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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!

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  • $\begingroup$ 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. $\endgroup$
    – Willie Wong
    Commented Mar 9, 2011 at 12:33

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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.

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