aligned}$$

and by Eq. (13)

$$\begin{aligned} ({{\,\mathrm{\mathbb {C}}\,}}\cdot e_2)\cap A_m\Omega = A_m \Big ( ({{\,\mathrm{\mathbb {C}}\,}}\cdot e_2)\cap \Omega \Big ) = \emptyset . \end{aligned}$$

Since

$$\begin{aligned} \left| f(0,z)\right| \le \epsilon _m \left| z\right| ^m \le \epsilon _m \left| w_m\right| ^m = f(0,w_m) \end{aligned}$$

when $\left| z\right| \le \left| w_m\right|$ we see that

$$\begin{aligned} f(0,w_m)i e_1 + (\left| w_m\right| {{\,\mathrm{\mathbb {D}}\,}})\cdot e_2 \subset \overline{\Omega }. \end{aligned}$$

Since $f(0,w_m)ie_1 \in \Omega$, convexity then implies that

$$\begin{aligned} f(0,w_m)i e_1 + (\left| w_m\right| {{\,\mathrm{\mathbb {D}}\,}})\cdot e_2 \subset \Omega . \end{aligned}$$

Thus

$$\begin{aligned} ie_1+{{\,\mathrm{\mathbb {D}}\,}}\cdot e_2 = A_m \Big ( f(0,w_m)i e_1 + (\left| w_m\right| {{\,\mathrm{\mathbb {D}}\,}})\cdot e_2\Big ) \subset A_m \Omega . \end{aligned}$$

Finally, by definition $(f(w_m)i, w_m) \in \partial \Omega$ so

$$\begin{aligned} (i,1) = A_m(f(w_m)i, w_m) \in \partial A_m \Omega . \end{aligned}$$

Thus $TA_m \Omega \in {{\,\mathrm{\mathbb {K}}\,}}_2$.

Since ${{\,\mathrm{\mathbb {K}}\,}}_2$ is compact in ${{\,\mathrm{\mathbb {X}}\,}}_2$, we can pass to a subsequence and suppose that $A_m \Omega$ converges to $\Omega _1$ in ${{\,\mathrm{\mathbb {X}}\,}}_2$. We claim that $\{0\} \times {{\,\mathrm{\mathbb {D}}\,}}\subset \partial \Omega _1$. Notice that

$$\begin{aligned} (A_m\Omega ) \cap (B \times B) = \left\{ (x+iy, z) \in B \times B: y > f_m(x,z) \right\} , \end{aligned}$$

where

$$\begin{aligned} f_m(x,z) = \frac{1}{f(0,w_m)} f\Big ( f(0,w_m) x, w_m z \Big ). \end{aligned}$$

In particular, if $\left| z\right| < 1$, then

$$\begin{aligned} f_n(0,z)= & {} \frac{1}{f(0,w_m)} f\Big ( 0, w_m z \Big ) = \frac{1}{\epsilon _m \left| w_m\right| ^m} f\Big ( 0, w_m z \Big )\\&\le \frac{1}{\epsilon _m \left| w_m\right| ^m} \epsilon _m \left| w_mz\right| ^m = \left| z\right| ^m. \end{aligned}$$

So

$$\begin{aligned} \left\{ (iy, z) : \left| z\right|< 1, \ \left| z\right| ^m< y < 1\right\} \subset A_m\Omega \end{aligned}$$

and thus

$$\begin{aligned} \left\{ (iy, z) : \left| z\right|< 1, \ 0< y < 1 \right\} \subset \Omega _1. \end{aligned}$$

In particular, $\{0\} \times {{\,\mathrm{\mathbb {D}}\,}}\subset \overline{\Omega _1}$. Since $({{\,\mathrm{\mathbb {C}}\,}}\cdot e_2)\cap A_m\Omega = \emptyset$ for all m, we also have $({{\,\mathrm{\mathbb {C}}\,}}\cdot e_2)\cap \Omega _1 = \emptyset$. Hence $\{0\} \times {{\,\mathrm{\mathbb {D}}\,}}\subset \partial \Omega _1$.

Now Proposition A.9 implies that ${{\,\mathrm{\mathcal {H}}\,}}\times {{\,\mathrm{\mathcal {H}}\,}}\in \mathrm{AL}(\Omega _1)$. Since

$$\begin{aligned} { \mathrm AL}(\Omega _1)\subset { \mathrm AL}(\Omega ), \end{aligned}$$

we see that ${{\,\mathrm{\mathcal {H}}\,}}\times {{\,\mathrm{\mathcal {H}}\,}}\in { \mathrm AL}(\Omega )$. $\square$