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Let $F$ be a non constant holomorphic map between two compact Riemann surface $X,Y$. Assume for each point $p\in X , mult_p(F)=1$. Then ,what can we say about the degree of that map (deg(F))?

Locally $F$ looks like a coordinate chart function. Can we say $F$ is an isomorphism? I am reading "Riemann Hurwitz formula " this question came in my thought but unable to get answer.

Infinity
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1 Answers1

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First off, the multiplicity is usually only defined for non-constant holomorphic mappings. So, let me assume that $F\colon X\rightarrow Y$ is a non-constant holomorphic mapping between compact Riemann surfaces.

We have that $F$ is surjective, since $F(X)$ is a non-empty clopen subset of the connected topological space $Y$. Thus, if we knew that $F$ were injective, we would be done. Now, since $F$ is unramified, by looking at its Local Normal Form, we see that $F$ is injective locally. However, in general, $F$ is not injective globally.

Let me give an example. By the Riemann-Hurwitz formula, any non-constant holomorphic mapping between complex tori is unramified. There are, however, holomorphic mappings between complex tori which are not biholomorphisms (i.e. that do not have degree one).

To see this, take you favourite (discrete) lattice $\Gamma \subset \mathbb{C}$, say the Gaussian integers $\Gamma=\mathbb{Z}+\mathbb{Z}i$. Pick your favourite complex number $\alpha \in \mathbb{C}\setminus\{0\}$, say $\alpha=1+i$. Then take a lattice $\Gamma'$ with $\alpha\Gamma\subsetneq \Gamma'$, say $\Gamma'=\mathbb{Z}\frac{1+i}{2}+\mathbb{Z}\frac{-1+i}{2}$. Next, consider the mapping $$f_\alpha\colon\mathbb{C}/\Gamma\longrightarrow\mathbb{C}/\Gamma'; [z]_\Gamma\longmapsto [\alpha\cdot z]_{\Gamma'}.$$ This mapping is well-defined since $\alpha\Gamma\subseteq\Gamma'$. Moreover, it is a non-constant group homomorphism. Additionally, it is holomorphic (why?). Finally, it is non-injective since $\alpha\Gamma\subsetneq\Gamma'$. Namely, pick a complex number $z\in\Gamma'\setminus\alpha\Gamma$. Then $f_\alpha([\frac{1}{\alpha}z]_\Gamma)=[0]_{\Gamma'}$, but $\frac{1}{\alpha}z\notin \Gamma.$ Thus, the kernel of $f_\alpha$ is non-trivial.

The example I gave is of minimal genus. Namely, by the Riemann-Hurwitz formula any unramified non-constant holomorphic mapping $f\colon X\rightarrow Y$ between compact Riemann surfaces satisfies $2g(X)-2=(2g(Y)-2)\operatorname{deg}(f)$, where $g(-)$ denotes the genus. If $\operatorname{deg}(f)\geq 2$, then $g(X)\geq 1$ and $g(Y)\geq 1$.

Note that Aphelli's example in the comments is a special case of the above construction.

Max Demirdilek
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  • If $T \subset \mathbb{C}$ is any (discrete) lattice, then $z \longmapsto 2z$ is such a map for $\mathbb{C}/T$. I think it’s worth adding that the maps satisfying the OP’s condition are exactly the holomorphic covering maps from $X$ to $Y$. – Aphelli Feb 02 '23 at 16:05
  • @Aphelli Yes, that is a valuable comment, thanks. In his Lectures on Riemann Surfaces, Forster (rather unfortunately) calls any proper non-constant holomorphic map between Riemann surfaces a (sheeted) holomorphic covering map — not only those that are unbranched. When you say holomorphic covering map, you of course refer to a covering map as defined in topology that additionally is holomorphic (i.e. an unbranched holomorphic covering map, in his terminology). I am writing this, simply to prevent someone mainly familiar with Forster from being confused. – Max Demirdilek Feb 02 '23 at 17:46
  • Sorry but I don't know the definition of unramified and unbranded cover. – Infinity Feb 03 '23 at 04:16
  • @Spectrum Firstly, a non-constant holomorphic map between Riemann surfaces is called unbranched iff it has no branch points iff it has no ramification points iff it is unramified iff its multiplicity at every point is one. – Max Demirdilek Feb 03 '23 at 06:55
  • Secondly, any non-constant proper holomorphic map between Riemann surfaces turns out to be a topological covering map between the underlying topological spaces iff one takes out all of its ramification and branch points. This motivates Forster's terminology: He calls a non-constant proper holomorphic map between R.S. a holomorphic covering map. In his terminology, a holomorphic covering map is a topological covering map iff it is unbranched. Unfortunate terminology, I agree! … – Max Demirdilek Feb 03 '23 at 06:56
  • … Now, any non-constant holomorphic map between compact Riemann surfaces is proper. Thus, in the compact case, any non-constant holomorphic map between R.S. is a holomorphic covering map a la Forster (so possibly branched).

    In summary, in this question you asked whether there exist unbranched holomorphic covering maps which are not isomorphisms. The answer is yes. Does that make things clearer? :)

     – Max Demirdilek Feb 03 '23 at 06:56
  • … and the degree of the covering map $f_\alpha$ is given by the index of the subgroup $\alpha\Gamma$ in $\Gamma'$. This might be worth adding also. – Max Demirdilek Feb 05 '23 at 06:56