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

Let $X$ be a non-compact complex manifold of Kähler type (i.e. there exists a Kähler metric on $X$ but it hasn't been endowed with one). For each $i \in \mathbb{N}$, let $f_i$ be a smooth function $X \to [0, 1]$ such that for every compact set $K \subset X$, there exists $N \in \mathbb{N}$ such that $f_i|_K = 1$ for all $i \geq N$.

Is it always possible to choose a Kähler metric on $X$ such that, for all $i$, $\|\bar{\partial}f_i\| \leq 1$ with respect to the norm on $\Omega^{0,1}(X)$ induced by the metric?

If the answer is no, can we do any better than a generic hermitian metric? That is, can we find a locally conformally Kähler, Hermitian-Einstein, or some other special metric which will give $\|\bar{\partial}f_i\| \leq 1$?

share|improve this question
You should add the hypotheses on the $\alpha_i$ that you have in mind, otherwise the answer is obviously "no": take the integer multiples of a non zero $\bar{\partial}f$. For instance, boundedness on every compact subset (in some/any hermitian metric) forbids this kind of stupid example, and is obviously necessary. –  BS. Oct 1 '12 at 10:45
@BS: Thanks for your comment. I have modified the question so that it more accurately reflects the situation I have come across. I think the hypothesis you proposed is now satisfied. –  Michael Albanese Oct 6 '12 at 8:06
The hypothesis is still not satisfied if the $f_i$ converge to, say, the indicator function of a compact set (which is possible, by convolutions). Then the $\bar{\partial f_i}$ are locally unbounded. –  BS. Oct 6 '12 at 8:56
@BS: Thanks, I didn't think of this. I've restricted the $f_i$ to be converging to the constant function $1$. –  Michael Albanese Oct 6 '12 at 9:55
add comment

Your Answer


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

Browse other questions tagged or ask your own question.