1
$\begingroup$

I have a very basic question on pluripolar sets. First remind their definition. Let $\Omega\subset \mathbb{C}^n$ be a domain. A subset $E\subset \Omega$ is called pluripolar if there exists a plurisubharmonic function $f\colon \Omega\to \mathbb{R}$, not identically $-\infty$, such that $E\subset \{f=-\infty\}$.

Question. What are the examples of pluripolar sets? For example, when a closed real analytic (or smooth) submanifold of $\Omega$ is pluripolar?

Remark. According to Wikipedia, pluripolar sets have Hausdorff dimension at most $2n-2$. Also the zero set of a holomorphic function is pluripolar.

$\endgroup$

2 Answers 2

3
$\begingroup$

Pluripolarity is a quite subtle property.

E. Bedford characterized pluripolar real-analytic submanifolds.

Added: Bedford (in Lelong-Skoda seminar, 1981) called a real submanifold $M$ of $\mathbb{C}^n$ a generating at a point $p$ if the tangent space $T_p M$ is not contained in a proper complex-linear subspace of $\mathbb{C}^n$. It is easy to see that a generating at any point submanifold is not pluripolar. Bedford prove that a real-analytic nowhere generating submanifold is pluripolar.

For example, all real analytic curves in $\mathbb{C}^n$ are pluripolar for $n>1$.

In opposite direction, it's useful to know examples of sets that are not pluripolar. There is a very surprising example due to K. Diederich and J. E. Fornæss (1982) of a non-pluripolar $C^\infty$ smooth curve in $\mathbb{C}^2$. This construction can be extended to $\mathbb{C}^n$.

$\endgroup$
2
  • $\begingroup$ Can you state please Bedford's result? $\endgroup$
    – asv
    Jan 28, 2016 at 15:41
  • $\begingroup$ @sva See the edit. $\endgroup$ Jan 28, 2016 at 17:47
3
$\begingroup$

As you said, analytic sets (other than the whole space) are pluripolar. By playing with subharmnic functions of the form $\sum u_j$ you can arrange some more complicated sets which are neither open nor closed (but the set where a plurisubharmonic function is $-\infty$ must be $G_\delta$). So you can have a dense pluripolar set, for example. Other non-analytic examples can be made by taking subsets of pluripolar sets. Pluripolar sets in (complex) dimension $1$ are easier to visualize: they are just very small, and sufficient conditions for a set to be pluripolar can be given in terms of Haurdorff-like measure terms (but only sufficient or necessary, there is a theorem that says that a complete characterization in metric terms is impossible). Then you can take products of these one dimensional sets times $C^k$.

But there is the following remarkable theorem:

If a pluripolar set in $C^{n+1}$ is a graph of a continuous function $f:C^n\to C$, then this function is analytic (thus the set is also analytic).

Nikolay Shcherbina, Pluripolar graphs are holomorphic. Acta Math. 194 (2005), no. 2, 203–216.

This answers your question about smooth hypersurfaces.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.