As far as I am aware, partitions of unity for smooth manifolds require the use of smooth functions with compact support (e.g. bump functions). However, for a complex manifold, the transition maps have to be not only smooth, but holomorphic. And by the Identity Theorem any holomorphic function with compact support is identically zero.

Question: How does one surmount this obstacle when working with complex manifolds?

Naively it seems to me that because of this the only possible complex manifold to define would be the open unit disk in $\mathbb{C}^n$, i.e. a manifold with only one chart, since Wikipedia says that the atlas of a complex manifold consists of charts to the open unit disk in $\mathbb{C}^n$.

  • $\begingroup$ Does the partition of unity have to be holomorphic just because the transition maps are? $\endgroup$
    – Arthur
    Commented Sep 12, 2016 at 20:44
  • $\begingroup$ @Arthur Honestly I'm not sure $\endgroup$ Commented Sep 12, 2016 at 21:00

1 Answer 1


Note that complex manifolds are in particular smooth manifold (holomorphic $\Rightarrow$ smooth). Thus one can define partition or unity as in the smooth case. We cannot have something like "holomorphic" partition of unity, of course.

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    $\begingroup$ In particular, partitions of unity are only used to glue smooth functions, differential forms, sections of bundles, etc., together, even though you're on a complex manifold. $\endgroup$ Commented Sep 12, 2016 at 20:50
  • $\begingroup$ How do we define "real smoothness" over the complex numbers though? Maybe the Wirtinger derivative? en.wikipedia.org/wiki/Wirtinger_derivatives Any function that is once complex differentiable is automatically holomorphic, I thought, so if we use complex differentiation, it seems like the notions of "smooth" and "holomorphic" are inextricable. I want to accept this answer but I just want to make sure I actually understand it correctly first. $\endgroup$ Commented Sep 12, 2016 at 21:02
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    $\begingroup$ There is a identification $\mathbb C^n \cong \mathbb R^{2n}$, thus all open sets in $\mathbb C^n$ are also open sets in $\mathbb R^{2n}$, and if $ \phi : U \to V$ is a biholomoprhism between $U, V \subset \mathbb C^n$, then it is also a diffeomorphism. @William $\endgroup$
    – user99914
    Commented Sep 12, 2016 at 21:09
  • $\begingroup$ Right but aren't holomorphic maps the problem? I get how complex derivatives can be identified with real derivatives plus the Cauchy Riemann equations, but only without the Cauchy-Riemann equations does the Identity theorem fail. So if we wanted to say a function from an open subset of $\mathbb{C}^n$ to $\mathbb{C}^n$ is "smooth" but not "holomorphic", what type of derivative are we referring to? Not the regular derivative that satisfies the Cauchy Riemann equations, right? That's why I am asking about Wirtinger derivatives. $\endgroup$ Commented Sep 12, 2016 at 21:15
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    $\begingroup$ Definitely the usual derivatives (in advanced calculus). For example $z\maspto \bar z$ is smooth but not holomorphic. @William $\endgroup$
    – user99914
    Commented Sep 12, 2016 at 21:18

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