Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

A complex manifold $X$ is said to be weakly pseudoconvex if there exists on $X$ a smooth plurisubharmonic exhaustion function $\psi$.

For example, Stein manifolds are weakly pseudoconvex (in this case $\psi$ can be chosen to be even strictly plurisubharmonic), compact complex manifolds are weakly pseudoconvex (take $\psi\equiv 0$), etc...

Now, let $X$ be a non-compact weakly pseudoconvex manifold. The question is the following:

Does there exist a smooth plurisubharmonic exhaustion function $\varphi$ on $X$ such that $d\varphi$ is never zero outside some compact set $K\subset X$?

Thanks in advance.

share|improve this question
    
Sorry Noz, I don't understand your question. –  diverietti Feb 4 '11 at 17:54
    
Forget my comment, I was out of my mind. I erased it –  Noz Feb 4 '11 at 18:08
1  
When you restrict a function to a submanifold you can get new critical points.Otherwise you will get into trouble with topology –  Mohan Ramachandran Feb 4 '11 at 18:22
2  
Amplifying Mohan's comment: Stein manifolds which have a psh exhaustion function with compact critical set are called "finite-type". They include affine algebraic manifolds but not, say, Riemann surfaces of infinite genus (because one can perturb the function near the critical set so as to make it Morse, but still proper and bounded below). –  Tim Perutz Feb 4 '11 at 20:32
    
Mohan and Tim, you are definitely right... $|z|^2$ is indeed a Morse function on $X$... I modify then my question... And I guess the answer is already contained in Tim's one then... –  diverietti Feb 5 '11 at 14:06
add comment

1 Answer 1

Expanded version of my earlier comment, which was directed at the unedited version of the question.

Stein manifolds are those complex manifolds $X$ which have a strictly psh exhaustion function, i.e., a proper, bounded below $C^\infty$ function $\psi$ such that the closed $(1,1)$-form $\omega:=-dd^c \psi$ is positive with respect to the complex structure $J$. (My convention is that $d^c f = J \circ df$ where $J$ is the complex structure acting on cotangent vectors.)

Those for which one can take $\psi$ to have compact critical set are called "finite-type". They include smooth affine algebraic varieties $X\subset \mathbb{C}^N$, which one can see by compactifying to a projective variety $\bar{X}=X\cup D$ and considering $\log \|s\|^2$ where $s$ is a section of a hermitian holorphic line bundle cutting out the divisor $D$.

Any open Riemann surface is Stein, but those of infinite genus (i.e. with infinite rank $H_1$) do not have finite type. If there were a psh exhaustion with compact critical set one could perturb it near the critical set so as to make it a Morse function. The downward gradient flow exists and converges to critical points, and so Morse theory bounds the rank of $H_\ast$ from above by the number of critical points.

share|improve this answer
add comment

Your Answer

 
discard

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.