# Extension of strictly plurisubharmonic functions on a Kähler manifold

I was wondering, suppose I have a non-compact Kähler manifold $M$ and suppose that outside some compact subset $A\subset M$, there exists a smooth function $f:M\backslash A\longrightarrow\mathbb{R}$ such that $i\partial\bar{\partial}f>0$. Is it always possible for me to find a smooth function $h:M\longrightarrow\mathbb{R}$ such that $h|_{M\backslash A}=f$ and $i\partial\bar{\partial}h>0$ on the whole of $M$? If not in general, are there sufficient conditions on $M$ that will allow me to do this? Many thanks!

-
Thanks for the responses below. Just 1 more thing. Do you think it is even possible to find a Kahler form $\omega$ on M so that $\omega|(M\backslash B)=i\partial\bar{\partial}f$, for some compact subset $B$ of $M$ containing $A$? I guess this is possible if for any compact subset $H$ of $M$, we can find a (1, 1)-form $\alpha$ such that $\alpha|H>0$ and $\alpha=0$ on $M\backslash K$ for another compact subset $K\subset M$ containing $H$. Then we can take $c\alpha+i\partial\bar{\partial}(\phi.f)$ for c>0 large and $\phi$ a bump function. But I'm not sure these (1,1) forms $\alpha$ exists. – user4317 Apr 19 '10 at 13:02
It might be possible to prove this, or something like it, using the regularized maximum of local Kaeher potentials; see Demailly's book. If $A$ is bounded by a contact-type hypersurface, the Gray stability theorem could be another option for patching your 2-forms. – Tim Perutz Apr 19 '10 at 13:45

It is instructive to conisder the case of Kahler metrics invariant under torus action. In this case your question becomes a certain (nontivial) question on convex functions.

Recall first, that Kahler metrics on $(\mathbb C^*)^n$ invariant under the action of $(S^1)^n$ have global potential that is given by a convex function $F$ on $\mathbb R^n$. Here $\mathbb R^n$ is identified with the quotient

$(\mathbb C^*)^n/(S^1)^n$ and we take coordinates $log|z_i|$ on $\mathbb R^n$. So we can translate your original question as follows

QESTION. Suppose you have a smooth convex function $F$, defined on $\mathbb R^n$ outside compact $\Omega$. Is it possible to extend $F$ to a smooth convex function on the whole $\mathbb R^n$?

It easy to construct an example of a non-convex $\Omega$ on $\mathbb R^2$, with convex $F$ defined on $\mathbb R^2\setminus \Omega$, so that $F$ can not be extended. For the moment I don't see how to make such an example when $\Omega$ is the unite disk, but it sounds plausible that such examples exist.

-

No, in general.

A trivial reason is that $f$ may not extend to a smooth function in a neighbourhood of $\overline{M\setminus A}$, but you can easily get around that by asking that $f$ agree with $h$ in a slightly smaller domain than $M\setminus A$.

A non-trivial reason is that the 2-form $\omega=i \partial \bar{\partial} h$ would be an exact Kähler form. Because of Stokes's theorem, a complex manifold can only support such a form if it has no compact complex submanifolds of positive dimension. For instance, $M=\mathbb{C P}^2 \setminus pt.$ does not, so there is no psh extension to $M$ of the Kähler potential $f$ for the Fubini-Study metric on $\mathbb{C P}^2 \setminus (\mathbb{CP}^1\cup pt.)$.