I'm currently trying to learn more about complex geometry and am working through Forsters book on Riemann surfaces. I'm having trouble with exercise 6.1 of this book.

It asks, why the presheaf of bounded, holomorphic functions on a general Riemann surface $X$ is not a sheaf.

The case $X=\mathbb{C}$ is clear to me. It's basically this argument about the failure of the gluing property.

For a general Riemann surface $X$ I have the definition: A function $f$ on an open set $U\subset X$ is bounded and holomorphic, if all the maps $f\circ\varphi_i^{-1}: \varphi_i(U\cap V_i)\rightarrow \mathbb{C}$ (where $(V_i,\varphi_i)_{i\in I}$ is a complex atlas of $X$) are bounded and holomorphic.

But now what? Do the patches $\varphi_i(V_i)$ cover all of $\mathbb{C}$? Then I could probably just use the result for $X=\mathbb{C}$. If not, where do I start then?

I don't feel super comfortable working with Riemann surfaces (or sheaves) yet, and my course on manifolds was a while ago. I would appreciate at least a little hint here. Thank you.

Edit: After thinking about it for a bit, the complex structure is probably irrelevant and this should also fail for topological manifolds and bounded continuous functions. An open set $U$ is covered its intersections with the charts of the atlas, but I'm still not sure about how to do the gluing.


1 Answer 1


Observe that the presheaf of bounded holomorphic functions on the unit disc $D(0, 1)$ is not a sheaf, e.g. by seeing that $\frac 1 {1 - z}$ is not bounded on $D(0, 1)$ but is bounded on $D(0, r)$ for $r < 1$. These constitute a collection of compatible sections but can not be glued together inside the presheaf of bounded functions so it is not a sheaf.

For the general case observe that every Riemann surface contains a disk, and on this disk the presheaf fails at having the gluing property.

  • $\begingroup$ i see. thank you, that makes sense. and, like i thought, the same argument ought to work for general topological manifolds with e.g. f(x)=1/|x|. $\endgroup$ Commented May 29, 2023 at 9:04
  • $\begingroup$ How is $1/z$ bounded on $D(0,1)$? As $z\to 0, 1/z$ is infinite. Don't you mean the function $1/(z-1)$ which is bounded on disks of radius less than one and unbounded on the closed disk of radius $1$? $\endgroup$
    – Hushus46
    Commented Feb 6 at 20:27
  • $\begingroup$ @Hushus46 you're right, I made an obvious typo here, thx. $\endgroup$ Commented Apr 4 at 16:50

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