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We know that if $X$ is a compact connected complex manifold,then every holomorphic function on $X$ is constant. Now,supposed that $X$ is not necessarily connect, then we can choose a connected component. We know that connected component is closed subset and every closed subset of a compact set is also compact. So the connected component is also compact, then we can deduced that every holomorphic function on the connected component is constant. Then We can deduced that every holomorphic function on $X$ is locally constant.

I think this may be not right but I can't find where is the problem in my proof in the above.

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This is correct. However, when people say "compact manifold" they almost always mean connected compact manifold. Rather, there is usually nothing to be gained by dealing with non-connected compact manifolds, since we might as well just look at each connected component.

(For non-compact manifolds, this is potentially trickier, because we have things like $(-\infty,0)\cup(0,\infty)$ which is a disjoint union of two manifolds, but they are sort of "touching," and in some sense inherently different from $(-\infty,-1)\cup(1,\infty)$, for example.)

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  • $\begingroup$ Ok,thanks a lot.May be I think too much. $\endgroup$ – Kevin Aug 18 '20 at 15:34
  • $\begingroup$ We all do.$\hspace{1pt}$ $\endgroup$ – Elliot G Aug 18 '20 at 15:34
  • $\begingroup$ Sorry,one more question.Dose that mean all holomorphic form on a compact complex manifold are closed? $\endgroup$ – Kevin Aug 18 '20 at 15:58
  • $\begingroup$ Yes because the derivative of $f\ dx^I$ is a sum of $\frac{\partial f}{\partial x^i}dx^i\wedge dx^I$, and the partial derivatives are necessarily zero since $f$ is constant. $\endgroup$ – Elliot G Aug 18 '20 at 21:03

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