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This question may be trivial, I did not think hard about it.

A friend of mine is looking for an irreducible (reduced) analytic subspace $C \subset \mathbb{C}^2$ with the following property. Let $f \colon C \rightarrow \mathbb{C}$ be the projection on the first factor. He wants that

1) All singular points of $C$ and all ramification points for $f$ lie in a limited set, so removing that set we obtain a topological covering from some open set of $C$ to $\mathbb{C}$ with a ball removed.

2) That covering should be trivial (even better if it is finitely-sheeted).

So the curve $C$ is connected, but only if one passes near the origin. Sufficiently far from that ther should be no way to jump between sheets. Is it possible to find such a $C$?

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  • $\begingroup$ Doesn't two copies of C meeting at a point work?? i.e. z^2+w^2=0. $\endgroup$
    – Kevin Buzzard
    Commented Feb 12, 2010 at 19:26
  • $\begingroup$ @Kevin. Sorry, I forgot to rule out the trivial case of a reducible curve with limited singular set. I will edit it. $\endgroup$
    – Andrea Ferretti
    Commented Feb 13, 2010 at 12:58
  • $\begingroup$ @Leonid. Let me see if I uderstand well. Your curve has two branch points. Turning around such a point makes you jump from one sheet to the other. So if you stay far enough from the origin you can only turn around both and end up on the original sheet, right? I believe you are thinking of a simpler argument using inequalities, but I cannot see it. $\endgroup$
    – Andrea Ferretti
    Commented Feb 13, 2010 at 13:00

2 Answers 2

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This is an extended version of my comment. Suppose we stay on the surface $z^2+w^2=1$ but away from the origin. The identity $|z^2+w^2|^2=|z|^4+|w|^4+2Re((z \bar w)^2)$  tells us that the square of $z \bar w$ has negative real part. The set of complex numbers $\zeta$ such that $Re(\zeta^2)<0$ has two connected components: it's the disjoint union of two open sectors. Finally, note that switch from (z,w) to (-z, w) involves going from one component to the other. 

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Let $C$ be a complex line in $C^2$, say $y=0$. Project it on $x$-line, all properties are satisfied:-)

If you really want "connected, but ONLY if one goes near the origin", take the set $\{(x,y): y^2=x(x-1)\}$ and project it on the $x$ coordinate. This is a non-singular curve. If you remove a compact set out of it, it becomes disconnected. There is no way "to jump between the sheets" of the square root away from the segment $[0,1]$.

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