A Class of Strictly Pseudoconvex Domains with Non-pluripolar Core

Abstract

We construct a class of strictly pseudoconvex domains in ${\mathbb C}^d$ whose core has non-empty interior. Consequently these cores are not pluripolar. This answers a question posed by Harz, Shcherbina and Tomassini.

A. Appendix

Proposition A.1

Let P(z, w) be a nonconstant complex polynomial. Then either

• (i) there are at most finitely many complex lines $L_1, L_2, \ldots , L_s$ in ${\mathbb C}^2$, that are disjoint from the zero set of P; or

• (ii) there is a complex line $L_0$, such that P is constant on every line $L'$ parallel to $L_0$, and so, every complex line L that is not parallel to $L_0$ must intersect the zero set of P.

Proof

Let ${\mathcal L}$ denote the set of all complex lines L in ${\mathbb C}^2$ such that P is not vanishing on L. Then P|L is constant.

Case(a) Suppose not all lines in ${\mathcal L}$ are parallel. We claim that in this case there are at most d lines in ${\mathcal L}$, where d is the degree of P.

Suppose this is false; then we can choose $(d+1)$ distinct lines $L_0, L_1, \ldots , L_d$ in ${\mathcal L}$, so that $L_0$ and $L_1$ intersect. Then every $L_i$ intersect $L_0$ or $L_1$, and so $S:=\bigcup _{0\le i \le d} L_i$ is connected, and $P|S = c$ where c is a constant. Consider any point $p\in {\mathbb C}^2 \setminus S$. Then there is a complex line $L^*$ through p that intersects every line $L_i$, for $i=0, 1 ,\ldots , d$, but does not pass through any of their intersection points. (We skip easy details here.) Then P|L has the same value c at each of the $(d+1)$ distinct intersection points of $L^*$ with S. Since degree of the polynomial $P|L^*$ is less than $(d+1)$, $P|L^*$ must be constant, equal c. We obtain P is constant on ${\mathbb C}^2$ contrary to the assumptions.

Case(b) Suppose ${\mathcal L}$ is a set of parallel lines, and there are more then d of them. Then P is constant on every complex line parallel to them, and every complex line transversal to ${\mathcal L}$ intersects the zero set of P.

To see this, consider $(d+1)$ distinct lines, say $L_0, L_1,\ldots , L_d$ in ${\mathcal L}$. Choose a complex coordinate system in ${\mathcal L}$, so that $L_i = \{(z,w)\in {\mathbb C}^2 : w =b_i \}$, with $b_0, b_1,\ldots ,b_d$ distinct constants. Consider, for any two constants $a, a'$ the polynomial $P(a, w) - P(a', w)$. Since it has $(d+1)$ distinct roots (i.e. $b_j$’s), it is identically zero. Hence, P is constant on every line parallel to $L_0$, and so, P being nonconstant, for every complex line L trannsversal to $L_0$, P|L is a nonconstant polynomial and so must have a zero.

Clearly cases (a) and (b) together imply the proposition. $\square$