The definition of holomorphic maps between Riemann surfaces confuses me a little. When we say a (continuous) map between two Riemann surfaces $f: X \to Y$ is holomorphic, we mean that for any pair of charts $\varphi: U_1\subseteq X \to V_1$ and $\psi: U_2 \subseteq Y \to V_2$, the composition $\psi \circ f\circ \varphi^{-1}: \varphi(U_1 \cap f^{-1}(U_2))\to \psi(f(U_1)\cap U_2)$ is differentiable at every point in its domain.

Now the word 'charts' here confuses me. Do we require the charts to be in the complex atlases of $X$ and $Y$? Or do we require the charts to be in the maximal atlases on $X$ and $Y$? Or do we just take arbitrary local homomorphisms $\varphi, \psi$ regardless of the atlases on $X$ and $Y$?

In the first case, will the definition of holomorphic maps depends on the specific atlases (not necessarily maximal) we impose on $X$ and $Y$?

It seems to me that the second case is somehow equivalent to the first one as we can simply write, for any charts $\varphi',\psi$ in the maximal atlases, $$\psi'\circ f\circ\varphi'^{-1}=(\psi'\circ\psi^{-1})\circ(\psi\circ f\circ\varphi^{-1})\circ(\varphi\circ\varphi'^{-1})$$ given some charts $\varphi, \psi$ in the given atlases. But I am not quite sure because even the same author may refer to different things when they mention the word 'chart' in different places. Thanks in advance.

  • $\begingroup$ The composition $\phi f\psi^{-1}$ should be holomorphic maps between open subsets of $\mathbb{C}$ not just continuous . What do you mean by a "complex" atlas? $\endgroup$
    – daruma
    Commented Jul 2, 2021 at 6:26
  • $\begingroup$ @daruma (1) see line 4 para 1. (2) an atlas with the compatibility condition; might be a vague terminology, but I think the point is clear. $\endgroup$
    – GoogleME
    Commented Jul 2, 2021 at 6:40

1 Answer 1


For example, $X=\mathbb{C}$ has two incompatible atlases.

The first analytic structure on $X$ is to just take the usual structure. That is the atlas consists of the charts where $U\to U\subset \mathbb{C}, z\mapsto z$.

The second analytic structure on $X$ is to take conjugates of everything. That is the atlas consists of all the charts where $U\to \mathbb{C}, z\mapsto \bar{z}$. You might think the transition functions are analytic but they are; $\phi\psi^{-1}: \bar{z}\mapsto \bar{z}$, if you follow the definitions. That is $\phi\psi^{-1}$ is an identity map when restricted to where it is defined.

Let's denote $X$ with the first analytic structure $X_1$ and the second structure $X_2$.

So observe that $f:X_1\to X_1,p\mapsto p$ is a valid holomorphic map of complex Riemann surfaaces but $f: X_1\to X_2, p\mapsto p$ is not. That is not to say there is no holomorphic map from $X_1$ to $X_2$.

What we see here is that no, you cannot take arbitrary homeomorphisms (as is case 2 of your question) in your definition of $\phi f \psi^{-1}$. Indeed, even the same space with different a different complex Riemann surface structure will give different answers as to whether this map $f$ is a holomorphic map or not.

  • $\begingroup$ I can see from your answer that the definition must depend on atlases. But do we need to consider maximal atlases? Or does it suffice to define the holomorphy only wrt. the charts in the not necessarily maximal atlases? $\endgroup$
    – GoogleME
    Commented Jul 2, 2021 at 6:52
  • $\begingroup$ Without an atlas, all you have is a topological space. A Riemann surface (or more generally other kind of manifolds) don't make sense without an atlas. $\endgroup$
    – daruma
    Commented Jul 2, 2021 at 15:11
  • 1
    $\begingroup$ Your atlas need not be maximal. Extending your atlas to a maximal atlas is like a compatibility condition. The map $f$ will remain holomorphic if you extend it to a maximal atlas. $\endgroup$
    – daruma
    Commented Jul 2, 2021 at 15:15
  • $\begingroup$ thanks for the explanation. $\endgroup$
    – GoogleME
    Commented Jul 3, 2021 at 6:23

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