MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In several complex variables , to determine the pseudoconvexity of a domain in $C^n$ is very important . There are various criterion to decide whether a domain is pseudoconvex . In particular ,if the domain is defined by a $C^\infty$ defining function $\phi>0$ , then 'Levi pseudoconvex ' is equivalent to the following matrix (which is called 'Monge-Ampere matrix') $$ \begin{pmatrix} -\phi & -\partial_\bar{k}\phi \\ -\partial_j\phi & -\partial_{j\bar{k}}^2\phi \end{pmatrix} $$ have precise one negative eigenvalue and n positive eigenvalues .

So my question is how to prove this ? Anybody knows? Thanks very much!

share|cite|improve this question
I fixed your math display. You probably should also include a definition of the notation $\partial_{\bar{k}}$ (I think I know what it is, but best to make sure), and perhaps (link to) a short description of Levi pseudoconvexity to refresh our memories. – Willie Wong Mar 9 '11 at 12:33
up vote 0 down vote accepted

Since $\phi$ is always nonzero inside $\Omega$ , so this matrix has precise one negative eigenvalue and n positive eigenvalues is equivalent to $-\partial\bar{\partial}log\phi$ is non-negative , but which means that $-log\phi$ is a plurisubharmonic exhaustion function for the domain $\Omega$ , i.e. $\Omega$ is pseudoconvex.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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