1
$\begingroup$

How to construct a surjective holomorphic map $F:X\to Y$ between (connected) complex manifolds with a real-valued function $u$ on $Y$ such that $u{\circ}F$ is plurisubharmonic on $X$, but $u$ is not upper semicontinuous on $Y$.

$\endgroup$
  • 1
    $\begingroup$ It is easy to construct an example when $X$ is disconnected. Let $Y$ be the complex plane, $u(0)=0$, $u(z)=1, z\neq 0$. Let $X$ be a disjoint union of two complex planes, and $F(z)\equiv 0$ on one of them, while on the other one $F(z)=e^z$. $\endgroup$ – Alexandre Eremenko Dec 12 '19 at 12:52
  • $\begingroup$ Yes, you are right. I think the difficulty of this question is when X is connected. $\endgroup$ – BLANK Dec 12 '19 at 13:45
  • $\begingroup$ When $X$ is connected, I see no difficulty: there is no such function. $\endgroup$ – Alexandre Eremenko Dec 12 '19 at 18:33
  • $\begingroup$ Dear Prof. Eremenko, could you kindly sketch a proof of your claim? I can give one when F is a proper map, but I can not do for general F. Thank you very much. $\endgroup$ – BLANK Dec 13 '19 at 0:20
1
$\begingroup$

Suppose that $X$ is connected. If $u$ is not u.s.c, then $u(a)<\limsup_{z\to a} u(z)$ for some $a$. Since $F$ is surjective, there is $b$ such that $F(b)=a$. Consider the set $E=\{ z\in X:F(z)=a\}$. This set is not a neighborhood of $b$, since otherwise $F$ will be constant in a neighborhood of $b$, thus, since $X$ is connected, constant on the whole $X$. So there is a sequence $z_k\to b$ such that $F(z_k)\neq a$. Then $F(z_k)\to a$ since $F$ is continuous, and thus $\limsup_{k\to\infty}u\circ F(z_k)< u\circ F(b)$ which the assumption that $u\circ F$ is u. s. c.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ It seems not to be enough for making a contradiction. Could we deduce that $\lim\sup_{k\to\infty}u\circ F(z_k)=\lim\sup_{z\to a}u(z)$? $\endgroup$ – BLANK Dec 13 '19 at 4:48

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.