2
$\begingroup$

Let $X$ be a complex manifold, and $\omega$ a K?hler form on $X$. A smooth function $\phi$ on $X$ is $\omega$-plurisubharmonic ($\omega$-psh for short) if the form $\omega+\sqrt{-1}\partial\bar{\partial}\phi$ is a K?hler form.

Question 1: Given an arbitrary smooth function $f$ on $X$, is it possible to find two $\omega$-psh functions $\phi$ and $\psi$ such that $f=\phi-\psi$?

Question 2: What if the regularity assumption is weakened? Is it possible for continuous $f$, $\phi$ and $\psi$? What can you say about the space of differences of general $\omega$-psh functions?

$\endgroup$
2
$\begingroup$

On your Question 1: the answer depends on the manifold.

For instance, if $X$ is the unit disk $D=\{z\in {\mathbb C}\vert |z|<1\}$, then, given $f\in {\mathcal C}^{\infty}(X,{\mathbb R})$, there exist two smooth plurisubharmonic functions $\phi$ and $\psi$ on $X$ such that $f=\phi-\psi$. Indeed, define $h$ a ${\mathcal C}^{\infty}$ function such that $hidz\wedge d\bar z=-i\partial\bar\partial f$ and it is easy to show that there exists $g:[0,1)\to [0,\infty)$ a ${\mathcal C}^{\infty}$ function such that $g(|z|^2)\geq h(z)$. Now, if you define $u:[0,1)\to {\mathbb R}$ by the formula $$u(t)=\int_0^t\left [\frac 1r\int_0^r(g(s)+1)ds\right ] dr$$ and set $\psi(z)=u(|z|^2)$. The function $u$ satisfies the equation $u'(t)+tu''(t)=g(t)+1$. Then it follows that $\phi:=f+\psi$ is plurisubharmonic.

On the other hand, on compact complex manifolds this is no longer true. For simplicity, let $X$ be a compact complex manifold of dimension $1$ and $\omega$ some Kahler metric on $X$. Suppose that any $f\in {\mathcal C}^{\infty}(X,{\mathbb R})$ can be written $f=\phi-\psi$, with $\omega_{\phi}:=\omega+i\partial\bar\partial \phi\geq 0$ and $\omega_{\psi}:=\omega+i\partial\bar\partial\psi\geq 0$. Pick $p\in X$ and let $z$ be local coordiantes centered at $p$, say $z$ is defined on the ball $B_2=\{|z|<2\}$ and choose $\chi$ a cut-off function with support in $B_2$ and which is $1$ on $B_1=\{|z|<1\}$. Set $$f_{\varepsilon}(z)=\frac {\lambda}{2\pi}\chi(z)\log(|z|^2+\varepsilon)$$ where $\lambda$ is a constant which is $>\int_X\omega$. Now denote by $\psi_{\varepsilon}$ and $\psi_{\varepsilon}$ the two $\omega$-psh functions such that $f_{\varepsilon}=\phi_{\varepsilon}-\psi_{\varepsilon}$. It follows that $$\omega_{\phi_{\varepsilon}}\geq \frac{\lambda}{2\pi}i\partial\bar\partial \log(|z|^2+\varepsilon)$$ on $B_1$. Since $\int_X\omega_{\phi_{\varepsilon}}$ is bounded (it is in fact constant), there exists a subsequence of $\omega_{\phi_{\varepsilon}}$ that converges weakly to a current $T=\omega+i\partial\bar\partial\varphi\geq 0$. The above inequality implies that $T\geq \lambda\delta_p$, where $\delta_p$ is the Dirac function at $p$. But this is impossible since $\int_XT=\int_X\omega<\lambda$.

Somehow related to your question, there is this notion of ${\it delta-plurisubharmonic}$ functions, which are functions that can be written as the difference of two plurisubharmonic functions. They appear in Cegrell's work, so if you are interested in this topic, you can start, for instance, with his paper "Delta-plurisubharmonic functions", Mathematica Scandinavica Vol. 43, No. 2 (May 2, 1979), pp. 343-352

$\endgroup$
  • $\begingroup$ Thanks a lot! Do you mean $\lvert z\rvert^2$ in the definition of $f_{\epsilon}$? $\endgroup$ – user135826 Feb 21 at 12:28
  • $\begingroup$ You're right, it's $|z|^2$ in the Poincare-Lelong formula. $\endgroup$ – user48958 Feb 21 at 17:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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