A finite branched holomorphic covering is a holomorphic map $f : V \to W$ between holomorphic varieties $V$ and $W$ such that

  • $f$ is a finite branched covering (in the topological sense)
  • There is a regular part $V_0 \subset V$ on which $f$ is a locally biholomorphic mapping

On the regular part, the behavior of $f$ is relatively simple since it is locally biholomorphic, and locally $V_0$ looks just like $W_0 = f(V_0)$. My question is: do we (can we) know anything about the nonregular part (maybe that's not the right term)? I.e., can we know the structure of the map $f:(V \setminus V_0) \to W$?

In particular I am interested in the case where $W$ is simply a open disc of $\mathbb{C}$ centered at the origin, and $W_0$ is the whole disk without the origin. How should I study the structure of $f^{-1}(0)$?


1 Answer 1


This is only a partial answer. Regarding the last part of your question:

Every (connected) finite branched cover $f: V\rightarrow \mathbb{D}$ of the disc is isomorphic to one of the form $p_n: \mathbb{D}\rightarrow \mathbb{D}$, where $p_n(z) = z^n$ and $n\in \mathbb{Z}^+$. In particular, $f^{-1}(0)$ consists of a single point since $z^n$ has a unique zero.

To get a better idea of why only the map $z\rightarrow z^n$ appears, note that every such branched covering arises from an unbranched covering $g:X\rightarrow \mathbb{D}^*$ which one can also show is of the form $p_k: \mathbb{D}^* \rightarrow \mathbb{D}^* $ up to isomorphism. The original branched covering $V\rightarrow \mathbb{D}$ is obtained by completing, i.e. "adding an extra point" to $\mathbb{D}^*$ and extending $g$ so that it takes the value $0$ on this new point. So $f^{-1}(0)$ consists of a single point (of "multiplicity" $k$) by construction. Now, if you don't require $V$ to be connected, you'll end up with possibly several disjoint copies of this situation, namely $N$ discs $\mathbb{D}_j$ equipped with maps $z\rightarrow z^{n_j}$. Then $f^{-1}(0)$ consists of $N$ points $v_j$ with multiplicities $n_j$. This is the situation you'll find when analyzing the preimage of a small embedded disc under a non-constant map of Riemann surfaces.

In higher dimensions, branched coverings are more difficult to understand. One useful fact is that such maps are proper (by definition). So if $V,W$ are smooth, then the Proper Mapping Theorem applies. In particular, $f(V\backslash V_0)$ is an analytic hypersurface in $W$ (since the ramification divisor $V\backslash V_0$ is an analytic hypersurface in $V$ - see Griffiths & Harris).

Edit: One nice reference for the basics of branched covers is the book "Holomorphic Functions to Complex Manifolds" by Grauert and Fritzsche.

  • $\begingroup$ Thank you so much for the answer. For the first part, what's the name of the theorem? I have heard different names like "local uniformization theorem" and "local canonical form", and I had trouble finding the original statement and proof. Why $f^{-1}(0)$ is just a single point: why can't a few sheets coalesce into one point in $f^{-1}(0)$ while the remaining sheets coalesce into a different point? $\endgroup$
    – ssquidd
    Jul 5, 2012 at 15:49
  • $\begingroup$ Hi ssquidd. I was going to reply with a comment, but it ran too long. I've edited the my answer to reflect your question. $\endgroup$
    – Kevin
    Jul 5, 2012 at 17:48
  • $\begingroup$ Miranda calls the local expression $z\rightarrow z^n$ of a holomorphic map between Riemann surfaces the "local normal form". That such maps have a this local form follows easily from power series manipulations. Check out his book "Algebraic Curves and Riemann Surfaces" for more. $\endgroup$
    – Kevin
    Jul 6, 2012 at 5:16
  • $\begingroup$ So, did you want to know more about the higher dimensional situation or is my answer satisfactory? $\endgroup$
    – Kevin
    Jul 20, 2012 at 18:31

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